Re: Boolean algebra

[chromatic]

Here's the guts of the reduction. It's not particularly brilliant, but it seems to work in my testing. If it were not Sunday night, I might get elaborate with error checking or bit vectors, which would really be the way to go with this.

There is room for someone to come along and to reduce my code's verbosity further. Welcommen.

#!/usr/bin/perl -w

use strict;

my $input = "(abdc)(adc-b)";

my @expr;
while ($input =~ s!\(([-a-z,]+)\)!!) {
    my @items = split '', $1;
    my (@and, @or) = ();
    while (@items) {
        my $elem = shift @items;
        if ($elem eq '-') {
            push @or, shift @items;
        } else {
            push @and, $elem;
        }
    }
    push @expr, [ \@and, \@or ];
}

my $first = shift @expr;
my @output = @{ @$first[0] };
while (my $next = shift @expr) {
    my (%keep, %drop);
    @keep{@{ @$next[0] }} = ();
    @drop{@{ @$next[1] }} = ();
    foreach (@output) {
        next if (exists $keep{$_});
        $_ = undef if (exists $drop{$_});
    }
}

my $output = join('', grep{ defined($_) } @output);
print "Result is; $output\n";

Update: I realized early this morning that this only handles the OR case in the second and subsequent input units. Back to the drawing board.