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Perl: the Markov chain saw

Re: Factors

by kvale (Monsignor)
on Jun 13, 2002 at 04:51 UTC ( #174101=note: print w/replies, xml ) Need Help??

in reply to Re: Re: Factors
in thread Vampire Numbers

Good observation! It looks like Fermat's method (the method I mentioned above) doesn't necessarily find all, or even any of the factors. It is just a heuristic. On the other hand, if you have an even number, you may extract factors of 2 until you get an odd number, in which case (x+y) and (x-y) must both be necessarily odd.

The generalization that gumby mentioned, x^2 = y^2 (mod p) may have more sucess, but I don't know if it is exhaustive. Here is code that implements it:
my $p = shift; my %residues; my %factors; foreach my $x (2..$p) { my $mod = ($x*$x) % $p; if (exists $residues{$mod}) { $factors{$x-$residues{$mod}}++ if $p % ($x-$residues{$mod}) == 0 +; $factors{$x+$residues{$mod}}++ if $p % ($x+$residues{$mod}) == 0 +; } else { $residues{$mod} = $x; } } foreach (sort {$a <=> $b} keys %factors) { print "$p = $_*", $p/$_, "\n"; }
For 54, this yields
1021% 54 54 = 2*27 54 = 6*9 54 = 18*3 54 = 54*1
But this code is unsatisfying because it takes longer than the simple trial by division. Thinking from a divide and conquer point of view, it is probably faster to use these methods to find a factorization, then recursively find factorizations of the factors untill all are prime.


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