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#34 | |
"Bob Silverman"
Nov 2003
North of Boston
11101010101002 Posts |
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I think you need to delete the 'not'.......... |
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#35 | ||
Aug 2006
135438 Posts |
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#36 | |
"Bob Silverman"
Nov 2003
North of Boston
22·1,877 Posts |
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The definition/wording in Wikipaedia is a bit vague: "given an integer N and an integer M with 1 โค M โค N, does N have a factor d with 1 < d < M?" If I give you M = [sqrt(N)] + 1, the answer is trivially yes. (or N is prime; which is in P) If M ~ (log(N))^k for any k, then the question can be answered in polynomial time. The meaning of "an integer M", is unclear, because it is clear that for some M depending on N the question is in P. |
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#37 | |
Aug 2006
176316 Posts |
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Sure, there are easy cases, but there are easy cases of SAT, too. So what's your view? How likely is it that FACTORIZATION is in P? (xilman, you're welcome to answer as well!) And if you're willing to go out on a limb, how likely to you think it is that:
I would have called all of these extremely unlikely, but now that I'm seeing people suggest that FACTORIZATION might be in P it seems like these should be reconsidered (since they are 'at least as likely'). * Either the one you gave above, or, say, "What is the m-th bit of the largest prime factor of N?" for 1 < m < lg N. |
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#38 | |
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
1165410 Posts |
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To be clear, I believe that P != NP. Paul |
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#39 | |
"Bob Silverman"
Nov 2003
North of Boston
22×1,877 Posts |
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#40 | |
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2·5,827 Posts |
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However, we could define it in another way, that used by gamblers who are not mathematicians. The probability is then the answer to the question: if you wished to place a bet, what odds would you negotiate before committing your money? My gut reaction, and it is only that, is that I'd settle for odds somewhere between 100:1 and 1000:1. That is a "fair chance" in my estimation. The reasoning behind my estimation that FACTORIZATION has a fair chance of being in P is entirely the result of analogy and wishful thinking. I don't know the answer --- which is pretty clear because you would have heard about it if I did. Anyway, history shows us that until recently factorization was as hard as primality proving and only exponential time algorithms were known for each. About forty years ago, the two problems were still of equal complexity but a subexponential algorithm was found for factoring. Dixon's algorithm, for instance, takes time which is in a loose sense half-way between exponential and polynomial. The looseness can be removed but I won't bother in this post. A decade or so later, GNFS factors in (heuristic) 1/3 exponential -- 2/3 polynomial time, again in the same loose sense. By that measure, we're already 2/3 of the way to finding a P-time algorithm. I emphasise that this is not at all rigorous and is almost entirely vigorous hand-waving. Turning to PRIMES, the factoring algorithms can still be used but asymptotically better ones have been discovered. ECPP and APRT-CL both run in (heuristically for ECPP) very nearly polynomial time. A few years ago, the polynomial time AKS algorithm was discovered. Paul Last fiddled with by xilman on 2010-08-16 at 07:57 Reason: Fix a couple of small bugs |
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#41 |
Aug 2006
5,987 Posts |
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Paul, thank you for your carefully-considered answer. (And yes, you have understood me correctly in terms of 'probability' -- think of it in terms of Bayesian knowledge sets if you must.
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#42 |
"Lucan"
Dec 2006
England
2×3×13×83 Posts |
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But nobody (except Xilman miscounting negatives) was trying
to suggest P = NP. I quoted Paul Seymour because I thought others here might find his criteria for plausibility (long and unintelligible) hilarious, especially coming from the chief editor of a major mathematical journal. (More or less the same as Bob's initial reaction). OTOH, the statement that P=NP would make much of his mathematical life pointless conveys the importance of the problem. Can someone explain simply why he said that? David |
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#43 |
Aug 2006
5,987 Posts |
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Not that I agree with the sentiment, but I think the theory is that if P = NP, proving theorem is easy, so who needs a mathematician? Just hook up the polynomial-time SAT solver computer.
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#44 |
"Lucan"
Dec 2006
England
145128 Posts |
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