Re: Hanoi Challenge

[fruiture]

in reply to Hanoi Challenge

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.

[tilly]

Good try. It works, but gains nothing in efficiency from using all pegs. Try it with 5 pegs and 4 disks. It takes 15 moves to finish - exactly the same as the 3 peg version. Try solving it by hand. You should have no problem finishing in 7 moves.

Can you find a way to have the program to benefit more from using all pegs?