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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.


In reply to Re: Hanoi Challenge by fruiture
in thread Hanoi Challenge by tilly

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