sub find_path {
  my ($start_at, $end_at, $been_there) = (@_, {});
  if ($start_at == $end_at) {
    return [$start_at];
  }
  else {
    my @try_these = grep { ! $been_there->{$_} } @{$adjacency_list{$st
+art_at}};
    return map {
      my @cdr_list = find_path($_, $end_at, {%$been_there, $start_at =
+> 1});
      map [$start_at, @$_], @cdr_list;
    } @try_these;
  }
}
[download]

Arghh stupid me!

Oh no, not a HW. In fact I am a little old for HWs *ashamed*. I am good with Java/C stuff but I am trying to get familiar with Perl for one of my projects so asking whatever comes into my mind. Perl's (incredible) shorthand expressions confuse me often and sometimes I spend needlessly long time trying to figure them out (else part in the code for instance). Though this time, I agree that, I did not consider enough. It was obviously too trivial. Sorry but thank you indeed!

I wrote this in very much a LISP style, which, if you're not familiar with it, can be pretty hard to follow. The else section constructs the non-trivial solution to the problem
1. Look at all the nodes adjacent to where we're starting. Weed out any we've already visited.
2. Call find_path for each of those nodes and gather up the results
3. Return the list made by prepending this node onto each of those paths

---

Caution: Contents may have been coded under pressure.