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Talk:Dixon's factorization method

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This is an old revision of this page, as edited by 149.156.82.207 (talk) at 20:36, 8 December 2010 (→‎Example is slightly faked: z=436). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

Comment

I cannot see a real difference to the Quadratic Sieve, but thar does not really matter.

Dixon's method is not a well-specified algorithm in that it doesn't really give a method for finding relations (presumably, brute-force search would be used). The quadratic sieve uses some mathematical facts and the sieving procedure to speed up the finding of relations. The number field sieve is yet another way of finding relations. Decrypt3 20:47, 24 March 2006 (UTC)[reply]

Question

I didn't understand this sentence: "This set of primes is called the factor base. Then, using the polynomial p(x) = x2 − n, many values of x are tested to see if p(x) factors completely over the factor base."

What is x? What is n? What does it mean to factor "over the factor base"?


n is the semi-prime which we are trying to factor x is one of many random numbers generated to try and find a p(x) which the prime factorization is contained in the factor base. for example if x=20 The prime factorization of 20 is: 2*2*5 If our factor base is [2,3,5] then p(x) factors completley or is "smooth" over our factor base.

I'm curious though is p(x)=x^2-n suposed to be p(x)=x^2 (mod n)

References

The references should be added to this page. Also, an expert is needed to say a few words about the history. Which Dixon is this one? Mhym 18:13, 24 June 2006 (UTC)[reply]

Overlap

This article has quite a bit of overlap with quadratic sieve, which also goes into quite a bit of detail regarding the linear algebra stage. This should be either in one or the other, or moved out to congruence of squares. Dcoetzee 02:47, 19 May 2007 (UTC)[reply]

Example is slightly faked

When testing N=84923 one actully find first b-smooth z=436, 436^2 mod N = 20250, which is 2*3^4*5^3.

Another important fact is that, we find in this example z=505 (less than "first" 513), is square 505^2 mod N = 256, which is 2^8. So whole procedure is somehow useless.

Maybe slightly better number N could be find to better ilustrate its application of Dixon's method.

--149.156.82.207 (talk) 20:36, 8 December 2010 (UTC)[reply]