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Here's my try. It's a brute force breadth-first algorithm. It will report all possible cycle-free paths from the start word till the end word. By default, it will not change the same letter in succession (so it won't do cone-code-core), but you can turn it on with --same. Use --dictionary to supply a dictionary of words (/usr/share/dict/words by default). Use --depth to give the maximum amount of words to be tried (0, the default, means no limit).
#!/usr/bin/perl use strict; use warnings; use Getopt::Long; my ($start, $end, $dictionary, $depth, $verbose, $same); $depth = 0; $verbose = 1; $dictionary = "/usr/share/dict/words"; GetOptions ('start=s', \$start, # Start word ($ARGV[0]) 'end=s', \$end, # End word ($ARGV[1]) 'dictionary=s', \$dictionary, # Dictionary, (/usr/share/d +ict/words) 'depth=s', \$depth, # Max depth (0 means infini +te) 'verbose=i', \$verbose, # Print progess 'same', \$same, # If set, allow the same le +tter # to be changed in successi +on. ); $| = 1 if $verbose; $start = shift if !defined $start && @ARGV; $end = shift if !defined $end && @ARGV; die "Need two different words of equal length\n" unless defined $start && defined $end && length $start == length $end && $start ne $end; my $L = length $start; # Read in the dictionary. Weed out words of inappropriate lengths, and # words containing characters that aren't lowercase letters. Add in th +e # start and end words (so you can use 'perl') and remove duplicates. my @words = do {my %seen; local @ARGV = ($dictionary); grep {$L == length && !/[^a-z]/ && !$seen{$_}++} map {chomp; $_} <ARGV>, $start, $end}; # Structure to keep track of a possibility: # 0: list of words used to get to this possition. # 1: hash of words used sofar. # 2: position of last letter change. my @tries = ([[$start], {$start => 1}, $L]); my $tries = 0; # How many positions have we tried so far? my @solutions; # List of solutions. while (@tries) { my $try = shift @tries; my $last = $$try[0][-1]; foreach my $c (0..$L-1) { next if $c == $$try[2] && !$same; my $re = $last; substr $re, $c, 1, '.'; my @new = grep /^$re/ && !$$try[1]{$_}, @words; foreach my $new (@new) { printf "\r%2d: %3d: %6d", 1+@{$$try[0]}, scalar @solutions +, ++$tries if $verbose; if ($new eq $end) { push @solutions, [@{$$try[0]}, $end]; } else { next if $depth && @{$$try[0]} >= $depth-1; push @tries, [[@{$$try[0]}, $new], {%{$$try[1]}, $new +=> 1}, $c] } } } } print "\n" if $verbose; foreach my $solution (@solutions) { print "@$solution\n"; } __END__

In reply to Re: Solving word ladder puzzles by Anonymous Monk
in thread Solving word ladder puzzles by bobf

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