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Here's my solution.

It's surely not optimal, but it's a very basic extension to the 3-peg solution, so the code is very short. It uses O(d1/(p-3)) moves O(2d/(p-2)) moves for the solution with d disks and p pegs (I'm too lazy to calculate the exact numbers).

(Updated fomula again. The O sign is meant if p is constant but d->inf)

Update: this is probably very suboptimal for more than 4 pegs.

#!/usr/local/bin/perl use warnings; use strict; $ARGV[0] =~ /(\d+)/ or die; my $pegs = $1; $ARGV[1] =~ /(\d+)/ or die; my $disks = $1; { my @pegnames = "A" .. "Z"; sub printmove { my($d, $f, $t) = @_; print $d, ": ", $pegnames[$f], " -> ", $pegnames[$t], "\n" } } sub rec { my($n, $s, $d, $t, @o) = @_; $n > 0 or return; my $k = @o < $n - 1 ? @o : $n - 1; #warn "[($n:${\($n-$k)} $s->$d]\n"; rec($n - $k - 1, $s, $t, $d, @o); printmove($n - $k + $_, $s, $o[$_]) for 0 .. $k - 1; printmove($n, $s, $d); printmove($n - $k + $_, $o[$_], $d) for reverse(0 .. $k - 1); rec($n - $k - 1, $t, $d, $s, @o); #warn "[)]\n" } rec($disks, 0, 1, 2 .. $pegs - 1); __END__

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

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