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Perl: the Markov chain saw

Re: Hanoi Challenge

by fruiture (Curate)
on Oct 22, 2004 at 18:13 UTC ( #401613=note: print w/replies, xml ) Need Help??

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.


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Re^2: Hanoi Challenge
by tilly (Archbishop) on Oct 22, 2004 at 18:26 UTC
    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?

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