- #1

JakeA

- 12

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I have some questions about the presentation of the P vs NP problem here. I'm quite new to this stuff, so pardon my ignorance if I make dumb statements.

http://www.claymath.org/millennium/P_vs_NP/Official_Problem_Description.pdf [Broken]

Specifically:

First of all this statement is confusing to me. Is he talking about 200 binary digit numbers? Decimal? It's a bit ambiguous to me.

By factoring is he talking about prime factoring?

Is he talking about any 200 digit number? I can prime factor 2

Finally I assume he's claiming that since you can verify factors of 200 digit numbers in a few minutes, if P = NP then you could factor numbers in equally bounded time. Is this correct? I'm not sure because I see 2 problems with it.

1) If the problem is bounded by n

2) If he's talking about prime factorization, he is effectively saying that you can verify a prime number in P time, a very dubious claim.

As far as I can tell with the state of the art in cryptography, new factoring algorithms are big problems, actually increasing the need for larger keys/certificates all the time. My view is that the reason cryptography works is not because P != NP, but because prime number verification is Non P, meaning that incremental increases in number size create exponential increases in computational time. Again, I'm no expert, just wanting more information.

Thanks

http://www.claymath.org/millennium/P_vs_NP/Official_Problem_Description.pdf [Broken]

Specifically:

The security of the internet, including most financial transactions,

depend on complexity-theoretic assumptions such as the difficulty of

integer factoring or of breaking DES (the Data Encryption Standard). If P=

NP these assumptions are all false. Specifically, an algorithm solving 3-SAT

in n2 steps could be used to factor 200-digit numbers in a few minutes.

First of all this statement is confusing to me. Is he talking about 200 binary digit numbers? Decimal? It's a bit ambiguous to me.

By factoring is he talking about prime factoring?

Is he talking about any 200 digit number? I can prime factor 2

^{200}in less than a second.Finally I assume he's claiming that since you can verify factors of 200 digit numbers in a few minutes, if P = NP then you could factor numbers in equally bounded time. Is this correct? I'm not sure because I see 2 problems with it.

1) If the problem is bounded by n

^{K}, who's to say what k is? It could be huge, much larger than the k for the NP problem.2) If he's talking about prime factorization, he is effectively saying that you can verify a prime number in P time, a very dubious claim.

As far as I can tell with the state of the art in cryptography, new factoring algorithms are big problems, actually increasing the need for larger keys/certificates all the time. My view is that the reason cryptography works is not because P != NP, but because prime number verification is Non P, meaning that incremental increases in number size create exponential increases in computational time. Again, I'm no expert, just wanting more information.

Thanks

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