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Your Favorite Math/Logic Riddles?
shma asks: "Whether you're involved in the Sciences, Mathematics, or Engineering, you undoubtedly enjoy finding simple solutions to seemingly difficult problems. I'm sure you all have a favorite mind-bender, and who better to share it with than the Slashdot community? Post your own problems and try to solve others. Just one request: If you have figured out the solution, link to it in a post, rather than write it out where anyone can see it." What brain benders tickle your fancy?
"Here's a sample to consider: You're in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it's too dark to tell what you're moving or flipping (no, you can't figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?"
They drive me nuts. Array and vecor logic. Fun
-B
You are making cupcakes for a party at which there will be 40 people. Half of them will be teenagers, a quarter of them will be adults, and the rest will be babies. Half of the babies don't like cupcakes, and one fifth of the babies left are too young to eat cupcakes. Half of the adults and three-quarters of the teenagers like chocolate cupcakes, and the rest of the people like cupcakes with sprinkles on them.
How many sprinkle cupcakes should you make?
What brain benders tickle your fancy?
Yeah, how did Bush get elected? Even more mind numbing is how he got re-elected. How'd that happen? Seriously, we have a man who is not the least bit curious, managed to squeak through school, managed to run two companies into the ground and yet still get elected as President of the United States where he is proceeding to run the country into the ground. Bush is cutting basic science research of all sorts (unless it is defense or energy related) and is pushing a religious agenda to the detriment of science and science education.
I have always been intrigued by the concept of infinity.
2=?
I wouldn't say I have a favorite problem but often when I'm bored I'll pen down the Pythagorean theorem and solve it manually. 0 = ax*x + bx + c. I'll work it out until I get the solution that (I hope) everybody knows and loves! It helps to keep my math skills alive during boring meetings.
x(n) = x(n-1)/2 if n is even
= 3*x(n-1)+1 if n is odd
All determinations of time presuppose something permanent in perception and that this permanent cannot be in the self, s
Is obviously 42
prove that a^n=b^n+c^n for any n.
What is the next line in the following sequence? 1 11 21 1211 111221 312211
How much wood would a woodchuck chuck, if a woodchuck could chuck wood?
Why yes, I AM a rocket scientist!
Turn a light on.
Slashdot - where whining about luck is the new way to make the world you want.
You are stranded on an island, on a path which splits in two directions. One direction takes you to "The Village of Death", the other path takes you to "The Village of Life." There are two tribes of people living on the island, one which ALWAYS TELLS THE TRUTH and one which ALWAYS LIES. A person is standing at the fork in the road. What is the ONE QUESTION (micro-variants don't count) you can ask this person which will ALWAYS get you to the Village of Life. Remember that you don't know which tribe the person is from.
Bill Gates is said to have solved the problem by memorizing the combinations first, the brute force approach.
It ones of those that requires a knack for seeing the simple things
"It is a greater offense to steal men's labor, than their clothes"
If you have a piece of paper, and you draw any quadrilateral of any size (rhombus, rectangle, or square) on that piece of paper. How can u divide that piece of paper in half so that it also evenly divides teh quadrilateral?
If you have a 5 gallon jug and a 3 gallon jug of water, and a hose so u can refill any as u please. What are the steps to get exactly 4 gallons of water?
SEND + MORE = MONEY What number does each word represent? The letters represent a single decimal digit. There is only 1 solution to this problem.
Beep. Boop. Beep. You have questions. I have answers and your home address.
I think the answer is either:
1. Turn the lights on, put 9 heads in each group
2. Cut all the coins in half and seperate...close enough to being split up
both may be technically correct, the best kind of correct...
cnidarian
The case of Sleeping Beauty:
We plan to put Beauty to sleep by chemical means, and then we'll flip a (fair) coin. If the coin lands Heads, we will awaken Beauty on Monday afternoon and interview her. If it lands Tails, we will awaken her Monday afternoon, interview her, put her back to sleep, and then awaken her again on Tuesday afternoon and interview her again.
The (each?) interview is to consist of the one question: what is your credence now for the proposition that our coin landed Heads?
When awakened (and during the interview) Beauty will not be able to tell which day it is, nor will she remember whether she has been awakened before.
She knows the above details of our experiment.
What credence should she state in answer to our question?
-------
Some people, "halfers", consider that the answer to this question is obviously "1/2". Others, "thirders", consider that it is obviously "1/3".
Fork universe as needed.
flip arbitrary number of quarters, seperate into two even groups.
If groups do not contain an equal number of heads up quarters, destroy universe.
The universe that remains will have the right number of quarters.
You integrate it.
We do not live in the 21st century. We live in the 20 second century.
a closed form solution to the Navier-Stokes equations? Quite a riddle I'd say.
a window |+|
Kind of annoying background music - link to an online Soduku game is here
Why do they call them slashdot editors?
Integrate cabin = log( cabin ) + c.
"Log cabin plus sea".
We do not live in the 21st century. We live in the 20 second century.
42
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make install -not war
Since it's an infinite sequence, you can separate the left-most X and rest still equals 2. Thus X^2 = 2, so X = sqrt(2).
How many Slashdot editors does it take to change a light bulb?
1. Grab a calculator (You won't be able to do this one in your head) 2. Key in the first three digits of your phone number (NOT the area code) 3. Multiply by 80 4. Add 1 5. Multiply by 250 6. Add the last 4 digits of your phone number 7. Add the last 4 digits of your phone number again 8. Subtract 250 9. Divide number by 2 Do you recognize the answer?
The gates in my computer are AND, OR and NOT; they are not Bill.
I haven't thought about a more efficient way, but it seems that the surest bet would be to split the coins into two equal groups, and then flip all the coins in each group. You're most likely to end up with 25 heads up coins in each group (being the maximum entropy state).
I have discovered a truly marvelous
Here is the little brain teaser I thought up-- see if you can solve
it...
In the following sequence:
1, 4, 8, 13, 21, 30, 36, 44...
What is the next number and why:
A. 48
B. 50
C. 53
D. 57
E. 61
F. There is no pattern
f(x)=x^(1/x)
It is not defined for negative values or 0. It is defined only for x>0. At 1, f(x)=1, then it peaks somwehere in 2.71
What I'm really interested in is the first derivative. Where f'(x) is 0, is the maximum of f(x). Just one catch: no limits in the formula. I don't want something that I need to calculate forever; I want a formula giving me a value I can calculate to an arbitrary precision.
The problem comes into play with the (1/x) in the exponent. All attempts to derive this result in a formula which contradicts itself, due to 0 being in either the base or the denominator of the exponent in the original f(x).
It is defeating all attempts of my college-intro-to-calculus. Searches at wolfram.mathematica.com don't help, either (or I'm just doing the wrong searches). Who can help?
If there are 23 people in a room, what are the chances that at least two of them will have the same birthday?
see http://en.wikipedia.org/wiki/Birthday_paradox for more info.
'Nuff Said
Just to get an easy, universally-agreed-upon one out of the way.... .9999 =! 1
Now we can move on to questions that can generate some real debate.
There's a good write up of this on MathWorld.
I met a man with seven wives. Every wife had seven sacks, and every sack had seven cats, and ever cat had seven kits. Kits, cats, sacks, and wives, how many were going to St. Ives?
Rank my idea: http://www.sinceslicedbread.com/node/531
Get over it.
http://www.websudoku.com/ is my sudoku fix of choice
In these days, bleeps and bloops mean something more
I picked the wrong night to log on to Slashdot after some heavy drinking.
And I thought my head was hurting before...
Google "same number of heads up" and read the first site at http://72.14.203.104/search?q=cache:ZqMemrb_jWUJ:w ww.creativepuzzels.nl/spel/speel1/speel2/munten2.h tm+%22same+number+of+heads+up%22&hl=en
Epimenides was a Cretan who made one immortal statement: "All Cretans are liars."
"The Epimenides paradox is a problem in logic. This problem is named after the Cretan philosopher Epimenides of Knossos (flourished circa 600 BC), who stated , "Cretans, always liars". There is no single statement of the problem; a typical variation is given in the book Gödel, Escher, Bach (page 17), by Douglas R. Hofstadter.
"Academicians are more likely to share each other's toothbrush than each other's nomenclature."
Cohen
My girlfriend figured it out, I guess that makes her hot.
Seperate 18 of the coins. The amount of heads remaining in the original pile will be equal to the amount of tails in the new pile. Flip over all the coins in the new pile to get the piles to have an equivilent number of heads.
Example
New Pile -- 5 heads, 13 tails
Old Pile -- 13 heads, 29 tails
flip new pile and get:
New Pile -- 13 heads, 5 tails
Old Pile -- 13 heads, 29 tails
I learned this trick in grade school many years ago; I am often surprised at how many people haven't heard of it:
.wellas6bydivisibleisitthen2byand3bydivisibleevenl yissumtheifAnd.well as3bydivisibleisnumberoriginalthethen,3bydivisible evenlyisnumbertheindigitstheof allofSUMtheIf
How can you *quickly* determine if an integer is divisible by 3? By 6?
Answer (with words reversed and spaces removed so you can think about it without seeing the answer too easily):
You find yourself before indistinguishable two doors, each with a statue. One door will lead to salvation, the other to death. The statue that guards the door to salvation always tells the truth, the statue to the door to death always lies. You may pose only one question to only one statue. What do you ask to determine which door is which?
Answer(ROT13): Nfx nal dhrfgvba gb juvpu lbh nyernql xabj gur nafjre. Gb qrgrezvar juvpu qbbe vf juvpu lbh arrq gb xabj gur eryngvbafuvc bs gur nafjre lbh ner tvira gb gur gehgu. Gur guvat V yvxr nobhg guvf evqqyr vf vg sbeprf lbh gb pbafvqre gur bcrengbe va gur ybtvpny fgngrzrag gb or gur inevnoyr. Nqqvgvbanyyl crbcyr nera'g hfrq gb nfxvat dhrfgvbaf jura gurl nyernql xabj gur nafjre fb gurl graq abg gb or noyr gb guvax bs n fbyhgvba evtug njnl. Gur jubyr guvat orpbzrf boivbhf jura lbh cbfr n dhrfgvba fhpu nf "Ner gurer gjb fgnghrf urer?"
First up, does Goedel's incompleteness theorem imply that computers will never be able to have human-like intelligence?
.99... = 1, or that formula that shows pi*log^-1 = 0 or whatever it is... is always struck me as the Grand Unification Theory of algebra and geometry. It's so simple and shows that these numbers, which are so hard for me to work with, combine in some fashion to show some property that is strikingly simple, like finding a beatiful crystal of clarity in a quarry of grey, difficult mathematics.
Other things I like are not necessarily problems, but things that just inspire awe, such as proving that
Computers are useless. They can only give you answers.
-- Pablo Picasso
1.Find problem on Slashdot
2.Solve problem
3.Email to generous stranger
4.Get $10 via PayPal
5.Profit!
See how easy that is! No tricks here...
* - by accepting the $10, you transfer all claims to any further monetary compensation from any party to Comatose51, including but not limited to $10 million from the Clay Institute
EvilCON - Made Famous by
- One room has three switches, labeled A, B, and C.
- Another room has three light bulbs, labeled 1, 2, and 3.
- Each switch is connected to one bulb, but you do not know which is connected to which.
- When inside either room, you cannot see the other room.
- You begin in the room with the switches and may turn the switches on and off in any way you choose.
- Once you leave the room with the switches, you may not reenter it. You may, however, go to the room with the light bulbs.
How can you determine which switch is connected to which light? Here is a hint and solution.I like this problem because people are ordinarily good at logic have so much trouble with it. I once had the pleasure of meeting Donald Knuth and stumped him with this puzzle.
Really simple. If you can't get this go back to calc 101 :p
....
From the Product Rule http://en.wikipedia.org/wiki/Product_rule
(u*V)' = u'*V + V'*u
Integrating both sides...
u*V = Integral [u'*V + V'*u]
Substituting u = x and V = 1/x, we have
1 = Integral (1/x - 1/x) = 0
1) Do something
2) ???
3) Profit!
Solve for ???
What is the degree of the angle between the hour hand and the minute hand when it is 2:15?
:14, = 67.5. The difference is 22.5
22.5 degrees.
Yes, you can do it iteratively until inifinity, but the minute hand is at 90 degrees off 12, and the hour hand is at 60 for 2, plus 30/4 for the
-- Is "Sig" copyrighted by www.sig.com?
.9999999999999999999999999999999999999999999 repeating = 1!
---In a time of Chimpanzees I was a Monkey.
Fill 3-jug
Pour 3g from 3-jug into 5-jug, leaving 2g empty in 5-jug
Fill 3-jug
Fill 2g in 5-jug from 3-jug, leaving 1g in 3-jug
Empty 5-jug
Pour 1g from 3-jug into 5-jug, leaving 4g empty in 5-jug
Fill 3-jug
Pour 3g from 3-jug into 5-jug for 4g in 5-jug
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make install -not war
This is one of my favorites - it has stumped many self-professed math geeks, yet high school freshmen have spotted the solution immediately.
x=y
x^2=xy
x^2-y^2=xy-y^2
(x+y)(x-y)=y(x-y)
x+y=y
2y=y
2=1
Every step uses perfectly valid algebra, yet something is obviously very wrong somewhere.
Enjoy...
An object at rest cannot be stopped!
Either the Monty Hall problem which serves as a meditation on the idiocy of expertese, or the mayday mystery which isn't a logic problem, but is certainly a more amusing way to waste time than most puzzles.
I'm sure this has been on slashdot before but nobody's mentioned it. http://www.ocf.berkeley.edu/~wwu/riddles/intro.sht ml Almost any logic problem you've heard of is described, and discussed on that site.
I just got that from my school listerv. Yeah, it's definitely a sad one.
Not really hard to prove, but it's cute.
Take any six-digit number that's of the form ABCABC where A,B,C are any integers (yes, they can be the same, yes they can be zero, although that might make it less than six-digits if A, or A and B, are zero), and that number is guaranteed to be divisible by 13.
A hunter follows a bear.
The hunter walks 1 km south, 1 km east, 1km north.
He's back at his starting point.
What is the colour of the bear?
PS: yes, it's math or logic!
Of course. I figured it would be a transcendent number, I just didn't know which one. And somehow, e never crossed my mind. Thank you to both, croto and sam_nead.
The rules are very simple.
o mments.JPG
1. place numbers 1-7 in each row.
2. place numbers 1-7 in each column.
3. place numbers 1-7 in each color group.
so that each number only appears once for each specified set.
By S. Viemeister
Here is the link:
http://vxweb.viemeister.us/math/7x7-puzzle-with-c
Enjoy!
Laffer curve: it's possible to cut taxes and increase revenues. Think about if the government taxed nothing. It would get nothing. If the government fully taxed everything, no one would do anything, so there would be no point, so the government wouldn't get any tax. If the government taxes between those bounds, it gets something. Thus, there is at least one local maximum, meaning lowering taxes can increase revenues.
Makes sense when you think about it. Punish people too hard and they won't produce anything - at least if they know you'll find out.
Rank my idea: http://www.sinceslicedbread.com/node/531
Just stack all the coins together on edge, and split however you like. Then neither set will have any head sides up.
1) No object can violate conservation of angular momentum.
2) To rotate an object one needs to give it angular velocity, hence angular momentum.
3) To have finite angular momentum, an object needs torque applied to it (or a force applied away from the center of moment).
4) Gravity acts on the center of moment and does not result in torque on any free falling object.
5) Cats dropped feet up manage to land on their feet.
6) Does this mean cats violate conservation of angular momentum; no wonder Egyptians worshiped them.
What is wrong with this discussion; no math involved from my Classical dynamics class.
That is a good one thats been around for ages, too bad it will never compile because of the 0 devide.
Dear aunt, let's set so double the killer delete select all
Farmer Jones has 12 sheep. All but 4 die. How many sheep does Farmer Jones have left?
You would be amazed how many people will get this wrong the first time they answer it.
I want a new quote. One that won't spill. One that don't cost too much. Or come in a pill.
The impolite formulation involves three college guys, who can only afford two condoms and the paid services of one person.
"She's a scientist and a lesbian. She's not going to let it slide." Orphan Black
what about this one guys:
there is a hunter who wakes up one day in his house, walks out, and sees the sun just above the horizon. So he decides to go hunting. He unavoidably walks three miles to the south, then four miles to the west. He finds a bear and kills it... (sigh). Well, the questions are: 1) how far is the hunter's house from the dead bear. 2) what date is it. 3) what is the dead bear's colour.
It's one of my favourite riddles.
Let's say I have a stack of sticks: all identical, inflexible, unbreakable. Sticks can touch only at their ends, not in between.
If I give you 3 sticks, you can make one triangle. If I give you 2 more sticks (5), you can make 2 triangles. If I give you another stick (6), how can you make 4 triangles?
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make install -not war
I can't take credit for these questions, but people have been doing them for centuries, and I enjoy playing with them. Prove that the square root of 2 is irrational. If that was fun and you want some more, prove that the square root of 3 is irrational. If you are still not satisfied, derive a proof for Fermat's last theorem.
"10001110101 - periodic table with a centerpiece of mind" -Clutch
Is ..:: Riddles ::... In has (amogst others) the famous "prison with a lamp" problem:
100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.
Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?
I first read about hypergame from Smullyan (I don't remember which book), but the mathworld reference is at http://mathworld.wolfram.com/Hypergame.html
This is representative of the problems I considered "cute" back in my high-school days: Show that, given any 6 points inside a circle of radius 1, some two are within distance 1 of each other. The proof is simple, but involves a cute trick.
Right here.
Incidentally, that page has been slashdotted in the past.
There is no problem that the Slashdot community can't solve. So what about that tiny 'P versus NP' problem?
DALMOZIAN if you don't have anything to say, type it
If you take two apples from three apples, how many apples do you have?
Most states have laws against siblings marrying, but is it legal for a man to marry his widow's sister in West Virginia?
These are pretty old and pretty easy...
Stand them all on end and end up with zero heads in both piles?
No matter how subtle the wizard, a knife between the shoulder blades will seriously cramp his style.
I think the one you mean is e^(i*pi) - 1 = 0. That's "Euler's Formula", athough there are many formulas that go by that name... An interesting excercise is to show why that formula is true.
Here's an excellent archive of puzzles going all the way back to 1996. These puzzles range from "interesting" to "amazingly clever".
http://www.greylabyrinth.com/archives.php
There is a king and there are his n prisoners. The king has a dungeon in his castle that is shaped like a circle, and has n cell doors around the perimeter, each leading to a separate, utterly sound proof room. When within the cells, the prisoners have absolutely no means of communicating with each other.
The king sits in his central room and the n prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners' rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers "yes" but is wrong, they all will be beheaded. If a prisoner answers "yes," however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
Here's one last monkey wrench to toss in the gears, though. The king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice. The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.
(written by a former roomate)
let's see... the problem goes roughly like this:
you have five hats (two red, three black) and three people. you queue the people up in order of height and have them face the same way (this way the tallest person can see the two people in front of him/her, the middle person can see the shortest person, and the shortest person can't see anyone). you put a hat on each person's head and instruct them that they are not allowed to take the hat off or turn around. you then ask them to tell you what color their hat is. after a while, the person at the front of the line correctly announces the color of his/her hat. how did the person at the front of the line know and what were the other hat colors?
This one is old school: we're dealing with pointers and very limited memory. Here's the set up:
You are given a pointer to the head of a linked list. Your job is to figure out if the linked list has a cycle (i.e. one of the nodes has a next pointer that points back to some earlier node in the list). The trick is that you only have enough memory on the stack for two node pointers. How do you do it?
I got asked this in a tech intterview as a "bonus question" and it drove me nuts for the better part of an hour. The answer is pretty simple but a little hard to see.
This is an interesting site with many math puzzles:
http://mathschallenge.net/
Teacher: So y = r cubed over 3. And if you determine the rate of change in this curve correctly, I think you'll be pleasantly surprised.
[The class laughs except for Bart who appears confused.]
Teacher: Don't you get it, Bart? Derivative dy = 3 r squared dr over 3, or r squared dr, or r dr r.
_______
2B1ASK1
A ten-inch hole is drilled through the center of a solid sphere. What is the remaining volume?
If all of your peers tell you not to succumb to peer pressure, what do you do?
I own itburns.net. What should I put there?
Q: Did you know they are serving free beer in the village of Life?
A truth-teller will say "no", and head to the village of Life. An
honest liar will say "yes", and head to the village of Life. A cunning
liar will say "yes", and may lead you to the village of Life -- or not,
in which case you can still claim victory since the liar will be haunted
by the suspicion that he or she may have missed out on the free beer.
Hint for the variant involving the robots:
Tourists don't always tell the truth!
One more variant, also due to Gardner or one of his readers: Suppose
that you speak the local language (shared by truth-tellers and liars)
perfectly, except you have forgotten if "pish" means yes and "tush"
means no, or vice versa, and that your question must be in a form
that requires a yes/no (well, pish/tush) answer.
"Why do some people confuse Halloween and Christmas?"
Answer
"Sometimes the only thing left to say is 'Oops'" -- debbers
File under 'M' for 'Manic ranting'
While I believe there is a mistake in the earlier steps, I can see immediately that if x+y=y then x=0. It is therefore doubtful that you can get 2y = y from that. It should rather be y=y which is true...
Rot13
K=L. Guhf K - L vf rdhny gb 0. jura lbh qvivqr ol (k-l) hc gurer lbh ner qvivqvat ol mreb juvpu lbh pna'g qb.
1
11
21
1211
filter: +3. Hey, look! all the trolls went away!
Make a second pile of 18 coins, and flip them all over. You take c heads, and 18-c tails for this second pile. the original pile now has 100-18 coins, of which 18-c coins are heads. Flip all of the second pile over, and now it has 18-c heads, rather than tails.
devide by zero, where your problem lies
8 something!
OK, it's not a riddle. But it's funny.
2 of my mathematicians friends have started a web site about enigma. But it only in french. http://www.lefactoriel.com/
You have 47 identical balls and one with a different weight. You also have a triangular balance with 3 plates.
One plate is heavier: The balance leans towards the heaviest
plate.
One playe is ligher: The balance raise the lightest one.
The 3 plates have the same weight: The balance stay flat.
How can you find out which ball is different and if it's heavier or lighter in 3 weighings ?
that question took me about 30 seconds.
my favorite math puzzle has always been take a three by three matrix and put the numbers 1-9 in the spaces such that if you sum any three boxes in a row it adds up to 15.
You are likely to get mugged by a Grue.
Not strictly math, but some mathematical aspects, some logic, and a lot of sleuthing.
http://www.notpron.com/
mostly SFW, though I have heard that some later levels link to (but don't show) porn.
--
"Extra Anus Kills Four-Legged Chick" -- Headline
I was going to say "duh, of course not, 0^0 is undefined, everyone knows that", since that's what I remember being taught all through school and university. However, a Google search tells me apparently it's not necessarily black or white. So I guess you have to go with what's useful for you. Anyway, it seems not even all software agrees; for example, both the built-in Windows calculator and Maple 8 say it's 1, but the PowerToys PowerCalc shows an error.
if x=y, then x+y=y can't work.
garble
There are an unknown number of people in a line, each with a black or white hat. The person in the rear can see the hat colors of everyone ahead, but not behind or his own. So, a person comes along and asks each person from the rear to the front in order what color hat they are wearing, and if they are wrong or have to guess, they are killed. They can only say a hat color, and everyone else in the line can hear what they say, but cannot know if they are killed. However, they collude beforehand to come up with a way to save the maximum number of them, which is n-1, since the first person asked cannot possibly get his hat color correct. So, what strategy will yield n-1 lives saved?
Bonus:
How is it done for m number of hat colors?
Good Luck!
here's an oldy MSFT asks on job interviews:
next to a door are 3 light switches. one of the switches are connected to an incandescant light bulb in the windowless room on the other side of the door. you may not open the door and flip the switches. you may only enter the room once. which switch lights up the room?
Solution
There are three doors: one has a brand new car behind it, and the other two have goats behind them. You pick one door, which is then marked, but not opened.
Now, one of the unmarked doors is opened to reveal a goat. You are given the choice of switching your first pick to the remaining unmarked, unopened door. Should you switch?
In Soviet Russia, backwards is everything.
An immortal and single-minded worm sits at one end of a 1km long rubber rope. The worm travels at a steady 1cm/sec toward the other end. The rope in infinitely and uniformly elastic. At the end of each second, the rope stretches (uniformly) by 1km. Does the worm ever reach the other end of the rope? If so, when?
Dividing by zero is not "perfectly valid algebra". Division is not closed on the set of real numbers. Its not really a riddle if you lie in the problem description. Otherwise the solution to the sample problem could be "Pull out 9 of the quarters into a seperate group. I was lying when I said you couldn't see any of them."
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
I always liked the pirates' booty problem (for those of you who giggled--oh grow up). There's a crew of 5, with a strict hiearchy from 1 down to 5. It's time to split the hoard of a 100 pieces up. The rules are: The highest rank proposes how to divvy things up and there is a vote, if it's 50% or more against, you kill that highest rank and then the next highest gives a plan, etc. Pirates will vote against unless there's a clear advantage to voting for (i.e. some pieces as opposed to strictly fewer). Pirates may be dastardly, but are also logical ;-)
So if you're the captain (numero uno) how do you dole the treasure out without getting killed?
Setup: There exists a function that accepts N points specified in cartesian 3D space and deterministically returns a plane, such that each set of N point has one and only one possible result value, and every plane can be specified by this function, given a proper set of points.
Question 1: What is the minimum value of N?
Question 2: In conventional language (ie not code or overly technical terminology) how does the function operate?
Hint: N is definitely less than 3, but if you were describing the function for N=3 one possible answer for Q2 would be "The function returns the unique plane which intersects all three points. When the three points are colinear or overlapping, the function returns False."
If you answer this question, please prefix "Answer" to the title of your [Reply] to avoid accidental spoilage. I'll post a link to the real answer in a day or two.
Kevin Fox
You have a set of DFA's which are linked to their neighboor. At each point in time their input is the state of their two neighboors. They are all identical, save for the two on the ends. They all (again, except for the ends), start at the same state. They transition simultaniously. An unknown number of them are hooked up, but they are hooked up linearly, with two end-points. Your goal is to arrange for them all to switch into a unique 'Firing' state at the same moment in time. Describe the DFA.
This is a non trivial problem. ...
...
Once upon a time, in a land far far away, there lived a knight who had
just rescued his first damsel in distress. The knight was called before
the king to receive a reward. The king told the knight that he had
written an amount of gold on a piece of paper and twice the amount of
gold on another piece of paper. He placed the two pieces of paper face
down in front of the knight, and told him he could chose either one.
The king would give the knight the amount of gold on the paper as a
reward. Or, the knight could opt to get the amount of gold on the other
paper instead.
This was the knight's first reward, so he had no idea what he was likely
to get. But the knight reasoned that no matter what amount he saw on the
paper he chose, he would take the other one because he had more to gain
than lose. For example, if the paper he chose had 16. He might win
another 16, and at worst he only loses 8!
The damsel points out that if the knight is going to end up with the amount
on the other piece of paper anyway, why not just choose it first and not
switch. His reward will be the same.
Is the damsel correct? Or is the knight's plan sound?
Refute the argument you disagree with. (Refuting the incorrect argument
is the challenge.)
You divided by zero between equations 4 and 5.
I pretend to know more than I really do by mooching off google and wikipedia.
The following two problems appeared in IMOs 1993 and 1994 (you can find the answers using Google, but I won't give a direct link).
A solitaire game is played on an infinite square grid. Initially, there are n^2 pieces in an n*n square formation. On each move, the player moves a piece either horizontally or vertically over an immediately adjacent piece into the square beyond, which must be unoccupied, and removes the piece that was jumped over. The objective is to end up with only one piece on the board. For which values of n is this possible?
Show that there exists a set A of positive integers with the following property: given any infinite set S of prime numbers, there are positive integers k>=2, n and m such that both n and m are the product of k primes in S, n is in the set A and m is not in the set A.
you've divided by 0 or infinity, depending on which way you go.
Here's a riddle.
It's Saturday night and some geek (sorry Cliff) is desperate and dateless so he spends 3 hours composing a submission, and YES!, it gets accepted by Slashdot just after midnight.
The riddle? How sad is that?
Assuming the earth is a perfect sphere, describe the solution set of points where you can go 1 mi south, 1 mile east, and 1 mile north and return to your starting point. Hint: the cardinality of the set is R cross Z + 1 (and yes, I know that's equal to R, but expanding it makes it a more effective hint). Feel free to email me for more hints.
U.S. War Crimes blog. Email for free Mandriva support.
You are a tourist, visiting a desert island just off the coast of South America. There's only one reason that you would be visiting this one-acre island, and that is that there is a tiny plateau reaching up a mile into the air, with the ruins of an Aztec temple on it.
As you walk along a path, you come to a fork. In the fork are two men, of which you know little, except that they must have come from one of the villages on the other islands nearby.
There are three villages--the Marqetteres always lie, the panguons always tell the truth, and the Shie'ep always do what everybody else is doing.
You may ask one question to one of the men. What do you do?
Answer(ROT-13):
Fvzcyr. Lbh vtaber obgu zra, naq jnyx fgenvtug gb gur zvyr uvtu cyngrnh, juvpu jbhyq or ivfvoyr sebz nal cbvag ba n bar-nper qrfreg vfynaq.
A barber has a sign that says he shaves all men, but only those men who do not shave themselves.
Who Shaves the barber?
Does he shave himself? He can't b/c he doesn't shave men who shave themselves. So somebody else must shave him, but that would mean that he doesn't shave himself so therefore he must shave himself.....
abcabc/13=(abc+1000*abc)/13=abc*1001/13=abc*77 nice!
What have I got in my pockets?
Ok, here is a progression of questions which require no special training. Make sure you only ROT13 one answer at a time if you're trying these yourself:
Assume Earth is a perfect sphere.
Q1) Where can you stand such that if you go 1km North, then 1km East, then 1km South, you're back where you started?
A1 rot13'ed) gur fbhgu cbyr. pregnvayl abg gur abegu cbyr, nf lbh pna'g tb abegu sebz gurer. naq vs lbh fnvq 1xz fbhgu bs gur abegu cbyr v'q fnl ab gbb, nf lbh pna'g tb rnfg sebz gur abegu cbyr, bayl fbhgu.
Q2) OK smarty. Where ELSE can you do it from, on the Earth's surface? No tricks are involved either, just a bit of thinking.
A2) n ovg bire bar xz fbhgu bs gur abegu cbyr: nsgre jnyxvat gur 1xz abegu, n 1xz jnyx rnfg pbzcyrgryl pvepyrf gur abegu cbyr, zrnavat lbh'ir qbar n ebhaq gevc. 1xz fbhgu gura ergheaf lbh gb gur vavgvny cbfvgvba. n srj crbcyr pbzcynva nobhg guvf bar, nf lbh nera'g jnyxvat va n fgenvtug yvar, rira gubhtu lbh'er nyjnlf urnqvat rnfg. lbh pna erzvaq gurz gung gurl jrera'g tbvat va n fgenvtug yvar va n1 rvgure. naq nfx gurz gb qrsvar 'rnfg' vs gurl fgvyy nera'g unccl.
Q3) You really think you're good don't you? OK, I want to know where ELSE!
(read this when you think you have it, before you read the real answer: gur nafjre vf abg nabgure cbfvgvba ba gur rnegu'f fhesnpr qhr rnfg (be jrfg) bs gur nafjre gb d2. jryy vg vf, ohg vg'f abg tbbq rabhtu, gurer'f fbzrjurer ryfr.)
A3) guvf nafjre vf nyzbfg gur fnzr nf gur ynfg, ohg vafgrnq bs cynpvat lbhefrys fb gung gur bar xz rnfgreyl jnyx vf n pbzcyrgr ybbc, lbh'er rira pybfre gb gur abegu cbyr, naq znantr gjb ybbcf! be, sbe gung znggre, lbh pna zbir rira pybfre, naq nf lbh nccebnpu gur '1xz fbhgu' cbvag sebz gur abegu cbyr lbh jvyy svaq zber naq zber fbyhgvbaf.
Enjoy.
If you do the population curve (takes a couple minutes, and somewhat tedious) it teaches an interesting lesson. The more money we donate to starving countries, like we do to various countries in Africe (both government and public donations), the more they reproduce. Thus they require more money, eventually dying out anyway. If we don't donate money, they drop to incredibly low numbers, and stabilize, but not as low as the other alternative. Thus the dilema. Either you donate and save some lives, causing more death in the end. Or you sit idle by, letting people die.
You have 9 brass balls and one set of scales (the old fashioned "justice" kind). One of the balls is very slightly heavier than the others. By only using the scales twice, how can you find out which one that is? That is, how would you weigh them?
Gur gbhevfg yvrq, naq jnf ernyyl tbvat gb Vgrebcbyvf. Fur jnf noyr gb qrqhpr gung gur svefg ebobg jnf gryyvat gur gehgu, naq gung ure gehr qrfgvangvba jnf gurersber gb gur yrsg, ba gur ebnq gur ebobgf unq orra geniryvat.
Gur xrl vf gur svefg ebobg'f frpbaq fgngrzrag ("Vs lbh nfxrq zr, V'q fnl Ovgobebhtu vf gb gur evtug"). Vs gur svefg ebobg vf n gehgu-gryyre, guvf fgngrzrag pna or npprcgrq ng snpr inyhr: Ovgobebhtu vf gb gur evtug. Vs gur svefg ebobg vf n yvne, ubjrire, gur fgngrzrag zhfg or n yvr; ohg jung vf gur yvr nobhg va guvf pnfr? N pnershy ernqvat fubjf gung vg pbapreaf ubj gur ebobg jbhyq nafjre n qverpg dhrfgvba nobhg gur ybpngvba bs Ovgobebhtu. Vs gur gehr ybpngvba bs Ovgobebhtu vf gb gur evtug, n ebobg gung yvrf jbhyq fnl "gb gur yrsg." Vs gur ebobg yvrf nobhg jung vg jbhyq fnl, vg jvyy fnl "gb gur evtug." Guhf gur npghny ybpngvba bs Ovgobebhtu zhfg or gb gur evtug va nal pnfr.
Fvapr Ovgobebhtu vf gb gur evtug, naq gur frpbaq ebobg qverpgyl pbagenqvpgf guvf, gur frpbaq ebobg zhfg or n yvne. Gurersber gur frpbaq ebobg'f svefg fgngrzrag ("Gung bar'f n yvne") vf snyfr, naq gur svefg ebobg vf n gehgu-gryyre. Guhf gur svefg ebobg'f svefg fgngrzrag ("Jr'ir whfg pbzr sebz Vgrebcbyvf") zhfg or gehr.
But bear with be.
There is a land populated by a lot of people and ruled by an insane but all-powerful tyrant. One day, that tyrant decides to play a very special game. He sends proclamations all over the land announcing the following upcoming "event":
"Some time in the future, I will summon one randomly selected peasant to my palace. He will come into my room, where I will be waiting with a pair of dice and a firing squad. I will roll the dice, and if I roll double sixes, he will be shot.
"If he is NOT shot, however, I will release him, but make him swear not to tell anyone else that he participated in this event. The next day, I will summon TEN new people to my palace, and repeat the procedure for the whole group: one roll of the dice, they all die if it's double-sixes, and they all go free otherwise. If they go free, the next day I summon in a group of 100.
"The cycle will keep repeating, each iteration with ten times as many subjects as the previous iteration, until I finally shoot someone (or a group of someones)."
Now, look at this and you'll realize that only about 10% of the people who participate in this experiment will survive. The math doesn't work out exactly, but since each iteration contains ten times as many people as the previous iteration, it's ABOUT that much. For example, if the tyrant finally rolls double-sixes on the fourth iteration of the experiment, 1000 people are killed while 111 people from previous iterations will have gone free.
On the other hand, it would seem that the people in any given iteration have a 35/36 chance of survival, since there's only a 1/36 chance that the tyrant will roll double-sixes for that group.
So let's say you live in this country. You get a knock on your door one day and find out you've been summoned to be a part of the current iteration. What is your chance of survival, 35/36 (~97%) or ~10%?
Does this chance of survival change when you're actually standing before the tyrant and he's about to roll the dice? If so, why?
(P.S.: If there is one "right" answer to this, I haven't heard it. I heard this puzzle in a philosophy class, and it has something to do with anthropic reasoning, if that's at all helpful.)
(P.P.S.: Assume that the dictator doesn't run out of people before the experiment is over. Really, this problem is less about the fiddly numbers than the basic probabilistic concepts involved.)
nt
This is one of my favorites...
...). It's currently 11:00 AM (or, t = 1 hour to noon).
;-)
There's a giant vat with a nozzle attached hanging over a basket. Next to the basket, there's a monkey. Inside the vat is an infinite number of balls (okay, it's a big vat), each labeled with a single natural number (1, 2, 3,
At t = 1 / k hour to noon (where k is a natural number) balls 2k and 2k - 1 fall out of the vat and into the basket. At the same instant, the monkey reaches into the basket, grabs ball k, and eats it. So, at 11:00, balls 1 and 2 fall into the basket, and the monkey eats ball 1. At 11:30, balls 3 and 4 fall into the basket, and the monkey reaches in and eats ball 2. This process continues ad infinitum.
The question, of course, is this: how many balls are in the basket at noon?
Now, if you've figured out that answer, here's a similar question that's both harder and easier. It's easier in that you're infinitely more likely to guess a correct answer, but harder since you're probably much less likely to be able to prove that it's correct.
Same problem, same vat, same monkey, same timing, same number of balls, except they're no longer labeled. So, at every t = 1 / k hour to noon, two balls drop into the basket, and the monkey reaches in and eats one ball. Now how many balls are there in the basket at noon?
I'll link to the solutions later, if nobody figures it out.
Just holding the defective lightbulb, and the world revolves around him. *ducks*
All those moments will be lost in time, like tears in rain. Time to die.
Given: two concentric circles of different sizes in a plane. There exists a line segment that is tangent to the smaller one and a whose endpoints lie on the larger one. The line segment is 10 units long.
What is the area between the two circles?
John von Neumann (1903-1957) [Hungarian/US mathematician and scientist] The following problem can be solved either the easy way or the hard way.
Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour. A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour. It does this until the trains collide and crush the fly to death. What is the total distance the fly has flown?
The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows:
Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it.
When this problem was posed to John von Neumann, he immediately replied, "150 miles."
"It is very strange," said the poser, "but nearly everyone tries to sum the infinite series."
"What do you mean, strange?" asked Von Neumann. "That's how I did it!"
http://www.ocf.berkeley.edu/~wwu/riddles/intro.sht ml/
Has nearly all the ones I've seen so far, plus oodles more. Keeps me distracted for hours on end. Warning: doesn't give solutions, can be frustrating.
When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl
A subject S knows that a proposition P is true if, and only if:
Achilles and the Tortoise In the paradox of Achilles and the tortoise, we imagine the Greek hero Achilles in a footrace with the plodding reptile. Because he is so fast a runner, Achilles graciously allows the tortoise a head start of a hundred feet. If we suppose that each racer starts running at some constant speed (one very fast and one very slow), then after some finite time, Achilles will have run a hundred feet, bringing him to the tortoise's starting point; during this time, the tortoise has "run" a (much shorter) distance, say one foot. It will then take Achilles some further period of time to run that distance, during which the tortoise will advance farther; and then another period of time to reach this third point, while the tortoise moves ahead. Thus, whenever Achilles reaches somewhere the tortoise has been, he still has farther to go. Therefore, Zeno says, swift Achilles can never overtake the tortoise. Thus, while common sense and common experience would hold that one runner can catch another, according to the above argument, he cannot; this is the paradox.
And of course figuring out identity is a fscking nightmare, damn metaphysics and damn Aristotle.
Every post I make begins with the assumption P=~P.
Given two points in 3-space, there is only one plane that is the perpendicular bisector of the line connecting the points.
The first prisoner should answer yes because they all passed through the room en route to their cells.
You are given an array of n x n integers. The goal is to end up with an array in which all entries are equal.
Four kinds of moves are allowed:
1. rotate a row
2. rotate a column
3. add 1 to all entries in a row
4. add 1 to all entries in a column
Show that the goal is achievable if and only if the sum of the numbers in the initial configuration is congruent to 0 mod n.
Yes, this one does have a solution. I got it from my logic professor, one of Quine's students.
You, the budding cultural anthropologist, have found yourself in yet another sticky situation. You have come across a village by accident during your travels, and the natives have forced you into a situation. Before you are two doors. Between the doors stands a man, and a sign written in your native language. The sign tells you that one door leads to a kingly feast, the other to certain death. You are allowed to ask one and only one yes-or-no question of the man standing there. However, the man either always lies or always tells the truth. Moreover, the man only speaks his own native language (but understands yours!): you know that 'yes' and 'no' are rendered as 'da' and 'na'...but you don't know which is which.
So. How do you get the kingly feast and avoid certain death? What one question do you ask?
Wow, you're a blockhead. For everyone else, just Google for it.
Here's a whole bunch. They're designed so that they must be solved by computer, but in order to do so you must understand the math behind them.
The only way to tell the difference between a hamster and a gerbil is that the hamster has more white meat.
An old riddle runs as follows. An explorer walks one mile due south, turns and walks one mile due east, turns again and walks one mile due north. He finds himself back where he started. He shoots a bear. What color is the bear? The time-honored answer is: "White," because the explorer must have started at the North Pole. But not long ago someone made the discovery that the North Pole is not the only starting point that satisfies the given conditions! Can you think of any other spot on the globe from which one could walk a mile south, a mile east, a mile north and find himself back at his original location?
There are 10 types of people in the world: those who know binary, and those who don't.
Oh, yes, the 'Laffer curve' is real to the extent that 0% of any number is 0, and 100% of 0 is also 0. However, arguments based on it presume that we are ABOVE the ideal tax level - a claim for which there is no supporting evidence. http://en.wikipedia.org/wiki/Laffer_curve
I don't know if this or a variation has been posted yet, so at the risk of being redundant, (hey, at least the information that it is my favorite puzzle would be new) here goes nothing:
Calling atheism and agnosticism a religion is like calling bald a hair color.
Yeah, I, for one, totally see your point. Everyone knows how much the wealthy elite feel punished at the mere sight of taxes. However they seem to have ways around the problem other than lowering productivity, such as creative accounting and political connections. It's the middle classes and below that governments know they can tax without complaint, with the occasional tax threshold increase that barely keeps up with inflation.
Here in Australia, Howard is a true genius at these games, convincing the masses that what is good for Australia is good for them and that tax cuts to the wealthy (and other less benign measures) are good for Australia. This offsets the Laffer curve with a large dose of unlikely optimism, a bubble that must eventually collapse as its premises are false.
Again:
"Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?"
This one makes you think... answer is simple!
x+1=x-1
Solve for x...
Sig (appended to the end of comments you post, 120 chars)... oops
Connect all nine dots using four straight lines drawn without lifting the pen from the paper:
. . .
. .
. .
. .
Stand all the coins on edge.
one of my favorite riddles:
the bridge: 4 men need to cross a bridge in the dark. The bridge can only take two persons at once, and to cross they need a light. The group only has one light, and after two persons crossed one of them need to return (if somebody rests on the other side) to bring the light back. The men are at different shapes, and it takes different times to cross the bridge for each of them.
1: 10min
2: 5min
3: 2min
4: 1min
If two men cross together, the faster one has to adjust its speed to the slower one: if #1 and #2 cross together it takes them 10min. Question: How long does it take them to cross the bridge. Bad example (as it goes faster):
1 + 4 -> 10min
4 5min
4 2min
--------------------
19min
It's possible to be faster than that (just by changing the order. A bruteforce algorithm would find the solution).
Solution (rot13):
guerr + sbhe
guerr
bar + gjb
sbhe
guerr + sbhe
friragrra zva
No, the validity of the Laffer curve itself does not depend on where we are. It is true, of course, that whether lowering taxes now will raise revenues, but first people like you have to stop scoffing at the novel idea that taxes can actually hurt the economy to the point of diminishing returns. Until that happens, debate about where we are on the curve can't proceed.
A little historical note: monarchs, being able to personally pocket the tax revenue, and thus historically had very strong incentives to maximize tax revenue, generally taxed about 8-15% of GDP according to most estimates. Just something to think about.
Rank my idea: http://www.sinceslicedbread.com/node/531
Wasn't this in the movie Die Hard?
-ted
"Here's a sample to consider: You're in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it's too dark to tell what you're moving or flipping (no, you can't figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?"
I turn the light on!
So where on the curve is your economy at the moment, below or above the local maximum? A point rather important to your theory that you seem to have left out.
this page has a graph which suggests that most countries lie below the maximum on the curve. Although to be fair, it is lacking in data.
By simply placing all coins on edge, you can be assured that both piles have exactly 0 heads.
You're actually making a better point than your toxic attitude would otherwise suggest. The rich can much more easily avoid taxes, legally or otherwise, than the average person, so raising their taxes really won't help revenue. (Also, shoving the tax load onto them removes any popular support for restraining government, but that's another issue.) If the rich see that one nation's attitude toward them has changed, they all just revise all future plans about investment. In the extreme case, since the rich, being rich, really don't need to work, if their taxes are too high, they just "consume" more leisure. You could, of course, just seize all their assets, but don't expect any more geese to be laying golden eggs where you live anytime soon.
I know you hate the rich and all, I'm just talking about the practical consequences of trying to milk more revenues out of them.
Plus, the premises of the argument behind the Laffer curve really are true. Again, you might not be past a Laffer point, but make no mistake there is a Laffer point. Denying this just makes wise people revise their estimation of the merit of your statements downward.
Rank my idea: http://www.sinceslicedbread.com/node/531
I amend my answer, and believe that it can be done with one point. The plane would contain the point, and be perpendicular to the vector pointing from the origin to that point. Good puzzle, and deceptive!
How quickly can you write out the full expansion for the following factorization:
(a-x)(b-x)(c-x)...(z-x)
Found at: notpron.com
/. crowd will love it. It has a ton of levels, but it is not in the finishing of notpron which is the payoff, but the journey. After playing it for awhile you'll grow addicted to the rush of seeing the new level pop up and being absolutely stumped as to how to finish it.
This is a series of riddles, with no payoff but the joy in moving on to the next level. Me and my friends (and some of my geek cousins) have found ourselves at times obsessed with this game. I've reached Level 33 and have been waiting for my friends to catch up ever since.
You'd be surprised how good these riddles can be. Ever need to edit pictures and look closer for clues? How about identifying musical notes? Having to decode Hex and ASCII values, image manipulation, even hidden JPG data? It's all inside.
Give it a try, I think the
Note: There are copycats to the original, including the recent frvade. I've played this one but definitely prefer notpron. YMMV. Good luck, riddlers.
My favorite math riddle: /joke
There are 10 types of people in the world. What are they?
I am scientifically inaccurate.
Sure it can.
But only if y = 0.
First we state that girls require time and money :-
:-
:-
Girls = Time x Money
And we know that time is money
Time = Money
Therefore
Girls = Money x Money
Girls = (Money)^2
And because 'money is the root of all evil'
Girls = (Evil)^2
Girls = the root of all Evil
Three people go to hotel. The hotel is $30 so they each pay $10. They're in their room and the manager remembers that actually the price is $25, not $30, so he gives the bellboy $5 and tells him to return it to the three people. The bellboy is lazy and doesn't feel like doing math in his head so he pockets $2 and gives a dollar each back to the three people. So now each of the three people paid $9.. $9*3=$27 + $2 that the bellboy pocketed = $29.
Where did the other dollar go?
It's stupid I know.
paid
I'm
worked
There are 5 friends Each has the number of dollars/other currency that corresponds to their friend number, i.e. ...
friend #1 has $1
friend #2 has $2
.
.
friend $5 has $5
So now, these blokes want to get into business, but they want to start
off on equal footing. Yet, being the farmer-descended Harvard grads
that they all are, they want to get to an 'equal footing' through a
business plan -- simply exchanging money amongst themselves would be a
gross misuse of their collective intellect.
One of these farm boys suggests that they get into wheat sales.
There's a weekly Farmer's Market that is organized every sunday in the
parking lot of the local super-mall, and there are plenty of
wheat-gorging locals in the neighbourhood. If they wait sufficiently
late into the evening to bring out their goods, they can name any
price since most of the other retailers should have run out of quality
wheat. Besides, he knows a place where they can get wheat @ $1 per 20
kilos, and they can make a pretty decent profit off of it.
The others like the idea, and just for kicks, they set up the following rules.
1. They will sell their wheat till it's all gone. Obviously.
2. They will all sell their wheat together, starting on the same weekend.
3. They must all finish selling off the last quantities of their wheat
on the same weekend, no sooner no later.
4. Individual amounts sold per sunday are irrelevant, but must be
non-zero. None shall abstain from selling on any weekend.
5. Each weekend they will decide on one selling rate and stick to it.
All quantities of wheat sold on that weekend must conform to the rate
they have decided upon.
6. On the last day, their collective earnings over all the weekends
must add up to be the same i.e. the original richest kid and the
original poorest kid (and all the rest) must end up with the same
amount of money in-hand.
On the first sunday, based on the money they had in-hand, each starts
out with the following amount of wheat...
friend #1 has 20 kg
friend #2 has 40 kg
friend #3 has 60 kg
friend #4 has 80 kg
friend #5 has 100 kg
In order to fulfil their objective, how many weekends will they take,
how much wheat will each of them sell on each weekend, and what rate
will they decide for each of the weekends that they sell their wares?
The post was a reply to spyinnzus's question (about the robot variant of the puzzle). If you found the original page using Google, you might enjoy following the links from there to some of my other puzzles.
Exactly solve the Schrodinger equation for Helium. Now _that's_ a riddle!!
Please ROT13 your answers. http://www.rot13.com/index.php
-ELiTe185
The whole concept of time travel to prior times is mind-bending, such as the "kill your father as a child" paradox. How can one exist if their father died before he was old enough to have fathered you?
Table-ized A.I.
The answer is always yes...
since they all had to enter the room to be locked in. they were all there, at a time before hand... i think.... i'm tired...bed time....brain hurt.
A bullet sounds the same in every language. So stick a fucking sock in it...
I remember one from fourth grade. It went something like...
You need 7 gallons of water, but you only have a 5 gallon and 3 gallon bucket, and a spigot from which to fill them.
Rules: You can not partially fill buckets from the spigot (eg. you cant fill the 3 gallon bucket up 2/3rds the way up).
The solution was having to pour from 5gal bucket into the 3, discard the 3, pour the 2 gallons left into the 3 gallon bucket and then fill the 5 and you have 7.
The Doormat
If you're not outraged, then you're not paying attention.
Just to make it more legible.
...
========
There are 5 friends
Each has the number of dollars/other currency that corresponds to their friend
number, i.e.
friend #1 has $1
friend #2 has $2
.
.
friend $5 has $5
So now, these blokes want to get into business, but they want to start
off on equal footing. Yet, being the farmer-descended harvard grads
that they all are, they want to get to an 'equal footing' through a
business plan -- simply exchanging money amongst themselves would be a
gross misuse of their collective intellect.
One of these farm boys suggests that they get into wheat sales.
There's a weekly Farmer's Market that is organized every sunday in the
parking lot of the local super-mall, and there are plenty of
wheat-gorging locals in the neighbourhood. If they wait sufficiently
late into the evening to bring out their goods, they can name any
price since most of the other retailers should have run out of quality
wheat. Besides, he knows a place where they can get wheat @ $1 per 20
kilos, and they can make a pretty decent profit off of it.
The others like the idea, and just for kicks, they set up the following rules.
1. They will sell their wheat till it's all gone. Obviously.
2. They will all sell their wheat together, starting on the same weekend.
3. They must all finish selling off the last quantities of their wheat
on the same weekend, no sooner no later.
4. Individual amounts sold per sunday are irrelevant, but must be
non-zero. None shall abstain from selling on any weekend.
5. Each weekend they will decide on one selling rate and stick to it.
All quantities of wheat sold on that weekend must conform to the rate
they have decided upon.
6. On the last day, their collective earnings over all the weekends
must add up to be the same i.e. the original richest kid and the
original poorest kid (and all the rest) must end up with the same
amount of money in-hand.
On the first sunday, based on the money they had in-hand, each starts
out with the following amount of wheat...
friend #1 has 20 kg
friend #2 has 40 kg
friend #3 has 60 kg
friend #4 has 80 kg
friend #5 has 100 kg
In order to fulfil their objective, how many weekends will they take,
how much wheat will each of them sell on each weekend, and what rate
will they decide for each of the weekends that they sell their wares?
A bunch of people standing outside a room in a random order wear each either a red or a green hat. They all don't like each other (i.e., they don't talk to each other or communicate in any way), and each person can't see what hat he himself wears. They managed to stand inside the room in a line, so that one side of the line is the people with the green hat, and the other is the people with the red hat: GGRRRRRGRRGGGGRGGGRR --> GGGGGGGGGGRRRRRRRRRR How did they do that?
http://en.wikipedia.org/wiki/Monte_Hall_Problem
New Orleans is an excellent example of how it works in practice. Dubya and his f[r]iends "saved" half a billion that was needed for flood prevention, and now the minimum estimates of the repairs is around a mere 200 billion dollars. How's that for a math puzzle?
Anyway, my larger point is that we're just fiddling while New Orleans sinks. Real scientists might be amused by mathematical puzzles, but America won't have many of those in a few more years.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
Bush doesn't initiate tax changes. They come from Congress and have to start in the House of Reps as set forth in Article I, section 7 of the US Constitution. You might wanna read Article II aswell, the limited powers of the presidency, U.S. Constitution.
The way these values are determined is by equating a function like f(taxation) to total GNP or GDP and graphing the results then finding the highest point on a curve where GNP+f(taxation) is a max. Historically large taxes tend to reduce the GDP and small taxes are insuffient for the government to perform its duties as required by the constitution.
The US currently collects about 3 trillion dollars in taxes every year.
How they do it and what their reasoning is is all freely available online
Maybe you can figure out where they're going wrong and run for office.
This must be my favourite personal quest - because it was difficult, because I found a solution, and because that solution beat the crap out of some research groups rather respected solution (in a very sepcific use case though, but still).
:)
The problem is to order rows and columns in a matrix so that a following QR factorization will be most efficient, computationally (the result of the factorization will not change with the ordering, but the amount of computational work changes as zeros and non-zeros are re-ordered by means of reordering rows or columns).
Tiny presentation of the optimization problem.
Page with pretty pictures of the original and re-ordered systems (a solution found by my optimizer).
And of course, you could just go ahead and read the whole deal - uh, did I mention the software is open source?
"The rich can much more easily avoid taxes, legally or otherwise, than the average person, so raising their taxes really won't help revenue." That's the worst argument for allowing the rich to be less taxed than the poor I have ever heard! The wealthy should be allowed out of taxes because the government is too lazy to make sure they aren't getting out of them regardless?? Your attitude leaves me incredulous! How about the government make an unpopular move and crack down on white collar crime? This would certainly increase revenue by a large amount as white collar crime is high among the top drains on the government budget. The government certainly does not seem worried about the effects of a crack down on welfare cheats, so why not the wealthy variant? "I know you hate the rich and all, I'm just talking about the practical consequences of trying to milk more revenues out of them." One wonders if it is in fact you who is dismissive of the poor, your attitude toward them in your posts here certainly is.
Some pictures of a real laffer curve.
obviously one step is not valid
--meh--
The Laffer Curve is a laugh and has been refuted
REALLY? Fundamental mathematical theorems have been overturned? You mean, it's possible for a continuous function to start at zero, increase, and then reach zero again without ever decreasing? Looks like I need to start hittin' the books again! Got a link to where this is established?
Rank my idea: http://www.sinceslicedbread.com/node/531
Here goes again with plain text formatting...
"The rich can much more easily avoid taxes, legally or otherwise, than the average person, so raising their taxes really won't help revenue."
That's the worst argument for allowing the rich to be less taxed than the poor I have ever heard! The wealthy should be allowed out of taxes because the government is too lazy to make sure they aren't getting out of them regardless?? Your attitude leaves me incredulous!
How about the government make an unpopular move and crack down on white collar crime? This would certainly increase revenue by a large amount as white collar crime is high among the top drains on the government budget. The government certainly does not seem worried about the effects of a crack down on welfare cheats, so why not the wealthy variant?
"I know you hate the rich and all, I'm just talking about the practical consequences of trying to milk more revenues out of them."
One wonders if it is in fact you who is dismissive of the poor, your attitude toward them in your posts here certainly is.
Two.
The function takes two points, a and b. Return the plane which contains a and which has the line between a and b normal to it.
Moving both a and b allows the plane to be located anywhere. Moving b while maintaining a allows the plane to be rotated in any way.
High-speed Road Trip (18.000KPH)
The coins are on fire?
This is not a turnip.
There is only one possible solution, with two potential end results:
Pick a random prisoner who has chosen "Yes."
End result:
1: The prisoner is correct and all prisoners will be let free.
2: The prisoner is incorrect but hits the king over the head with the chalice and frees all the prisoners.
Raising taxes often boosts the economy. And until you folks get over the idea that lower taxes stimulates the long term health of the economy this will never happen.
"initiate" is such a tricky word. for instance if he were to run on a platform of tax reduction and order his bitches in the house to write a bill that lowered taxes, would that not be initiating them?
Although the moon is smaller than the earth, it is farther away.
Start with a 80, 80, 20 isosceles triangle. Starting from each large angle draw a line connecting with the opposite side. One line will make a 60 degree angle with the small side and the other will make a 70 degree angle with the small side. Now draw a line from each intersection of those lines with their sides to each other. The problem is simple. Find all the angles. It looks deceptively simple but it's really difficult.
" The answer is always yes..."
You're the 3rd person to suggest that. You might want to try reading the problem again.
How about we decentralize the leader. In other words, the King *has* to call on everyone an arbitrary number of times. Who cares who the leader is? Make everyone the leader. You always flip the same way, that way being the way you flipped the chalice the first time you went into the room. Since k n*k, as long as someone has flipped it n*k +1 times, he knows that every other prisoner has gone in (mind's a bit mushy, so that math may be off). The only problem left then is that the King might decide to call everyone in once, in which case they'll never go free....but its closer...
Yes, it is undefined and you don't need a calculator. Actually, you shouldn't. But the bottom line is that it approaches 1, in the limit: lim(x->0) x^0 = 1, and for x^x this holds as well; you can try both on a calculator with very small values for x, if you want. In some "practical" problems, such as fitting a polynomial on a set of points, you really need 0^0 to be 1.
However, if you're going to try the other limit, lim(x->0) 0^x, the answer would seem to be 0 (but since this function is not continuous, there is no need to make it equal 0).
The theoretical answer? it depends on the function you're looking at.
The pratical answer? Try 1.
There's nothing clever at all about that, it's just your basic 5th-8th grade, Iowa Proficiency Exam (they still have those?) question. Riddles should be clever. Your "riddle" was as clever as: what's two plus two?
Although the moon is smaller than the earth, it is farther away.
Three salesmen are late to town for a conference. Eventually late that night they find a hotel with one room left. (It's a room for three). The bell boy says "The room is $30, please". The salsemen are delighted with such a reasonable rate and each fork over $10.
When the bell boy gives the cash to the night manager, the manager says "No, no, no no. This is not right - The room is only $25 dollars - not $30. Here are five $1 notes. Please give them back to the guests".
On the way back back, the bell boy thinks, "I have five dollars and three guests. I can't divide this evenly. So I'll just keep two dollars for myself."
The salesmen take their cash and turn in for the night.
So.......
Each Saleman has effectively paid nine dollars (ten dollars minus one returned).
The bell boy has two dollars in his back pocket.
$9 + $9 + $9 + $2 = $29.
Where has the last dollar gone???
Suck on that, my fellow brains-on-stilts.
Heh heh heh.
One wonders if it is in fact you who is dismissive of the poor, your attitude toward them in your posts here certainly is.
...RIGHT?
*sigh*
I'm trying to explain to you the practical consequences of what you're proposing. I was just describing aspects of reality. You'll notice I never said anything was good or bad, just what will or won't happen. If you can find the place where I was dismissive of the poor, I'd really like to know where it is.
Now, I already explained enough so you could understand the difference between cracking down on the rich vs. the poor. The rich can easily scurry away and/or stop producing. The poor can neither easily scurry away nor stop producing. Again, this is not to say anything is "good" or "bad", just that it "will" or "won't" raise tax revenues. Contrary to your staunch refusal to dispassionately analyze the topic, there really are relevant practical considerations in raising taxes.
In fact, I'd like nothing more than to test out your ideas. Check out the link in my sig. I submitted an idea to a policy site. The idea is that basically, in one state, we do what you propose: high taxes on the rich, high minimum wage, good workers protections and workplace safety requirements, etc. In the other state, do the oppose: no min. wage, low taxes on the rich, no safety requirements, etc. If you're really serious about your views, you'd leap at the chance to do this and see who's right based upon which state people flock to.
You do think you're right, right?
Rank my idea: http://www.sinceslicedbread.com/node/531
Isn't that assuming that all but the 18 are tails? I don't read the problem as specifying that. It say that 18 are heads, not that only 18 are heads.
"You're in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it's too dark to tell what you're moving or flipping (no, you can't figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?"
By switching on the friggin light. Sheesh.
Seriously though: If you remove the cause of a problem, you solve it *for real* and needn't spend your life patching it.
Visit http://ringbreak.dnd.utwente.nl/~mrjb/growingbettersoftware to download your free copy of the book
Tonight's homework assignment: a one page essay on what the words "leadership" and "bully pulpit" mean. Extra credit for describing how government in theory differs from government in practice, or why everyone refers to the recent rounds of tax cuts as "the Bush tax cuts".
I don't care if it's 90,000 hectares. That lake was not my doing.
I always liked this one: parallel train tracks appear to converge in perspective. what angle must they be laid at to appear to be parallel in perspective?
must... stay... awake...
My simplest proof
Starting at the north pole certainly works.
There is a circle some amount north of the south pole, concentric to the axis of the earth's rotation, with a circumference of 1 mile. Starting at any point 1 mile north of any point on this ring will work too. Note that this set of points is itself a circle concentric to the axis of the earth's rotation.
Similarly, for any positive integer n, there is a ring some amount north of the south pole with a circumference of 1/n. Starting at any point 1 mile north of any point on this circle will work too. Similarly, this set of points is itself a circle concentric to the axis of the earth's rotation.
None of these rings overlap with any other, or with the north pole. Any other point near enough to the south pole will send you around the axis of the world's rotation a non-integer number of times, and you won't end up where you started.
And of course, "going a mile south" when you start at or within a mile of the south pole is an undefined operation, so presumably we exclude those points too.
You will know the anmswer is correct when you find it! Note "1" is _NOT_ a prime
Take this statement: "All rules have an exception".
If this statement is true, then this rule itself must have an exception, in which case the statement is not true. However, if the rule itself is the only one without an exception, then the statement is making itself true by not having an exception.
What is a paradox? A simple demonstration that human logic is flawed.
lucm, indeed.
The only person I ever knew who could solve the riddle was an accountant, everyone else (engineers, computer scientists, mathematicians, etc.) get's terribly flummoxed.
just a ghost in the machine.
One more variant, also due to Gardner or one of his readers: Suppose that you speak the local language (shared by truth-tellers and liars) perfectly, except you have forgotten if "pish" means yes and "tush" means no, or vice versa, and that your question must be in a form that requires a yes/no (well, pish/tush) answer.
"If path A is a path to your village, say 'pish'; otherwise, say 'tush'."
No questions required.
A Psychotic Mathematician (The Professor) has kidnapped 20 of his students.
They are being held hostage, and the Professor has announced to them that he will provide a thought puzzle to determine whether he will let individuals go free.
The students will be placed in a single room in single file and must always face forward. Students can clearly see all students in front of them.
Every student will be given either a red or blue hat, but no one knows the color hat on their own head.
To be let free, a student must announce the color of the hat on their head.
All students can announce only one color, and all other students can clearly hear the announcement. Announcing anything but red or blue will result in all students being brutally subjected to absurd math problems indefinitly.
They are allowed to discuss how to get out of the situation before being placed in line, but the Professor, after going through years of intense mathematic study, will pessimize (force the situation into the worst possible outcome) the situation for a given solution the students come up with.
How many students can be guarenteed freedom in the optimal solution?
What is the optimal solution?
What principle in the real world does it reflect?
I have tried to be very explicit in this problem. If someone notices an issue or has a question, please mention it.
I wouldn't consider the mad hatter mad. Just reality impaired. He sure can make a mean cup of tea.
This sounds very much like a similar problem I solved in college, involving a light switch instead of the chalice and omitting the king's ability to manipulate the communications tool of the prisoners. I have the following questions:
:)
1. Do the prisoners know k?
2. Define "arbitrary number of times."
These are my sticking points. Based on my questions, I'm sure you can guess the strategy I'm going after from the problem I solved before. My other question is this: Am I on the right track by iterating that strategy k+1 times, or is it completely different?
you've never heard of integer overflow? ;)
I think I get the reasoning, but from the given data I guess it could be either 1231 or 111221, depending on the specific logic used.
just try proving something like the binomial theorem. Or trying to understand what -2! means, and why. Or even convincing yourself that when you add 32 and 549 by the common method you get the same answer you would if you put 32 strokes next to 549 strokes and counted them all up together. Of course there are people who understand these things, but for every one of them there are a hundred more who have simply accepted them. Mathematics is a wonderland right in front of our faces.
And who would be foolish enough to suggest that the economy is described by a continuous function?
The chalice cannot store any information because they can't know if the king manipulated it. The king knows their rule, and they can't know if he is interfering or not.
The fact that, as you claim, only two people "know the answer" shows that not a whole hell of a lot of eyeballs have looked at this problem. It's easy to be confident of the solution when only two of your friends have checked the answer.
I look forward to debunking your solution.
Rank my idea: http://www.sinceslicedbread.com/node/531
One problem, what if there are 3 prisoners and lets just say K is 2...
What if the king knowing this picks 2 of the prisoners and alternates them (and makes sure one is an upper, and one is a downer...).
he then calls just those two prisoners alternating, 7 times each....
I think this is on the right track though. You just also need to factor in taht the king will call every one the same number of times and can't stop until someone says 'yes'.
I'm not interested in arguing with fools like you. Or maybe you're just incredibly naive and ignorant and you are optimistic that I'll take the time to educate you on the reality? Whatever. Please mark me as a foe and we shall eagerly ignore each other forever.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
This looks like it is on the right track. Looking at the ways a king can mess with the signal:
1. trick the leader into thinking that someone signalled to them but turning cup back up and calling the leader out. To fix this may need to wait for at least n+k affirmations. This by itself falls down though because the king may choose not to signal.
2. trick a non leader into thinking they have successfully signalled but the king invalidated their signal. To fix this each non leader can signal k+1 times.
Bringing this together. Since everyone is going to signal upto k+1 times the leader should wait for n*(k+1) cups. Assuming the king doesn't mess with signals
Now if the king hides signals from the leader - leader may received k signals less. So they should wait for just n*(k+1)-k. This is less than signals from all but one user (n-1)*(k+1). Algebra works out that if n>1 this holds. If n=1 then there is just one prisoner and they say 'yes' on being called out first.
Now if the king hides tries to block all signals from one non leader - leader may receive k signals less again. Same logic as above
So i think the solution is all non leaders signal upto K+1 times, and the leader waits for n*(k+1)-k signals.
It doesn't say the "economy is described by a continous function". I don't even know what that means, and I'd be surprised if you did.
Rank my idea: http://www.sinceslicedbread.com/node/531
Interesting point but the president can't order congress to do anything. He has hardly any power.
All bills for raising revenue shall originate in the House of Representatives(Article I, sec 7.)
The house often has a substantial number of reps who oppose the president, But if the house passes the bill it goes to the Senate. And anywhere in the process the bill can be filibusted, killed in judicial committees, etc. by either house. If the Senate passes the bill it goes to the president for signature. Sure he signs it into law but I posted the above because I get the impression many people think the president raises or cuts taxes autonomously and that's not the case. Congress has all the power.
If you're interested, oyez is a great website that has many recordings of supreme court decesions. I like to get the names of the plaintiffs and defendants and try to guess who will win. The outcomes and the reasoning behind them are sometimes quite informative.
You claim that the Laffer curve is a good economic model. You also claim that the Laffer curve is a continuous function.
Hence you claim "economy is described by a continous function".
It's called logic.
Here's a small one:
Each character represents a different digit (the dots are just to preserve the formatting). No two characters represent the same digit. Happy solving :-)
He knows everyone has been in there when it gets to (n-1)(k+1) or he will be waiting for him to flip it as well.
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
The king can only flip the chalice k times, so you just need a solution that can be messed with up to k times. So if you have one guy flipping it rightside up each time he sees it upside down and all the other guys flipping it down the first k+1 times they find it up, the first guy can just count the number of times he has to flip it. Once he gets to (n-1)(k+1), I believe he knows everyone else has been in there.
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
Switch. (I've seen this question before; the first time I saw it, I think I was one of those people who messed up.)
The BART is the SF Bay area's excuse for a subway / mass transit. To ride, you buy a ticket at a kiosk full of $x worth of "BARTness". When you get board the train, you stick your ticket into a turnstile, and it hands it back to you. When you reach your desgination and get off the train, you again stick your ticket into a turnstile. It deducts the cost of your trip (based on how far you traveled) and gives your ticket back. You keep using the ticket until $x is used up. BUT: Suppose you walk toward the train, put your ticket in the entrance turnstile. You pick up your ticket, then you change your mind and leave, putting your ticket in the exit turnstile to get out. The cosmic BART megamachine will charge you the maximum possible fare, even though you haven't gone anywhere. For a good reason. What's the reason?
My favorite these days is very simple (good for kids), and actually quite practical:
By what procedure can two people split a single piece of cake (or slice of pizza, or bowl of soup, etc.) such that both agree that the outcome is fair? In other words, how do you avoid the "No fair! His piece is bigger than mine!" problem?
The most rabid believers in American Exceptionalism are the exact same people whose policies are destroying it.
I just want to note that, at least in U.S./Canada, there is no tax on being rich. The taxation occurs on the first derivative of wealth, assuming it is positive. That being said, using "the rich" in an argument doesn't make any sense, as your tax rate is determined solely on your instantaneous increase in wealth.
The Laffer Curve thing is not a fundamental mathematical theorem. It is a cualitative stamenent of an economical theory, from which cualitative consequences can be drawn, and, if hard data is provided, maybe cuantivative ones too (but that is much, much harder). Those consequences are as valid---assuming the derivation of those consequences is sound, which is a very big if,---as the hypothesis that supports the original economical theory, which of course cannot be of a mathematical nature.
The truth (*) of an economic theory does not follow from the soundness of mathematical arguments, but from the validity of its hypotheses and the concordance of its predictions with reality. I'll leave it as an exercice to check whether these two are in favor of Laffer's theory.
The fact that the theory can be stated mathematically does not imply its truth, as there is plenty of examples of false but mathematicallt expressed theories around, in economy and in other subjects.
(*) Of course, this usage of the workd "truth" is both vague and incorrect. Yet it is clear that with big enough doses of philosophy of science this statement can be made correct.
My favorite math riddle?
The Condom Problem
Can't analyze it graphically. Making a function would be much too complex (involving understanding every person indifference curves according to their work:money-earned [after tax] ratio) and estimating on empirical evidence would require changing the tax rate a bunch of times and seeing how it effects revenue in both the public and private sector. Then you'd need to make sure changing the tax rate did not change the structure of the economy so that the data gained from that tax rate is still valid.
Had to do it.
"Sure there's porn and piracy on the Web but there's probably a downside too."
Imagine you have got 12 marbles. They all have exactly the same weight, except one. Using a balance, try to find this marble with only 3 weightings and also determine if it is heavier or lighter!
:-)
Easy as it looks, it took me two days to find out the solution to this problem. And don't give up, it is possible without any tricks
Open Source Alternatives
Allow me the ego to be freudian with you - LnxAddct.
;-) :-) Go linux! :-) :-) :-) ;-)
* You write you name in a terse, leet form, vowels being for the swaety prole masses no doubt
* You grew up on bash scripting, weened onto perl, and now suck python out of the can while looking for chicks.
* You post a python script for solving methematical puzzles
* You post a python script for solving methematical puzzles, and worry about its bug freeness
* You post a python script for solving methematical puzzles, and worry you might have not done it pythony enough.
* You post a python script for solving methematical puzzles, encoded to get past lameness filter (instead of using code tags? ca you do that??)
* You post a uuencoded, compressed file of a python script for solving methematical puzzles, and instructions on how to get it working
* You post a uuencoded, compressed file of a python script for solving methematical puzzles, and instructions on how to get it working instead of posting the file to any hosting account, and posting a coral cache link to it.
#hostfile 0.0.0.0 primidi.com 0.0.0.0 www.primidi.com 0.0.0.0 radio.weblogs.com
Dubya and his f[r]iends "saved" half a billion that was needed for flood prevention, and now the minimum estimates of the repairs is around a mere 200 billion dollars. How's that for a math puzzle?
Right, it's all Dubya's and the Republican's fault. Forget the fact that the local (N.O.) and state government (LA) have been controlled by the Democratic party for a long time. Forget the fact that spending destined for the levees has INCREASED in the years since the Republicans took office. Forget the fact that this money was misused by the local government. Forget the fact that the exact outcome of the levees breaking and the flooding during a storm similar to Katrina has been forecast for DECADES and N.O. didn't do shit to prepare for it.
All I said was that all bills for raising revenue must originate in the House of Reps as set forth in Article I, sec, 7 of the USC. Is that not the case? If that's not the case, enlighten me. It's more productive to impart knowledge than to accuse one who lacks it of being a fool.
Maybe I'm completely wrong. Help me out.
He just showed you how all five of the possible answers are correct given only the information you supplied.
How is it that you have any solution more "real" than the solutions he just provided? Could it be that you haven't really thought through your riddle so well?
Fill the 3 gallon and empty into the 5 gallon, do this once more to fill the 5 gallon. Hold your finger at the water line on the 3 gallon jug, which should now be a third full. Empty the 3 gallon and fill the 5 gallon, and then empty the 5 gallon into the 3 gallon until the water level reaches your 1 gallon mark.
In these days, bleeps and bloops mean something more
So sorry, I read this statement:
"The rich can much more easily avoid taxes, legally or otherwise, than the average person, so raising their taxes really won't help revenue."
And I took it as twisted apologia for lowering taxes on the rich.
I understand now that it was not. Perhaps it was your assumptions that I hate the rich and that I denied the Laffer curves existence, both assumptions without merit and agressively misleading, that led me to believe you were in opposition to my arguments and a far right supporter.
You will find my arguments to be based on the Laffer curve, but introducing the idea that deception may increase the maximum point on the graph, much like the price of shares on a stockmarket may be increased through similar means, but creating a bubble that must eventually burst.
This is the point that you refuse to accept, seeing it as incompatible with your precious curve and therefore a denial of its existence on my behalf.
And on the policy thing, you think there could be a socialist state in the USA? Watch out, the FBI are knocking down your door, the red peril must be stopped.
Well, It was certainly interesting to see some of the questions/ riddles and problems posted here. And it's been a while since we posted comments to an article that was certainly meant for us nerds. But most of these problems can be solved with math formulas or other processes mostly because the input allow you to furmulate an output result - no matter how tricky.
But the one logic (?) riddle that will puzzle man probably for all eternity is this: Understanding the behaviour and thinking processes of a woman. We can argue about it for years and some of us think we have an answer and some are close to understanding them, but no man will EVER be able to fully understand a woman.
That my friends, is the ultimate riddle in the universe: Understanding those weird people with vaginas.
"I used to have that really cool,funny sig
***Please read this post!!!***
/. editors, can you consider putting that part in bold in the main story? Because a lot of people are missing it. A lot.
"Just one request: If you have figured out the solution, link to it in a post, rather than write it out where anyone can see it."
***Read the Above!!!***
I've noticed a lot of people missing this point, so I'm trying to make this post stand out as much as possible. I know you're not supposed to, but this is a special case.
"When the atomic bomb goes off there's devastation...but when the atomic bong goes off there's celebraaaaation!"
Probably old, but sometimes true (at least since I got married).
(Ladies, please don't bother to flame. If I had been smart, and married myself a geek-chic, I would probably be able to disprove the following statements)
----
We know that women like two things, your time and your money,
therefore...
women = time x money
Now time equals money, so
women = money x money
Money is the root of all evil, so
money = square root of evil
By squaring both sides of the equation,
money squared = evil
Now go back to the equation
women = money x money
See where I'm headed?
women = money squared = evil
A problem with this "proof" occurred to me immediately: Since evil is
negative, the square root of evil must be imaginary, which would mean
that money is imaginary, and therefore, by definition, so is time.
Then I realized that, in my life at least, that's pretty much true.
Because it gives some insight into the role of sexual reproduction in evolution.
A long time ago, there was a kingdom of 10,000 people. Half were male and half were female. The king decreed that males were intrinsically superior, and he wished to increase the ratio of males to females in his kingdom without shedding blood. To accomplish this, he dictated that each couple must stop having babies after giving birth to a girl. As long as they had *only* boys, though, they could continue to bear children (but didn't have to).
In other words: Any couple that gave birth to a girl was forbidden to have more children.
This policy was enforced for 1000 years. At the end of this time, what was the (approximate) ratio of males to females in the kingdom?
Assume that the base ratio of giving birth to a boy vs. a girl is 1:1. That is, each outcome is equally likely.
The most rabid believers in American Exceptionalism are the exact same people whose policies are destroying it.
First puzzle: Given a rectangle 2 units high and 200 units long, it's trivial to pack 400 unit-diameter circles into it, in a rectangular arrangement. The question is, how can you fit the 401'st circle? (There is a way, but it's non-obvious; this is not a trick question!)
Second puzzle (given to me by my discrete mathematics professor to solve over spring break, which I did): Sam and Polly are logicians. A Martian comes down in a spacecraft and announces, "I'm thinking of two integers x and y, where 3 <= x <= y <= 97. I'll tell their sum to Sam, and their product to Polly." He does so, and leaves. The following conversation ensues:
Sam: "You can't possibly know what x and y are."
Polly: "I do now."
Sam: "Now I do, too."
Find x and y.
Weeks of coding saves hours of planning.
That's where the "arbitrary" number of times comes in. The prisoners can wait until they have all been in at least f(k,n) times, and the king will eventually comply. So what's f(k,n)?
Everyone is entitled to his own opinions, but not his own facts.
A man can sire children by multiple wives, and vice versa, as long as any given couple stops after giving birth to a girl.
The most rabid believers in American Exceptionalism are the exact same people whose policies are destroying it.
Wrong try again
Answer
Each non leader must signal 2k+1 times
Leader must wait for N*(2k+1)-k signals
WhereN = n-1 = number of prisoners less 1. (Don't count the leader)
k = number of signals king can mess with
logic:
Total number of signals that can be received by leader including kings false signals where all but one non leader prisoner is allowed out. i.e. the king does his best to trick the reader into thinking that all prisoners have been let out but they haven't:
LHS = (N-1)*(2k+1)+k
Total number of signals that can be received by leader where all non leaders signal less max number of signals a king can muffle. i.e. the king does his best to make the leader wait forever in indecision:
RHS = N*(2k+1)-k
Invariant that needs to hold is:
LHS < RHS
Although the submitter asked you not to reveal the answer in the thread.
The most rabid believers in American Exceptionalism are the exact same people whose policies are destroying it.
My favorite "riddles" aren't exactly riddles, but games requiring a certain amount of logical problem solving nonetheless.
Some example games are:
(1)The words I, me, my, and mine are forbidden. Using any of these words will result in a demerit, penalisation, loss, or similar. Creativity counts. Creative cheating counts extra.
(2)Think "What did I just think?" As soon as you forget what you just thought, you lose.
(3)Think "What did I just think?" As soon as you remember what you just thought, you lose.
And so on. In (1), some fundamental words are removed from the list of words available in speech, forcing the player to keep constantly on the toes while reforging verbal associations. In (2), a winner will have some characteristics of a befuddled butterfly, while a lose will be unaware of their loss. In (3), a winner may set "What did I just think?" spinning, place it in the corner of the mind, and check up on it every now and again. When such games are participated in by multiple parties great parts of the fun and attraction lie in the inevitable arguments about who's won, who's lost, who's the most points, and "why don't we add this rule," or "doesn't that rule we added five minutes ago have this implication?"
exegene refugee memories in hiding
|| /\ || /\ /\ /\
() || () () ()
fixed art fixed art fixed art fixed art
fixed art fixed art fixed art fixed art
da da dad a da d a jf kjrtfi34rtnfijrvnjrv
-schwal "Hanging is too good for punners, they should be drawn and quoted"
Last week my friend proved with a few steps that 1 = -1.
To my surprise I could not find any error in his proof..
For now,can any of you prove that.
Why does yahoo do this
This is the shorted puzzle I've ever created.
It is linguistic, mathematic and a touch cryptic.
Just fill in the missing digit:
1_
EMail: 0110001101100010010000000110001101110010 0110000101111010011011100110000101110010 0010111001100011011011110110
This is one of my favorite teasers-- not especially hard but it might make you think for a bit...
Suppose someone is paying you with six sacks of gold. Each sack contains 100 coins and you have examined all the coins and verified the count in each sack. However, you have it on excellent authority that one whole bag of gold is counterfeit! Further, you have been tipped-off that the counterfit coins weigh exactly 1.1 ounces while the real gold coins weigh only 1.0 ounces. When you heft the bags they feel about the same so you're not really sure which one is the fake. You do have a scale, but it will only work ONE time. You can put ONE set of items on it and get ONE reading. You can't trick it by putting stuff on and off while its reading, so choose your items carefully! Assume for the problem that the tare weight of the bags is inconsequential. Using only these items in the manner described, is it possible to determine which bag has the counterfeit gold, and if yes, which one is it-- how would you know?
I could buy two tickets from the kiosk. One, I always use to get on/off at the stop nearest my house, the other, I always use to get on/off at the stop nearest my office. The exploit would be that, since both cards will always show zero distance moved, if BART didn't charge me anything for zero distance moved I would ride for free. By charging full price for zero distance moved, they prevent this exploit (and in turn exploit people who actually travel zero distance like in the problem statement).
One from my youth...
"What is the beginning of eternity,
the end of time and space,
the beginning of every end,
and the end of the human race?"
Kyle King
IT Consultant
kyleking.com...Obedience training...for your PC!
I love these puzzles:
http://www.logicmazes.com/theseus.html
Just referencing the Buffon's noodle article.
Buffon's needle is a way to calculate pi: you throw a needle of length l on a grid with lines spaced D>=l apart. The probability of the needle crossing the grid is related to pi, and Buffon used integral calculus for deriving it.
"Buffon's noodle" is a generalization where you can throw an arbitrary geometric line shape of length l on such a grid, and derive the expected number of grid crossings without even knowing the shape. This leads to a solution of Buffon's original problem which works without integral calculus.
There was an episode of Dr Who with this problem, it was a Baker episode, it had something to do with Mars and pyramids or something.
Ask one what the other would say, and choose the opposite door. The truthful will tell the liars answer, which is the wrong door, and the liar will lie about the truthful answer.
I knew knowing that would pay off some day!!! See, plot and story over special effects anytime.
--
Reload to see this sig
Ok, I came up with this brain tickler, or whatever you call it. It's a little bit different than the others, because it's more about the process of answering than it is about the actual answer. And I'm going to blatantly break the rule about "no answers in the post"... because I'm lazy. So don't read the whole way if you don't want to find out.
...
OK, here goes: If you're launching a satellite into space on a rocket, what direction do you launch it in? The next paragraph is a hint.
The first part of the problem is figuring out what kind of answer is expected, but whoever is asking can help out the askee at this point by specifying that, in addition to "up", a compass direction is expected (North, South, East, West).
Don't read beyond this point if you don't want the answer: if you just think about it, you can figure it out.
OK, the first partial credit is assigned if people realize that the Earth's rotation can give a boost to the rocket (that is, if you launch in the direction in which the Earth is already rotating, you don't have to fight the rotational speed). The second bit of partial credit is assigned if the person says, thinking alound, "Okay... well, the Sun rises in the East..." (Surprisingly, most people do say this). Then, finally, full credit is given if they correctly deduce from this that the Earth is rotating towards the east, and a rocket should then be launched up and to the east (and at the equator, to get the biggest boost).
Incidentally, this also explains the location of Cape Canaveral and other such establishments: it's near the equator, and there's a large body of water to the east (so the parts that fall to the ground - hopefully not everything! - land in water).
Does anyone else find this a fun puzzle?
At least, it's unsolvable if the king has any memory at all and doesn't simply interfere at random k times.
He can call someone the minimum number of times all in a row, and revert the chalice to the state it was in before he called them. There is no evidence for anyone that he was ever called, nor can there be. It's CERTAINLY true if k > n, but it's probably true for smaller numbers of k all the way down to at least 2 and probably 1. With a single flip and given a set minimum amount of calls, it's possible for the king to completely eliminate evidence that one of the prisoners got to go at all.
...and you chop down a tree with it. But when you're doing it, the handle of the ax breaks.
So you replace the handle. A week later, you're using it to chop up some firewood. This time you strike a stone and the head of the ax breaks. You buy a new head.
A neighbor comes to borrow the ax. He says, "so, is this the same ax you chopped the tree down with?"
Well? Is it?
Here's a longer version of the riddle. MUCH longer.
Not really a riddle but showing that 0.9999... is equal to one seems to bug a lot of people.
x = 0.9999....
10x = 9.9999...
9x = 9
x = 1
0.9999... = 1
My favorite puzzles are what you might call meta-puzzles. That is, sometimes a puzzle will have properties that you can analyze in a non-brute force way.
For instance, consider a board with the following shape (1 represents usable squares, and 0 unusable squares):
0 0 1 1 1 0 0
0 1 1 1 1 1 0
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
0 1 1 1 1 1 0
0 0 1 1 1 0 0
in a puzzle similar to HI-Q you start off with pieces on all of the "1" square except for one, and then you jump piece over each other, taking the jumped piece and try to end up with a single piece on the board.
Unlike HIQ, this puzzle isn't solvable when you start with the hole in the center of the board.
The first meta puzzle is to prove this.
The second meta-puzzle is to discover a general rule that lets you know what starting holes can lead to a solution, and what squares the final piece can end up on.
There is such a rule, and it doesn't require brute force to find it.
Once you understand the mathematics of this game, it's trivial to prove that the HI-Q board (which looks like this):
0 0 0 1 1 1 0 0 0
0 0 0 1 1 1 0 0 0
0 0 0 1 1 1 0 0 0
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
0 0 0 1 1 1 0 0 0
0 0 0 1 1 1 0 0 0
0 0 0 1 1 1 0 0 0
is ALWAYS solvable, no matter which piece is the first hole.
I'll give you one hint. The trick is to find a game with rules that are much easier to analyze than "jump over a piece and take it" but where "jump over a piece and take it" is a legal move. If you can prove that there's no solution in this game, then there can't be a solution in the more restricted game.
Another hint that may be more malicious than helpful is to consider an infinite board - if you can simplify that, then understanding a finite board is easier. Of course if you can do that, then you already know most of the answer.
find all combinations to obtain all integers from 0 to 100. The 6 operations being +, -, /, *, ^ and !. You can merge the 4s, too (ie, 44 is possible).
:p
First stumbled upon this math game at a school in Exeter while doing a language school during the summer at the age of 13. Back home I tried it out and succeeded, but it took me a whole month!
For example, 0 = 44-44, 1 = 44/44, 2 = 4/4 + 4/4, etc etc.
Very fun
From a Marilyn Vos Savant column. Google for explanation:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
I love puzzles, and most are just a bit of a push to work out, but I've had a ton of people argue to no end with me on this one.
And it should be called Petals Around the Roses, as there may be more than one.
let me first reword the problem to the way I think of it. Let us think of all the prisoner having "tokens" that they are passing around via the cup. Flipping the cup up means you leave one of your tokens in the center room. Flipping it down means you take the token. Obviously only one token can be left in the center room at a time.
m l#100prisonersLightBulb if you are interested in this type of problem and want to figure out the "fastest" solution. That whole forum is full of really hard math puzzles.
Now the problem just becomes one of having the leader collect enough tokens. The cup can start up or down, so there may be an extra token in play. Allowing the king to flip the cup out of sight of the prisoners allows him to add or remove a token from the system with each flip. Thus if the prisoners start with a total of x tokens among all of them, over the course of the game that total may change to between x - k and x + k + 1 (+ 1 there because the cup might start sitting upright).
Now, the solution is that all prisoners except the leader start with 2k + 2 tokens (so the prisoners start with a total of (n - 1)(2k + 2) tokens. The leader will say yes once he has collected (n - 1)(2k + 2) - k tokens. We need to show that the leader can always collect this many tokens, and that he cannot get this many tokens unless ever prisoner has left at least one token in the center room.
We can see that the leader can always collect this many tokens because the king can take at most k tokens out of play leaving only (n - 1)(2k + 2) - k tokens. Since all players will be called an arbitrary number of times, they will all get a chance to leave all their tokens in the center room and the leader will get a chance to pick them all up (the ones that the king doesn't take).
Now consider the case where one prisoner is not called out for a very long time and all other tokens are allowed to be transferred to the leader first. In addition we will assume that the cup starts up and that the king adds k tokens to the system. In this case, there are (n - 2)(2k + 2) + k + 1 = (n - 1)(2k + 2) - 2k - 2 + k + 1 = (n - 1)(2k + 2) - k - 1 tokens in play. This is one less that the required number of tokens for the leader to say yes, so he can't possibly say yes until that last prisoner comes out and gets to transfer one of his tokens to the leader.
That is a bit hand wavy, but I think it is a good proof. I have no idea if this is the best possible solution, though. I have good reason to believe that it is not the fastest (if the king chooses who to bring out arbitrarily). Check out http://www.ocf.berkeley.edu/~wwu/riddles/hard.sht
-- corygwilliams AT google's mail place
Similar to your clock puzzle: Imagine having a clock where the hour hand and the minute hand are identical, and there is no way you can see which is which. But you can measure their angles exactly. How many times in 12 hours will it be impossible to see what time it is?
We got this puzzle in a "creative maths" class I took once. Took me surprisingly long to solve it...
You're at a T intersection trying to get home. You don't know which one to take, but there are two men standing there who know. One always tell a lie and one always tell a truth.
You can ask only one question (same question to each man) to find out how to get home.
What question do you ask?
eTrade SUCKS
Problem:
Although I found this puzzle very difficult (I may not be very clever), I did eventually solve it on my own.
What is the largest sofa (by area; any shape) that can be moved round a right-angle corner in a corridor of unit width? (This is still unsolved.) See, for example, http://www.studyworksonline.com/cda/content/articl e/0,,EXP1755_NAV2-95_SAR1800,00.shtml
Just a kind notice to anyone who decides to play that - make sure you've got Javascript enabled.
I spent damn near two hours before realizing my mistake...
Good answer.
The only thing I can add to this is that the number of meddlings the king can do is actually k+1, as he gets to set the default chalice position (up / down). So, much like N = n - 1, K should equal k+1.
The ______ Agenda
A pound of gold or a pound of feathers?
And, no, the answer isn't that they weigh the same.
I don't understand why you have to mess so much with this. First, let's drop the part where the king can interfere. Clearly, the simple solution (not time-efficient, but who cares) is to choose one leader, who waits for 99 'ups' and whenever he sees an 'up' turns to 'down'. The other 99 are only allowed to turn 'down' to 'up', and only one time. Now, since the king can interfere K times (I assume K is known to the prisoners as well, otherwise it can't be solved) , the startegy should be for the leader to wait for, say, (2K-1)*99 'ups', and the other 99 must turn 'down' to 'up' in the first 2K chances they get. It is 2K times longer than if the king could not interfere, but it works and the king has nothing he can do about it, no matter what order he chooses for inviting the prisoners into the room.
the "rose" is the pip in the middle...3's and 5' are the only faces with a middle.
count the pips in the corners
Do not taunt Happy Fun Ball.
What am I missing as this seems simple:
K+N = Yes
Each prisoner only changes the state of the chalice once. Each prisoner keeps track of the amount of times the chalice changes state. Once the number of times the chalice changes state is greater than K+N times (k, the number of times the king can flip the chalice, and n, each prisoner flipping it once) the prisoner can answer yes.
So why did I just get everyone killed?
Cheers, Greg.
Let's say you have two variables a and b, such that a = b.
Thus ab = a^2,
ab - b^2 = a^2 - b^2,
Factoring the right side,
ab - b^2 = (a + b)(a - b)
Factoring the left side,
(a - b)b = (a + b)(a - b)
Divide both sides by (a - b),
b = a + b
By this we see that
b = 2b
and thus
1 = 2
Subtracting 1 from each side,
0 = 1
How can this be?
This is not necessarily a problem that you sit down and solve. It's a problem in which you come up with the method to solve the problem.
One day, a really advanced alien comes down to earth and finds the 2 smartest guys on the planet, Tom and Harry. The alien tells them, "I'm going to destroy Earth, unless you guys can solve my riddle. I'm thinking of 2 numbers between 3 and 99. The sum of the 2 numbers is... " and he whispers the sum into Tom's ear. "And the product of the 2 numbers is..." and he whispers the product into Harry's ear.
"what are the 2 numbers?"
Being such smart fellas... They sat there and thought about it for awhile...
Tom suddenly says, "Ah ha! you can't possibly know what the 2 numbers are!"
Harry then responds, "Ah Ha! you can't possibly know what the 2 numbers are either!"
Then Tom, says, "Ah ha! Now I know what the 2 numbers are!"
Then Harry says, "Ah ha! Now I know what the 2 numbers are also!"
what are the 2 numbers, and more importantly, how did they find out what they were?
(GP AC here)
Yes, I believe that's the "elegant" solution -- Well, actually you have to substitute "K+1" for "K" everywhere if the state of the chalise isn't known at the beginning. When the master prisoner sees the cup flipped he won't know if he's the first prisoner and it started like that OR if it started right-side-up and another prisoner already used one of his flips. Therefore it is the same as if the king had one more "flip" worth of interfernce. The only way this isn't true is if the first prisoner KNOWS he's the first prisoner and can elect himself master, but that doesn't seem to be in the problem.
I was just modeling it as a pure information-theory problem which isn't a good way to get an optimal solution but it's a good way for producing an existence proof)
When the numbers become meaningless look at the meaning behind words.
The setup is a central room (the king room) with cells around it. Basically a lot of the story keeps pointing out that the room with the king is central to the setup with the cells surrounding it. Since cells are typically only build with 1 door it then becomes logic to assume that each prisoner as he was led to his cell has had to pass through this central chamber.
So every prisoner simply has to say "yes".
Some discount this simple answer but if you read the problem carefully I think it is the only real answer. The math solution can't work. Not when you got no limits, you can only calculate X if X is the only unknown.
So then ask yourselve why spend all this time discussing setting up the layout when it is not needed. The king could have set himself up in an empty cell or have the prisoners brought from the dungeon to his throne room. But no, he sits in a central room with the cells around its perimeter.
MMO Quests are like orgasms:
You may solo them, I prefer them in a group.
Ok, this is regarded by many as THE puzzle to solve... you have been warned...
Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are "da" and "ja", in some order. You do not know which word means which.
good luck!
Does the level of the lake fall, rise, or stay the same?
Exigo spamos et dona ferentes
The problem with that is King doesn't need to turn the challice himself, he only needs to call a prisoner (a non watcher) out to turn it over. Since he can call them an infinite number of times there is no K cap with that strategy.
"2) Everyone else, upon finding an upright chalice, turns it upside down. Otherwise, they leave it. They each must turn it over exactly 3k+1 times, where k is the number of times the king can move the chalice."
I've only had one coffee, so I'm not going to work the numbers, but if each non watcher only turns the challice right side up ONLY IF THEY HAVE NOT PREVIOUSLY DONE SO. That small change may make your strategy work.
You must use each number exactly once. The only operations permitted are addition, subtraction, mulitplication, division and parenthesis. Nothing tricky, this is straight forward.
What is nice about the riddle is that it would seem that it is either trivial or not doable. It turns out that it is doable but surprisingly difficult. Don't give up, it really is doable without any "cheats".
S = 9
E = 5
N = 6
D = 7
M = 1
O = 0
R = 8
Y = 2
9567 + 1085 = 10652
Work (not overly rigorous, includes some logical guesses):
The problem can be expressed as such
(1000*S + 100*E + 10*N + D) + (1000*M + 100*O + 10*R + E) = 10000*M + 1000*O + 100*N + 10*E + Y
where
0 <= S != E != N != D != M != O != R != Y < 10
Simplify
9000M + 900*O + 90N + Y = 1000S + 91E + 10R + D
Due to the size of the coefficient on M, M must be 0 or 1.
Assume M = 1
9000 + 900*O + 90N + Y = 1000S + 91E + 10R + D
The only solution that can balance the large value on the left is such that S = 9.
Assume S = 9
9000 + 900*O + 90N + Y = 9000 + 91E + 10R + D
900*O + 90N + Y = 91E + 10R + D
O must then be 0 for the same reason (as the 1 has been used).
Assume O = 0
90N + Y = 91E + 10R + D
Play around with (mathematical term) your remaining choices. R must be 8 to bring 91E and 90N close in value. N = E + 1 for the same reason.
Assume R = 8, N = E + 1
90(E + 1) + Y = 91E + 80 + D
90E + 90 + Y = 91E + 80 + D
10 + (Y - D) = E
Using a little guesstimation, D must be large and Y small, so
Assume D = 7 (largest remaining number), Y = 2 (smallest remaining number)
10 + 2 - 7 = E = 5
=> N = 6
Having been told only one solution exists and having found a solution, DONE.
This took a lot longer to type than to solve.
1 12 1112 3112 132112 1113122112 311311222112 what's the next line?
Fleur de Sel
I have a solution, but the prisoners have to know the number k:
They choose one of them as a leader.
When he finds a chalice "wrong side up" he flips it over. If it has the right side up, he doesn't touch it. He keeps track of the number of times, he has flipped it.
The others flip it only when it has the right side up. If it has the wrong side up they don't touch it. They also keep track of the the number of times they flip it. They each only flip it (2k+1) times.
When the leader is about to flip it back for the ((n-1)*(2k+1) - k) time, he answers yes to the question.
Proof that it will always work:
If only (n-2) of the non-leaders have been in the room, the max times the leader can have flipped it is:
(n-2)*(2k+1) + k = (n-1)*(2k+1) - k - 1, so he cannot possibly answer yes too early.
Every time one of the non-leaders flip it, it has to be reset. The king will reset it no more than k times, the leader will reset it no more than ((n-1)*(2k+1) - k -1) times, combined they will max reset it (n-1)*2(k+1) - 1 times. But the non-leaders will flip it (n-2)*(2k+1) times, so we can't get "stuck" either.
I've thought of puzzles like this being like search trees. It seems that we navigate possible solutions like a depth-first search, because one thought leads to the next, leads to the next... Smarter people can search deeper, but then they can take longer to find the answer to puzzles like this because they follow wrong paths longer. The solution: mentally switch to a breadth-first search. Try to throw you your last thought and come up with a completely fresh approach. Then do it again. After you have a few different approaches, then think about extending them.
i.e. 3 is 2 points, 5 is 4 points, all the rest are 0.
I'm sorry, this is just dumb. I'm only repeating what others have said so far but maybe that will help drive in the point that this is Worst. Puzzle. Evar.
With the king able to manipulate the chalice, there is absolutely no way for them to communicate to eachother, because the king can always turn the chalice back to the way it was before the prisoner entered the room. So every prisoner will always see the exact same thing. Stupid, stupid.
Joseph?
My variant on solution #2 was apparently a little more violent than yours: Flip the switch rapidly until you burn out the bulb. That should take a lot less time than waiting for it to burn out. Static charge might also help.
The problem with logic puzzles like this is that logic puzzles presuppose a lot of things. In the real world, you could pretty easily take a switch out of the wall and add a resister so that one of the bulbs was dimmer than the others. Or you could get to the other room and the bulbs could be high up in the ceiling and out of reach, and your solution is screwed. If you're fast enough you could theoretically throw the switch and make it to the other room in time to see the bulb change states. Considering all of the walls these days are a variant on carbonized newspaper, you could pretty easily kick a hole into the wall to see the lights change. You could pull the wire attached to the switch until you yanked back the bulb in the other room. Even without getting violent, every light switch seems to have a state in the middle which causes the bulb to be much louder than the rest.
Heck, take off your clothes, cut / tear them into thin strands. Loop one strand around one of the switches in the "on" position, and hook that around something low to the ground, such as the bottom of a door frame. Walk to the other room. Pull the switch. Or cut the clothes and soak it in a liquid before anchoring it snugly to the switch just barely in the on position and something on the ground. Walk into the other room, and wait for the clothes to dry and shrink.
Do you even know which way is the on position? I've walked into a lot of old houses where swithches were mounted with "down" as the on position, and without any labeling at all. Newer houses frequently use multiple switches, whereby the XOR of two switches determines whether or not a light is on. What about switches that are mounted Left / Right? Are you assured that all of the lights work? All of the switches work? Does the house have power?
The only way that NONE of the light from the other room can reach the current one is if none of the switches did anything. If there was any light variance, you could measure that and get a relative position based upon strength.
It also says that you can't see the other room. Are you blind?
Logic puzzles, basically, are a nice way of communicating a simple state problem. And if you want it to work as a puzzle, you agree to this limitation / logical purity. Light switches only have two states. Walls are impermeable. No light gets through. That way you can permutate through the simplified logical consequences of each potential action and the ramifications thereof. Actual real-world solutions are usually much easier. A light in the real world may have a tremendous number of states, but in a puzzle it is on or off. A chalice in the real world has angle and position information, can be scratched into, fashioned into a weapon, stolen, etc, but in the puzzle it is either up or down. Otherwise it doesn't work as a puzzle.
If you're proposing a logic puzzle, but you use a real-world side-effect that you don't make explicitly clear in the puzzle , you're violating how logic puzzles work. That doesn't make you smarter than people who solve lots of puzzles, it just means you've written a bad logic puzzle.
The ______ Agenda
(based on an AC comment above and my bug fix to it)
They choose a watcher
If the watcher is called and the cup is upside down he flips it right side up
If anyone else is called and the cup is right side up they flip it to be upside down, unless they've done it previously in which case they leave it alone.
When the watcher has flipped the cup n+k+1 times he says yes.
Since the King can only interfere k times, you have to out wait him, hence +k.
The +1 is because the cup may not be right side up on first time.
Being quite lazy and uncoffied, it may be +-1, so I'd add +16 because I'm a programmer and do stuff like that. But that strategy would work.
Once a traveler got into terrible weather and seeked shelter at a monastery with high walls and a heavy gate. He knocked, but no reaction. He tried and tried till finally he was granted access only under the condition, he would never talk to anybody nore stare into peoples faces and leave the monastery early the next day.
...
Well, Mr. Traveler was a stupid dumbass and while sitting at the table for supper, he asked loudly: "WHY DO AT LEAST TWO OF YOU HAVE SPOTS ON YOUR FOREHEAD?"
The 100 monks were not happy at all and one explained to him:
"We know that some of us might have spots on their forehead. That's the reason why we don't have mirrors here and don't speak a word. If we get to know to have a spot, we die the very next night."
42 days later, all the monks with spots die. How many are they?
(The monks see all others at supper every day and thus know if one is missing. They will not stop that habit now. Ignoring the facts or stop thinking does not help them. A monk either has a spot or not. They don't disapear.)
I read this in a novel (Mark Hadden's excellent The Curious Incident of the Dog in the Night-time). You are a contestant in a game show. The host tells you that behind the three doors in front of you are two goats and one car (one item behind each door; the idea is to win the car, by the way...). You pick a door, and then the host opens one of the other two doors to show you a goat. The host now offers you one last chance to change your mind and pick the other remaining unopened door. Should you? (In other words, which strategy is better overall: stick or change?)
[A man condemned to be hanged] was sentenced on Saturday. "The hanging will take place at noon," said the judge to the prisoner, "on one of the seven days of next week. But you will not know which day it is until you are so informed on the morning of the day of the hanging."
... He is convinced, by what appears to be unimpeachable logic, that he cannot be hanged without contradicting the conditions specified in his sentence. Then on Thursday morning, to his great surprise, the hangman arrives. Clearly he did not expect him. What is more surprising, the judge's decree is now seen to be perfectly correctly. The sentence can be carried out exactly as stated.
The judge was known to be a man who always kept his word. The prisoner, accompanied by his lawyer, went back to his cell. As soon as the two men were alone, the lawyer broke into a grin. "Don't you see?" he exclaimed. "The judge's sentence cannot possibly be carried out."
"I don't see," said the prisoner.
"Let me explain They obviously can't hang you next Saturday. Saturday is the last day of the week. On Friday afternoon you would still be alive and you would know with absolute certainty that the hanging would be on Saturday. You would know this before you were told so on Saturday morning. That would violate the judge's decree."
"True," said the prisoner.
"Saturday, then is positively ruled out," continued the lawyer. "This leaves Friday as the last day they can hang you. But they can't hang you on Friday because by Thursday only two days would remain: Friday and Saturday. Since Saturday is not a possible day, the hanging would have to be on Friday. Your knowledge of that fact would violate the judge's decree again. So Friday is out. This leaves Thursday as the last possible day. But Thursday is out because if you're alive Wednesday afternoon, you'll know that Thursday is to be the day."
"I get it," said the prisoner, who was beginning to feel much better. "In exactly the same way I can rule out Wednesday, Tuesday and Monday. That leaves only tomorrow. But they can't hang me tomorrow because I know it today!"
- the income=f(tax ratio) is not a smooth function.
- it is by all account not even a function, eg
you can have more than one value of revenue for one tax ratio.
The usual parabol-like curve is based on a very simplistic model which is useless except for pushing an ideological agenda. The only sound reasonning in this is that high tax rates increase the incentive for escaping taxes, which is commmon sense. No need for half-baked maths to get that.4 8 15 16 23 42 Explain
"They each only flip it (2k+1) times."
I like this one because it's simple but the result is surprising (to me at least).
You have a rope tied around two trees that are 200 metres apart. A 10 metre tall truck has to pass under the rope by driving through the centre of the gap between the trees. How much slack do you have to put into the rope to allow the centre of it to be lifted over the 10 metre tall truck?
Input error. Replace user and press any key to continue.
Actually, I'd say it's indefinate, not undefined... just as 0/0 is indefinate, it could be every number (0*n=0, thus 0/0=n for any n) n/0 for any non-zero n is undefined (there's no real number you could multiply zero by to get n) 0^0 is either 1 or 0 depending on how you want to look at it: n^0 = 1 or 0^n = 0, so it has a finite result set (which is a bit easier to deal with than 0/0's infinately infinate result set...) Ok, here's a mind bender: you say that there are infinately many integers, and there are infinately many non-integers between each pair of integers, does this mean that there are more non-integers than there are integers? infinity * infinity is still infinity... My answer would be that, yes, it is still infinity, it's just an infinately larger infinity... You could even say that it's a significantly larger value of infinity... What's infinity divided by infinity? :)
The Diamond Question
You are an intelligent but greedy treasure hunter. You and other 4 equally intelligent and greedy treasure hunters have obtained a chest containing 100 identical diamonds. The 5 of you come up with the following plan to split the treasure:
- First, the order for the treasure hunters to speak is decided with a draw;
- The first person to speak has to tell everyone his scheme of how the diamonds should be split among the group;
- After the person described his scheme, the other treasure hunters will vote;
- If over half of the group (i.e. > 50%) vote for the scheme, it will be used;
- Otherwise, the scheme will be discarded and the person evicted from the group. He will have no say and will not get any of the diamonds. Then, the group will go back to step 2 and the next person will describe his scheme, the others will vote, and so on.
Luckily, you have won the draw and get the chance to speak first. How will you design your scheme? Obviously, you cannot split the diamonds evenly (i.e. each person get 100/5 = 20 diamonds) because the other greedy treasure hunters will simply vote against your scheme (and also evict you in the process) for a chance to get a bigger share. Is there a scheme that give you the maximum number of diamonds and getting over half of the group to support it?
Note: you need to get over 50% of support including yourself, which means you need at least 2 votes for a total of 3 for, 2 against
Clearly the barber is a Cretan. :)
The barber problem is not really a problem in that there is no paradox. Your ackward wording makes it unclear, but the common phrasing of this problem is "the barber shaves all men who don't shave themselves." That means that of the subset of all men who don't shave themselves, the barber shaves them. Don't shave themselves -> Barber shaves them. However, the opposite is NOT specified by the rules of logic. Shaves themselves does not -> Barber does not shave them. It does imply something known as the contrapositive, the negative of both sides of the equation with the order of implication reversed... Hence if the barber does not shave them, then they must shave themselves.
None of this precludes the barber from shaving someone that shaves themselves. Or precludes the barber from not shaving someone who shaves themselves. This says NOTHING about people who shave themselves. It only makes a logical statement about the subset of people who do not shave themselves.
Hence, there is no problem if the barber shaves themself.
Of course, I'm using the common form of the problem not the one from the poster above. The one from the poster above seems to be trying to close that "loophole" in the equation, but just winds up saying the equivalent of "I've got a dollar, but I've only got twenty-five cents."
The ______ Agenda
You have 24 coins, one of which is measurably heavier than the others, but otherwise indistinguishable. You have a pair of scales. Find the heavy coin in the minimum number of weighing operations.
I like this one as an interview question for potential developers. The first attempt leads to a situation that, if extrapolated on, reaches the correct solution.
vi is my shepard, I shall not font.
Guy asked me for a quarter for a cup of coffee. So I bit him.
The Place
There is a circular abbey, inhabited by the most devoted but intelligent monks. Each monk spends its entire life in its own cell. The cells are line up one next to the other, around the atrium. Each cell has one balcony looking over that round atrium from where each monk can inspect the rest. All the monks visit that balcony every day of their life, at the same time in the morning and look at each other. The monks are not allowed to communicate with anyone by any means (words, signs, telepathy or whatever) at any time.
The Voice of God
One day, the monks hear the voice of God whispering them the following words:
"Tomorrow, some of you will be chosen to abandon this vain life. This night, at your sleep, an angel of death will sneak into your cells. This angel will paint a black spot on the forehead of the chosen ones. It is your duty to obey my will, so, those among you currying the spot should within, a finite period of time, end your life and come to join me!.
The Conundrum
The next day, there were indeed monks with spots. Of course since there were no mirrors inside the cells, and the spots weren't relief, the monks couldn't tell by themselves whether they had the spot. But finally, the chosen monks... committed suicide!
How they did it?
Try to partition a 4x4x4 cube into 8 pieces, each of 8 cubelets.
Each piece must hold together as a physical object.
ie cubelets must be joined face-to-face.
So far this is very easy.
The extra twist is that the center of gravity of each piece must lie
outside of the piece.
Example 1: the 0,0,0 and 0,0,1 cubelets share a face, but the center of gravity,
at 0,0,1/2 lies on the face of 0,0,0, so it is not allowed
Example 2: the 0,0,0,and 2,2,2 cubelets have center of gravity 1,1,1.
That is outside both cubelets, but they don't touch so that is not allowed
either.
The origin of the puzzle is in statistical pattern recognition: come up with
a gnarly but fair test for recognition methods aimed at twisty shapes in low
dimensions.
There might not be a solution. I imagine it as a computer search problem.
So the puzzle is the meta-puzzle: how to write a program to solve this?
Is that the adherents to this are living oxymorons. They are simultaneously being taken in by this tripe theory, but for the theory to hold water, it has to capitalize on an extreme cynicism -- the idea that any money taken by the government is completely wasted and does not in and of itself stimulate the economy (through contracts, enabling services, and quality of life.)
So they are cynical fools.
Someone had to do it.
... unless there are other constraints, such as the king having to turn the chalice k times, eventually. The description specifically states that the king may do so, though.
If there only is a solution that relies on the king flipping the chalice k times, then the prisoners are boned, since the king is not obliged to flip it at all.
If there is a solution that does not rely on the king flipping the chalice k times, then the king (knowing the solution and the complete state of the game) can simply wait until the cup indicates a success, flip it, hide the smug grin on his face, and call in the appropriate prisoner.
If construction was anything like programming, an incorrectly fitted lock would bring down the entire building...
I write out a check to you for some amount, $X, unknown to you, and then I write out another check for $(2*X). I put each in an envelope, and put the envelopes in a hat.
.5 * Y/2 + .5 * 2 * Y = 1.25 * Y. Therefore, switching gets you an expected gain of 25%, so you should switch.
You choose an envelope from the hat. After opening it, if you like, you may instead take the other envelope. Either way, you get to cash one of the checks. So - what is your best strategy?
Here are two seemingly airtight arguments, which conflict with each other.
1. You gain no information when you open the envelope, since you have no idea what X was, so switching clearly makes no difference - you are still getting one of the checks at random.
2. Call the value of the first envelope you drew Y. Now, there's a 50% chance it was the larger check, and a 50% chance it was the smaller check. Therefore, the expected value of the other check is
They can't both be right. But where is the flaw in the reasoning?
Extra credit - believe it or not, both are wrong! There is a strategy by which you can do better than just taking the envelope you drew, but strategy 2 is not it. What is this strategy?
One. You can do this by using just one point. A point has three arguments, the X,Y and Z value, now imagine a plane through the (0,0,0) point. The X argument specifies the 'pitch' of the plane, the Y argument the 'roll' and the Z argument translates the plane over the Z-axis. For example, when you define the Cartesian space as real values from 0-1, the point (0,0,0) defines a plane spanned by the X and Y axis. The point (0.5,0,0) defines the XY plane tilted at 45 deg up. The point (-0.5,0,0) defines the XY plane tilted down at 45deg. The point (-0.5,1,0) defines the plane perpendicular to the previous plane. Now when the plane does not intersect the z-axis at the (0,0,0) point just use the z-argument of the position where it does intersect the Z-axis. While this function already lets you define most planes there are a few exceptions, these are planes parallel to planes described by 2 of the basis axis which do not cross the z-axis. A way to still describe these planes is to define an exception that whenever a plane is parallel to the X or Y axis, ie (1,0,0) or (0,1,0) the z-argument does not translate the plane over the X-axis but over respectively the Y or X axis. I hope this makes some sense. -- Nicholas Piël
In case it hasn't been posted, yet, here's a traditional logic puzzle:
;-)
You're in a room with three light switches. They control three lights in another room. Your task is to play with the switches, then go to the other room and determine which switches control which lights. It can be done in one trip. That's the easy one.
Here's the hard one...
It's easy to get out of a maze if you have a magic-marker in your hand, right? You mark doors with an X as you go through them, and rooms with an X on the floor when you first enter them, and remember the door you came through with an --> or some such mark. When in a room, go through an unmarked door. If you come to a room that's been visited, go back. If there are no unmarked doors left, take the one with the arrow. Basically, just do a depth-first search, with a linear time solution.
What if you introduce one-way doors between rooms? Assume that you can't peek and look into the next room, but instead must commit to going through the door before knowing what's on the other side. All you have is your magic marker, and you can remember a few numbers in your head. You have no sense of direction, and can't make a map.
There's lot's of interesting conclusions, but no simple solutions that I know of. If I ever get around to it, I'm going to build a giant one-way door maze in my basement. I'll put bad kids in it and let them starve to death
Beer is proof that God loves us, and wants us to be happy.
the following list seems easy to continue. find the other solution 3,1,4,1,5,....
Computers are like air conditioners.
- They stop working when you open Windows.
This is an oldie (but a goodie).
First, suppose that there is just one marked monk. Then the next day, when they all go to the balcony, he will see that nobody else has a mark. Therefore, since he knows that somebody is marked, he must be marked, and thus he should commit suicide.
Now suppose there are two. Each of them sees just one other marked monk. Now, the next day, they *still* see one other marked monk. But, if there were only one, he would already have commited suicide. Therefore, each then knows that he must be marked as well, so both should then commit suicide.
In general, if N are marked, they will each see N - 1 other marked monks. If, after N - 1 days, they have not committed suicide, then they all know there must be N, and they should then commit suicide.
perl -e 'fork||print for split//,"hahahaha"'
You are through to the final round of a game-show and have a chance to win a car which is hidden behind one of three doors. Behind the other two doors are goats.
You must choose one of the doors. The host then opens a different door to the one you selected to reveal a goat, leaving only two doors. You then have the option to stick with your initial choice, or switch to the other unopened door.
So, assuming you prefer cars to goats, what is the best strategy to adopt? :
a.) always stick with your initial choice
b.) always switch
c.) it doesn't matter
What I really like about this puzzle is that just about everyone who sees it knows the answer staight away, is absolutely certain that they're right, and is completely wrong.
(the subject line is taken from the title of the book where I first read this.)
N is the set of all integers. M is the set of all integers except 0. Given the associativity and commutativity of addition, removing one member from an added set of integers must of course be equivalent to subtracting the same number from the sum of the set.
Never let logic and reality get in the way of a good, liberal hate-on for Republicans.
(-1: Post disagrees with my already-settled worldview) is not a valid mod option.
Good riddle.
Can I ask all three questions of the same god?
I asked a friend of mine this the other day, and he instantly came back with the answer... "12 arbitrary units".
I was suitably impressed, especially since it answers practically any quantitative question, and always correctly.
(Actually it also reminds me of the old "How long is a piece of string?"... "Twice as long as half its length.")
Everything in moderation, including moderation itself
* Solution may only work if you're really really committed to it, or are Wowbagger the Infinitely Prolonged
"A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
e^(i*pi) + 1 = 0
What do e, i and pi have in common?
While it is true people are greedy and will continue to work hard, even as more and more gets filched by the government, at some point revenue will decrease because people will stop working so hard for diminishing returns.
It's very cynical to tout as good a government that lays like a 2000 pound pig on top of the population, and then shouting joy about it as the population barely moves it around, said movement, if nothing else, caused by the economic activity necessity to stay alive.
But see, if we take taxes, i.e. confiscate money that is not ours, to use for projects that we approve of then it doesn't feel like stealing to us.
Which I find as a disgusting attitude, and why I have little long-term hope for humanity. They preform all the age-old tricks of a dictator, but laundered because it's at the behest of power hungry, charismatic individuals who launder it via "see, the voteres elected me and want it!"
(-1: Post disagrees with my already-settled worldview) is not a valid mod option.
How many days can Snowhite cook - if every day she cooks twice as many dwarves as the day before?
I doubt that we will ever figure out - and I suspect that even if we did figure out we couldn't do much about it
> It's more productive to impart knowledge than to accuse one who lacks it of being a fool.
Hint to those who resort to namecalling: When you namecall, you are angry. You are angry because, subconsciously, you recognize the truth in your opponent's argument, and cannot come up with a legitimate response, and you know it.
So when calling someone a "fool", think to yourself: "He may very well be right, and I have no real response." Now there may be one, you just don't know it yet. And your argument is now on thin ice.
What's that saying from that idiotic "make yourself more erudite" commercial? Ah, yes. "Better to reinforce your argument than raise your voice." (Ok, granted, that attempt at alliteration was somewhat strained...)
(-1: Post disagrees with my already-settled worldview) is not a valid mod option.
Galley's Rule of Nines demonstrates a simple mathematical rule that I discovered about twenty years ago. You can take any positive whole number with at least two digits, and with a few manipulations, the end result is always nine! So how does it work? There are two parts to the rule: Part One: If you find the sum of the digits of any whole number (larger than nine) and subtract that sum from the original number, the result will always be a multiple of nine. Part Two: Every multiple of nine's digits will add up to nine (or another multiple of nine, who's digits will add up to nine, etc.) In other words, no matter what number you being with, the end result will always be nine! Download it FREE at http://www.galleytech.com/
"I'm not a cool person in real life, but I play one on the Internet". Galley
...richie - It is a good day to code.
Every even integer greater than 2 can be written as the sum of two primes. (The same prime may be used twice.)
Now prove this !
a = b
+a
2a = a + b
*a
2a^2 = a^2 + ab
-2ab
2a^2 - 2ab = a^2 - ab
factorise
2(a^2 - ab) = 1(a^2 - ab)
divide out
2 = 1
Points if anyone works out where the problem lies.
...but this man's father is my father's son.
Ignorance is curable, stupid is forever.
I've never understood why people consider this an interesting paradox:
The common formulation is "All Cretans are liars". In this formulation, there is actually no paradox, since liars can sometimes tell the truth. (So the Cretan saying "All Cretans are liars" may be telling the truth, or a lie. In both cases we don't have any contradiction).
Seeing this, people change the formulation to "All Cretans always tell lies". We want to determine whether the Cretan saying this is telling the truth, or not. Let's consider these two hypothesis:
If the Cretan is telling the truth, it means the sentence is true, so according to it he's telling a lie, hence a contradiction.
If the Cretan is telling a lie, it means the sentence is false, ie. "it's not true that all Cretans always tell lies". Which means they at least sometimes tell the truth (but may also tell lies at other times). This is compatible with the hypothesis that the Cretan is telling a lie. This time there's no contradiction, so this hypothesis is the right one.
So the answer to the problem is that the Cretan is telling a lie.
Am I missing something?
Another version: you have 13 coins, but do not need to tell if the fake is heavier or lighter, just point out the fake coin.
my first programming teacher would start first day of class (when i was a freshman in HS) with......if anyone can answer my next question in less than 3 seconds you get a 100 in the class. this only works when saying it....doesnt have the effect when seeing it typed but he would say.. whats 2 plus 2 times 2. most people would say 8 right off the bat. others wouldnt say anything. -Rob
that the bear probably is, in fact, a penguin
You sir are an idiot.. No wonder you are on my Freaks list......
A witty saying proves nothing. Voltaire (1694-1778)
A variation is to say upfront that the answer can be determined in three goes. The puzzle is to work out how, and that's very tricky.
2*0=0
http://www.phobe.com/s_cat/s_cat.html
Join the Slashcott! Feb 10 thru Feb 17!
A community of people meet everyday in the town square. However, they cannot communitate with one another because they are deaf and don't know sign language. They wander admist each other looking at everyone else in the town. After doing that they go home. It is a rule that if you have a special mark on your forehead you must commit suicide. A certain number of marks are on a few of the townspeople (obviously no more than 1 per person). After three days everyone with marks is dead. Assuming that they are all perfect logicians, how many people had marks on their forehead?
Do your stuff.
Somewhere down the road, you(Dark Lord Seth) will be banned again. Or you will grow up.
There is a cube which houses cube crawlers. The corners of the cube are labeled A to H, where G is furthest away from A (that is, three edges away from A, in the opposite corner). All cube crawlers are born in A, and each day they travel along one of the edges, to rest in the next corner. They will never turn around and travel back along the edge they traversed the previous day, but will always choose one of the two edges they did not traverse the previous day, with a chance of 50% for each of them. In corner G, the Cube Crawler Predator is situated, who devours all cube crawlers that end up in G. The question is: what is the average life span of a cube crawler?
Obviously, the minimum life span of a cube crawler is three days, since G is three edges away from A. Theoretically, the maximum life span of a cube crawler is infinity, because it can travel in circles from A to B to C to D and back again to A.
It took me two days to solve this problem when I first encountered it, using a purely mathematical approach. However, when I stated this problem to a physicist he provided me with the correct answer in a few minutes (without proof, that is).
You're wrong in step 6 already. 1/i = -i.
Step 4 is not valid. a^2=b^2 is not equivalent to a=b. a=b implies a^2=b^2, but the reverse is not true. Therefore, the only thing you've proved is that if 1=2, 1=1.
It should go:etc.
Note that none of the digits will ever exceed 3.
I have a truly marvelous proof for this, which, unfortunately, this post is too small to contain.
Those who sacrifice security to condemn liberty deserve to repeat history or something. - Benjamin Santayana
We're angry because it's exasperating trying to reason with people such as you, who appear to be brain damaged at best, not because you're correct. I read over your posting hisory, and its a hoot.
That being said, WTF is "Impiuos"? Did you mean "impious"?
Hint for you: When you mispell your own, self-chosen, nickname, you're a moron.
But, the rest of us have a good laugh everytime we see it, so thanks.
even better, for the fingerless:
fill the five, then fill the 3 using the 5. The 5 will have 2 left in it - dump the 3, then pour the 2 gals from the 5 into the 3. now, fill the 5, then top off the 3 with it, and since there's only 1 gal of room remaining in the 3 you'll be left with 4 gals in the 5.
Easy!
I'd have to say that one of the cooler one's I've seen is the integral of sec^3() d. Not only would you need to do a trig substitution, but it ends up being (what I think of as) one of those cyclical type problems.
actually, seeing simple things is an act of intelligence, as a specialized skill or tactic. Habit will sometimes lead you down the wrong path of seeing more than what is there.
"It is a greater offense to steal men's labor, than their clothes"
===========START=============
... posted again due to hyperlinks hiding the '_'.
1_
============END==============
That's it.
EMail: 0110001101100010010000000110001101110010 0110000101111010011011100110000101110010 0010111001100011011011110110
I loved this on in grade school. Doesn't apply prefectly to a math problem, but was a fun riddle.
MAKE TEN BY ADDING THESE FIVE LINES TO THE LINES BELOW: | | | | |
| | | |
Of course all of you know the answer. But I'll post it in a reply to this.
hmm, it would appear that the theta isn't appearing.. Sorry, it's supposed to be sec^3(theta) d(theta)
I believe this one originated from that wacky Lewis Carrol.
A man decides that he wants to become a lawyer, but he doesn't have the money to go to law school. His friend who is already a lawyer offers to teach him for $10,000. The friend says "pay me half up front, and half when you win your first case. If you lose your first case, then you don't have to pay me the second half."
The man agrees, pays the $5000 and proceeds to study law. After some time he passes the bar and officially becomes a lawyer. But then he becomes interested in other things and never tries his first case.
His friend wants the other $5000, so he sues the man for it. Now, since he is a lawyer, he decides to represent himself, and thus this becomes his first case.
Now suppose you are the judge in the suit. If the man wins the suit, then he does not owe the $5000, but since he won his first case, he owes the $5000. If he loses, the he must pay the $5000, but since he lost his first case, by the agreement he does not owe the $5000.
By the perception of illusion, we experience reality
One person will always flip the cup from up to down. The other prisoners will only flip the cup from down to up. If it's not in the state they want on entering, they leave it be.
There's two main problems here:
1. You don't know the initial state of the cup.
2. You don't know what the king has been screwing with.
These are actually the same problem, in that the state of the cup might have been tampered with and you have to route around it.
In the simplest problem (uncertain initial state only), you could have each prisoner only flip the cup from down to up twice only. This means that after it has been flipped 2(n-1)-1 times, the original prisoner knows that everybody has been in the room. He just keeps a count of how many times he's had to flip it himself.
With the king being an ass about it, they have to count for his weird ass states too. In order to do that, the prisoners must each flip the cup from down to up an additional k times, to overcome the king's fucking with them.
-So one prisoner flips up to down and keeps a count.
-The other n-1 prisoners each flip the cup from down to up k+2 times (k times for the king, one time to account for the initial uncertainty, once for themselves).
-When the counter prisoner gets a count of (k+2)(n-1)-1, he knows everybody has been in the room at least once.
- Give a man a fire and he's warm for a day, but set him on fire and he's warm for the rest of his life.
In chemistry we are taught that when we combine two chemicals in a reaction, each with their own unique set of properties, a new compound is formed with a new set properties, distinct from those of the component chemicals. My question: Where do the new properties come from and where do the old properties go? Bill
Fill the 3G jug and empty it into the 5G jug, repeat without emptying the 3G you have in the 5G jug, you are left with 1G that doesn't fit in the 3G. Empty the 5G, put th 1G into that jug, and top it up with another 3G. There you go, 4G!
What can I say, it sounded simpler in my head...
---
Computer Support in Sydney
a a a d a e a g a h a i a l a m a n a r a s a t a w a x a y b a b e b i b o b y d a d e d o e f e h e l e m e n e r e s e t e x f a g o h a h e h i h m h o i d i f i n i s i t j o k a l a l i l o m a m e m i m m m o m u m y n a n e n o n u o d o e o f o h o m o n o p o r o s o w o x o y p a p e p i r e s h s i s o t a t i t o u h u m u n u p u s u t w e w o x i x u y a y e
The solution seems pretty difficult.
Stephan
http://stephan.sugarmotor.org
A simple 'duh, no shit' answer, once you figure it out...
Always switch. There is a 1/3 chance your first choice is right and a 1/2 chance the remaining door is right.
k is not undefined. k is a number that the king and prisoners know ahead of time. This should be obvious, as the problem does indeed say there is a solution.
How many licks does it take to get to the center of a Tootsie Pop? (And how many does it take to get sued for sexual harassment?)
I might know what I'm talkin' about, but then again, this is Slashdot...
Finally, someone who understands the English language.
Too simple for the slashdot crowd, but nice to stumble anyone else with:
Q: Whats your age?
A: In twelve years I am twice as old as I was six years ago.
You can easily replace the numbers accordingly so the solution is your age.
The name of the game is significant.
To figure out what the name of the game has to do with it, you have to look at the dice.
Don't play with the numbers.
Look at the dice.
And, remember, the name of the game is "Petals Around the Rose".
It took me about 10-15 minutes to figure out, and I figured it out graphically, not mathematically.
Those who sacrifice security to condemn liberty deserve to repeat history or something. - Benjamin Santayana
You have two people in front of you. One always tells the truth, and believes what he says is true, and will state so. The other one always lies, and believes in the opposite of the truth, and will state the opposite. Can you tell them apart by asking yes/no questions? I remember that confusing the hell outta me when I was younger in the movie Labyrinth I think..
In undeveloped countries, the consumer controls the market. In capitalist America, the market controls you.
29 - 1 = 30
Prove it.
Ron Gage - Westland, MI
n people are put in n different glass rooms, so that every person can see everyone else. Each person is then given a hat, which is one of n different colors. (However, everyone's hat may be the same color, or different colors, etc.) No one can see their own hat, but they can see everyone else's. Everyone then tries to guess the color of the hat on his or her head. If at least one person guesses correctly, everyone wins. What is a winning strategy?
I hate the one hundred and twenty character limit for signatures with an all-enveloping, all-destroying, incredible pass
sin x
----- = (simplifying n) = six!
n
Parent gives a good solution to the problem.
See subject. ;-)
My UID is prime is yours?
How do you tell two Thevenin-equivalent sources apart: one with a voltage source in series with a resistance, the other with a current source in parallel with a resistance?
I'd say you can't do step 4: dividing by (x-y) to lead to x+y=y since (x-y)=0 [can't divide by 0]
There's an error when going from 4 to 5. I think it's that when going from (a/b)^c to a^c/b^c , c must be >= 1.
The answer is 4 8 15 16 23 42
This doesn't work because everone will just flip the chalice once. If the king flips it back before the watcher gets to see it, they will not be able to confer the information again.
I suppose you could have the watcher flip the chalice at some point to tell the other to report gain, but the lack of limits on the number of times someone can be called makes it hard to specify any strategy.
I'm still trying to figure out what people mean by 'social skills' here.
I am reminded of a short bit in the Journal of Irreproducible Results, where they had proof of -1 being the largest integer. Rough rephrasal:
The defining property of the largest integer is that there is no larger integer. But if you take any integer and add one, you get a larger integer. So the largest integer must fulfill x = x + 1, and taking it one step farther for good measure, x = x + 2.
x = x + 2
Square both sides.
x^2 = x^2 + 4x + 4
Subtract x^2.
0 = 4x + 4
Subtract 4x.
-4x = 4
Divide by -4.
x = -1
Checking this result, -1 is indeed an integer and nothing is one more than -1. QED.
The teller is going to St. Ives, while the destination or origin of the man, wives, etc is not told. It is reasonable to assume, in this case, that unless there was a huge discrepancy of speed between the teller and the man/wives party, they must have been going in opposite directions if they were to meet.
Thus, if the teller is going to St. Ives, the man and his wives/etc must have been going in the opposite direction (or, perhaps, perpendicularly, to an undetermined destination).
While as given, this cannot be solved due to lack of information, my answer in this case is none of them are going to St. Ives except for the teller of the story. The teller isn't included in the final querry, so I submit an answer of "zero".
Some people think the answer is obviously 1/2, while others think it must be 1/3. Which is it?
How to algebraically get an answer of '7' with only four numerals to choose from, all of them '2'.
Example: To make '20' with three '2's = 22 - 2
I have reasoned three ways to solve the '7' problem, and a friend found a fourth. Absolutely stumped me and kept me up for a few nights....
Before I part with'em: two pennies weigh ~4.996+/-0.014g, have a zinc core, and the face of Lincoln. You can keep 'em.
Interesting point but the president can't order congress to do anything. He has hardly any power
Ha! If you've read America: A Citizens guide to democracy inaction, by Jon Stewart, you'd know that the president does indeed have a little power over congress. The president can make the congressmen a nice sandwhich, and this can get the congressmen to sway his way sometime.
That's all well and good, but Laffer didn't stop there -- he argued that we're at a position where the slope of that curve is negative -- that is, if we tax more, our revenue decreases. The Laffer Curve is the central argument to supply-side economics.
What's amazing is the degree to which supply-side economics resembles religion: because it is almost entirely devoid of evidence. Supply-siders are the creationists of the economics world. It is fairly rare to find any government, much less our own relatively low-tax government, which has ever received lower revenue from marginally higher taxation (or vice versa). Even during the Reagan years, when the supply-siders ran free like herds of buffalo in the White House, the tax cut resulted in dramatically lower revenue.
Well, duh. The problem with Laffer's argument is that, if the government was trying to maximize revenue, it'd just hill-climb right up to the top of the Laffer Curve and stay there. But there is a second force at work: voter anger. If you raise taxes too much you get voted out of office. This force acts to pull taxes down, and hence to the left (positive-slope) side of the curve. The stronger that force, the more on the positive-slope side. To counter this, you need some force to actually move to the negative-slope side, but Laffer, nor any other supply-siders, have ever really proposed one.
It's baloney.
Here's a hard way/easy way problem:
A machinist makes a solid metal sphere. Then he drills a hole in it, dead center, all the way through.
The hole is six inches long. How much metal remains in the ball?
rj
Don't you mean "skillz"?
#I feel like pwning n00bs...#
News for merdes. Shit that matters.
Ask me about my sig.
Very nice! Nicely written and explained. I don't think it's hand wavy. If anyone is paranoid about whether you might be off by one anywhere in your count, then fine, give each non-counter more tokens...
I love this problem because it shows how unrealistic these logic puzzles are.
The 'correct' answer is always something like 98 0 1 0 1, where the head dude keeps 98 gold.
To see the flaw, simplify it to 3 pirates. The 'correct' answer here is 99 0 1. The reasoning is that if there were 2 pirates, the head dude would just keep it all since he wins the vote. So, with 3, the reasoning is that the head dude tosses a coin to the last dude, because if the head dude dies, the second one takes everything and the last one gets nothing. OK, the flaw is that THE HEAD PIRATE WILL NOT LET HIMSELF DIE!! He's not going to die, just to prove 3 wrong. The head pirate has no power. The 'optimal' solution where the main dude takes 99 and tosses a coin to the last guy is stupid. Why on earth would 2 and 3 accept one gold between them when they have the power to kill the first guy whos taking 99?
So now what? Given that 'optimal' solution, 2 and 3 decide to split the pot 50/50 and force the head dude to fork it all over in return for their lives. The head dude responds by offering one of them more than 50. A biddin war ensues, and likely ends with the head dude keeping 49 and either 2 or 3 keeping 51. Whatever happens, there is no reasonable solution where either 2 or 3 get less than 50 gold.
Help me take back Slashdot. When did 'News for Nerds' become 'FUD and Conspiracy Theories for Extremist Nutjobs'?
Oh, woe is me. I have a perfect logic puzzle, but was unlucky enough to be otherwise engaged when this story was posted. (By the way: a soft couch, a carefully selected DVD, half a bottle of rum, and a girl. Guess which element to this excellent scenario was fucking ruined by copy protection? I'll give you a hint: I may have just switched sides in this movie piracy debate. Fuck the RIAA. It was a perfectly legal store-bought DVD. Fuck them all.)
But anyway, logic puzzles. This logic puzzle is excellent. I've had it up on my site (http://www.xkcd.com/blue_eyes.html), and after I got boingboing'ed I got a lot of email about it, so I've been able to tweak the wording to get rid of most of the confusing stuff, leaving only the logic. It's extremely subtle; I've never seen anything like it.
Here's the puzzle:
A group of people live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight.
On this island live 100 blue-eyed people, 100 brown-eyed people, and the Guru. The Guru has green eyes, and does not know her own eye color either. Everyone on the island knows the rules and is constantly aware of everyone else's eye color, and keeps a constant count of the total number of each (excluding themselves). However, they cannot otherwise communicate. So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes, but that does not tell them their own eye color; it could be 101 brown and 99 blue. Or 100 brown, 99 blue, and the one could have red eyes.
The Guru speaks only once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone with blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb, It is not a trick question, and the answer is logical. It doesn't depend on tricky wording, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
xkcd.com - a webcomic of mathematics, love, and language.
Hmmm, let me guess.
Low taxes, no expensive workers protection, no minimum wage will move business to the 2nd coutry.
Business in the first country will not be able to compete with business from the 2nd country. Since workers protection
is good they will be afraid to hire people since it will be costly to fire them.
This will lead to much higher unemployment in the first country.
Now the answer depends on the unemployment benefits. If they are low - people will flock to the 2nd country faster.
If they are high - it will create another drain on the 1st country resources. In 100 years it will be significantly poorer than
2nd country and people will flock to the 2nd country then.
It is already happenning: Europeans moving to the US, I even know a few Europeans who moved to China.
Just compare France and the US.
If opportunity came disguised as temptation, one knock would be enough.
3^2 * 67^1 * 977^1
Hi - I like that famous story about the man sentenced to die within the next week but where the judge says he will be surprised on the day it happens - there was even a book called "The Unexpected Hanging" based on it.
Given an unbounded square grid ( ie an infinite square grid ) of one ohm resistors, What is the resistance between two adjacent nodes?
As I was going to St. Ives,
I met a man with seven wives.
Each wife had seven sacks.
Each sack had seven cats.
Each cat had seven kits.
Each kit had seven mitts.
Mitts, kits, cats, sacks, and wives,
How many were going to St. Ives?
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
3 1 2 2 1 1
1 3 1 1 2 2 2 1
1 1 1 3 2 1 3 2 1 1
3 1 1 3 1 2 1 1 1 3 1 2 2 1
continue the series!
Ah, arrogance and stupidity, all in the same package. How efficient of you. -- Londo Mollari
No, I claimed the relationship between tax revenues and tax rates is continuous, no that the entire economy can be described (predicted?) by a continuous function (of what? and predicting what?)
Of course, you're right I was imprecise, there can be jump discontinuities, but if the number of actors is large, these are small enough to be ignored. What is most important is that you assume it doesn't behave like -1/x near the origin: increase to a vertical asymptote and then increase from negative infinity back to the axis. But I think that's a safe asssumption.
But go ahead, keep believing that you can increase taxes indefinitely while increasing revenues. It doesn't correspond with reality, though.
Rank my idea: http://www.sinceslicedbread.com/node/531
The king asks: "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?"
I'm guessing any passage through the central chamber must've happened BEFORE they were locked up, no?
Cool, so how can a positive function increase from zero, and go back to zero without even decreasing? This I'd like to know!
Rank my idea: http://www.sinceslicedbread.com/node/531
Here's a similar apparently paradoxical result:
-1 = sqrt(-1) * sqrt(-1)
-1 = sqrt(-1) * sqrt(1 / -1)
-1 = sqrt(-1) * sqrt(1) / sqrt(-1)
-1 = sqrt(1)
-1 = 1
Can you spot the flaw?
Whoa, where are you in your argumentation? Claiming that you disagree with the Laffer curve is "aggressively misleasing" but the point you're trying to make is "incompatible with [my] precious curve"? Let's be a little less schizo here.
And in case you hadn't noticed, California and Massachusettes exist. FBI's actually a pretty big fan of them.
Rank my idea: http://www.sinceslicedbread.com/node/531
Reminds me of the famous interview question "Why are manhole covers round"? There are several possible solutions to the problem, but my favorite is the way I (naively) answered this when I first heard the question - "Because the manholes are round"
And thinking about it more, in the 3 pirate case the answer will be 0 0 100, where the third pirate gets everything and the head pirate lives (assuming that they don't start fighting during negotiations). The head pirate can't trust pirate 2 with any deal, because 2 just needs to vote no and he gets everything. So, the head dude needs to dump it all on 3 for his vote. If the head dude tries to keep anything, 3 can just say no, knowing that the head dude will not die just so that 3 doesn't get anything.
Haven't noodled through the 5 pirate problem, but I do know that the 'correct' answer found using induction is wrong given the problem.
Help me take back Slashdot. When did 'News for Nerds' become 'FUD and Conspiracy Theories for Extremist Nutjobs'?
Nice try, but wrong on about every sentence:
...RIGHT?
the income=f(tax ratio) is not a smooth function.
Not necessary, just at at some point a higher tax rate will come with lower revenues.
it is by all account not even a function, eg you can have more than one value of revenue for one tax ratio.
It is made with ceteris paribus assumptions, so just changing the tax rate can only do one thing. Okay, free will, non-determinism and all that. But I don't think that's what you were arguing.
The usual parabol-like curve
Does not require a parabolA-like curve.
is based on a very simplistic model
based on the concept of elasticity, which is far from simplistic insofar as leaving things out.
which is useless except for pushing an ideological agenda.
It's not just the right-wing that wants to raise revenues, right? And that's the purpose of taxation, right? You don't advocate taxation solely to hurt people *without* raising revenue, right?
The only sound reasonning in this is that high tax rates increase the incentive for escaping taxes, which is commmon sense.
No, that's not the main reason the Laffer Curve holds, except in very high-tax regions. High taxes also mean - in violation of your ideological agenda - that people will stop producing the things the government gets tax revenue from. This is just an application of the concept of substitution, elasticity, and deadweight losses. Think about that for a second: taxes can actually hurt! Novel concept there!
No need for half-baked maths to get that.
Yeah, claiming that returns have to fall at some point to get back to zero is "half-baked". I'd like to know what you deem non-baked.
Rank my idea: http://www.sinceslicedbread.com/node/531
Another way is to fill the 3G jug and pour it into the 5G jug. Fill the 3G jug again and fill the 5G jug. 1G remains in the 3G jug. Empty the 5G jug. Pour the remaining 1G from the 3G jug into the 5g jug. Fill the 3G jug one last time and pour those 3Gs into the 5G jug. There are 4Gs in the 5G jug.
I saw this one first in Scientific American (a long time ago). As I recall the answer was:
e^(10000 - y), where y is Euler's constant.
I think it works out to 10^43429
What you should have stated was that each step seems valid.
Those who sacrifice security to condemn liberty deserve to repeat history or something. - Benjamin Santayana
If you count highly-unintuitive functions as well, you don't even need a point: A single real value is enough to specify a plane.
The proof goes roughly like this:
The set of 3D planes has the same cardinality as R^3, as can be seen by the representation of planes via half-space functions (they have four coefficients, but since you can renormalize without changing the plane, only three are really relevant).
R^3 has the same cardinality as R (the set of real numbers). The idea behind the proof of this is that you can write a real number as its decimal expansion, put the digits into three groups (take every third digit), and re-assemble three real numbers from those three groups. The proof itself is slightly more complicated because you have to work around the fact that 0.9999... = 1.0.
I have omitted some fine detail here, but what this all means is that there are as many real numbers as there are planes in 3D cartesian space - there is a bijection between the respective sets, and this bijection is the function the Parent was asking for. It operates by splitting the real number into three real numbers, interpreting these as the coordinates of a point (or alternatively yaw/pitch/distance from zero) and constructing a plane from that as explained in other posts.
You take a solid sphere and drill a hole along the diameter of the sphere. The hole is 6 inches long. What volume of the sphere is left over after drilling?
or a more simple way of stating a similar concept:
2 != 1
0(2 != 1)
0 != 0
I just broke algebra, without even dividing by zero.
DYWYPI?
My fav math puzzle is:
There are two mathematicians in a room. The product of two integers >=2 is given to the first mathematician, and the sum of the same two integers is given to the second one. Hence, the first mathematician only knows the product and the second only knows the sum of the two integers. However, both are aware that the first knows the product and the second knows the sum of the two integers.
The first mathematician is asked whether he can determine the two numbers, and he answers no.
The second mathematician is then asked whether he can determine the two numbers, and he too answers no.
The first mathematician is then asked once again whether he can determine the two numbers and this time the answer is yes!
What are these two numbers?
Not an easy problem. Using only two NOT gates, and as many AND gates and OR gates as you want, construct a circuit with three inputs and three outputs. If the three inputs have values X Y and Z, the outputs will be not-X not-Y and not-Z. This will be true for all possible values of X Y and Z. Not a trick problem. Two valued logic.
spoiler i guess if it works.. not sure if it was posted yet..
.. and since I decide the minimum number of visits every prisoner gets I am going to say it's a gazillion willion fuckillion to the k ..fckit i love exponentiation). It does not matter if the king calls him first out of the prisoners or tenth. The point is, he is the first to try to transmits his name.
.. so if his binary name is 1001010101010010101001001010010.. When he first enters he will point the chalice "up" (one). The second time, he will point it "down" (zero).
.. it will probably takes hundreds of n's of transmission. Once the prisoner who is second in the sequence sees the complete unmanipulated name of prisoner 1 twice. Then it is his turn to transmit his name twice. And so on until every prisoner has transmitted their name two times. When prisoner number "n" has seen everyone's name he decides to say "yes".
.. multi bit gates work by doing the manipulation on the second turn, or third etc turn) based complex circuit that basically has some stable solution that a prisoner looks for. Note we aren't restricted to boolean gates. In fact NOT gates may fuck up the whole thing. That is, we can have a prisoner who sums 7 bits. etc. Given that the king can manipulate the order in which prisoners are called, I am doubtful a solution like this even exists (given a known set of gates, can arbitrarily reordering them produce predictable results at a location)? What I mean is would a prisoner say Bob, who decided to act as a simple X gate .. given that he is always a X gate, can he expect to get a sequence say 1010111010101 in a row regardless of how the "gates" are ordered? He just continues to do his job until he sees that sequence. This would involve the king to not have access to a large amounts of time and computational power. If he had that, and without knowing k, you are left with a situation of prisoners unable to decide on a suitable trigger situation.
.. got u a headache?
I decided on seeing what is the most complex solution and got these two.
Half ass solution 1 (given the prisoners know k):
If they know k, the following solution may work, I think. Try to optimize it. Actually i dont even know if this will work.
The prisoners decide on names for each other. The names has to be long like say n to the k to the something. The point is each prisoner's name is longer than k, and the names are unique from each other to the point where k manipulations cannot make a prisoner's name look like the others.
Then, each prisoner is given an order/sequence id. That is from 1 through n. The person who has the number 1, when he gets in the chamber he transmits his name (it will take multiple visits
When I say transmits his name, I mean "spell out his name in binary"
Before he has transmitted his name twice if he sees the chalice not left the way he had originally left it, he calls the king a shithead TAMPERER and re-starts his transmission of his name bit by bit. After his second transmission is completed, if he waits k turns and still sees no manipulation, he starts retransmitting his name. Yes it sucks to do it this way
If k is unknown and large enough, the king can fake to prisoner number "n" that everyone transmitted and foil the plan.
Note, this may take a zillion years for the prisoners to be freed, but the point is they'll be freed. Maybe the king will let them out early on the planning night just for coming up with a solution.
Half ass solution number 2 (given the prisoners DO NOT know k):
The problem doesnt say whether the prisoners know k. If prisoners do not know k, the problem becomes really hard. Because to decide on a trigger to say "yes", you have to ensure that it is not the king transmitting it to you. There may be a "solution" that involves a form of the satisfiability problem, that is each prisoner makes themselves into a logic gate (ie, a subset of prisoners act as reversible gates etc
Actually the second solution is utter nonsense
LOL.
Five sailors are shipwrecked on an island. On the island they find alot of coconut trees and a monkey. The sailors decide to spend an entire day collecting coconuts and then the next morning to split up the coconuts evenly between them and go their separate ways on the island.
The first day they spend gathering coconuts and place them all in one huge pile.
In the middle of the night one of the sailors wakes up and decides to take his share now. He splits the coconuts into 5 piles. One coconut won't divide out so he gives it to the monkey. He takes his 1/5 share, hides them on the island and puts all the remaining coconuts back into one pile. The second sailor wakes up and does the same thing - splits the pile into five, one coconut doesn't divide out and gives it to the monkey, takes one pile away and puts the remaining coconuts back into one pile. And so it goes for all five sailors. In the morning the sailors wake up make no comment about the smaller pile, divide the remaining coconuts between them with one coconut left over that they give to the monkey and go their separate ways.
How many coconuts were there?
There is a room with n people in it, where all their birthdays (only the month and day; year is disregarded) are randomly distributed. For what value of n is P(at least one pair of people have the same birthday) > 0.50 and for n-1, the same operation yields a probability of less than 0.50?
IWARS.
People, in general, disappoint me. Politicians even more so.
Thats not the method I have.
1^2 = (-1)^2
Take root on both sides.
1=-1
Now whats wrong in this proof...I couldnt find any.
Why does yahoo do this
A very small town is being planned. This town consists of just 6 buildings - 3 houses, and 3 service buildings - a gasworks, a power station, and a waterworks.
Each house must recieve all 3 services - gas, water and electricity.
To recieve a service, the house must be connected to the service building by a pipe*. The pipes can be as long as you like, and have as many corners or junctions as needed. However, the pipes from the different services must not cross at any point. The pipes can also not be routed through the houses, the houses are an endpoint only.
So in super-lameo-ASCII vision (ignore the dots, they're just there because Slashdot doesn't seem to like strings of spaces), the design for a 2 house town would look like this:So the question is, can you find an optimal design for a 3 house town?
*the truly observant amongst you will note that electricity tends to travel down wires rather than pipes. Nothing to say the wire's not inside of a pipe though, is there?
Answer in ROT-13. Paste here to decode
Ybbx pnershyyl ng gur dhrfgvba, "pna lbh svaq na bcgvzny qrfvta?" Ab, lbh pnaabg, fvapr lbh pna'g svaq n jbexvat qrfvta ng nyy. Vs lbh jnag gb trg ernyyl fznegnefr nobhg vg lbh pna chapu n ubyr guebhtu lbhe cncre naq pbaarpg lbhe cvcr gb gur bgure fvqr, be jenc nebhaq gur rqtrf gb onfvpnyyl znxr vg 3 qvzrafvbany naq nibvq pebffvat lbhe yvarf, ohg ernyyl, gung'q or nobhg nf purnc nf.. jryy, nf zr nfxvat guvf dhrfgvba va gur svefg cynpr.
Curiosity was framed. Ignorance killed the cat.
Oldie but goldie, and certainly a riddle that has made most people I've mentioned it to think rather hard. Consider the following situation: we play a game where I present you with two envelopes, each of which contains a certain amount of money; you may choose one, open it, and keep the money in it. One of the envelopes contains twice as much money as the other, but outside of that, you don't know how much they contain (I do, though, because I put the money inside).
:) If you haven't heard of this riddle yet and think you know the solution, be sure to also explain it. :)
Now, there's a twist: after you choose one envelope but *before* you open it, I offer you a chance to back out of your decision and take the other envelope instead. You can do this or not; it's completely up to you. No matter what you do, the envelope you choose now is final, and you get to keep the money in it, but you don't get the money in the other envelope.
Naturally, you want to maximize the amount of money you get.
Now, the problem is as follows: after you choose one envelope, which contains a certain amount of money n, you know that the other envelope contains either n/2 or 2n. In other words, the other envelope contains, on average, 1.25n, which is more than the envelope you currently have - so you should trade yours for the other one. Okay.
However, you could also look at it the following way: call the smaller amount of money n, then the envelope you have contains either 2n or n. On average, it thus contains 1.5n, and the same goes for the other envelope, too - it also contains either 2n or n, and thus, on average, 1.5n. So it doesn't matter whether you trade yours or not. Okay.
Unfortunately, it's obvious that not both these things can be correct - it either is better for you to trade, or it's not, but not both at the same time. What's going on here?
If you heard about this before and know the solution already, please don't post it so the riddle's not spoiled for everyone else.
quidquid latine dictum sit altum videtur.
In math lab, they had this problem in brownies. I said: split the brownies into a top half and a bottom half. Not sure I could do it with paper, but I could do it with paper-thick mica.
Given :
Twelve billiard balls, numbered 1-12, otherwise identical
except for the fact that one (and only one) differs in
weight from all the others. It is unknown whether it is
lighter or heavier.
The Problem :
Part 1:
Using only a balance scale (capable of determining
heavier/lighter or equal), what is the smallest
number of trials necessary to determine (over all
cases) which ball is the odd one and whether it is
heavier or lighter than the others ?
Part 2:
Demonstrate the answer to Part 1
I actually got this one at a job interview and was unable to solve it. When told the answer I was very upset and explained why I would never use such a shoddy solution (and yes, I got the job :). You see, a lightbult is defined as a light-producing device; the specification says nothing about heat. If you rely on heat to make your solution work, you're asking for trouble in the long run. As lightbulb technology improves, there is absolutely no guarantee that lightbulbs will continue generating heat. In fact, it is almost a certainty that eventually they would generate only light and no heat at all, at which point your solution will break and you'll have to spend weeks hunting through your code to figure out why it isn't working. Add to that the fact that any programmer worth his salt wouldn't even think of undocumented features like this (especially because in his time all lightbulbs produce no heat), the chances of the bug being found go down significantly. The moral of the puzzle should be: "thou shalt not use undocumented features".
No you didn't. You broke the rules of algebra. You just demonstrated a rule you didn't know about. While there is a zero identity of equality, there is no zero property of inequality.
In other words, your second step is invalid; you're not allowed to do that.
Mod me down and I will become more powerful than you can possibly imagine!
Here's another nice one, courtesy of Raymond Smullyan.
Suppose there are three people, called A, B and C. Each of these is a "perfect logician"; that is, given some information, they all are able to immediately draw any and all conclusions that can possibly be drawn from this information. Furthermore, suppose there are four red and four green stamps.
Now, all three of them close their eyes, and two stamps are glued to their foreheads, each; the remaining two stamps are put away. Now, they all open their eyes again.
Then, the first, A, is asked whether he knows the colours of the stamps on his forehead. He says he doesn't. Then B is asked the same thing, and also says he doesn't, and afterwards, C is asked and says he doesn't, too. Now, A is asked a second time, and he still says he doesn't know. But then, when B is asked a second time, he now says he does know.
The question is: how?
quidquid latine dictum sit altum videtur.
You have a large bag and a set of numbered coins from 1 to n. You start by added coins with the numbers 1,2,3,4,5 into the bag. Then you take out the coin with the number 1 and set it aside. Now there should be 4 coins in the bag. You then add the numbers 6,7,8,9,10 into the bag and remove the 2 coin. There should be 8 coins in the bag. If you could repeat this procedure an infinite number of times, how many coins will be in the bag?
These puzzles seem to fall into a limited number of classes, when you view how you get the solution. ...etc
ONe class is the strict logic, like the old chestnut where you get some set of statements like "1, joe lives next to the geen house" "The blue house is next to joes best friend"
Another class involves a play of words, like I came to town on monday and left two days later on friday, where friday is the name of the airplane
A third class includes physical situation outside the language, like the early post in this thread on the MS puzzle of a closed room with a lightbulb and three switches.
IN any event, it seems like these puzzle fall into a small number of classes, and onece you know that, you can develo simple heuristics to solve them. Of course, it may be that the puzzle requries math or some other specialized skill you don't know, but if the heuristic tells you that, you have solved the puzzle, because you now know where to get the answer frm
All you seem to care about is how taxes relate to government income. Sure, when Reagan cut taxes in the early 80's tax income dropped. But the economy improved by leaps and bounds.
Unemployment went from 11% to 6%. Incomes rose, and more people bought homes.
Who the fuck cares about tax revenue. I only care about how citizens are doing.
2 bad u r such a jerk, because the idea that math is obsolete for a large fraction of the population is an interesting idea..since a caclculator costs about a dollar to make, and since 90% plus of hte population can solve all of their math needs with a calculator, why do we bother teaching math before the second year of college ?
I think you could make a serious argument about this, if one wnated to get into a serious discussion of what "education" consists of, and how many people in the US really NEED to know algebra, or trig, or something ike that. I will bet dollars to donuts that surveyors, a profession traditionally associated with strong math, are BETTER if they don't know math because the computer makes fewre erros !!!!
So there isn't enough information.
04 08 15 16 23 42
Mmmmm.. Feed me more tv.. Must consume... Must consume...
-- If we don't stand up for our rights, now, there will be no right to stand up for them later.
I don't understand your explanation. Take this example and please explain what I'm doing wrong:
Split the 50 coins into 2 piles of 25. Assume the 18 heads are randomly distributed, so you get:
Pile 1: 4H, 21T
Pile 2: 14H, 11T
Randomly select 18 coins from Pile 1 to flip:
Pick 3H and 15T from pile 1 and flip them
Pile 1 now has 16H and 9T
So the end result is:
Pile 1: 16H, 9T
Pile 2: 14H, 11T
Either I'm doing something wrong, or your explanation is unclear, or it just doesn't work.
I think there's a fault (now that I'm caffinated I can tie my shoe laces too!). He doesn't need to keep calling them. The puzzle is that he only needs to call them the same arbitrary number of times. So he could call them only once each and that would defeat this.
It's a nice solution otherwise.
"The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose. "
If you really get one of 'those' meetings or classes, you can try this. It is so boring, you have already made another Tic Tac Toe crossed set of lines. Take all 10 numeral digits and put them in the Tic Tac Toe so that all horizontal, all vertical and all diagonal sums each add up to ... 15
I give no hints.
First, here is a proof by counterexample that there is no general solution to this problem. The problem has 3 parameters, n the number of prisoners, k the number of allowed flips, and a, the arbitrary number of times the king must call each prisoner.
Proof: no solution for k > n * a
Let k > n * a
This means the king can flip the challice once for every time a prisoner is called in to see him. So he can, for example, always turn the challice up before each prisoner is called in. He will select a permutation S of the prisoners, and call each of them in a times before calling the next. Each prisoner will see a string of a UP challices. They will have no way of knowing whether they are the first or last in the permutation S.
Proof there is no "watcher" solution:
let w be a prisoner who watches for a solution and answers yes when everyone has been in the room.
The king overhears the prisoners planning, and so knows w. He selects S, a permutation of the prisoners, such that it maps w to 1. He calls each of them in a times before calling the next. The watcher, w, can never answer yes because no other prisoners have ever been in the room when the watcher is.
In fact, for any number of "watchers" up to n-1 the king can beat the strategy by knowing who the only non-decider is and showing that prisoner the challice last.
So here are some hints for you:
First, get Brian to either give you adequate constraints on k before starting or rephrase and expand the question so that he also asks what constraints on k need to be placed.
Secondly, the algorithm used by the prisoners will be symetrical, though they might assign themselves an asymmetry in the form of keys used to labels the prisoners.
So, basically, you're saying that you encode the specification for a plane in another number?
That seems to go against the problem statement that the N points are specified in cartesian 3D space. If their encoding can be arbitrary, then they can't very well be called N points in cartesian 3D space.
Why are you letting these clowns ruin our country?
Is if he really would have broken if she hadn't said anything.
No sig for you!!
It's been know to cause fights and make people's brain hurt.
III.IIVIVIXIIVIVIIIVVIIIIXVIIIXIIIIIIIIVIIIIVVIII
Duh
3.243F6A8885A308D313
ok, you are in a factory with 14 machines, producing coins. All the machines are supposed to produce exactly the same coin. But by weighing some random coins, you find out that there is one machine that produces a coin that is exactly one gram heavier than the others. Find the correct machine by doing ONE measurement.
:)
It's not that difficult, but it took me a few minutes to figure it out the first time
Doolittle :
Bomb no.20 : To explode of course.
It is not directly clear from the initial description, but brian0918 has clarified that the sequence in which the prisoners are called satisfies the condition that for all i in N there is a finite but arbitrarily long prefix of the sequence after which every prisoner has been in the room i times. This is a requirement for "watcher" solutions, but it is also a requirement for this to be solvable at all. If you had to set a fixed number a of times that each prisoner has to be called, then the king could just call one of the prisoners a times in a row, then the next one and so forth. That would reduce the problem to one where each prisoner gets one bit of information from his predecessor, possibly manipulated by the king. I don't think I have to prove that with no information about the sequence and just one untrustworthy bit of information from your predecessor, you cannot decide in one shot whether everybody has been in the room by now.
k is an arbitrary, but fixed and known, number. Given the sequence description, k can be any number at all without rendering the problem unsolvable.
No, c 1 is valid. The problem is that that's only valid when the denominator of the fraction is positive.
The darkness... controls the music. The music... controls the soul.
Three salesman are attending a small town sales convention. With only one hotel, the options are limited. Even worse, the hotel only has one room remaining. When the men enter to stay for the evening, the hotel owner decided to make a few extra bucks on the last room. Instead of charging the usual $25 for a room, he ups the price to 30 buck even. Each of the salesmen forks over $10 each. The bellboy escorts them to their room, but is stiffed for a tip. Later that evening the hotel owner realized that taking advantage of the salesmen during their time of need was not the proper way to do business and gives the bellboy the $5 he overcharged to return to the salesmen. As the bellboy makes his way up the stairs, he thinks (a) he didn't get a tip and (b) the salesman can't split $5 evenly, SO he only returns $3 and keeps $2 for himself. Here's the question. After the refund, each salesman paid $9 for the room, and the bellboy kept $2... where is the remaining $1 from the original $30.
Bob asks Alice how old her three sons are, and Alice replies that the product of their ages is 36 and the sum of their ages is the number of windows on the building they happen to be in front of. After a moment of thought, Bob replies that that is not enough information to determine their ages. Alice realizes her mistake and adds, "the youngest in years has red hair."
[Def: youngest in years means floor(age of child A) floor(age of child B) and floor(age of child A) floor(age of child B) where child A is the youngest in years and age is given in years.]
Since a solution was already posted, i don't feel bad posting another, differently stated.
split m coins with h heads into 2 groups of size n and (m-n).
group (1) contains h(1) heads and t(1) tails. correspondingly, group (2) has h(2) heads and t(2) tails.
consider group (1) of size n.
h(1) = h - h(2)
t(1) = n - ( h - h(2) ) = n + h(2) - h
now, apply f() to flip over all the coins in group (1).
f(h(1)) = t(1) = n + h(2) - h
f(h(1)) = h(2) when (n - h) = 0 , i.e. when n = h .
so make a group equal in size to the number of initial heads then flip them all.
x^2 = x * x = x + x + ... + x (x terms)
d(x^2)/dx = 2x
d(x^2)/dx = d(x+x+...+x)/dx = 1+1+...+1 (x terms) = x
2x = x
2 = 1, provided x != 0.
"But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose. "
I can see what you mean, but only if I ignore this sentence:
"and he can call and recall each prisoner as many times as he wants"
Split the coins into two groups of twenty-five. Put the coins on their edges. 0 coins in each pile are heads-up.
I talked to a college graduate that went to an interview at Microsoft. Apparently they throw a ton of these at you. The sad part is that like 90% of the people know this and get guides (online or something) to study the puzzles beforehand, so they know all the answers and don't actually do any thinking on their feet. Anyways, here's one of them ...
You have 2 strings. Each string takes 1 hour to burn from end to end. You do not know the burn rate of the strings, it is completely random, and the burn rate varies along the string (ie 90% of the string may burn in the first 2 minutes and the last 10% may burn in the last 58 minutes). And the burn profile for each string is different (ie you can not flip the strings to have them burn symmetrically against each other).
Use the 2 strings to measure a 45 minute time period.
So I started over with the 9 in a non-corner edge box, and 5 or less in the middle. 8 has to go in a corner or else you end up with nowhere for the 6 again. So 8 goes in a corner not involved with the 9. This leaves only one possible place for the 7, in a non-corner edge. At this point we know the middle has to be 5, because 1-4 quickly leads to multiple uses of numbers in the grid. So putting 5 in the middle leaves the rest easy to fill in:
492
357
816
or any rotation of n*pi/2, n is an integer
I was introduced to this in high school at the MS Governor's School camp (hi everybody!) under the name "Polar Bears." The rhyme is:
/Was the guitarist for the band in the talent show.
Polar Bears
They Come In Pairs
They sit around a hole [1] in the ice
Like petals on a flower.
The game was great and swept the camp even though the hints are stronger in this version, newly promoted "BearMasters" are then given the question:
"Now, how many fish?"
It's sad how strong that memory is.
[1] MSGS'ers say it with me: "hoooooooooooole in the ice"
The title itself tells you that the hole must be ten inches long rather than ten inches wide, since there would not otherwise be sufficient information for a unique solution. So there really was enough information.
SPOILER WARNING!
6/(1-5/7)
Petals AROUND the rose meant nothing to me. I looked at them for about 20 minutes and had no clue.
Only until I read this "bears in holes" thing did it suddenly make sense.
I guess I am really weird.
Spoon not. Fork, or fork not. There is no spoon.
No.
Here's the answer:
You ask each person, "which direction will the other guy point me to if I asked him which way will take me home?"
eTrade SUCKS
Lying awake at night, you hear a motorcyle drive by. As it passes the house, the pitch of the motor drops by 1/2 step on a standard piano scale. How fast was the motorcycle traveling?
The Whole Time I was thinking that flipping a coin would give a random result. Trying to figure out how the heck to do it that way is a nightmare.
12 marbles, 1 balance scale, 3 chances to use the scale.
One marble is different in weight (lighter or heavier).
How do you find which one?
(more clues to answer below)
3 places to put the marbles,
Balance Scale [left side, right side] , and on the table.
3 states for the balance Scale: left side goes up, left side goes down, no movement.
The full answer is very elegant.
Very fun to write down.
Then more questions come up, like: how many trials are necessary for 13 balls? 16? 30? any random number?
You, kind sir, suffer from wishfull thinking.
> No, I claimed the relationship between tax revenues and tax rates is
> continuous,
That's the first time you've said that. And in an argument with someone else I find:
him: " The trouble with the Laffer curve is that the income=f(tax ratio) is not a smooth function."
you: "Not necessary, just at at some point a higher tax rate will come with lower revenues."
So you state here that the function isn't necessarily continuous. I don't know why you think self-contradiction and pure invention are good arguing techniques.
> no that the entire economy can be described (predicted?) by a
> continuous function (of what? and predicting what?)
You claimed that the Laffer curve is a good economic model. You also claimed that the Laffer curve is a continuous function.
Hence you claimed "economy is described by a continous function". It's still called logic, and it's still a correct use of the verb "to describe".
The problem most of us have with the laffer curve is that it's simplistic pseudoscience that ignores the many factors at play in a modern economy, as characterised by your king example. Yes, you are trying to distill a highly complex subject down to one formula.
You try to defend this with latin "ceteris paribus assumptions, so just changing the tax rate can only do one thing". Newsflash, changing the tax rate won't only change one thing. Scientists actually try to keep all other things equal through the use of control experiments, economists just talk shit.
> Of course, you're right I was imprecise, there can be jump discontinuities,
> but if the number of actors is large, these are small enough to be ignored.
There are jump discontinuities in experimental data, but the laffer curve is a theoretcial abstract, and I've never seen it drawn with discontinuities by anyone (other than Martin Gardner). Do you understand the difference between theory and data?
> What is most important is that you assume it doesn't behave like -1/x near the
> origin:
Your favourite trick: make all sorts of bizarre claims as to what other people think when they've never said any such thing. Quote me.
> increase to a vertical asymptote and then increase from negative
> infinity back to the axis. But I think that's a safe asssumption.
Personally I follow the standard of plotting the independent variable along the horizontal axis.
> But go ahead, keep believing that you can increase taxes indefinitely while
> increasing revenues.
Your favourite trick: make all sorts of bizarre claims as to what other people think when they've never said any such thing. Quote me.
The king can only change it k times. The value of k influences how long it will take before the prisoners can be confident of a yes answer. This is a thinly disguised scenario for a "zero-knowledge" cryptographic authentication protocol.
*** Possible Spoiler***
I have two possible solutions. The first involves the prisoners electing a starting prisoner. once he is selected he places the chalice at the edge of the table nearest to his door. This would assume the king only changes the orientation of the chalice and not the position on the table. Then the other prisoners are called and if the chalice is placed one position counter clockwise to their cell then they move its position. Once all the prisoners have been called the original starter will notice the chalice in the position in the cell counter clockwise next to his. He can then answer the question yes. My other solution would be as others have described in the orientation of the chalice, as others have said if the king can change it then you can't possible calculate anything from it. But I think that there is a chalice and a cup on the table. reading the problem it mentions the word cup when describing that the king can manipulate it.
Ah I got it. Took 16 rolls (written down) and almost hour. The fact that I am ignorant about roses didn't help :)
:)
Anyway I read that story and it didn't appear to me that he was trying to solve it by memorization, but rather that after an hour, seeing hundreds of rolls, he remembered many of them, which isn't all that surprising. What I got out of the story, is that he persistently kept at the problem trying many different ideas until he finally got it, even after everyone else in the group had solved it or quit.
I don't think that how quickly someone solves any one particular problem is much of an indicator of how smart they are. We were doing brain teasers on an ACM trip my freshman year of college, and I was one of the first to get most of them. Then there was this one that everyone got right away and I couldn't get it. When I finally did a day later I was kicking myself because it was so obvious - I just wasn't looking at it the right way.
This story showed my that Bill Gates is a very persistent and determined person which is probably a big reason that he was so successful. That and he needed to get a girlfriend at that point, as do I apparently
This is a math question which I've asked many people to test them for job positions ranging everywhere from mathematicians requesting jobs as encryption engineers to sales people selling embedded applications. Their first and second responses to the questions can provide me with a tremendous amount of information regarding the persons' intelligence as well as their social skills. I don't expect sales people to come up with the right answer, but I do expect that after a quick guess, they should admit their guess was wrong, not try to sell me their answer no matter how wrong it is. Only once have I ever seen anyone answer the question without using paper, and he was a Ph.D. in math from Chambridge. Often I like to simply hear peoples suggestions on how best to solve the problem or at least find out what the first thing they'd look up would be to solve the problem.
Give it a try :
Given two circles of equal size (in other words, both are in fact circles and have equal radii). To simplify the question we'll assume the radius of the circles is 5. What is the distance between the origins of the two circles so that the area of the intersecting part is equal to the nonintersecting area of one circle. In other words "How far apart are the centers of the two circles so that if you compare the overlapped region of the circles so that it is equal to one half the area of one of the circles)
The following bit of information is what I tend to look at for answers.
First, with the exceptions of math freaks, people immediately answer 5 without thinking about the fact that this is completely wrong. I've had a sales person spend 10 minutes trying to convince me that his answer is in fact correct. He did no get the job.
Shortly after, I typically am hoping to hear "Well, it must be less than 5 since the circles would have to have a greater overlap in order to achieve half coverage". For programmers, I hope to hear this observation before they reach for a pencil. If this conclusion isn't reached or they've gone the wrong direction, the programmers typically don't get the job either.
I typically allow a few minutes from this point before I see a bunch of work on paper and if they're on the right track, I sometimes let them go ahead until they solve it or give up. Alternately, if they are not on the right track, I ask them "What's the first thing you'd look up to solve the problem?". What I'd like to hear is something like "I'd find out how to calculate the area of the intersect".
The beauty of the problem is that the question itself (especially when you can draw two circles and describe it) sounds so incredibly simple, but in fact even most programmers can't solve it since this is the type of math they learn an later forget.
As a note, I am a programmer and I've taught myself all my math, well at least post high-school math, I just tested out of the college level stuff since I didn't have the money to pay the whole credits. It took me 3 months on and off solving this problem. All together I think I had 45 hours invested it (not counting the endless hours laying in bed thinking about it)
Of course not. Since the description of a set of numbers changes when the number changes, you'll never have a number sequence of "1111" or "2222", so therefore, you'll never describe a sequence longer then 3 of the same number.
"That's so plausible, I can't believe it!" - Leela
You're assuming the density of the bowling ball.
If the ball is heavier than water, there is one answer.
If the ball is lighter than water or the same as water, you have another.
In no case does the water level rise.
I'm a little confused....
let's just say for god A, 'da' means yes. Then does 'da' also mean yes for gods B and C? or do each speak there own langauge so if both god A and god B said 'da', it could be they were both answering yes/no or it could also be that they didn't agree wtih each other
A once well-known study at Stanford used a cohort of 1,000 students, half mathematicians, half engineers. The students were introduced one at a time into a small concrete bunker. Inside was a bare concrete room. In one corner was an old bucket of water and in the opposite corner was an old tin bowl. The task was to get the water from the bucket into the bowl.
First time round 100% of the students picked up the bucket and poured the water into the bowl.
Then they were fed into the room a second time. This time there was Wilton carpet on the floor, a window with a view across campus, Laura Ashlet wallpaper and drapes, a fine rosewood table with a linen tablecloth, and a Waterford Crystal pitcher of iced water on it. In the corner was the (empty) bucket, and in the other corner was the tin bowl. Same task: get the water into the bowl.
The 500 engineers picked up the pitcher of water and poured it into the bowl.
The mathematicians picked up the pitcher and poured the water into the bucket, thereby reducing the problem to the terms of Part I, and we already have a solution to Part I...
Mathematics, sorry, "math" is a thinking skill. If you don't know "math" you are an inferior product. I believe that America is used to pumping out inferior products. So please, feel free, stop fitting your human resource units with skills. No one wants your overweight units anyhow. It's no real loss. ;}
what is the square root of 69?
ate something
(its 8.306623862918)
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to take the switch?
I love the proof that .9999... repeating to infinity is equal to 1. People never believe it until they see the simple demonstrative proof:
1/3 = 0.3333333...
3*(1/3) = 3*(0.3333333...)
3/3 = 0.9999999...
1 = 0.9999999...
Before this goes off the front page, I'd like to thank everyone who wrote in with their puzzles. They'll certainly be keeping me busy for a while.
-Shma
I came here for a good argument
Two puzzles - with answers I'm afraid. 1) Two doors, one leads to instant death, the other to safety. Two guards, one always tells the truth, the other always lies. How do you escape? 2) Sherlock Holmes and Dr Watson arrive at two doors, one yellow and one red, Sherlock says "We take the yellow door Watson". Why? 1) Ask either guard, which door _the other guard_ would say leads to safety. Then take the door which he did not indicate. (Which ever guard you ask, the answer will contain one truth and one lie, so the answer will be the wrong door) 2) "It's a lemon entry my dear Watson!"
***You learn something Every day. And then you die.***
How many people must you invite to a party to be certain (probability = 1) that at least 3 of the attendees either all knew each other beforehand, or, of the three, none of them knew either of the others?
Here's one I came up with myself. It's not hard, but the answer is surprising.
What is the angle at the point of the "wide" version of the seven pointed star? This is the figure you would create if you start with a circle and draw 7 equally spaced dots, then connect every dot to the 2nd one following. Express the answer in degrees as a mixed fraction (e.g. 23 1/3 degrees).
The number seven is considered lucky, and you might find the answer surprisingly lucky!
It's worth pointing out that at the time when monarchs could personaly pocket tax income (and it's pretty much never been that straightforwardly simple, even when "tax" meant "tribute"), social infrastructure, including the ability to assess and collect tax, was far, far more primitive. It's not apples and apples.
This is old and it's just a mathematical trick, but it's still fun figuring it out. You'd be surprised how many people will never get it.
http://digicc.com/fido/
So you state here that the function isn't necessarily continuous.
True, the continuity deal was too restrictive. The Laffer Curve holds in *more* instances. Thanks!
You claimed that the Laffer curve is a good economic model. You also claimed that the Laffer curve is a continuous function.
Hence you claimed "economy is described by a continous function".
The Laffer Curve is not a model of the economy. It does not purport to describe an array of economic activity, just that at some point, revenues diminish despite tax increases.
The problem most of us have with the laffer curve is that it's simplistic pseudoscience that ignores the many factors at play in a modern economy, as characterised by your king example. Yes, you are trying to distill a highly complex subject down to one formula.
Nope, no formula. And claiming that raising tax rates eventually diminishes revenues is not pseudoscience.
You try to defend this with latin "ceteris paribus assumptions, so just changing the tax rate can only do one thing". Newsflash, changing the tax rate won't only change one thing. Scientists actually try to keep all other things equal through the use of control experiments, economists just talk shit.
I said it will do one thing, not change one thing. Sure, raising taxes could raise revenues, increase unemployment, lower interest rates, and so on. But the person was claiming that the same change in tax rate (with all else held equal as per the curve's ceteris parib assumptions) could BOTH increase AND decrease revenues and is thus irrelevant to revenues. This is false.
And there are economists that dismiss conclusions grounded in empirical data for the precise reasons you gave: you could never control enough variables to get a good result.
There are jump discontinuities in experimental data, but the laffer curve is a theoretcial abstract, and I've never seen it drawn with discontinuities by anyone (other than Martin Gardner). Do you understand the difference between theory and data?
Okay, fine, if you raise taxes, revenues will JUMP down, rather than decline smoothly. Happy now?
> What is most important is that you assume it doesn't behave like -1/x near the
> origin:
Your favourite trick: make all sorts of bizarre claims as to what other people think when they've never said any such thing. Quote me.
I never attributed anything to you. I was using "you" in the sense of "one". What's important is that one not allow that revenues can increase asymptotically and then increase from negative infinity... yeah, not a tough assumption to make.
> increase to a vertical asymptote and then increase from negative
> infinity back to the axis. But I think that's a safe asssumption.
Personally I follow the standard of plotting the independent variable along the horizontal axis.
Me too. Your point?
> But go ahead, keep believing that you can increase taxes indefinitely while
> increasing revenues.
Your favourite trick: make all sorts of bizarre claims as to what other people think when they've never said any such thing. Quote me.
Okay, you don't believe you can increase taxes indefinitely while increasing revenues. Congratulations: you agree with the concept of the Laffer Curve. Which is all I was trying to show. What a waste of time, when you already agree with me!
Rank my idea: http://www.sinceslicedbread.com/node/531
Sure it is. Monarchs tried to maximize revenues. If building a new bridge meant he could get more in taxes, accounting for the interest rate, he would do it. If some monarchs systematically guessed wrong on the utility of bridges, the ones that didn't would beat them out (war, immigration). The evidence I gave is actually more conclusive: that monarchs did not systematically engage in social spending and internal improvements probably means they don't increase tax revenues, which mean they probably don't increase the health of the economy, else they would lead to more tax revenues.
Rank my idea: http://www.sinceslicedbread.com/node/531
Assumption "2) To rotate an object one needs to give it angular velocity, " is wrong. For an explanation, see e.g. "H. Esséna t=application/pdf&identifier=oai%3AarXiv.org%3Aphy sics%2F0401146
The Cat Landing on its Feet Revisited or Angular Momentum Conservation and Torque-free Rotations of Non-rigid Mechanical Systems
American Journal of Physics 49 (1981) pp.756-758." The "Cat problem" is also mentioned in ch.3 in http://citebase.eprints.org/cgi-bin/fulltext?form
You can actually get an idea of this on an ordinary office swivel(sp?) chair. Sit on it, stick your arms straight out-left and right, as quick as you can move one arm forward and the other backwards. Your body will turn in the opposite direction as a reaction. Now slowly lower your arms-still pointing forward and backward. When they are back along your body (front and back) move them slowly so they are along your side. Now your body posture should be the same as you startet with, but rotated a small angle. Repeat as many times you want to e.g. rotate 360.
The rich have economic power, by virtue of controlling the medium of economies, and can manipulate economic policy. Thats not really news. I think it might be interesting if you did something like add a 100% tax cap after $500,000 of income. Nobody would have any incentive to make that much money any more - but the money that goes to extreme (CEO level) salaries and from huge investments doesn't just go away. Think of it as the accelerated version of trickle down economics.
I had a conversation with Brian0918 on AIM this morning, in which he revealed he's really trolling when I pointed out to him there is no solution (see my other posts) on this topic. Here's a little tidbit: ". i usually just post the problem to get people into big disputes, which so far has worked 2 out of 2 times". If you want the full conversation, email me at sbartaNOSPAM_at_MAPSONgmail.com.
Rank my idea: http://www.sinceslicedbread.com/node/531
It can't, but that doesn't meant the Laffer curve is infallible. It makes certain assumptions, like there are no external actors it doesn't account for. A 100% controlled economy can easily exist, with all wealth and production controlled by the state, and still have greater than 0 productivity.
Am I the only one who doesn't understand or accept the "one" answer? Ok, I get that we're supposed to assume that the man and his wives (and cargo) were going in the opposite direction because this is the only explantion for meeting a fellow traveler. Pretty weak, but it's halfway logical so I'll let this one pass.
HOWEVER, you guys need to pay attention to the final question: "Kits, cats, sacks, and wives, how many were going to St. Ives?" This cannot be answered "one" unless we also assume that the speaker is a wife. Or a talking cat. Or a sack given life by a mad wizard.
Is it really reasonable to assume any of these things? I suppose the traveler *could* be a wife, but there's nothing in the riddle that would hint at such a thing. The narrator didn't ask how many PEOPLE or BEINGS were going to St. Ives; we're only concerned with the "Kits, cats, sacks, and wives"; ostensibly the wives mentioned in the previous line and not referring to the speaker at all. If he meant wives and cats in general as opposed to these specific wives, I suspect that the answer would be unknowable unless you could somehow account for every wife and cat and kitten and sack traveling/being taken towards St. Ives from all directions. And if "How many were going to St. Ives?" was meant to refer to all travellers in the vicinity, then "Kits, cats, sacks, and wives," has no business being attached to that sentence. Riddles that lie aren't really riddles at all; they're what I like to call "pointless drivel."
So yeah, it's a pretty dumb riddle, but the commonly accepted answer is supendously retarded. I can only surmise that readers are too distracted by the rhyming to notice. An equivalent mathematical puzzle would go something like this:
"What's "i" squared?"
"Um. Negative one..."
"HAH! Wrong! You thought I meant the imaginary number didn't you?! In this equasion the variable i=5, so the real answer is 25! Haha, you fell for it."
Yeah, real fucking hilarious...
The way this works is each | represents one tick mark and you have to move one and only one | mark to make the problem true. This is ideally suppossed to be done with match heads so you can think of V and X as two seperate match heads combined together. You can't change the = sign and make it a less than symbol. I will post the answer in the next reply.
XX||
|| = ------
V|||
My answer is that they do weigh the same, and if you try to claim "oh no, I really meant TROY pounds" I will sue the crap out of you for false advertisement :-p
for a second there, I thought you were saying "I reprogrammed my genetic code in order to temporarily boost my brainpower" and I was like whoa, that's hardcore.
sarcastic putdowns are the last refuge of the arrogant and stupid, to paraphrase a famous phrase
With three questions there are eight possible cases you could distinguish. Since you're only asking for the order of the gods (six possible permutations), it seems momentarily plausible. But then we remember that Random's answers don't tell us anything, so if (da,da,ja) maps to (False, Random, True), then (da,ja,ja) must also). Similarly, there must be two sets of answers that map to each possible permutation. But we don't have twelve possible outcomes from the questions. Parent is a troll. QED.
However, this can easily be solved by increasing the number of times each non-leader prisoner flips it by k. If they each flip it 2k+1 times, then the leader can wait until he gets to a count of (2k+1)(n-1)-k.
Proof that it is reachable: Since the prisoners will flip it (2k+1)(n-1) times and the king can only cancel out k of those with his flips, this count will be eventually reached.
Proof that everyone will have visited the room at this time: The worst case scenario would be if every non-leader prisoner but one has flipped it all 2k+1 times and the king has used all k of his flips in the same direction. This will cause the count to reach (2k+1)(n-2)+k. I believe if you do the algebra, you will find this is one less than the required (2k+1)(n-1)-k. In order to get that last flip, the last prisoner will have to go into the room.
If you also consider that you don't know which way the chalice will be initially, you can just assume it will be one way. If it is not, you can assume the king has used one of his now k+1 flips to make it that way, and adjust the formula as needed.
Any objections?
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
switch 1, leave alone
switch 2, flip, wait 5 minutes, unflip.
switch 3, flip at the unflip.
Run to room, check temps/states of all bulbs.
1 is either hot/on or cold/off.
2 is either warm/on, or warm/off
3 is either Hot/off, or Cold/on.
Storm
Random isn't truly random - you just don't know whether or not he's telling the truth. However, if you ask a suitable question:
If I were to ask the X god what the word for no/yes is, what would he say?
I won't give the exact phrasing away, but if you can narrow down what the words are, then you can easily figure everything else out...
If that's true then you should be able to prove it.
The poster is /not/ posting a problem without a solution, or one that only his close buddies have confirmed. I have no idea who the poster is, and have come across a slight variant of the problem before; it's a great problem, and does indeed look insoluble at first, but the given solution works, and if you take a little time to check the numbers, you'll see that.
Concrete example: n = 2. There are two prisoners, just the counter (Alice) and one other (Bob). Each time Alice comes in, she turns it upside-down if it wasn't already. Each time Bob comes in, he turns it right-way-up if it wasn't already.
First claim: If Alice has ever turned it over k+2 times, she knows then that Bob has been in.
Justification: She can know this because between any two , the cup has been turned back somehow at least once; so in total it's been turned right-way-up at least k+1 times. Since the king can only flip it k times, at least one of the times it was turned back up must have been by Bob. So Bob must have been in.
Second claim: Sooner or later, whatever the king does, Alice will have turned it over k+2 times.
Justification: The statement that each prisoner gets called in arbitrarily many times (as long as they're not set free first) can be restated as that no prisoner has ever been called in for the last time. So starting from any given time, we can find some time after it that Bob is called in, then some time after that that Alice is called in, then another that Bob is called in, and so on as far as we like. In particular, we can find (2k+2) pairs of times alternating in this way.
Each time that Bob goes in, he ensures the chalice right-way-up. Each time Alice leaves, she leaves it upside-down. So between each time Bob goes in and the next time in our sequence that Alice leaves, the cup gets turned upside-down at some point, either by her or the king. But we found (2k+2) such pairs of times, and the king can have turned it upside-down at most k of times, so Alice must have done it at least k+2 times.
Thus sooner or later Alice will have turned it over k+2 times, and so can confidently say that both she and Bob have been in.
This is, I think, a watertight proof that the strategy works in the case where there are just two prisoners; the same proof can be extended to cover the general case with more prisoners, but it'd take a better writer than me to set it out in a way that's both pedantically rigorous and reasonably clear to read. Your argument that the state of the cup can't carry enough information does sound convincing - it's roughly what I thought when I first heard the problem - but if you try to formalise it information-theoretically it doesn't quite go through.
> True, the continuity deal was too restrictive. The Laffer Curve holds in
> *more* instances. Thanks!
You can keep trying to redefine what you think the laffer curve is supposed to indicate, but you'll continue to be wrong. Partly because the theory itself is wrong, and partly because you lack the ability to think and argue coherently.
> The Laffer Curve is not a model of the economy.
*uses LeonGeeste anime school of arguing trick* So you agree with me that the laffer curve is not a viable economic model. Thanks.
> It does not purport to describe an array of economic activity, just that at
> some point, revenues diminish despite tax increases.
Pathetic backpedalling on your part. Most people agrees that "at some point, revenues diminish despite tax increases". "At some point" is the crucial part of that phrase that you're not dealing with.
> And claiming that raising tax rates eventually diminishes revenues is not
> pseudoscience.
I'm calling the laffer model in its entirity pseudoscience. Here's a hint, boy, the laffer curve does not state that "raising tax rates eventually diminishes revenues", that's one of the assumptions used in the laffer model. The two are not the same thing. Try and work this out.
I surely agree with two if the assumptions, that 0% of x is 0 and that 100% of 0 is 0. All that stuff in the middle: wrong. To then proclaim that decreasing taxes *will* increase revenue without a further assumption (hopefully backed up with some empirical data) as to where you are on the curve: double wrong with wrong sprinklings and lashings of wrong sauce.
Mmmmm. What a big feast of wrong you're trying to jam down that throat of yours. As before, here's hoping you choke.
Now, if you're trying to rewrite the laffer model so that all it is saying is that "at some point [...] raising tax rates eventually diminishes revenues", let's call that the lesser-laffer model, then almost no-one will disagree with you.
> I said it will do one thing, not change one thing.
So what is this "one thing" that changing tax rates will "do"? And don't say increase revenue, because it depends on where you are on the curve. Is it that it will change total revenues? Wow, big news.
> Sure, raising taxes could raise revenues, increase unemployment, lower
> interest rates, and so on. But the person was claiming that the same change in > tax rate (with all else held equal as per the curve's ceteris parib
> assumptions) could BOTH increase AND decrease revenues and is thus irrelevant
> to revenues. This is false.
Nope they didn't say that. They were probably slightly misled by one of the sideways-drawn economists' graphs; they most certainly didn't say that if something can increase and decrease revenue then it is irrelevant to revenue, as it is obviously affecting revenue how can it be irrelevant?
They did say that it's shit economics. Even for economics. And I heartily concur.
> And there are economists that dismiss conclusions grounded in empirical data
> for the precise reasons you gave: you could never control enough variables to > get a good result.
Which is why it is pseudoscience. It oftend pretends to be science but doesn't follow any of its strictures, such as falsifiabilty or empirical data. And then it's used as a reason to fuck with people's lives.
Thanks for proving us right. Again.
> Okay, fine, if you raise taxes, revenues will JUMP down, rather than decline
> smoothly. Happy now?
No: "will". It depends, according to the theory, on where you are on the curve. You keep trying to ignore that one rather salient fact. What the curve in its entirity does show, is that if you adjust the tax rate, then tax revenues will either increase or decrease. Wow, that's big news to everyone here I'm sure.
> > > What is most important is that you assume it doesn't behave like -1/x near
>
http://www.math.hmc.edu/funfacts/ffiles/30002.8.sh tml
Magic Sudoku is a variation of the original Sudoku. It's adds some new restrictions:
* Every main diagonal has the numbers 1 to 9 without repetition. (the same as original sudoku but applied to diagonals).
* Every Quadrant has only a number.
* There is some special rows (filled with blue color). The value of the numbers who are in this cells is lower or equal as the number of coloured cells in every quadrant.
checkout at PrintSudoku.
You didn't quite get it, did you?
The laffer Curve is all about tax revenues, it has nothing to do with state expenses. On the contrary, since it helps to establish maximum tax revenue, it would actually help New Orleans by giving more funds for securing the area.
www.beholder.co.uk/planetarium/ is my favourite mathematical/logical/linguistic puzzle. it takes a bit of solving, mind.
I toyed with all sorts of verifiable communication systems, and there is something in having messages that can withstand more error than all previous message sizes combined plus something bigger than k, but there is difficulty in transitioning between pairs of talking prisoners. There needs to be some sort of "bell ringining" (a broadcast by constantly flipping the challice) which is dificult to negotiate responses to. So here's something unbelievably simple that works:
Let's add one simplification, the king gets k = (k in original puzzle) + 1 flips because he gets an extra flip from the challice's undefined initial state. We'll be generous and assume this could cause an error the king desires anywhere and not just at the start.
Prisoners are numbered p0 through pn. (Here n is one less than in the original puzzle.) They choose a number f discussed below. Here are their strategies:
p0: sees a challice up - turns it down
counts turns, once he or she has flipped it more than the numebr of time discussed bellow, tells the king "Yes"
p1 through pn: sees a challice down - turns it up
counts turns, once he's done it f times he stops (no longer turns challice)
Notice that these messages can not interfere with each other.
The king's flips can either add to the number counted by p0 (by flipping it down before showing p0) or subtract from it (by flipping it up before showing any other prisoner). All of the flips will be eventually recorder because every prisoner is shown the challice after any other prisoner and thus after any transition (thanks AC). The number of up flips fated to be made by prisoners is fn. The king can adjust this as low as fn - k or as high as fn + k. The most that the challice could possibly be flipped without all the prisoners entering the room is f(n-1) times. So when not all the prisoners have been in the room the total number of flips is 2k. Now the first prisoner, once he or she flips the challice more than fn - f + k times, knows that all the prisoners are accounted for and can tell the king "Yes". Don't be surprised if I'm off by +/- 1 total; the prisoners would be wise to use a bigger f to absorb any fencepost errors.
The question: "What would someone from your tribe say if you asked them which way I needed to go at this fork to reach the Village of Life?"
If this is the truth-teller, so is his tribe member, so you get the correct answer. If it is the liar, his tribe member is also a liar, so the first guy would point towards the Village of Death, and the second guy lies about what the other guy would say, pointing you to the Village of Life.
This assumes honest liars, but does not assume which Village each tribe is from (indeed, they need not be from either for this to work).
I agree with the statement about Americans teaching themselves to become inferior. We keep abstracting technique out of our teaching system and replacing it with computers and tools. The large majority of us may be literate, but that doesn't mean we know what the hell we're doing any more. People who really take the time to study why things work the way they do are still getting jobs. I think a major issue is the thought everyone in America needs a College Education to get a job. People need to wake up. College Educations are not valuable if you're not using them to learn why what you're doing works.
The reason math and science are obsolete is even a topic of discussion is we're becoming overly dependant on our technology to do work for us.
What would happen if we had no electricity? A major blackout?
What about when people do allow math and science into obsolecence. Are we going to allow the Foundation to come true?
Who will fix the tools when we've become so dependant on them that we no longer can?
I wouldn't consider the mad hatter mad. Just reality impaired. He sure can make a mean cup of tea.
That's quite interesting. Could you then explain how is it possible, that in 2004 in Poland after CIT decrease from 27% to 19% (sic!) total tax revenue increased?
And who's saying that...
Three travelling salesmen are stuck in a small town due a road being washed out. The town is rather small and has only one hotel. The three salesmen arrive at the hotel at the same time and all are interested in renting a room. Unfortuantely there is only one room available. The desk clerk says the room is $30.00 for the night and if they each pay $10.00 they can share the room. The salesmen agree this is the only reasonable option so they each pay $10.00.
A short time later the desk clerk realizes he made a mistake and over-charged the salesmen. He calls the bellhop over and says, "I overcharged those three guys in room 9, I should ahve only charged them $25.00 for the night. Please take the gentlemen $5.00 and offer my appologies". The bellhop takes the $5.00 and starts heading for room 9.
On the way, the bellhop realizes it will difficult to split $5.00 between the three gentlemen. Not being the most honest person around, the bellhop decides to pocket $2.00 and return $3.00 to the gentlemen in the room.
Now the salesmen have each paid a total for $27.00 for the room ($10.00 ((the origianl amount paid)) minus $1.00 ((the dollar returned to each by the bellhop)) times 3). The bellhop has kept $2.00. $27.00 plus $2.00 is $29.00.
Where is the missing dollar?
If VISTA is the answer, you didn't understand the question
So what you're saying is that you wrote the problem ambiguously, and to resolve ambiguities we ought to choose the interpretation that yields a solution?
Your description of the problem does not say when k is fixed. A perfectly valid reading of the problem is to suppose that the prisoners are told that the king will decide k when the game begins.
You ought to admit that the problem was unclear, instead of insulting everyone who interprets your ambiguity the wrong way.
I knew breasts would make it into this posting somehow.
Intelligent Design: because MATH is HARD.
And what about the remaining 1/6? Is that the chance that you are fooled and that there is no prize at all?
You have two planks: 3m and 2m long. They are placed in a X pattern across a corridor, each with one end on the floor touching one wall and leaning against the other wall. They cross each other at 1m from the floor.
And no, I don't know the answer.
ID: the nose did not occur naturally, how would we wear glasses otherwise? (apologies to Voltaire)
"I lied"
so is it a mistake in the original problem that, if each prisoner will be brought out X times, the king can bring out the first prisoner all X times before ever going back to the next prisoner? because if he can do that, he can bring out the counter all X times, and foil the prisoners' plan.
You can't add the probabilities. They change when the guy opens the door.
Sorry if this is a dupe - I've tried searching the thread with likely keywords without finding it.
The most elegant puzzle I've ever heard follows. I've heard it attributed to various people, but haven't yet come across any well documented attribution.
Consider a standard 8x8 chessboard, and a set of rectangular tiles that are exactly 1x2 squares in size. It's obvious that you can cover every square on the chessboard with 32 tiles.
Now, remove the squares from two opposite corners (eg. top left, bottom right). Is it possible to cover every remaining square using 31 tiles?
(Hint: the solution can be given in one or two short sentences and requires no formal math.)
Unfortunately, your proofs are both flawed:
> The problem has 3 parameters...a, the arbitrary number of
> times the king must call each prisoner.
There is no such "a". In mathematical parlance, "an arbitrary number of times" doesn't correspond to any particular number - it means that all numbers must be considered. In other words, "a" is not a parameter of the game, and hence your equation---k > n * a---makes no sense (unless k is infinite).
The problem is simple if k is known to the prisoners, and verifiably impossible if k is unknown to them.
The solution (when k is known) has been given by others (hint: counting). The refutation (when k is unknown) is information-theoretic. Any valid prisoner-decision algorithm must be finite, and so must terminate after receiving at most A bits of information. All bits of information must be transmitted through the chalice. The king can invalidate up to k bits of information by cleverly flipping the chalice (i.e., since he knows the prisoners' algorithm, he can just randomize the chalice every time it's about to transmit a bit). Thus, if k >= n*A, the prisoners' algorithm must terminate before any valid information has been received by any prisoner, and hence cannot guarantee the required result. Since k is unknown when choosing A, no algorithm can be devised which will always succeed.
An excellent site with puzzles: Thirty Puzzles for Mathematicians and Computer Scientists
For example:
Bigger or Smaller: Alice chooses two distinct real numbers between 0 and 1, writes them onto two chits of papers and places the chits in a jar. Bob gets to select one of the chits randomly and open it. He then has to declare whether the number he sees is the bigger or smaller of the two. Is there any way he can be correct more than half the times Alice plays this game with him?
f(f(x)) == -x? Is it possible to write a function int f(int x) in C that satisfies f(f(x)) == -x? Without globals and static variables? Is it possible to construct a function f mapping rationals to rationals such that f(f(x)) = 1/x?
30 Coins: 30 coins of arbitrary denominations are laid out in a row. Ram and Maya alternately pick one of the two coins at the ends of the row. Could Maya ever collect more money than Ram?
I always loved this one, because it is so counter-intuitive (at first): There is a worm which starts to crawl along a rubber band. The rubber band starts with a length of 1 meter, with the worm at one end. The worm crawls 1 centimeter per second. At the end of each second, the rubber band expands with another meter. Suppose the worm will live forever and the rubber band never breaks, will the worm ever reach the end of the band? If so, when?
We used to do 5 Minute Mysteries. My favorite was "Four men are found dead in a cabin in the woods, all sitting in seats facing the same direction. What happened?"
It's not really a riddle because there is supposed to be a question-answer discussion leading eventually to the solution, but all the details to solve the problem are in fact already given.
I am sorry, but you can ALWAYS add probabilities. There are three doors. One is already open, so it has a probability of zero that the car is behind it. The other two are still closed, and if there is indeed a car behind one of them, the probability that the car is behind one of them is 1. The probabilities of the two remaining doors therefore must add up to 1. And, indeed, the correct answer would be that the "other door" now has a probability of 2/3. If you still don't believe me, I really would like to do some betting with you.
>>> he has to call every prisoner in an arbitrary number of times.
>
> the king can call any nominated leader/counter the maximum number of times
False
There is no "maximum number of times" - arbitrary means just that, arbitrary. You want 50 times? 100? 1000? 6023*10^23? The king must give you any of those.
In particular, that means each prisoner will be called in after each other prisoner has been called in an arbitrary number of times, leading to the simple counting solution most people come up with. (Proof: suppose otherwise---i.e., there exists a prisoner A and a prisoner B such that A is not called in after B has been called in P times. Once B has been called in for the Pth time, A has been called in N_a times. However, A must be called in an arbitrary number of times; in particular, he must be called in N_a+1 times. Thus, A must be called in again. Contradiction. Therefore, there cannot exist such a pair of prisoners A,B---i.e., there are no two prisoners such that the first only sees the second called in a fixed number of times.)
The key is to have a correct understanding of what "an arbitrary number of times" means in math-speak.
Mathematical Puzzles: A connoisseur's Collection by Peter Winkler: at Amazon. The first chapter is readable at Amazon! With wonderful puzzles.
Yes, please be a little less schizo.
If California and Massachusettes are already progressive states then why are you trying to get another one established? Some half of you must see your experiment as already complete.
Algebra is actually a very good problem-solving tool. I can't answer for trig or calculus, as I've never studied those. It's not enough to have the calculator figure everything out, though; you really need to understand how to calculate the answer yourself.
If math and science truly become obsolete (never!), I sincerely hope that people like you double up on your spelling and grammar lessons.
its logic guys, quit trying to re-invent the wheel.
/.ers just wanted a multimeter? 8^)
turn on switches one and two, and wait about a minute.
then, turn off switch two and immediately go into the room.
if the bulb is on, its switch one. if the bulb is off, but still warm, it was switch two. if the bulb is off and cold, then it was switch three.
man, how many
Well, the answer is the one I gave you, as you can look up on any relevant website. If you need help to internalize the theory behind it, I'm afraid I'm not your guy. Good luck.
I have a truly marvelous proof for this, which, unfortunately, this post is too small to contain. Please share, Mssr. Fermat.
http://alternatives.rzero.com/
but it's well on the way to the solution. just design it so that every prisoner can act as the leader somehow -- distributed leadership if you will, all that's missing from the abovce solution is a way to beat the king if he decides to call the leader in n times first thing, eg.
that is, if the king called the leader in n times before he called in anyone else, the leader would never collect any tokens . . . someone else would have to become the leader.
it gets quite complex.
mr c
"Physics is like sex. Sure, it may give some practical results, but that's not why we do it." - R. Feynman
By the way, if it doesn't exceed your bravery quotient please mark me as your foe so we can happily ignore each other in perpetuity. I have no use for foolish cowards, with or without the bravery of their anonymity.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
One of my favorites is from Martin Gardner, which I paraphrase below:
One day two mathematicians, Igor and Pavel, meet in the street. "How are you? How are your sons?" asks Igor. "You have three sons as I remember, don't you? But I have forgotten their ages."
"Yes, I do have three sons," replies Pavel. "The product of their ages is equal to 36." Looking around and then pointing to a nearby house, Pavel says, "The sum of their ages is equal to the number of windows in that house."
Igor thinks for a minute and then responds, "Listen, Pavel, I cannot find the ages of your sons."
"Oh, I am very sorry", says Pavel; "I forgot to tell you that my oldest son has red hair."
Now Igor is able to find the ages of the brothers. What are their ages?
A great puzzle, easy to figure out, no need for advanced math, and a true "Aha!" moment when the trick is figured out....
Best Buy can have you arrested
Before running off to church this morning I wrote 50 and 18 on my hand and by the time I decided to give up on the service and start solving it I couldn't remember what the goal was. I thought it was to make two piles, one filled with nothing but heads up and the other filled with nothing but heads down.
:P
/.
Needless to say, the puzzle is much easier to solve when you know what the goal is.
On a different note, I started chasing down a possible solution but decided that this was significantly too much work to go through to impress the readers of
Ever-so-respectable Tripod "webmasters" simply cannot call anybody a fool and expect to be taken seriously.
1 11 21 1211 111221 312211 13112221 .....
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
{Yawn}
It's the quadratic formula. And it's really hard to derive in my experience, so I'm surprised that you both a) can do it and b) forget what it's called.
"I don't care about the Constitution!" --Bill O'Reilly, November 17, 2009
It only works if the prisoners know k ahead of time. Unfortunately as written, the riddle does not say whether they do. It's clear that they are given the *rules*--that the king has k opportunities to flip the chalice. But it's NOT stated (read it carefully) that they are given the value of k ahead of time. If they know the king is limited, but they don't know that limit, your solution does not work, because the value of k+2 is unknown to the prisoners. They will follow their stategy for infinity without ever Alice ever knowing when she hits her limit.
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
Well, I'm not sure that I'd accept the conventional wisdom as being evidence of the best course, especially in an area like economics where true expirimentation is essentially impossible. A common fallacy of economic theory is assuming that all actors are fully informed and will always act in thier best interest. Even if you did, it only follows as being sound economic policy if the maximation of tax revenue is your only goal, as opposed to things like what you do with that tax revenue. And yet one final time, you're making the assumption that these monarchs had the ability to improve tax collection methods more than they did, which is unsupported.
As the riddle is written, the value k is not known to the prisoners. The rules are (i.e. the existence of k), but the riddle is silent as to whether the actual value of k is communicated to the prisoners. This may be just be a case of shitty writing, but if we're going to play riddles, we should at least pay attention to their details as presented. After all a huge part of the fun of riddles is in their (carefully crafted) details.
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
One more favor. Please designate me as one of your foes. It will save time in ignoring you in the future.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
The fact that an increasing function which increases away from zero cannot go back to zero is a triviality once you have set up basic properties of inqualities, and has been known since, well, the beginning of time. It has not much to do with the correctness of the Laffer Curve theory, though.
The king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you...
As this reads, the prisoners have a complete understanding of the game as you have explained it to me--i.e. they know of the existence of k but not the value. As written, you do not specify the value of k nor do you state that the value of k is specified to the prisoners. Since a logical solution depends on the prisoners knowing the value of k, and you do not state that they have that knowledge, there can be no logical solution.
From your reaction here is seems we were supposed to know that k is specified to the prisoners. Since that is nowhere in your original write-up, I can logically conclude that you suck at writing up riddles.
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
Your original write up only states that the prisoners know of the existence of k, not the value.
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
Well, the issue as I see it, is that there needs to be a point where no number of changes is ambiguous. So one person gets designated the leader, and keeps a running total of how many times the chalice is right-side up. The other people flip it right-side up, if and only if they haven't already flipped it x times.
So, we know that the minimum number of times that the chalice will be seen as right-side up by the leader is x(n-1) - k, because the king can swap right-side up chalices up to k times. The minimum number of times that the chalice needs to be flipped to guarantee that everyone has flipped it, assuming it starts down and no interference is x(n-2) + 1. However, the number that must actually be worked with is x(n-2) + k +2, since the king can add k right-side up chalices that weren't there, and it could start right-side up as well.
Next, we equate the two to find the lowest possible number allowing for an overlap. So, x(n-1) - k = x(n-2) + k +2. This simplifies to x = 2k+2. Plugging this into x(n-1) - k, we get 2kn +2n - 2k - 2. This becomes 2(kn - k + n - 1). So, when the chalice has been seen right-side up 2(kn - k + n - 1) with each person flipping it a maximum of 2k+2 times, the leader can honestly answer that everyone has been in the room. That being said, it could take a really, really long time.
If the Government taxes everything (100% tax rate) and supplies 100% of "All services you will ever want." Then you get something like socialism.
A Capitalistic system would not work, but a system could be made to work with everyone being an employee of the state and everyone receiving all they needed from it.
OK it might not be "The best system" but it show the paradox up for what it is.
A sig is placed here
To display how futile
English Haiku is
you have 12 seemingly identical balls, but one of them is odd, in that it is lighter or heavier than the others. Using a pair of scales, determine, in three weighings, which ball is the odd one and whether it is lighter or heavier. Solution can be found here: http://www.curiouser.co.uk/puzzles/12ballssol.htm Destroyed a perfectly good Saturday trying to figure this one out
This is an interesting idea but I'm going to reject it for two reasons:
1. The question asks how many points, not real numbers.
2. (more importantly) If the slope of the plane and (X|Y|Z)-translation of the plane are both irrational numbers -- sqrt(2) and e for example -- there isn't a single rational number that can be passed to the function to specify this plane, hence not every plane can be specified by the function.
Kevin Fox
Mighty K says: I amend my answer, and believe that it can be done with one point. The plane would contain the point, and be perpendicular to the vector pointing from the origin to that point. Good puzzle, and deceptive!
;)
Sjampoo replies: While i do believe it should be possible with just one point i dont think your solution is the correct one. You won't be able to describe planes that intersect the O-point (0,0,0). Maybe better luck next time
Frankir clarifies: Easy. Just define your function returning a plane shifted by some constant towards (0,0,0).
This is exactly correct. Nicely done, guys!
Kevin Fox
I see the idea behind your solution, but I don't see how you get from case 1 (one spotted monk) to case 2 (two spotted monks). If each monk sees another with a spot, he knows that guy is slated for death, but how can either be sure about himself? On day N + 1, isn't the state the same as on day N?
My idea (and this is not at all my thing, so I'm sure I'm missing something) is this: Each morning, each monk looks out at the others. One at a time, if a monk sees another with a spot, he goes back inside (and, presumably, waits for tomorrow). The last monk left knows he has a spot and does himself in (perhaps by leaping off the balcony). We should have a complete solution in N days, right?
The one problem I can see is if there are just two monks left, both with spots. One goes in, but the next day, he'll never know if he has a spot or not. I suppose this could be solved by each of them sitting down if he sees a spot. If one sees the other guy sitting, he knows it's his turn.
Like I said, I suck at these things, but it seems like it should work. My only question is, who cleans up the bodies?
What if life is just a side effect of some other process and God has no idea we exist?
But the monks don't know how many there are to start.
If there are 5 marked to start, but a monk sees 4 others with dots, how is he to know that it wasn't 4 that were originally marked?
You frigging anonymous cowards are so tedious. Just mark me as a foe and we'll ignore each other. I'm not at all interested in "discussions" with intellectually dishonest fools.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
Has anyone yet discovered a valid purpose for the existence of anonymous cowards on /.?
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
The problem states that the king must call each prisoner an arbitrary number of times. He can't call someone n times and then never call them again. At any point in time each prisoner will be called in again at some point in the future unless someone says yes.
turn on the light
http://www.npcgaming.com Dedicated Gaming Servers
Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
So that the hole was just six inches long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!
(the answer isn't hard to calculate, but try proving that you don't really need to know the information that wasn't provided)
If k is known to the prisoners, then the answer posted by notshannon is the correct one.
The key to the solution is that each non-counter prisoner can flip the cup 2k+1 times. This effectively nullifies the king's influence.
Another way to think about it is with tokens. Say there are n prisoners. One of the prisoners has no tokens, he just collects them. Call him the collector. The other n-1 prisoners, each have 2k+1 tokens. Call them depositors.
At any one time, the table in front of the king either has a token or it doesn't. If a depositor sees the table empty, then he deposits a token on the table. If a depositor sees a token on the table he leaves it there. If the collector sees a token on the table, he takes it. If the collector sees no token on the table, he does nothing.
The analogy is: Putting a token on the table is like flipping the cup to face up. Taking a token from the table is like flipping the cup to face down.
The king (with his k flips) can effectively steal k tokens or add k tokens. This does not influence the decision of the collector guy when he is counting the tokens.
When the collector has collected (n-2)(2k+1) + (k+1) tokens, then it means that each of the depositors has deposited at least one token. If the king was allowed only to steal k tokens and not add k tokens to the table, then each prisoner would just need k+1 tokens, not 2k+1. And in that case, the collector would just need to collect (n-2)(k+1) + 1 tokens to say "yes".
But because the king *can* add k tokens, we need to give extra tokens to the prisoners so that in case the king does not add those k tokens, the prisoners can add them, and the collector can collect enough tokens to make the "yes" decision, without having to wait for the king to add those k tokens.
I would like to refute the claim that we cannot have a collector prisoner (counter/leader prisoner). Someone said we cannot have a collector prisoner because the king can just call out the collector at the end, after all the depositor prisoners have been called out. The puzzle says that all prisoners will be called out as many times as you choose. So say that first all the depositor prisoners were called out p times. Then the collector prisoner was called out p times, I can always choose a number q > p. And so the depositor prisoners will have to be called out q times, as well as the collector. Therefore, the depositor prisoners will have to be called out even after the collector prisoner has been called out p times.
I don't see you complaining that there might be infinitely many prisoners, so you clearly didn't have any trouble working out that 'n' (the number of prisoners) is some particular finite number that doesn't change throughout the course of the problem. But you seem unable to apply the same brain power to 'k'. I can only conclude that you are deliberately being obtuse to cover up the fact that you can't solve the problem?
A man gets to a train station at exactly 5pm everyday. His wife immediately picks him up and drives him home. One day, he arrives early at 4pm. Instead of waiting, he begins walking home on the route his wife drives along. His wife picks him up somewhere along the route from the train station to his house. He gets home 10 minutes early. How long did he walk?
As with many problems that stump people for a long time, the fault is in the first step. You are squaring a quantity without taking care to consider the direction. Think of the diffrence between performing math on a scalar versus performaing math on a vector.
If you do this proof with a unit vector in the positive x direction (aka 1) and a unit vector in the negative x direction (aka -1) the math will NOT say that they are equal, no matter how you manipulate them.
You find yourself alone on an unknown island. Some natives capture you and tell you that for trespassing on their sacred ground they MUST kill you. However, their laws demand that they kill you in a very specific way. They tell you to make a final statement ... and based on your words, they MUST slowly torture you to death if you lie or they MUST execute you quickly and painlessly if you speak the truth (silence is a golden truth) ... what four words can you say to save your life ... ???
A warrior's greatest weapon is wisdom, therefore, keep your mind even sharper than you keep your sword.
On day N + 1 the state isn't the same as on day N. It's proof by induction. We have to assume that all the monks are perfect logicians, but that's given.
The theorem they each deduce is that if there are N marked monks, then after N days the marked ones will know who they are, and will kill themselves.
The base case is for N = 1, which I think is clear.
The inductive step is, suppose it is day N. Then the marked monks all see N - 1 marked monks. (The unmarked ones see N.) By induction, if there were only N - 1 marked, they would have committed suicide after day N - 1. But they didn't. Therefore, they all now know they are marked as well, and will then commit suicide, before day N + 1.
Which I think is pretty much what I said before... but hopefully I made it a bit more clear. 2 is just a special case of N.
If, after 4 days, the 4 marked monks he sees are still alive, he knows he must be marked as well - otherwise they would all have deduced they were the marked ones already.
The 18 coins one was brilliant. This is less so, but fun.
Consider all pairs of prime numbers whose difference is 2 EXCEPT 3/5. For instance, 11/13 or 29/31 etc.
The numbers sandwiched between all such pairs is divisible by 6. Without resort to fancy proofs... why is this, and why is the 3 and 5 pair different from the others?
tone
tone
From what I can tell, this is correct. I it also refered to in another post...
IE citizen are treated like any other renewable resource; the problem
is the same for vampire victim management and for fishery management --
extract the maximum from the citizen. If the government tried to get
more, they'd get less.
Just say, "I will die slowly."
A warrior's greatest weapon is wisdom, therefore, keep your mind even sharper than you keep your sword.
What about the one titled "complex solution" .. got an opinion on that one?
a + b = c
=> 4a-3a + 4b-3b = 4c-3c
=> 4a + 4b - 4c = 3a + 3b - 3c
=> 4(a + b - c) = 3(a + b - c)
=> 4 = 3
Thanks to the internet and its myriad of contributors of all shapes and sizes for this wonderful piece of logic
Gur nafjre vf sbhegl-guerr. Hfr vaqhpgvba. Vs gurer ner rknpgyl gjb zbaxf jvgu fcbgf, naq lbh frr bayl bar, lbh ernyvmr lbh zhfg unir bar naq ner qrnq gur arkg qnl. Vs gurer ner rknpgyl guerr, gura rnpu fcbggrq zbax frrf gjb fcbgf. Vs gubfr gjb ner abg qrnq nsgre gur svefg qnl, gura rnpu bs gur guerr ernyvmrf gurl zhfg unir gur znex. Rgp., rgp.
Then use the proposed solution.
Isn't a valid answer to any of these problems 'there is no solution'? And then perhaps proving this?
It's actually a pretty good exercise in lateral thinking.
Exactly. It's the principle of use it or lose it. Luckily there are still people who can deal with most technical issues. However there are now a lot fewer people who can repair their own car.
You're an arrogant prick that refuses to acknowledge ideas set forth by anyone else unless they are agreeing with you. How about you mark everyone as Foe and set your comment limit to only those comments already modded +5. Would save the majority of us from you bullshit posts and your completely ignorant and arrogant attitude.
Simply put. Fuck off.
http://www.griddlers.net/
They're fun because of the pictures that are formed when you complete them. But staring too long at those 50x50 ones could be bad for the eyes.
I agree that my solutions bends the posed problem quite a bit, and I understand that you cannot accept it for that reason.
However, I cannot agree with your second objection. Neither your original problem nor my answer contains anything about rational numbers at any point - we've always been talking about real numbers. In the context of real numbers, the problem with irrational numbers simply doesn't exist.
I know (from my own personal experience) that it can be hard to accept that a single real number contains enough "information" (whatever information may mean) to specify finitely many real numbers, but it's a result that you'll invariably encounter when studying set theory. A proof of this fact is illustrated nicely in "Proofs from the Book" by Martin Aigner and Günther Ziegler - I don't have an online link handy, I'm afraid.
When you take the square root of a number you have 2 solutions. You have to consider both.
If you disagree with me on social issues, then it's pretty clear that you are a narrow-minded bigot.
No, I know about induction. I just didn't read the problem carefully enough. I understand the argument now.
I don't think it holds if there are more than two monks with marks, though. If there are three monks and two of them have marks, and all that they know is that some of them have marks, they don't have enough information to make a decision.
Actually, the problem as stated implies that more than more than one not but not all of the monks will be chosen, so if N is more than 2 and there are more than two with marks, it won't work.
What if life is just a side effect of some other process and God has no idea we exist?
If the initial state was unknown then the counter must count to n*k+2 and each pawn must likewise flip the cups k+(k+2)/2 times...
Here's one that I think came from a Martin Gardner column:
A man is taken prisoner and told that he will die sometime next week, Monday through Friday, but he won't know which day until they come to take him away to be executed.
He thinks for a minute, then is comforted, thinking they can't possibly kill him under those constraints, reasoning thus:
I can't be executed on Friday, because if Monday through Tursday pass, and I haven't been killed yet, then I will know that I will die on Friday, but they said I wouldn't know, so Friday is out.
Therefore, I can only be executed on one of the days Monday through Thursday. But it can't be Thursday for the same reason that it can't be Friday, because by then I would know.
Each of the rest of the days are likewise eliminated.
However, on Wednesday, they came and executed the prisoner. And sure enough, he didn't know what day he would be executed until they came for him.
Where did his reasoning go wrong?
If you disagree with me on social issues, then it's pretty clear that you are a narrow-minded bigot.
No, not the Linux distribution, but a puzzle that was put to me in the 80's ...
There's a game called the Red Hat Game. Each person has a coloured hat on their head, and they don't know what colour it is. The players are put into a room, where they can see each others hats. The purpose of the game is to work out the colour of one's own hat. The rules are:
1. if you see somebody with a red hat, you have to put up your hand.
2. when you've worked out the colour of your hat, you leave the room.
In one particular instance of the game, each player has a red hat. The players go into the room, and after awhile the most intelligent player leaves the room. The question is: how did he know he had a red hat?
The problem is one of logic - there's no trick solution.
Wasting time to try to educate such a bloody twit, but here's the /. catch. There is a limit to how many friends and foes you can designate. There are constructive reasons to designate friends, and my slots are mostly full of thoughtful and interesting people.
There are a rather large number of worthless twits like yourself. You're not even worth a foe slot. It's rather a shame that /. doesn't offer an effective way to deal with such vile and mindless abuses of anonymity. If /. offered the configuration option, fools like you would not exist in any visible sense, even when they are cowering behind their anonymity.
I can wish you nothing worse than to have another one of your endless stream of terrible days. And it would still be nice if you designate me as a foe. I would not at all object to being the official foe of a large number of such spineless "people".
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
This reminded me of a seemingly logical and valid proof that 0.999... = 1 that our high school teacher showed us. Every once in a while I contemplate investigating what the fallacy is but to this day I haven't and it's been many days since that day in high school ;-). Here it goes.
1) x = 0.999... (0.9 with bar above the 9)
2) 10x = 9.999... (9.9 with bar above the fractional 9)
Subtracting the first equation from the second we get:
9x = 9
Therefore, x = 1!
Seemingly innocent but there must be a mathematical fallacy. Any one care to explain what is wrong with the above "proof"?
But I'd still appreciate the foe setting just in case you ever develop the courage to post under less than complete anonymity. It will save time in ignoring you.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
Ummm... First, turn on the light?
If you disagree with me on social issues, then it's pretty clear that you are a narrow-minded bigot.
People were looking for such answers, not word game answers because those answers aren't acceptable in this context. Shame on you for wasting peoples time with a phony riddle.
Preserve old classics: copy your collection onto all hard drives.
No it is not, as you can look up on any relevant website yourself, if you take the time. If you can find a website which says that the probability for the other door is 1/2, please point it out to me, because that site is wrong. And I think you will find that either no such site exists, or that it is about a different problem.
You are imprisoned together with 2 other prisoners. 2 of you have been chosen to be executed at random, and the guard knows which two. Your chance of being executed is thus 2/3.
But you ask the guard that, since at least one of the others will die as well, if he can't tell you who it will be. The guard reasones that indeed this doesn't matter, and tells you the first guy has to die.
But now you know the first has to go, so the second one that will die is either you or the other guy, and your chance of being executed decreases from 2/3 to 1/2...
or not...
Where is the flaw, or is this correct?
int main(void) {while(1) fork(); return 0;}
Let me guess, you're one of the few who actually likes the taste of troll-bait.
Yes, well done :)
Guy asked me for a quarter for a cup of coffee. So I bit him.
Finally, a suggested answer that fits the structure of a real solution! (although I might have missed one) Let others look at it and see if they find is a problem.
This has the proper structure of the solutions I've seen, which only 1 other person's had that I noticed (although I probably missed some). Assuming you didn't make a mistake in the minimum requirements, it would be a valid solution.
He's just one of the few that seem to understand the problem's intricate wording (I didn't write it)
That phrase you complain about is a phrase that is often used in math/logic problems, as other posters have pointed out to other people complaining about the wording of that phrase, "or twenty times, or any number you choose". That means "finite, but unbound".
What you fail to realize is that 1/6th of the probability is consumed by the Feyerabend Principle that balances out Bell's Theorem, that would otherwise lead to a probability surplus in the universe.
t ml
Okay, you're right. Here's a good page:
http://mathforum.org/dr.math/faq/faq.monty.hall.h
You have a fishbowl.
Hence: You have fish.
Hence: You like animals.
Hence: You like people.
Hence: You like women.
Hence: You like sex with women.
Your friend does not have a fishbowl.
Question: Is your friend gay?
Note: This is only a joke, don't flame me about it. >_>
I found it very easy cos I learned this at school (20 years ago!). I was about to try and explain it but thought better of it and googled for it instead.
what happens if the counter is called out the required number of times before any other prisoners? That is, the king calls the assigned counter out the required amount of times in a row that he has to call a prisoner out prior to calling the rest of the prisoners out. Remember, he heard them make plans. Who's going to call out "yes" in that case?
So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
I am speaking about this constraint. The way stated here, the number I "choose" no matter how high, can result in a single designated counter's turn in the chamber being used up and denied communication. So it seems any solution must get rid of the notion of an elected individual. No? And then by the way after using that individual's turns up, given a nonzero "k", the king can choose to flip his chalice. Erasing all traces of his existence.
When will you post your solution?
A man lives on the 17th floor of a building. Each morning he gets up and rides the elevator down to the lobby and goes off to work. Every evening, he gets in the elevator in the lobby, rides up to the 11th floor and then walks up the last 6 floors to his apartment. Why?
Before you ask, it's not for exercise.
How about this one:
2 men standing on a hill, one points to the other and says: "Brothers and sisters I have none, but that mans father is my fathers son".
How are the 2 men related (if at all)?
Puppet Master
The day Microsoft creates a product that doesn't suck, it will be known as the Microsoft Vaccuum Cleaner!
Was my favourite riddle book when I was in school; it's by Yuri B Chernyak and Robert M Rose.
It apparently comes from a Russian tradition of maths and science riddles.
chicken from minsk or chicken from minsk (UK)
Eg, from chapter 6:1
"A plumbing problem"
A faucet / tap has been left slightly open for some time, and a gentle stream of water flows downward. Why does the stream become thinner as it gets farther away from the faucet / tap.
There are lots of simple problems like this that can really make you wonder why you'd not looked at water streams as closely in the past! There are some pretty brutal problems too. You get hints and full answers (usually with equations).
The Brit lives in the red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the left of the white house.
The green house owner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the house right in the center drinks milk.
The Norwegian lives in the first house.
The man who smokes Blends lives next to the one who keeps cats.
The man who keeps horses lives next to the one who smokes Dunhill.
The man who smokes Bluemaster drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blends has a neighbor who drinks water.
Who owns the fish?
(written by Albert Einstein in the late 19th century)
unless I made a mistake or misunderstood the question.
My solution is based on the condition listed "eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose."
I define the cup when right-side-up called "full", when it is upsidedown called "empty". Turning the cup from "empty" to "full" I call "filling the cup", the other way I call "drinking the cup". These naming is to make the solution easier to understand.
One of the prisoner will be the "producer", he is the only one who will fill the cup. All other prisoners are "consumers", they will only empty the cup.
Set number M = 10k (i.e. ten times k). Whenever the producer's turn, he will fill an empty cup, but ignores a full cup. For consumer's turn, he will empty a cup but ignores an empty cup, until he personally have emptied M cups, then he always ignore the cup.
Everytime the producer comes out and see the cup empty (after he filled it last time), he counts 1 cup consumed. When he counted (n-1)M + k + 1 cups consumed (or (n-1)M + 2k to be safe), he can be sure everyone has come out at least once.
Oliver.
... the solutions given thusfar (here - I haven't looked elsewhere - require that the prisoners have some way of communicating and can elect a leader or a counter. How would they do that if they're in soundproofed cells and only one of them is able to leave at a time?
---- You are fully entitled to my opinion.
"...but make no mistake there is a Laffer point."
This should really say "...but make no mistake there is at least one Laffer point."
You can't really assume it's a smooth function, let alone with a single point that is optimal. Hell, you can't even assume the curve doesn't change over time.
"Pulling together is the aim of despotism and tyranny! Free men pull in all sorts of directions" -- Havelock Vetinari
Careful with that argument. It seems to have as an implicit assumption that monarchs know what's best. Absolute monarchy - is that the kind of government you favour?
Real Daleks don't climb stairs - they level the building.
Actually, if you check the numbers, despite revenue being lower as a portion of GDP, absolute revenue was HIGHER by the time reagan left office.
GDP Data
Revenue as a portion of GDP
GDP per capita
Taking the simple method of multiplying the revenue as a % by the gdp per capita, the revenue per capita went from $4306.54 to $4980.34 in chained 2000 dollars.
"Pulling together is the aim of despotism and tyranny! Free men pull in all sorts of directions" -- Havelock Vetinari
From 1980 to 1988 that is.
"Pulling together is the aim of despotism and tyranny! Free men pull in all sorts of directions" -- Havelock Vetinari
This should really say "...but make no mistake there is at least one Laffer point."
Hm...good point. To clarify that issue, my statement probably should have read:
"Again, you might not be past a Laffer point, but make no mistake there is a Laffer point."
Oops, it already did.
Note I referred to being past "a" Laffer point, not "the" Laffer point, so obviously I didn't assume there was only one.
You can't really assume it's a smooth function,
Not necessary. So there exist tax rates above which revenues jump up, or tax rates above which the revenues jump down. Doesn't change the fundamental insight.
let alone with a single point that is optimal.
Never said so or even hinted. See above.
Hell, you can't even assume the curve doesn't change over time.
Of course it will change over time; the theory never says otherwise, and neither did I. It's phrased under a ceteris paribus clause. The claim is that it's possible that at some specific point in time, with all its policies, if you just lower taxes, you will have higher revenues that if you had those policies but not altered clauses. The circumstances can of course change.
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My apologies, I didn't read your original statement as carefully as I thought I did. I picked out the '...there is a laffer point' due to the italics I guess.
"Pulling together is the aim of despotism and tyranny! Free men pull in all sorts of directions" -- Havelock Vetinari
Because, your cow-orker may have a psychological obsession that causes him to murder female offspring, (like Kronos, who used to eat his sons, for example) but clearly he does not have any such obsession that applies to males.
Or vice-versa, right?
Fanatically anti-fanatical
The issue is not what I know, it is what the prisoners themselves know. The problem is stated with variables and can be solved in the abstract with variables. But the solution is only valid from the prisoner perspective if the integer values of those variables are known by the prisoners prior to the development of their group strategy.
The prisoners will of course know how many their total number is (n), since it's clearly stated that they are given a chance to all meet and strategize prior to the game beginning. What is NOT stated is that the king will share the value of k with them.
Any solution depends on counting by the prisoners. This implies that they must have numerical value that they are counting toward. But if they do not know the value of k, they will never know when they have counted to the limit of their strategy, such as ((n-1)*(2k+1) - k) to quote one proposed solution. If k is unknown to the prisoners, that equation will never produce an integer value for the prisoners and thus they will never know when to stop counting and answer yes.
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
Careful with that argument. It seems to have as an implicit assumption that monarchs know what's best. Absolute monarchy - is that the kind of government you favour?
Check out Democracy: The God that Failed (Transaction, 2002) by Hans-Hermann Hoppe (on amazon and wikipedia), who argues that while democracy and monarchy are morally bankrupt forms of government, monarchy is clearly preferable for essentially the reasons I gave: the monarch owns the capital value of the country, while a democratic leader is just a caretaker and only owns temporary usage rights. A monarch will maximize capital value, which means taking into account long-term effects of his actions, while a democratic leader will maximize his current takings at the expense of the future - get while the gettin's good (think in terms of the Senator Bilkins Highway to Nowhere). This is not to say all monarchs are foresighted, and all presidents are short-sighted, just that the former's incentive structure encourages long-term thinking, while the latter's discourages it. Again, it's a ceteris paribus statement: the same person will be more reckless with a country if it does not cost him the capital value than if it does cost him the capital value.
Nor does it assume monarch's know what's best, merely that they have more incentive to search, and because there were more of them (over 1000 in the Holy Roman Empire) they have more data to go on. A monarch might have heard from 20 other princes that they raised taxes to 20% and saw a loss in revenue, whie 6 raised taxes to 20% and saw a gain. He could then adapt appropriately. Likewise, another could talk with other monarchs who made massive internal improvements and find out if that increased the number of people wanting to live there and thus the taxes they could get away with charging and then see if that offset the cost of making that improvement. Further, unlike the average voter, a monarch feels the full impact of his policies and doesn't face the "what does your vote matter?" problem, so he is more likely to research what actually works.
Again, these effects could all be minor relative to other effect, but they nowhere rely on monarchs as monarchs knowing what's best.
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Very cool. I'd never seen that solution before (obviously).
This was an old Higher School Certificate problem in Australia as I recall ...
Envision a circular building of radius r. There is a guard dog for this building, chained to it somewhere on the perimeter of the building. The chain is exactly 1/2 the circumference of the building.
So the dog is able to reach the exact opposite point of the building by going either way around it.
Question: What is the area the dog can guard ?
The lower part is easy - it's a semi-circle. The upper parts are trickier - the path of the dog traces a path reminiscent of a cardioid.
94TT
You apparently don't understand induction :)
The knowledge of how many monks have spots is being communicated by the monks not killing themselves. (Well, at least assuming they can't just commit suicide without having a dot.)
1 dot == 1 day
2 dots == 2 days
3 dots == 3 days
The monks can see between n-1 and n dots. What is unknown is if they have a dot. If they wait more than n-1 days then they know that there must be a dot they can't see, and that has to be their own.
I didn't see an answer posted yet. I must not have looked hard enough, I suppose. I don't think it's that hard, although I did solve a similar puzzle years ago from William Wu, which means I had already gone through this kind of process (http://www.ocf.berkeley.edu/~wwu/riddles/medium.s html).
It's a cool puzzle, and the steps you have to go through to figure it out are fun to find. The meataphysical implications of the statement kicking off the process, even though everyone knew the statement to be true before it was said, is by far the most interesting part.
On the 100th night, all blue eyed people leave the island.
I have misplaced my pants.
Vector....Its been a long time since I learnt it in school,am not able to understand much about it.
In simple maths term [ instead of advanced maths like vector..] that involves just 1,2,3.... + - / * , does it appear to be reasonable?
However , I do accept what another poster has said: both roots of a sqr.root should be considered.However Iam not entirely satisfied....coz for one of the roots it does satisfy...
Why does yahoo do this
"make no mistake there is a Laffer point" This, of course, is false. It assumes a 'smooth' tax code. Ours is anything but. There maybe two Laffer points, many, or an infinite number. Martin Gardners Scientific American article from the 80's was a definitive demolition of the case for a single Laffer point. Surprising that the author was not aware of it.
You made these assumptions, which are checkable:
1) probability of the sex of a newborn baby is 50% girl 50% boy
wrong.
2) one child being of a particular sex does not affect the probability of the next child in any way
wrong.
This is why math majors need to base their examples on subjects they actually know something about (which, in this case, would not include anthropology, biology, or medicine).
However, your answer is correct. The births of the two children are independent events because without knowing a great deal more than you can possibly know (about the parents' family histories, basically) you cannot possibly make any estimation more accurate than "there are two choices, so by assigning them equal possibility we can estimate 50% probability".
I said a positive function, not a positive slope function. Try reading my posts. Or the construction job ads.
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Thanks for reading this post of mine:
o ld=1&commentsort=0&tid=228&mode=thread&pid=1380916 4#13809477
http://slashdot.org/comments.pl?sid=165444&thresh
before firing off on subjects you don't really understand.
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You have an evil and paranoid enemy, who you wish to poison. Fortunately, You have a circulare cake in the centre of which a microscopic drop of deadly poison has been placed. You have to cut the cake, using vertical, but possibly curved, cuts, to comply with your enemies paranoic requirements while ensuring that they eat a poisoned piece and you do not. You therefore have to divide the cake into some number of pieces such that:
a) they are all the same shape (or your enemy might take a liking to the "wrong" piece)
b) at least one piece avoids the centre (so that you can eat it)
c) no cake is left over (wasting cake is punished by a horrible death)
More formally, you have to divide a circle into a finite number of congruent pieces so that at least one piece avoids the centre by some non-zero distance.
You're probably right...
I think I get it now. My problem was with getting to the more general case. The idea is that each monk sees N marks, and knows that the ones with marks see either N or N - 1 marks, so they have more information than he does. On each successive day, they each know that the previous day's scenario didn't happen, so after N days, they have it figured out, right?
I still get hazy with more than 4 marks, but I'll just trust that it works. It makes my brain hurt too much to follow it otherwise.
What if life is just a side effect of some other process and God has no idea we exist?
Maybe, but the discussion is to the validity of the Laffer curve which deals only in government revenue :)
No, real world examples of this show that the economy shuts down and is replaced by a black market economy which works via capitalism, leaving the government stuck with obligations to the people with little income, leaving the people who "play by the rules" to fight over the scraps while the people who illegally shift to capitalism become wealthy.
There is an outside chance I do understand it. Gardner's point is that since the curve is of unknown shape, with mulitple 'laffer points' and possibly even sections in which changes in tax policy produce no change in revenue (a straight line) it is a useless policy device: 1. You cannot tell where you are on the curve. 2. You cannot assume that at what ever point you are on will be improved by a change in either direction 3. You cannot assume that you will not pass over some discontinutiy and get unexpected results. Now, simplfiy the tax code first, then you might find Dr. Laffer's work useful for policy making. As it is you might as well roll dice.
I agreed from the outset, and in the post you didn't read before firing off your mouth, that there can be multiple Laffer points (though I deem it unlikely). Also (~8th time): I never claimed the Laffer curve guarantees that cutting taxes can increase revenues, just that it's a possibility. It is useful for policy making, because ~95% of politicians don't understand, or act as if they don't understand, its fundamental insight: taxes can kill the goose that lays golden eggs. Taxes can hurt the economy, even without benefiting the government. People won't admit this very plain insight because they don't like the implications: that the politicians they have been supporting have been hurting more than helping.
That's why they hem and haw about it, but when presented with its unassailable logic, they revert to "Well it's still a stupid curve... because we might not be past a Laffer point... uh, and the curve is wrong too." Well, sorry, it's not. And if people stopped claiming it's not, we could actually start finding out where we are.
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The guru would not be the one to leave. It is impossible for the guru to know the color of her own eyes unless somebody communicated it to her. So when is it spoken, the word "green" for it to be known to the guru that she has green eyes?
bite my glorious golden ass.
Its also not stated that the prisoners will ever be fed in their cells, does that mean that any solution needs to found within a few days before the prisoners are fed? Neither is it explicitly stated that the prisoners have an understanding of a number system, does that mean we cannot assume that they know how to count? Stop nitpicking. The problem is obviously both clear and solvable as several people have now found apparently independent solutions to it.
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
What's 69 in Roman Numerals? I recall this from a Rudy Rucker novel, I forget which one.
Tag lost or not installed.
with an long and interesting history. Fortunately, someone has written it up so I don't have to do so here:
http://en.wikipedia.org/wiki/Monty_Hall_problem
Tag lost or not installed.
I really don't feel like going into the boring game of explaining what you said, what I understood, and the rest: no amount of that will change the fact that whatever statement you had in mind was still completely irrelevant: Laffer's Curve Theory is not a fundamental mathematical theorem, it is not a mathematical theorem, and it is not a theorem.
As for your fun attempt at a come back: I am a professional research mathematician, and I'm mostly reading invitations for post-docs right now: when I get the time, I promise to look into those ads you mention. Cheers.
(Sorry -- just answering in case you thought nobody noticed your post and felt bad :)
The squares you removed are both white, but each rectangle covers one white and one black square. QED.
I didn't say it was a fundamental theorem; math doesn't talk about taxes. It does, however, have a lot to say about which functions have maxima, and the tax revenue/tax rate curve is one of them (positive on [0,100], zero at 0 and 100, no vertical asymptotes). Don't try to bluff your way out of this. I said the function was positive. You took that to mean strictly increasing (positive slope). You were wrong. Sorry, no two ways about it. If you stick your neck out that far on a mistake that big, I'd really hate to know what your "research" looks like, unless that's "research" for a good employer at the day labor site.
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1 = 1
10 = 2
100 = 3
10110 = ?
I hope this is not too easy...
k is obviously a known constant, the question is known to whom.
As you noted the original poster has had to clarify this several times throughout the thread. If it was so obvious, why all the clarifications?
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
I happen to agree with the broad principles of the Laffer Curve. Surely it is correct at the limits (t = 0 and t= 1) and provably the government can increase or decrease tax revenues by broadly increasing or decreasing tax rates. As a policy tool, however, it is lacking robustness. Because of the complexity of the tax code one cannot use the curve to predict the impact of changes to the code; for small changes in t one cannot predict even the direction of the change in revenues. So while it may be an interesting strawman for entertaining legislators, it is not a serious economic tool. Simplify the tax code, and perhaps it could be made meaningful.
I don't see how it's the tax-payer's business to protect places which are clearly too dangerous for human habitation. The fact that they need huge levees just to stop the place constantly flooding might suggest that there are better places to live.
If anything, only locals should have to pay for it, not anyone else.
I looked at the reasoning by others, but they all seem to miss something. Namely: "if it's 50% or more against, you kill that highest rank" So, if the problem were simplified to just 2 pirates, the lower ranking pirate gets everything by default by just voting against whatever the other pirate proposes. The lower ranking pirate has all the power since he makes up 50% of the vote with only 2 people. Thus, it should be assumed that pirate #5 will always vote against, in hopes of being put in the final situation. He has nothing to gain by not voting against, since his life is not in danger. Now, knowing this, how would the other pirates react? (Ack! I'm out of time! I'll come back later!!) *signs out*
Tell you what. Mark me as foe and let's ignore each other. I have no use for fantasy-based fools, and you have no use for reality.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
Obviously to everyone, or there would be no point in mentioning it.
"As you noted the original poster has had to clarify this several times throughout the thread. If it was so obvious, why all the clarifications?"
I believe he has also had to clarify multiple times that the king was asking if they had been all in the room since they were origionally locked in their cells, even though that was explicitly stated in the origional problem. You have no idea the crap I have had to explain to people on /. All that means is that the world is full of idiots who like to type.
The vast majority of people who posted got it. If it was so confusing, how were they able to do that?
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
Dude: you said:
in responding to someone's "The Laffer Curve is a laugh and has been refuted".
I am sorry if I understood that you were claiming that the Laffer Curve [theory] is a fundamental mathematical theorem. If you weren't, then I have no idea what you were saying in the second sentence in that post.
If you used the word increase in that post without meaning that the function was increasing (strictly or non strictly,that's rather unrelated), well, again, I have no idea what you meant in your third sentence in that post.
As for your comments on my research... Nah, I have nothing to say about that: I really don't care.
Well: I'd imagine without such a big stretch of imagination that there were other factors concurrent with the tax cut.
In Argentina, after a massive economic crisis, taxes on exports were raised to unheard-of levels and, guess what?, total tax revenue has ever since been at their highest point ever in the history of the country, by a margin which surpassed every expectation (even of those who proposed and implemented the raise).
The only thing that one can conclude from examples is that taxes do not happen in the void.
I was claiming that the only thing you need to prove the existence of the Laffer Curve is some very fundamental mathematical theorems and some very trivial assumptions (if you credibly claim you'll take everything, you get nothing... no one has disputed that, here or elsewhere). Then, it's a straight shot to proof. Ergo, if you claim that the curve itself has been refuted, you're claiming that some fundamental mathematical theorems have been overturned (or, you're claiming something even stupider, but I err on the side of overestimating people). Understand now?
8 06454
I don't know what's so hard for you to understand about "increase". Think about the function f(x)=x^2 defined on [-2,2]. At -2 it's 4 and at 2 it's 4. So I could say, on this bound, it starts a four, and decreases, and then reaches 4 again. That wouldn't mean I'm claiming that x^2 is strictly increasing, which is exactly how you interpreted me! Look at your post again:
http://slashdot.org/comments.pl?sid=165444&cid=13
"The fact that an increasing function which increases away from zero cannot go back to zero is a triviality once you have set up basic properties of inqualities,"
You thought the function was strictly increasing so it could not go back to zero! Yet that was obviously not what I said. I said in some region after 0, it increases, not that it always increases.
Comprende?
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What's the matter? Did your "copy protection" have a hole in it and now she wants you to marry her?
Here's one a friend put to me a couple months ago that I quite enjoyed.
Riddle:
You step off an airplane in a strange city about which you know nothing, and you immediately hail a taxi cab. The taxi has "licensed cab #n" written on the side of it.
If your goal is to guess exactly the number of licensed taxi cabs in the city, what number should you guess?
If your goal is to guess the number of taxi cabs in the city to within plus or minus m, what number should you guess? What if you're trying to get the answer to within some fraction of the actual value?
Commentary (with no detailed spoilers, but probably not interesting until you've solved the problem already):
Within the group in which this question was posed (consisting entirely of slightly intoxicated physics grad students), we arrived at a thoroughly convincing answer in minutes, and then spent the next hour coming to the conclusion that the question is a lot more complicated than it sounds and our initial answer was incomplete.
You have to assume something about the possible distribution of numbers of cabs in the world in order to arrive at an answer. That assumption is so obvious, at least to those trained in frequentist statistics, that it's easy not to recognize that you're making it. Consider if, for example, the question had been posed on a planet on which every city was required to have exactly 1000 licensed cabs. In that case, the winning strategy will be very different from the straightforward answer. On our world, you'd really have to know a lot about the distribution of cab fleets in cities in order to answer the question properly. (At some level this is intuitive - if you step off a jumbo jet and get into taxicab #1, it's unlikely any real person would guess they'd landed in a city with a major airport but only one cab.)
BushCo (super clever man) is pushing for zero taxes? Wow, I am still paying 35% plus FICA+Medicare of roughly 12.4%(self-employed) + 1.45% so I pay 48.85% of my income in taxes by the end of the year. People at the other end pay 10% plus 6.2%(assuming the are employees)+1.45% so 17.65% in taxes for the year.
Why again is it fair that I have more taken from me becuase I produce more? Without me I know of at least 50 less jobs that would be out there. BTW, the top rate was reduced 1% the bottom by 5%. Sounds like a tax cut for the poor. 1% is less than 5% lat time I checked.
Let us not forget local and state taxes as well. We work past July just to pay taxes.
It stopped being profitable because of what, exactly? I'm asking because I do not know. Did the cost of greyness not change (in the way of more credible threats, burocratic impediments, whatever, on cheaters, for example)?
The Laffer Curve Theory can be refuted without even a hint of overturning any theorem, fundamental or otherwise (theorems cannot be overturned: at most, statements mistakenly believed to be theorems are shown to be false). The Laffer Curve Theory rests on reasoning of economical nature, on hypothesis of economical nature, and is only sensibly discussed in economical terms. You can make all kind of models of the same situation which can be refuted in economical terms. A very simple example is the following:
If you think for a minute about this alternative theory of the relation between taxation and revenue, you'll see that pretty much any mathematical reasoning you can do with the Laffer Curve can be done on the Suarez-Alvarez curve mutatis mutandi, just taking care of reversing some signs, and replacing maxima by minima and so on.
You will clearly agree that this Suarez-Alvarez Curve theory is quite inadecuate. Pretty much the same fundamental theorems are used in dealing with it as with Laffer's. The key difference (at least, so supply-side economists try to convince us of this), the Laffer Curve theory is supported by economical reasoning and econometric data, while the Suarez-Alvarez Curve Theory apparently is not.
How many fundamental mathematical theorems have been overturned because of the tragic fall of the Suarez-Alvarez Curve Theorem?
As for the rest of your comment, forgive if I ignore it.
Nice one! I think I have the answer though:
There's 4 distinguishable situations:
I hit the 'submit' button, consider the last post a good hint. Here's the rest of my answer:
Let's call the 'life expectancy' in 'situation P' P, in 'situation Q' Q, etc.
Then the following holds:
A bit of algebra yields that P equals 6, so my answer is that on average the crawlers live 6 days.
Okay, I'm back.
Another thing that does not seem to be clear in the problem is whether voting against would cut the pirate off from getting his share (whatever it may be) if the majority voted for. It seems others are assuming that by voting against, the pirate forfeits the proposed share of the treasure for himself. The problem doesn't state this anywhere, so I don't assume this (although I do assume that all the pirates want to live, don't care at all if the others die, and know that the others think the exact same way).
I was never good at these types of problems, so I don't have a solution thought up yet. But using the above info should provide an adequate solution.
You are MacGyver (that's hint #1).
You are running on a jungle, a hungry lion behind you, and you reach a river, which has a deadly crocodile waiting for his meal.
You have 1 binoculars, 1 (empty) match box, and 1 small clamp.
If you stay, the lion eats you, if you swim the crocodile eats you.
How do you cross the river?
Hint #2: You can't call Steve Irwin
What, do I need a sig now?
Obviously to everyone, or there would be no point in mentioning it.
d =13802205d =13804515d =13805037d =13804553d =13801559d =13801707d =13801933d =13805810d =13801427d =13804138d =13802845
d =13806048
d =13805847
You must be baiting me, or just exaggerating to make a point. Many word problems revolve around information that is known to one party in the "story" but not another. The devil is in the details.
I believe he has also had to clarify multiple times that the king was asking if they had been all in the room since they were origionally locked in their cells, even though that was explicitly stated in the origional problem.
Those are cases of poor reading--as you say the problem text is explicit about when the game starts. I challenge you to quote the text where the OP states explicitly that k is known to all parties in the story prior to the start of the game. Sure, yes, it's easily assumed and the answer is only valid if you do. But my point is that if you're going to offer up a math/logic problem, you should at least be able to state your starting conditions clearly.
The vast majority of people who posted got it. If it was so confusing, how were they able to do that?
They did it by either asking about k or by stipulating in their own post that k must be known to the prisoners.
http://ask.slashdot.org/comments.pl?sid=165444&ci
http://ask.slashdot.org/comments.pl?sid=165444&ci
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Look I'm not saying that it's an insolvable problem, I'm just saying it's lazily and inexactly written. I'm not alone.
Consider this comment:
http://ask.slashdot.org/comments.pl?sid=165444&ci
Or this one:
http://ask.slashdot.org/comments.pl?sid=165444&ci
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
I'm not blaming the victim, I'm just saying it's not anyone else's fault. Natural disasters happen. If you live below sea-level in a hurricane zone, and the shit hits the fan, don't cry to other people for not protecting you.
What 'leader' are you on about? I don't care about Bush, I'm not even American. But then it makes it easier to attack people when you put them into boxes.
Do you think I'm going to argue with a fool? As far as I'm concerned, your only purpose in life is to mark me as your foe.
Freedom = (Meaningful - Coerced) Choice != (Speech | Beer^2), and sad sock puppets' bad mods avail them naught.
You must have been a real pain in the ass to your friends when you were playing games as a kid.
l e.htm
Yup, I was the one who always probed the details. Sometimes it was just annoying but sometimes it was the key to the solution. What can I say, I grew up a stickler.
Here's a good example of a riddle where the details of the telling are essential to solving it...I used to love these as a kid. I think someone else posted this somewhere in this thread too.
http://www.greece.k12.ny.us/taylor/topics/grypuzz
Build a man a fire, he's warm for one night. Set him on fire, and he's warm for the rest of his life.
You have a persecution complex.
You're missing the argument completely. The whole point is that the overall tax rate was lower at the end of reagan's term (as determined by revenue as a % of GDP), but that the absolute revenue was higher. It is possible to lower taxes and increase revenue. Using the numbers I located, this occured even if you account for population growth and correct for inflation. The GDP per capita increased between the start and the end of his term, and per capita revenue grew 15%.
Why do you consider it inherently a bad thing that revenue fell as a portion of GDP? To use a weak analogy, the goal is to take a smaller slice but enlarge the pie. In other words, (20% of 100) > (21% of 90).
Incidentally, you should use 1980 as a start, but only as a comparison to before Reagan. He didn't take office until 1981. Considering 1989 has some usefulness if you assume that Bush Senior's policies didn't really impact things until later, but I agree that CATO should have left it out. This is why I ran the numbers from 1980 to 1988. They discuss this very point in the linked article however. They did use inflation corrected dollars. All amounts were given in 1992 dollars, but they did not factor out population growth.
Now, I will concede that we can't know for certain whether the tax rate modifications were a cause of the improved GDP. Nonetheless, your original statment that "Even during the Reagan years, when the supply-siders ran free like herds of buffalo in the White House, the tax cut resulted in dramatically lower revenue." is provably false, unless you redefine revenue as a percentage instead of an amount. Since that isn't the standard defintion, using it that way in an argument is misleading.
"Pulling together is the aim of despotism and tyranny! Free men pull in all sorts of directions" -- Havelock Vetinari
I'm confused. Your sig leads to a left wing website, but you're advocating for lower taxes. Are you a troll on that website?
Okay, 100th time: I never claimed the math alone proves the idea. Of course you need to add economic postulates like no tax yields no revenue, full tax yields no revenue, and some tax gives positive revenue. However, no one is really stupid enough to dispute these assumptions. That' why I said you have to refute the math if you claim the Laffer Curve is "refuted". Get it now? If not, work it out tomorrow night after you get home from brick-laying.
Now, let's look at the example you constructed specifically to show a case where the Laffer theory would not hold up despite the validity of the math. Keep in mind, this is a case where you dug to the very depths of your intelligence, and came up with the best possible example. So it should be pretty devastating if I can show that Laffer's claim holds up even under your economic model, don't you agree.
In the example that you specifically constructed to show how I'm wrong and how bright you are, there still exist regions in which raising taxes causes revenues to decline!!! Oops. Back to the drawing board.
By the way, I'm not so sure about this whole idea that you're a math professor. Math professors don't make dumbass mistakes like interpreting "positive function" to mean "positive slope function" and then sticking their necks out on it. For this reason, I'd like to know where you're a math professor so I can tell your colleagues about what went on here. If on the other hand, you're been lying this whole time, quietly slipping out would be the bright thing to do, m'k?
One final mistake: I don't know why you called it the Suarez-Alvarez Theory. Shouldn't it be the Suarez-Alvarez de Lopez de Castillo de Gonzales de Madrid de Saxe-Coburg-Gotha de Rodriguez-San Felipe Theory or something? Just a suggestion.
Rank my idea: http://www.sinceslicedbread.com/node/531
I am sorry, but I really do not care about you that much. While writing the previous post, I considered better options, but I really did not feel like writing them up.
Heh. Now, there also exist regions where raising taxes lowers revenues. Of course, that is true also of Laffer's curve. That is pretty much irrelevant to anything, too.
Go to Google and search for my name. It should be pretty easy to filter out the non-mathematical stuff. Now, in fact, I am not a professor of Mathematics: I'm right before becoming one in Buenos Aires. I am a professor in another university, but of Computer Science. My research is only in Mathematics, though.
Well, Suarez-Alvarez is my last name.
I have to confess I do not follow. All interpretations I seem to find of this are way too stupid, I'll just assume I have not found the correct one.
You are not very entertaining to write to, you know?
I am sorry, but I really do not care about you that much. While writing the previous post, I considered better options, but I really did not feel like writing them up.
Like hell you didn't! You had to construct an example to "prove" (hah!) me wrong. You picked that one. If you had a better one, you should have used it. You're just as careless creating examples as you are interpreting "positive function" to mean "positive slope function".
Heh. Now, there also exist regions where raising taxes lowers revenues. Of course, that is true also of Laffer's curve. That is pretty much irrelevant to anything, too.
Everybody agrees raising taxes can raise revenues. Laffer presented the obvious yet forgotten insight that raising taxes can also lower revenue. It's not irrelevant: it shows we need to keep in mind that taxing too high can defeat the purpose, something many on the left are unwilling to even consider. Laffer's insight is correct, and even your carefully constructed example validates it.
About the name I proposed for your theory: I thought that was your full name. Usually people who use hyphenated last names have a whole host of names part of their full names, and that includes the names of royalty (Saxe-Coburg-Gotha). Whatever reason you use a hyphenated last name, it's probably stupid. Just use your full name: Suarez-Alvarez de Lopez de Castillo de Gonzales de Madrid de Saxe-Coburg-Gotha de Rodriguez-San Felipe.
I hope I can find you on Google so I can link them to certain posts of yours. I also hope it doesn't cost you your job, although it probably will.
Rank my idea: http://www.sinceslicedbread.com/node/531
I'm reading the various posted "solutions" and there's one obvious (to me) thing missing. Why would the lowest ranked pirate ever agree with any distribution by the others? See below:
"if it's 50% or more against, you kill that highest rank"
So, if the problem were simplified to just 2 pirates, the lower ranking pirate gets everything by default by just voting against whatever the other pirate proposes. The lower ranking pirate has all the power since he makes up 50% of the vote with only 2 people.
Thus, it should be assumed that pirate #5 will always vote against, in hopes of being put in the final situation. He has nothing to gain by not voting against, since his life is not in danger. Plus, nowhere in the problem does it say that by voting against that a pirate loses his share of the bounty.
Given that pirates 1) put a slightly higher priority on living rather than getting gold, 2) don't give a damn whether the others live or die (so if they can get the same amount of gold either way, they'll choose to do their buddy(s) in anyway), 3) know that all the other pirates think the exact same way.
Now, knowing this, how would the other pirates react?
Nevermind, you're correct with 97-0-1-0-2 and 97-0-1-2-0. The unexplained (read: assumed by the smart ones) bits threw me off on the logic. But there's a reason why I don't accept 98-0-1-0-1 as an answer.
The following is my attempt to have it make sense to even laypeople like me.
Working backwards from pirates #4 and 5:
No matter what #4 says, #5 will always vote NO because he will receive everything (50% or more rule). Thus, #4 wants to avoid this scenario.
[#5 = 100]
Pirates #3, 4, and 5:
#3, knowing that #4 will vote YES no matter what to stay alive, keeps all 100 pieces to himself since the vote will be, #3&4 = YES, #5 = NO.
[#3 = 100, #4 = 0, #5 = 0]
Pirates #2, 3, 4, and 5:
Knowing that the above will happen if #3 gets his way, #2 gives one gold piece each to #4 and #5. Since #4 and #5 want to get at least something, they will both vote YES. So the vote will be #2&4&5 = YES, #3 = NO.
[#2 = 98, #3 = 0, #4 = 1, #5 = 1]
Pirates #1, 2, 3, 4, and 5:
Now things get interesting. #1 must get three votes (himself and two more) to survive. Given the above, #2 will always vote NO. To ensure the other votes, #1 will only have to give one piece of gold more than what the above scenario will provide to two others. Since he only needs two more votes, one piece will be given to #3 (to save on cost) and one more piece to either #4 or #5 (no need for both). So the vote will be #1 = YES, #2 = NO, #3 = YES, #4 = YES/NO, #5 = NO/YES.
[#1 = 97, #2 = 0, #3 = 1, #4 = 2 or 0, #5 = 0 or 2]
98-0-1-0-1 (or 98-0-1-1-0) will not work because neither #4 nor #5 will be given a reason to vote YES. Remember, pirates will only vote YES if there is an advantage for themselves, not for the greater good.
Okay, so in the first situation he votes against it, and the first pirate gets thrown overboard.
In the second situation he votes against it, and the second pirate gets thrown overboard.
But in the third situation, there are only three pirates left (3, 4 and 5). #3 knows that #5 will vote against it, but he only needs two votes to survive. #4 will vote for this proposal whatever he's offered, because he doesn't want to propose something that #5 might reject.
So #3 proposes that he gets all 100 gold pieces. #4 agrees, because he doesn't want to die. #5 rejects it, but it's two votes to one, and he gets nothing.
Whatever he's offered in the first proposal is better than what he would get offered in the second proposal, which is better than nothing (what he would get from the third proposal).
Self-referential sigs do not a humourous poster make.