XP is just a number PerlMonks

### Factors

by gumby (Scribe)
 on Jun 11, 2002 at 18:19 UTC ( #173606=note: print w/replies, xml ) Need Help??

in reply to Vampire Numbers

A more efficient algorithm (of Gauss and later Kraitchik) would be to find solutions to the congruence x^2=y^2 (mod p) e.g.

38^2=3^2 (mod 1435)

It is trivial to then check if the factor (38+3 in this case) is a permutation of n/2 digits from the original number.

Replies are listed 'Best First'.
Re: Factors
by kvale (Monsignor) on Jun 12, 2002 at 21:01 UTC
I think what is being suggested here is that instead of creating all possible pairs of numbers (x,y) from the digits of a number p and testing if x*y == p, it is more efficient to factor the number p first and and then test all possible factorizations to see if they are just a permuation of the digits of p.

Finidng a quadratic residue x^2=y^2 (mod p) is useful, because then x^2 - y^2 = (x+y)*(x-y) = 0 (mod p) and so (x+y) or (x-y) may be factors of p.

In practice, one starts with an x that has the smallest square larger than p, then computes x^2-p to see if it is a perfect square. If not, increment x and repeat.

This is pretty fast relative to testing by division from 2 to sqrt(p), but there exist even faster methods based on the quadratic residue. The Quadratic Sieve, invented by Pomerance in 1981 is a direct derivative and the Number Field Sieve is a related method.

Getting back to Vampire numbers, what are the relative efficiencies for a number p with 2d digits? For the first method, a naive approach generates all permutations of 2*d digits and splits each into two d-digit numbers to test. Thus each 2d digit number requires about (2*d)! == (2*log(p))! tests, or (2*log p)^(2*log p) by Stirling's approximation. For the second method, a quadratic sieve has running time of order exp(sqrt(log(p)*log(log(p)))) to find a single factor of p. So finding a single factor using QS is more efficient that testing all factors using the first method. But I don't know the running time for finding all the factors using the QS and I don't know the typical multiplier for the scaling result of the QS method.

Anyone?

-Mark
The Quadratic Sieve has running time

O(e^(sqrt(1.125ln(n)ln(ln(n)))))

But with certain improvements (such as using Wiedemann's Gaussian Elimination algorithm over GF(2)), it's running time is given by

O(e^(sqrt(ln(n)ln(ln(n)))))

The Number Field Sieve is better for numbers over 100-digits and has running time

O(e^(1.9223((ln(n))^1/3(ln(ln(n))))^2/3))

Excuse my notation (but TeX would be overkill).

Note: O(stuff) means the magnitude of the running time is < A * stuff for some constant A and all values of n.

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);
}
[download]```
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";
}
[download]```
For 54, this yields
```1021% factor.pl 54
54 = 2*27
54 = 6*9
54 = 18*3
54 = 54*1
[download]```
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
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;
}
}
[download]```

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