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### Re: Re: Factors

by YuckFoo (Abbot)
 on Jun 12, 2002 at 22:30 UTC ( #174047=note: print w/replies, xml ) Need Help??

It seemed like a good idea so I cooked it up according to your description. Correct me if I'm wrong but it seems quadratic residues won't account for all factors.

Program correctly factors 24 to 4*6 and 2*12 but misses 3*8. Program finds no factors for 54.

YuckFoo

Update: With a little thought one can see the program will only find factor pairs that are both even or both odd. x+y * x-y = p.

```#!/usr/bin/perl

use strict;

my (\$num) = @ARGV;

my \$x = int(sqrt(\$num)) + 1;
my \$y = sqrt(\$x * \$x - \$num);

while (\$x - \$y >= 2) {

if (\$y == sprintf("%d", \$y)) {
print "Ok \$num = ", \$x-\$y, " * ", \$y+\$x, "\n";
}
else { print "No x = \$x, y = \$y\n"; }

\$x++;
\$y = sqrt(\$x * \$x - \$num);
}

Replies are listed 'Best First'.
Re: Factors
by kvale (Monsignor) on Jun 13, 2002 at 04:51 UTC
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% factor.pl 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.

-Mark
Re: Re: Re: Factors
by gumby (Scribe) on Jun 13, 2002 at 15:43 UTC
If you want to find a square root of a (mod p) use the Shanks-Tonelli algorithm.
```# Calculate the least non-negative remainder when an integer a
# is divided by a positive integer b.
sub mod {
my (\$a, \$b) = @_;
my \$c = \$a % \$b;
return \$c if (\$a >= 0);
return  0 if (\$c == 0);
return (\$c + \$b);
}

# Calculate a^b (mod c), where a,b and c are integers and a,b>=0, c>=1
+
sub mpow {
my (\$a, \$b, \$c) = @_;
my (\$x, \$y, \$z) = (\$a, \$b, 1);
while (\$y > 0) {
while (\$y % 2 == 0) {
\$y = \$y / 2;
\$x = \$x**2 % \$c;
}
\$y--;
\$z = mod(\$z * \$x, \$c);
}
return \$z;
}

# Shanks-Tonelli algorithm to calculate y^2 = a (mod p) for p an odd p
+rime
sub tonelli {
my (\$a, \$p) = @_;
my (\$b, \$e, \$g, \$h, \$i, \$m, \$n, \$q, \$r, \$s, \$t, \$x, \$y, \$z);
\$q = \$p - 1;
\$t = mpow(\$a, \$q/2, \$p);
return 0 if (\$t = \$q); # a is a quadratic non-residue mod p

\$s = \$q;
\$e = 0;
while (\$s % 2 == 0) {
\$s = \$s/2;
\$e++;
}

# p-1 = s * 2^e;
\$x = mpow(\$a, ++\$s/2, \$p);
\$b = mpow(\$a, \$s, \$p);
return \$x if (\$b == 1);

\$n = 2;
RES:  while (\$n >= 2) {
\$t = mpow(\$n, \$q/2, \$p);
last RES if (\$t == \$q);
} continue {
\$n++;
}

# n is the least quadratic non-residue mod p
\$g = mpow(\$n, \$s, \$p);
\$r = \$e;

OUT:  {
do {
\$y = \$b;
\$m = 0;
IN:   while (\$m <= \$r-1) {
last IN if (\$y == 1);
\$y = \$y**2 % \$p;
} continue {
\$m++;
}
return \$x if (\$m == 0);
\$z = \$r - \$m - 1;
\$h = \$g;
for (\$i = 1; \$i < \$z; \$i++) {\$h = \$h**2 % \$p;}
\$x = (\$x * \$h) % \$p;
\$b = (\$b * \$h**2) % \$p;
\$g = \$h**2 % \$p;
\$r = \$m;
} while 1;
}
}
Here's a more 'reference' style implementation. Basically, this is the algorithm you'll find in the paper 'Square Roots, From 1; 24, 51, 10 To Dan Shanks' by Ezra Brown, but at the cost of using Math::BigInt.
```#!/usr/bin/perl -w

use Math::BigInt ':constant';
use strict;

sub tonelli {
my (\$s, \$e, \$n, \$x, \$b, \$g, \$r, \$m, \$t);
my (\$a, \$p) = @_;

die "\$a has no square roots (mod \$p)" if \$a**((\$p-1)/2) % \$p == -1;

\$s = \$p-1;
\$e = 0;
while (\$s % 2 == 0) {
\$s = \$s / 2;
\$e++;
}

for (\$n = 2; \$n >= 2; \$n++) {
last if \$n**((\$p-1)/2) % \$p == \$p-1;
}

\$x = \$a**((\$s+1)/2) % \$p;
\$b = \$a**\$s % \$p;
\$g = \$n**\$s % \$p;
\$r = \$e;

\$t = \$b;
while (1) {
for (\$m = 0; \$m <= \$r-1; \$m++) {
last if \$t % \$p == 1;
\$t = \$t**2;
}

return \$x if \$m == 0;

\$x = \$x * \$g**(2*(\$r-\$m-1)) % \$p;
\$b = \$b * \$g**(2*(\$r-\$m-1)) % \$p;
\$g = \$g**(2*(\$r-\$m)) % \$p;
\$r = \$m;
}
}

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