Well, no idea if this is very efficient, but it seems to solve all situations (the hanoi-driver does not complain) and is based on your code, just generalized.
#! /usr/bin/perl -w
use strict;
my $peg_count = shift;
my $disk_count = shift;
my @pegs;
{
my $p = 'A';
push @pegs , $p++ for 1 .. $peg_count;
}
solve( [ @pegs ] , reverse 1..$disk_count);
sub solve {
my ($pegs, $first_disk, @rest) = @_;
my $from = shift @$pegs;
my $to = shift @$pegs;
my $hold = shift @$pegs;
return unless $first_disk;
solve( [ $from, $hold, @$pegs, $to ] => @rest );
print "$first_disk: $from -> $to\n";
solve( [ $hold, $to, @$pegs, $from ] => @rest );
}
Update: Well, it is absolutely NO more efficient than using 3 pegs anyway, but at least it works.
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