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Everything posted by Rainman
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Every Natural Number can be Unambiguously Described in Fourteen Words or Less
Rainman replied to BMAD's question in New Logic/Math Puzzles
I might be necro-posting here, but I find this interesting. The problem with the "proof" is that it tries to pass off the English language as a formal mathematical language. By what standards do we decide if a sequence of words unambiguously describes a natural number? Are those standards semantic or syntactic? Let's break it down. If certain sequences of words unambiguously describe natural numbers, let S be the set of those sequences. Define the function f : S → N that maps each such sequence of words to the natural number that it describes. Your claim is that f is surjective. Your "proof" is as follows: 1. Suppose f is not surjective. Then it misses a non-empty subset of N. 2. Any non-empty subset of N has a smallest element. 3. Thus we can define n as the smallest natural number that f misses. 4. But now f(the smallest natural number that f misses) = n. So f did not miss n, a contradiction. 5. Since assuming that f is not surjective leads to a contradiction, that assumption must be false. 6. Hence f is surjective. This rewording of the OP should make it clearer. Notice how the phrase "the smallest natural number that f misses" is used semantically in step 3, but syntactically in step 4. If we did invent a formal mathematical language that allowed such syntax, then we have only used that piece of syntax once in our "proof". So we could just replace it with any other piece of unused syntax by argument of symmetry. Our language, our rules: 1. Suppose f is not surjective. Then it misses a non-empty subset of N. 2. Any non-empty subset of N has a smallest element. 3. Thus we can define n as the smallest natural number that f misses. 4. But now f(Mynd you, møøse bites Kan be pretti nasti...) = n. So f did not miss n, a contradiction. 5. Since assuming that f is not surjective leads to a contradiction, that assumption must be false. 6. Hence f is surjective. By now it is made clear that step 4 is technically just a second assumption rather than a justified conclusion. -
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I'd hate to disappoint them, but...
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I probably shouldn't even be doing math now because I'm quite sick and feverish, but I maintain my stance that we don't have enough information to solve this problem either. We would have to know the probability distribution for French coins among all Euro coins. Imagine for example that French coins would be very rare, with only one Euro coin in a million being French. It is then much more likely that you only have one French coin and simply happened to pull it three times, than that all three coins are French which would be a one in 1018 anomaly. On the other hand, imagine that 90% of all Euro coins were French instead. Now the base probability that all three coins are French, even before you pull out any of them, is 0.93 = 0.729. Doing the test of pulling coins out and finding them French would only increase the probability that all are French. On a side note, in my last post I claimed that we could calculate P(A|B), which upon further thinking seems to be false. We can't even do that without the probability distribution.
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What part of the body was measured for temperature? Different parts of the body cool at different rates.
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Still, I did assume that a+b+c+d = 40. Indeed 12+11+9+8 = 40. My counterexample holds.
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What are the allowed operations? Can we only use the standard operations (connect two points or extend a line segment) or may we use other tricks, for example align the straightedge along the diameter and use the other side of the straightedge to draw a parallel line?
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Hats on a death row!! One of my favorites puzzles!
Rainman replied to roolstar's question in New Logic/Math Puzzles
1. Nobody needs to sacrifice himself. The first prisoner to guess has a 50/50 chance of survival, regardless of whether he adheres to the strategy or not. Nothing he can do can improve his chances of survival, so he might as well help the others (not assuming any grudges). 2. Well, yes, the average prisoner most likely is more stupid than the average non-prisoner. But this being a math/logic puzzle, aimed at people who are more intelligent than average, the general assumption is that the prisoners would be intelligent enough to understand the solution proposed by the puzzle solver. And anyway, in this fictional setting with a king who likes to toy with his people, it is plausible that these prisoners have survived through similar challenges before and thus are far more intelligent than regular prisoners, through the survival of the fittest. 3. I could apply the same argument as for point 2, that this is a math/logic puzzle and we should not assume distrust unless explicity told to do so. And I could also argue that in this fictional world with an oppressive king, the people might have developed more trust towards each other. 4. The OP's description of the procedure seems to imply that each prisoner's judgment is carried out before they move on to the next prisoner. So unless you have heard a prisoner die before you (except the first one who might die due to random chance), you would have no reason to assume that a mistake has been made. And again, if we were supposed to account for the possibility of mistakes, we would have been explicitly told to do so, and possibly even have been provided with the exact probability of a mistake being made. 5. Finally, even with all that being said, all it takes for one mistake/betrayal to be canceled out is another mistake/betrayal. Maybe the first prisoner intended to betray you, but was too stupid and accidentally gave the "good" answer. So even if the prisoners were indeed selfish, stupid, distrusting, and prone to betrayal, it would boil down to a crap shoot anyway, even if you did follow the strategy. So once the strategy has been proposed, there is really no reason to deviate from it (unless you have heard people dying before you as I mentioned earlier). -
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How do you arrive at this answer? If I'm not mistaken, the answer is approximated by running the following program indefinitely, given a full list of primes in order P1 = 2, P2 = 3, P3 = 5, and so on: 1: set S = 0, i = 1 2: set S = S + (1-S)/Pi2 3: set i = i+1 4: return to step 2 As this program keeps going, S should tend to the desired probability. Not being a programmer, I haven't been able to run this for large numbers, but it seems to me that it would converge closer to 0.4 than 0.6. I would be very surprised if it converged at exactly 6/pi2, as this would indicate a clear correlation between pi and the set of primes.
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Monochromatic rectangle on a two-colored plane
Rainman replied to bonanova's question in New Logic/Math Puzzles
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Monochromatic equilateral triangle on a 2-colored plane
Rainman replied to bonanova's question in New Logic/Math Puzzles
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Equilateral Triangle: Color and distance
Rainman replied to BMAD's question in New Logic/Math Puzzles