Ever been in an office with 25 people all talking on the telephone at once? I have - and it seems to work just fine. Nobody on either end of the 25 lines ever gets confused. Just need to make good mic selections - heck you could do it all with just whispers.
Of course, it might get annoying to be in an office with more chatter than you're used to. But "give it a few months. You'll get used to it... or you'll have a psychotic episode." Either that or wear headphones.
Three of the monks have red eyes, here's Monk "A"'s thought process:
I see 2 Red-Eyed Monks, that means they both see:
1 Red-Eyed Monk (if I have Brown Eyes)
2 Red-Eyed Monks (if I have Red Eyes) Assume I have Brown Eyes:
Each of the 2 Red-Eyed Monks I see therefore see only 1 Red-Eyed Monk. Now then, each of them must be thinking:
Either there is 1 Red-Eyed Monk (him)
Or there are 2 (him and me)
Assume I have Brown Eyes
Therefore there is only 1 Red-Eyed Monk (him)
So HE must be thinking:
Either there are 0 Red-Eyed Monks
Or there is 1 Red-Eyed Monk (Me)
Tourist says there's at least 1
Therefore, I have red Eyes
I will kill myself tonight (night 1)
But, DAMN, he didn't end up killing himself
Assumption INCORRECT
I have Red Eyes
I will kill myself tonight (night 2)
But, DAMN, he didn't end up killing himself
Assumption INCORRECT I have Red Eyes I will kill myself tonight (night 3)
Monks A,B,C are indistinguishable - they all act exactly in the same way. Therefore A,B,C kill themselves all on the same night.
Midnight isn't key, you right, but because they always kill themselves in synchronization. Midnight just emphasizes this fact and makes for a better story
The easy answer: Think of a number between 0.999... and 1 Can't do it? Don't feel bad, no one can. Since there is no number between 0.999... and 1, they are not distinct from one another and must be the same number (think of how you would order them, neither is less than the other...).
The harder way: Didn't everyone do this at some point? Say I want to represent the number above the "^" on the following numebr line going from 0 to 1.
|----|----|----|----| 0---.25-^.50--.75---1 [I hope that came out right...]
One way to do this is to divide the line in half. If the number falls below this division (left) write a "0" in the first place after the "decimal" point, if it's above (right), write a "1" So... ".0"
Now divide this new subinterval in half again. Again write "0" if the number is in the left half, and "1" if its in the right So... ".01"
Keep going to represent the numebr in a sort of binary format (I'm not a CS person, is there a real name for this?). So... ".0110..."
But what happens if the number you want happens to fall exactly on a dividing line, say on the line corresponding to.25 So... ".0?"
Is it "?" a "1" or a "0"? To take care of this you can define your procedure a bit better: "0" if in [Left_End,Mid_Point), "1" if in [Mid_Point,Right_End]. So... ".0100000000..."
But that's just as arbitrary as: "0" if in [Left_End,Mid_Point], "1" if in (Mid_Point,Right_End]. (Mind the difference between "(" and "["). So... ".0011111111..."
Since the procedural difference is arbitrary, the numbers are EXACTLY the same. Extending this into real decimal is trivial: just use 10 divisions instead of 2 every iteration. Thus, 0.999...=1.000...
--Not that anyone will see this since my Karma is officially "BAD"
Close - but you're a night off if I'm thinking right. As someone pointed out above, everyone kills themselves on the Nth Night (where there are N Red-Eyed Monks). I'm, of course, assuming that since these monks have nothing better to do all day, they'll think about this a lot and therefore be 100% rational (as in Game Theory).
Case 1: 1 Red-Eyed Monk (A) _Day 1_ A: "I See No Red-Eyes Monks!" _Night 1_ A KILLS HIMSELF
Case 2: 2 Red-Eyed Monks (A, B) _Day 1_ A: "I See 1 Red-Eyed Monk (B). B sees 0 or 1(Me)" B: "I See 1 Red-Eyed Monk (A). A sees 0 or 1(Me)" _Night 1_ NOTHING HAPPENS
_Day 2_ A: "B did not kill himself. B saw 1 Red-Eyed Monk(Me). I have red Eyes" B: "A did not kill himself. A saw 1 Red-Eyed Monk(Me). I have red Eyes" _Night 2_ A, B BOTH KILL THEMSELVES
Case 3: 3 Red-Eyed Monks (A, B, C) _Day 1_ A: "I See 2 Red-Eyed Monks (B,C). B sees 1(C) or 2(Me,C). C sees 1(B) or 2(Me,B)" B: "I See 2 Red-Eyed Monks (A,C). A sees 1(C) or 2(Me,C). C sees 1(A) or 2(Me,A)" C: "I See 2 Red-Eyed Monks (A,B). A sees 1(B) or 2(Me,B). B sees 1(A) or 2(Me,A)" _Night 1_ NOTHING HAPPENS
_Day 2_ A: "If B saw 1(C), C saw 1(B) and I have Brown Eyes. Therefore they will kill themselves tonight." [See Case 2:] B: "" C: "" [I'm getting damn tired of writing this. Someone could write a teeny little program that does all this] _Night 2_ (3 Red-Eyed so...) NOTHING HAPPENS
_Day 3_ A: "B, C did not kill themselves. Therefore each saw two(Me,C|B). Therefore, I have Red Eyes." B: "" C: "" _Night 3_ A,B,C ALL KILL THEMSELVES
Case N:... _Night N_ A,...,"N" ALL KILL THEMSELVES
--Not that anyone will see this since my Karma is officially "BAD"
Slashdot Mirror
It was recently determined that Adam Bressen's lawyer is, in fact, timothy's father...
Ever been in an office with 25 people all talking on the telephone at once? I have - and it seems to work just fine. Nobody on either end of the 25 lines ever gets confused. Just need to make good mic selections - heck you could do it all with just whispers.
Of course, it might get annoying to be in an office with more chatter than you're used to. But "give it a few months. You'll get used to it... or you'll have a psychotic episode." Either that or wear headphones.
The Tourist Is Critical
Consider _Case 3_:
Three of the monks have red eyes, here's Monk "A"'s thought process:
I see 2 Red-Eyed Monks, that means they both see:
1 Red-Eyed Monk (if I have Brown Eyes)
2 Red-Eyed Monks (if I have Red Eyes)
Assume I have Brown Eyes:
Each of the 2 Red-Eyed Monks I see therefore see only 1 Red-Eyed Monk. Now then, each of them must be thinking:
Either there is 1 Red-Eyed Monk (him)
Or there are 2 (him and me)
Assume I have Brown Eyes
Therefore there is only 1 Red-Eyed Monk (him)
So HE must be thinking:
Either there are 0 Red-Eyed Monks
Or there is 1 Red-Eyed Monk (Me)
Tourist says there's at least 1
Therefore, I have red Eyes
I will kill myself tonight (night 1)
But, DAMN, he didn't end up killing himself
Assumption INCORRECT
I have Red Eyes
I will kill myself tonight (night 2)
But, DAMN, he didn't end up killing himself
Assumption INCORRECT
I have Red Eyes
I will kill myself tonight (night 3)
Monks A,B,C are indistinguishable - they all act exactly in the same way. Therefore A,B,C kill themselves all on the same night.
Midnight isn't key, you right, but because they always kill themselves in synchronization. Midnight just emphasizes this fact and makes for a better story
As Above, 0.999...=1
.25
The easy answer:
Think of a number between 0.999... and 1
Can't do it? Don't feel bad, no one can. Since there is no number between 0.999... and 1, they are not distinct from one another and must be the same number (think of how you would order them, neither is less than the other...).
The harder way:
Didn't everyone do this at some point?
Say I want to represent the number above the "^" on the following numebr line going from 0 to 1.
|----|----|----|----|
0---.25-^.50--.75---1
[I hope that came out right...]
One way to do this is to divide the line in half. If the number falls below this division (left) write a "0" in the first place after the "decimal" point, if it's above (right), write a "1"
So... ".0"
Now divide this new subinterval in half again. Again write "0" if the number is in the left half, and "1" if its in the right
So... ".01"
Keep going to represent the numebr in a sort of binary format (I'm not a CS person, is there a real name for this?).
So... ".0110..."
But what happens if the number you want happens to fall exactly on a dividing line, say on the line corresponding to
So... ".0?"
Is it "?" a "1" or a "0"? To take care of this you can define your procedure a bit better:
"0" if in [Left_End,Mid_Point), "1" if in [Mid_Point,Right_End].
So... ".0100000000..."
But that's just as arbitrary as:
"0" if in [Left_End,Mid_Point], "1" if in (Mid_Point,Right_End]. (Mind the difference between "(" and "[").
So... ".0011111111..."
Since the procedural difference is arbitrary, the numbers are EXACTLY the same. Extending this into real decimal is trivial: just use 10 divisions instead of 2 every iteration.
Thus, 0.999...=1.000...
--Not that anyone will see this since my Karma is officially "BAD"
Close - but you're a night off if I'm thinking right. As someone pointed out above, everyone kills themselves on the Nth Night (where there are N Red-Eyed Monks). I'm, of course, assuming that since these monks have nothing better to do all day, they'll think about this a lot and therefore be 100% rational (as in Game Theory).
...
Case 1:
1 Red-Eyed Monk (A)
_Day 1_
A: "I See No Red-Eyes Monks!"
_Night 1_ A KILLS HIMSELF
Case 2:
2 Red-Eyed Monks (A, B)
_Day 1_
A: "I See 1 Red-Eyed Monk (B). B sees 0 or 1(Me)"
B: "I See 1 Red-Eyed Monk (A). A sees 0 or 1(Me)"
_Night 1_ NOTHING HAPPENS
_Day 2_
A: "B did not kill himself. B saw 1 Red-Eyed Monk(Me). I have red Eyes"
B: "A did not kill himself. A saw 1 Red-Eyed Monk(Me). I have red Eyes"
_Night 2_ A, B BOTH KILL THEMSELVES
Case 3:
3 Red-Eyed Monks (A, B, C)
_Day 1_
A: "I See 2 Red-Eyed Monks (B,C). B sees 1(C) or 2(Me,C). C sees 1(B) or 2(Me,B)"
B: "I See 2 Red-Eyed Monks (A,C). A sees 1(C) or 2(Me,C). C sees 1(A) or 2(Me,A)"
C: "I See 2 Red-Eyed Monks (A,B). A sees 1(B) or 2(Me,B). B sees 1(A) or 2(Me,A)"
_Night 1_ NOTHING HAPPENS
_Day 2_
A: "If B saw 1(C), C saw 1(B) and I have Brown Eyes. Therefore they will kill themselves tonight." [See Case 2:]
B: ""
C: "" [I'm getting damn tired of writing this. Someone could write a teeny little program that does all this]
_Night 2_ (3 Red-Eyed so...) NOTHING HAPPENS
_Day 3_
A: "B, C did not kill themselves. Therefore each saw two(Me,C|B). Therefore, I have Red Eyes."
B: ""
C: ""
_Night 3_ A,B,C ALL KILL THEMSELVES
Case N:
_Night N_ A,...,"N" ALL KILL THEMSELVES
--Not that anyone will see this since my Karma is officially "BAD"
Out of curiosity...
Is there anyone out there with a Slashdot Subscription that saw this story on the front page...?
Also, how would you respond to the claim that "newfangled" digital technologies cheapen the original art form of film?
Always sorta been curious but never bothered to take any initiative and actually look it up myself:
How precise is the frequency of AC from a typical power company? And how much does it change over time (if any)?
Clockless Chips
Facts:
1. Hackers are mammals.
2. Hackers hack ALL the time.
3. The purpose of the hacker is to flip out and phreak the system.