All,
A problem at Project Euler led me to try and find an extremely fast way of determining the number of unrestricted partitions of an integer. I was suprised to find that nothing on CPAN did this already (or at least I couldn't find anything). So once I solved the problem, I decided to throw together some code and perhaps upload it.
Before I do, I want to solicit your input.
package Integer::Partition::Unrestricted;
use strict;
use warnings;
use Carp;
use Math::BigInt;
our $VERSION = '0.01';
sub new {
my $class = shift;
croak "Incorrect number of parameters" if @_ % 2;
my $self = bless {}, $class;
$self>_Init(@_);
return $self;
}
sub pos {
my ($self, $new_pos) = @_;
return $self>{POS} if ! defined $new_pos;
croak "The 'pos' argument must be a nonnegative integer" if $new_
+pos =~ /[^09]/;
croak "Out of range 'pos'" if $new_pos > $#{$self>{PART}};
return $self>{POS} = $new_pos;
}
sub last_pos {
my $self = shift @_;
return $#{$self>{PART}};
}
sub val {
my ($self, $pos) = @_;
my $last = $self>last_pos();
$pos = $last if ! defined $pos;
croak "The 'pos' argument must be a nonnegative integer" if $pos
+=~ /[^09]/;
return $self>{PART}[$pos] if $pos <= $last;
$self>_Add_Partitions($last + 1, $pos);
return $self>{PART}[$self>last_pos()];
}
sub next {
my $self = shift @_;
if ($self>pos() < $self>last_pos()) {
$self>pos($self>pos() + 1);
return $self>val($self>pos());
}
$self>_Add_Partitions($self>last_pos() + 1, $self>last_pos() +
+1);
$self>pos($self>pos() + 1);
return $self>val($self>pos());
}
sub prev {
my $self = shift @_;
$self>pos($self>pos()  1) if $self>pos();
return $self>val($self>pos());
}
sub gen_iter {
my ($self, $target) = @_;
$target = $self>pos() if ! defined $target;
croak "The 'gen_iterator' argument must be a positive integer" if
+$target =~ /[^09]/;
my @part = (0, (1) x ($target  1));
my $done = undef;
return sub {
return () if $done;
my $min = $part[2];
my $total = $part[0] ? 0 : 1;
my $index = 0;
for (0 .. $#part  1) {
if ($part[$_] > $min) {
$total += $part[$_];
next;
}
$index = $_;
last;
}
$part[$index]++;
$total += $part[$index];
if ($total > $target  $part[$index] > $part[0]) {
@part = ($index ? ++$part[0] : $part[0], (1) x ($target 
+$part[0]));
}
else {
@part = (@part[0 .. $index], (1) x ($target  $total));
push @part, 1 if $part[0] == 1;
}
$done = 1 if $part[0] == $target;
return @part;
};
}
sub _Init {
my ($self, %opt) = @_;
my $n = 1;
$n = delete $opt{init} if exists $opt{init};
croak "The 'init' option must be a positive integer" if ! $n  $n
+ =~ /\D/;
croak "Invalid option provided" if %opt;
$self>_Set_Defaults();
$self>_Add_Partitions(1, $n);
return;
}
sub _Set_Defaults {
my $self = shift @_;
$self>{PART} = [ Math::BigInt>new(1) ];
$self>{PENT} = [ 0 ];
$self>{POS} = 0;
return;
}
sub _Add_Partitions {
my ($self, $min, $max) = @_;
$self>_Add_Pent($max);
for my $i ($min .. $max) {
my $sum = 0;
for my $j (1 .. $self>_Find_Idx($i)) {
if (($j % 4) % 3) {
$sum += $self>{PART}[$i  $self>{PENT}[$j]];
}
else {
$sum = $self>{PART}[$i  $self>{PENT}[$j]];
}
}
$self>{PART}[$i] = $sum;
}
return;
}
sub _Add_Pent {
my ($self, $max) = @_;
my $need = int(sqrt($max + 1)) * 2 + 1;
return if $need <= @{$self>{PENT}};
my $last = $self>_Pent_2_N();
for ($last + 1 .. ($need  1) / 2) {
push @{$self>{PENT}}, ($_ * (3 * $_  1) / 2), ($_ * (3 * $
+_  1) / 2);
}
return;
}
sub _Pent_2_N {
my $self = shift @_;
my $c = $self>{PENT}[1];
my $x = (1 + sqrt(1  (4 * 3 * 2 * $c))) / 6;
return abs($x) if $x == int($x);
return abs((1  sqrt(1  (4 * 3 * 2 * $c))) / 6);
}
sub _Find_Idx {
my ($self, $tgt) = @_;
my $min = 0;
my $max = $#{$self>{PENT}};
my $mid;
while ($min <= $max) {
$mid = int(($min + $max) / 2);
my $val = $self>{PENT}[$mid];
return $mid if $val == $tgt;
$tgt < $val ? ($max = $mid  1) : ($min = $mid + 1);
}
return $self>{PENT}[$mid] < $tgt ? $mid : $mid  1;
}
"This statement is false";
__END__
=head1 NAME
Integer::Partition::Unrestricted  Work with unrestricted integer part
+itions
=head1 VERSION
Version 0.01 developed on 20060227
=head1 SYNOPSIS
use Integer::Partition::Unrestricted;
my $part = Integer::Partition::Unrestricted>new(init => 100);
my $last = $part>last_pos();
while ($part>pos() <= $last) {
my $pos = $part>pos();
print "$pos = ", $part>val($pos), "\n";
$part>next();
}
my $next = $part>gen_iter(10);
while (my @part = $next>()) {
print "@part\n";
}
$part>pos(0);
print $part>prev();
=head1 DESCRIPTION
This module allows you to work with unrestricted integer partitions
=head1 SYNTAX
The new() constructor can be called with a hash of additional options
my $part = Integer::Partition::Unrestricted>new();
or
my $part = Integer::Partition::Unrestricted>new(init => 42);
=head1 OPTIONS
=over 4
=item init
This option allows you to specify an initial number of integer partiti
+ons
my $part = Integer::Partition::Unrestricted>new(init => 1000);
=back
=head1 METHODS
=head2 pos
This method returns the current iterator position if no arguments are
+specified.
You can set the position any where between 0 and L<"last_pos">
my $curr_pos = $part>pos();
# Set iterator to end of currently calculated partitions
$part>pos($part>last_pos());
=head2 last_pos
This method returns the position of the last calculated parition
my $calculated = $part>last_pos();
=head2 val
This method returns the number of partitions of the integer provided a
+rgument.
It will default to using L<"pos"> if no argument is provided.
If the specified value exceed L<"last_pos">, as many new partitions wi
+ll be
calculated as necessary.
# The number of integer partitions of 100 is
print $part>val(100), "\n";
=head2 next
This method increments L<"pos"> by 1 and returns the number of partiti
+ons
for that value. It will increment L<"last_pos"> by 1 and calculate th
+e next
integer partition if necessary.
# Set position to end of calculated values
my $n = $part>last_pos();
$part>pos($n);
# Print number of positions for next integer
print $n + 1, " = ", $part>next(), "\n";
=head2 prev
This method decrements L<"pos"> by 1 and returns the number of partiti
+ons
for that value. It will always return 1 at position 0.
my $prev_val = $part>prev();
=head2 gen_iter
Sometimes it is necessary to know more than just the number of partiti
+ons.
This method will return an code ref to iterate over the partitions of
the integer argument specified. It will default to L<"pos"> if no arg
+s are supplied.
$part>pos(42);
my $next = $part>gen_iter();
while (my @part = $next>()) {
print "@part\n";
}
=head1 AUTHOR
Joshua Gatcomb, <Limbic_Region_2000@Yahoo.com>
=head1 ACKNOWLEDGEMENTS
This module was inspired by a problem in Project Euler
(L<http://mathschallenge.net/index.php?section=project&ref=about>)
=head1 BUGS
None known. Bug reports, fixes, and feedback are desired.
=head1 PERFORMANCE
While this module is pure perl and is OO, it is fairly fast and attemp
+ts not to
waste memory. I am sure there could be plenty of improvements and I w
+elcome
suggestions (preferrably as patches).
=head1 COPYRIGHT
Copyright (c) 2006 Joshua Gatcomb. All rights reserved.
This program is free software; you can redistribute it
and/or modify it under the same terms as Perl itself.
=head1 SEE ALSO
The following links may be helpful if you don't know what an integer p
+artition is
L<http://en.wikipedia.org/wiki/Integer_partition>
L <http://mathworld.wolfram.com/Partition.html>
=cut
Beyond needing a test suite, I also considered adding a couple of methods to peek at values without changing the position. Feel free to comment on that or any other new methods you think might be useful.
Re: RFC: Integer::Partition::Unrestricted by brian_d_foy (Abbot) on Feb 27, 2006 at 23:45 UTC 
That's very cool!
Recall, however, that pos is a Perl keyword, so you might want to use something else. In general, I like to spell out the words (and ambs has an article on this in the next issue of The Perl Review).
 [reply] [d/l] 

 [reply] 

You could make that argument, but it fails when you need to use those builtins inside the package that defines those methods. And, just because Unix is voweldeficient doesn't mean that you need to be. :)
 [reply] 


 Re: RFC: Integer::Partition::Unrestricted by polettix (Vicar) on Feb 24, 2007 at 14:46 UTC 
I think you probably don't go and look for feedback for problem 78 on Project Euler, so I decided to give you one here :)
Your generator is quite impressive but it is a bit overkill in this case. The problem statement requires divisibility by 1_000_000, which means that you can do all sums and differences modulo 1_000_000 and still get a good answer; moreover, you can cache values of p(n) modulo 1_000_000. As a consequence, you can work with "regular" integers and get rid of bigint, which slows you down.
In my solution, I am a bit less optimised than you, because I'm calculating pentas all the time; in any case, it runs in about 19 seconds on Centrino 1.6GHz. If you're interested, I've put it into Project Euler forum entry #78 (I'm polettix).
I also saw that you didn't publish this code, but I think it could be useful  who knows? :)
Flavio
perl ple'$_=reverse' <<<ti.xittelop@oivalf
Don't fool yourself.
 [reply] 

frodo72,
Thanks for the feedback. I lost interest in Project Euler a long time ago. Since my intent, as noted in the forum of problem 78, was to write code reusable for others  I ignored optimizations specific to the problem posed. I have not given up on the idea of publishing it but it needs to be part of a larger puzzle and right now I don't have time for that.
 [reply] 

