Re: character generator

[tmoertel]

Here's one way to compute the combinations that can be created by drawing a single element from each of a list of given sets:

    #!/usr/bin/perl -l

    use warnings;
    use strict;

    use List::Util qw( reduce );

    sub combinations {
        no warnings qw( once );
        reduce { outer_r($a,$b) } [[]], reverse @_;
    }

    sub outer_r {
        my ($ys, $xs) = @_;
        my @product;
        foreach my $x (@$xs) {
            foreach my $y (@$ys) {
                push @product, [$x, @$y];
            }
        }
        return \@product;
    }

The combinations function takes a list of sets (each represented as an arrayref) and returns the combinations that can be created from them:

    use Data::Dumper;
    $Data::Dumper::Terse = 1;
    $Data::Dumper::Indent = 0;

    print Dumper( combinations( [1..3], ["a","b"] ) ), "\n";
    # [[1,'a'],[1,'b'],[2,'a'],[2,'b'],[3,'a'],[3,'b']]

With this function, we can turn to your question of how best to represent your character sets. I would just use strings to keep things simple. A helper function will convert strings into the form needed by combinations and then convert the results back into strings:

    sub charset_combinations {
        my @charsets = map [split//], @_;
        map join("", @$_), @{ combinations( @charsets ) };
    }

Let's use our new helper to find all of the 3-character combinations that can be made from the charset "abc":

my @abcees3 = charset_combinations( ("abc") x 3 );
print "@abcees3\n";
# aaa aab aac aba abb abc aca acb acc\
# baa bab bac bba bbb bbc bca bcb bcc\
# caa cab cac cba cbb cbc cca ccb ccc

We can even draw successive characters from different character sets:

    my @charsets = qw( abc 123 !@$ );

    foreach my $string_length (0 .. @charsets) {
        my @genstrings =
            charset_combinations( @charsets[0..$string_length-1] );
        print "$string_length: @genstrings\n";
    }

    # 0:
    # 1: a b c
    # 2: a1 a2 a3 b1 b2 b3 c1 c2 c3
    # 3: a1! a1@ a1$ a2! a2@ a2$ a3! a3@ a3$\
    #    b1! b1@ b1$ b2! b2@ b2$ b3! b3@ b3$\
    #    c1! c1@ c1$ c2! c2@ c2$ c3! c3@ c3$

I hope this gives you some helpful ideas.

Cheers,
Tom