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I am just a beginner of computer science. I learned something about running time but I can't be sure what I understood is right. So please help me.

So integer factorization is currently not a polynomial time problem but primality test is. Assume the number to be checked is n. If we run a program just to decide whether every number from 1 to sqrt(n) can divide n, and if the answer is yes, then store the number. I think this program is polynomial time, isn't it?

One possible way that I am wrong would be a factorization program should find all primes, instead of the first prime discovered. So maybe this is the reason why.

However, in public key cryptography, finding a prime factor of a large number is essential to attack the cryptography. Since usually a large number (public key) is only the product of two primes, finding one prime means finding the other. This should be polynomial time. So why is it difficult or impossible to attack?

zyl1024
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  • What you first said is checking if the number is prime or not. – Emil Sep 28 '12 at 09:53
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    Because the numbers are HUGE! – nullpotent Sep 28 '12 at 09:55
  • If you believe that algorithm X has polynomial complexity, try to write the polynomial which expresses its complexity. If you succeed then X has polynomial complexity, if you fail you may want to console yourself with the thought that X does not have polynomial complexity, which will be more comforting than the thought that you have failed to find the (or a) polynomial. But, more seriously, try writing an equation for the complexity of integer factorization in terms of the number of digits in the integer and study its form. – High Performance Mark Sep 28 '12 at 11:40

3 Answers3

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Casual descriptions of complexity like "polynomial factoring algorithm" generally refer to the complexity with respect to the size of the input, not the interpretation of the input. So when people say "no known polynomial factoring algorithm", they mean there is no known algorithm for factoring N-bit natural numbers that runs in time polynomial with respect to N. Not polynomial with respect to the number itself, which can be up to 2^N.

Ryan Culpepper
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  • Isn't the size of the input, the interpretation of the input? And why do we use bits to represent the size of the input rather than the interpretation of the input? – Gerald Oct 07 '15 at 15:04
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The difficulty of factorization is one of those beautiful mathematical problems that's simple to understand and takes you immediately to the edge of human knowledge. To summarize (today's) knowledge on the subject: we don't know why it's hard, not with any degree of proof, and the best methods we have run in more than polynomial time (but also significantly less that exponential time). The result that primality testing is even in P is pretty recent; see the linked Wikipedia page.

The best heuristic explanation I know for the difficulty is that primes are randomly distributed. One of the easier-to-understand results is Dirichlet's theorem. This theorem say that every arithmetic progression contains infinitely many primes, in other words, you can think of primes as being dense with respect to progressions, meaning you can't avoid running into them. This is the simplest of a rather large collection of such results; in all of them, primes appear in ways very much analogous to random numbers.

The difficult of factoring is thus analogous to the impossibility of reversing a one-time pad. In a one-time pad, there's a bit we don't know XOR with another one we don't. We get zero information about an individual bit knowing the result of the XOR. Replace "bit" with "prime" and multiplication with XOR, and you have the factoring problem. It's as if you've multiplied two random numbers together, and you get very little information from product (instead of zero information).

eh9
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If we run a program just to decide whether every number from 1 to sqrt(n) can divide n, and if the answer is yes, then store the number.

Even ignoring that the divisibility test will take longer for bigger numbers, this approach takes almost twice as long if you just add a single (binary) digit to n. (Actually it will take twice as long if you add two digits)

I think that is the definition of exponential runtime: Make n one bit longer, the algorithm takes twice as long.

But note that this observation applies only to the algorithm you proposed. It is still unknown if integer factorization is polynomial or not. The cryptographers sure hope that it is not, but there are also alternative algorithms that do not depend on prime factorization being hard (such as elliptic curve cryptography), just in case...

Thilo
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