I ignored assumption #2, but the results of the following code do not violate that assumption (for this example anyway). I get two possible groups of five letters, for the sake of time, I'm not scanning any dictionary for these combinations. Even though it's probably easier to brute force a dictionary, I went the "let's find valid letter groups" route :)
use strict;
use warnings;

my $word = "aaaaa";
my @words;
{
 push @words, $word
 if
   $word =~ tr/bumps// == 2 and
   $word =~ tr/seam//  == 2 and
   $word =~ tr/domes// == 3 and
   $word =~ tr/shake// == 3 and
   $word =~ tr/pokes// == 3 and
   $word =~ tr/dukes// == 3;
 last if $word eq "zzzzz";
 $word = inc_letter($word, 4);
 redo;
}

print "$_\n" for @words;

sub inc_letter {
  my ($word, $i) = @_;
  if (substr($word, $i, 1) eq "z") {
    $word = inc_letter($word, $i-1, 1);
    substr($word, $i, 1) = substr($word, $i-1, 1);
  } else {
    substr($word, $i, 1)++;
  }
  $word;
}

[download]
Update: The puzzle changed; this solution has not. If a general solution is desired, then more work would need to be done here.

In reply to Re: Challenge: Mystery Word Puzzle by runrig
in thread Challenge: Mystery Word Puzzle by Limbic~Region

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