Hey, before you changed the history, I had a Nobel prize, owned a few billion dollars and was in the process of winning the elections to the president of the world. Thanks to your changes I don't even have an idea what I should have gotten the Nobel prize for, my bank account is rather low, and there's not even the position of the president of the world. Fix that!
The key to immortal computers is to attach a doomsday mechanism to it, which triggers if the computer fails. Since the failure of the computer would ultimately be a quantum event, this doomsday mechanism would implement a quantum suicide. And since a quantum suicide mechanism is supposed to never trigger for the one who would get killed, the computer cannot fail.
Now, chances are that quantum suicide doesn't really work, and the mechanism would trigger anyway. But then, after the doomsday mechanism is triggered, there's nobody left who could care about it.:-)
But you don't need to check the vast majority of those paths. We know that the optimal path doesn't cross itself (because if the path crosses itself you can construct a shorter path by simply changing those crossing lines so that they don't cross any more). We know that the points lying on the convex hull are passed through in order (because otherwise the lines would cross). I don't know how to calculate the number of paths one would have to consider in the worst case (the best case is easy; one path, if all cities lie on the convex hull), but it's definitively much less than (500!).
Computers in the real world cannot even solve every P problem with current algorithms. That's because they can only address a limited amount of memory, and they only work for a limited time before they break down.
Everyone knows how to choose up sides, right? The captains alternate picks. We're so used to doing that we just assume that is the fairest way to divide up players, but experience shows that one team often ends up much stronger than the other when teams are picked this way.
You don't need experience to get to that result. Simple logic tells you that the one starting will inevitably get the stronger team. The argument is easy: First, group the players in pairs: The first pair contains of the two best players, the second pair gets the next best player, and so on. Now it's obvious that in the first round, the two people of the first pair are chosen. The captain who gets the first choice of course will pick the better one. But the captain who has the first choice will also have the first choice on the second pair, and on each following pair. Therefore he will inevitably get the better team.
It's also easy to give a strategy which is more fair: Instead of the commonly chosen pattern ABABABAB..., let them choose according to the pattern ABBAABBA... That way, while A gets the first pick on the first pair, B gets the first pick on the second pair, and so on. Or if you group the players in four, captain A will get the best and the least good from each group of four, while captain B will get the middle two.
Of course it still isn't guaranteed to give the optimal teams, but it at least doesn't guarantee captain A the better team. IN your example, captain A would get the total score 88+12+6+1 = 107, and captain B would get the total score 50+25+3+1 = 79. So in this case captain A would still get the much better team, but at least not as pronounced as with the ABAB... method. And if we take instead e.g. the scores 85, 83, 81, 70, 52, 44, 32, 15, then the usual ABAB... method would give the total scores 250/212, but the ABBA... method would give 222/240, so in this case B would get the better team, but his advantage would not be as pronounced as the advantage A has with the usual ABAB... method.
Polynomial time vs. Non-polynomial time? That's what I thought it meant. Ok. So its Quick solvability vs. Quick checkability and if P=NP its both.
P means polynomial time; more exactly, it is the set of all problems which can be solved in polynomial time by a (deterministic) Turing machine. But NP doesn't mean non-polynomial time. It means nondeterministic polynomial. The set of NP-problems is the set of problems which can be solved with a non-deterministic Turing machine in polynomial time. Especially all P problems are also NP problems. However the reverse is only true if P=NP.
Agreed, but I had to think of something to say and that's the only field I know enough about to comment. Im sure there are better examples. It basically reduces something that requires ~ base^N computations or higher to only requiring ~ N^power computations. That's like reducing hours to seconds for sufficiently large N.
It might mean reducing hours to seconds. But it also might mean reducing millions of years to mere centuries.
Certainly? Web pages have a nice tree structure. Even if you take into account that e.g. the same image may be used twice, you still get a DAG. It should be trivial to get memory management right.
There's nothing wrong with 'Eye Pee' law! Nothing. Next week, I'm going to copywrite the words 'the', 'it', 'and' and the letters 'E' and 'S'. You are all on notice! I'm open to licensing, say 20 dollars per use, with only 2 dollars per use for book publishers and newspapers. Internet sites will be on a per-click basis. Piracy will not be tolerated! Pay me or be a crimminal. Also, you won't be allowed to use these words in speech. You will all have to find alternatives.
No problm. I will not uth thi mintiond words ath wll ath thi lttrth. Yith, thith will mak my txtth a littl bit hardr to rid, but that'th OK. Thi only problm ith that pipl now will think that I'm lithpling.
Trademarks should be restricted to what they were invented for: Identifying products. If I sell a soft drink and claim it's Coke when it isn't, that's bad. This is what trademark infringement was invented against. But if I make a button saying "I like Coke" or "I hate Coke" or "I didn't have a Coke today", or if I write a book with the title "My first Coke", then this should not be a trademark infringement.
HOWEVER, the Tolkien estate is still LEGALLY REQUIRED to sue anyone they know who is using their mark without their permission.
I know laws can be pretty stupid, but I'm sure it cannot be that stupid. I bet they only need to sue anyone whose use is infringing.
Otherwise the trademark "Coke" would be long lost because they didn't sue any of the millions of people who used that word daily to refer to, well, Coke.
Slashdot Mirror
When Faraday was asked by the finance minister what this electricity was actually good for, he answered: "One day you will tax it."
Also, geochemistry? Shouldn't that be hermechemistry?
Hey, before you changed the history, I had a Nobel prize, owned a few billion dollars and was in the process of winning the elections to the president of the world. Thanks to your changes I don't even have an idea what I should have gotten the Nobel prize for, my bank account is rather low, and there's not even the position of the president of the world. Fix that!
This proves that P!=NP right there. Go get your $1M!
Wow, you get 1 million factorial dollars for that? :-)
Which is then consulted as source by the traditional media.
Depends. It might be that the best algorithm for breaking our public key cryptosystems is O(n^1000). That's polynomial, but still safe.
The key to immortal computers is to attach a doomsday mechanism to it, which triggers if the computer fails. Since the failure of the computer would ultimately be a quantum event, this doomsday mechanism would implement a quantum suicide. And since a quantum suicide mechanism is supposed to never trigger for the one who would get killed, the computer cannot fail.
Now, chances are that quantum suicide doesn't really work, and the mechanism would trigger anyway. But then, after the doomsday mechanism is triggered, there's nobody left who could care about it. :-)
But you don't need to check the vast majority of those paths. We know that the optimal path doesn't cross itself (because if the path crosses itself you can construct a shorter path by simply changing those crossing lines so that they don't cross any more). We know that the points lying on the convex hull are passed through in order (because otherwise the lines would cross). I don't know how to calculate the number of paths one would have to consider in the worst case (the best case is easy; one path, if all cities lie on the convex hull), but it's definitively much less than (500!).
Computers in the real world cannot even solve every P problem with current algorithms. That's because they can only address a limited amount of memory, and they only work for a limited time before they break down.
Well if you prove that A = Not A, you're certainly going to give mathematicians and physicists everywhere a heart attack.
Not if you use fuzzy logic. There A is simply half-true.
You don't need experience to get to that result. Simple logic tells you that the one starting will inevitably get the stronger team. The argument is easy: First, group the players in pairs: The first pair contains of the two best players, the second pair gets the next best player, and so on. Now it's obvious that in the first round, the two people of the first pair are chosen. The captain who gets the first choice of course will pick the better one. But the captain who has the first choice will also have the first choice on the second pair, and on each following pair. Therefore he will inevitably get the better team.
It's also easy to give a strategy which is more fair: Instead of the commonly chosen pattern ABABABAB..., let them choose according to the pattern ABBAABBA... That way, while A gets the first pick on the first pair, B gets the first pick on the second pair, and so on. Or if you group the players in four, captain A will get the best and the least good from each group of four, while captain B will get the middle two.
Of course it still isn't guaranteed to give the optimal teams, but it at least doesn't guarantee captain A the better team. IN your example, captain A would get the total score 88+12+6+1 = 107, and captain B would get the total score 50+25+3+1 = 79. So in this case captain A would still get the much better team, but at least not as pronounced as with the ABAB... method. And if we take instead e.g. the scores 85, 83, 81, 70, 52, 44, 32, 15, then the usual ABAB... method would give the total scores 250/212, but the ABBA... method would give 222/240, so in this case B would get the better team, but his advantage would not be as pronounced as the advantage A has with the usual ABAB... method.
Not totally. The one-time pad will remain secure.
Well, O(n^1000) algorithms are not exactly what I'd call useful.
If you can solve it in one Planck time for some problem size, solving it for twice that size will need about 10^240 times the age of the universe.
Well, as long as you don't break an ordinal rule ...
Polynomial time vs. Non-polynomial time? That's what I thought it meant. Ok. So its Quick solvability vs. Quick checkability and if P=NP its both.
P means polynomial time; more exactly, it is the set of all problems which can be solved in polynomial time by a (deterministic) Turing machine. But NP doesn't mean non-polynomial time. It means nondeterministic polynomial. The set of NP-problems is the set of problems which can be solved with a non-deterministic Turing machine in polynomial time. Especially all P problems are also NP problems. However the reverse is only true if P=NP.
Agreed, but I had to think of something to say and that's the only field I know enough about to comment. Im sure there are better examples. It basically reduces something that requires ~ base^N computations or higher to only requiring ~ N^power computations. That's like reducing hours to seconds for sufficiently large N.
It might mean reducing hours to seconds. But it also might mean reducing millions of years to mere centuries.
OK, then replace the quad-processor DOS computer on steroids with a quad-processor MacOS computer on steroids. :-)
Certainly?
Web pages have a nice tree structure. Even if you take into account that e.g. the same image may be used twice, you still get a DAG. It should be trivial to get memory management right.
Of course not. IE6 shops tend to have no problems with Firefox breaking plugin compatibility. :-)
Well, the tools will be restricted to non-free countries, because after all, in free countries you don't need them anyway ...
There's nothing wrong with 'Eye Pee' law! Nothing. Next week, I'm going to copywrite the words 'the', 'it', 'and' and the letters 'E' and 'S'. You are all on notice! I'm open to licensing, say 20 dollars per use, with only 2 dollars per use for book publishers and newspapers. Internet sites will be on a per-click basis. Piracy will not be tolerated! Pay me or be a crimminal. Also, you won't be allowed to use these words in speech. You will all have to find alternatives.
No problm. I will not uth thi mintiond words ath wll ath thi lttrth. Yith, thith will mak my txtth a littl bit hardr to rid, but that'th OK. Thi only problm ith that pipl now will think that I'm lithpling.
Joke's funnier when you spell it right.
He didn't want to get sued by T*lk**n Estate.
As if the author would see a single cent of the movie revenues anyway.
Trademarks should be restricted to what they were invented for: Identifying products.
If I sell a soft drink and claim it's Coke when it isn't, that's bad. This is what trademark infringement was invented against.
But if I make a button saying "I like Coke" or "I hate Coke" or "I didn't have a Coke today", or if I write a book with the title "My first Coke", then this should not be a trademark infringement.
I know laws can be pretty stupid, but I'm sure it cannot be that stupid.
I bet they only need to sue anyone whose use is infringing.
Otherwise the trademark "Coke" would be long lost because they didn't sue any of the millions of people who used that word daily to refer to, well, Coke.