The 11/23/2003 NPR Weekend Edition Sunday puzzler was (roughly) as follows: take an English word that ends in -m. Replace the 'm' with -rn, and you get the antonym of the original word. What is the pair? As I was just learning Perl at the time, I decided to test out my skills, and this is what I came up with. (The idea behind both versions is that @solutions is small enough that I can visually pick out the true solution visually. And indeed, using words-2-16mdk, I get only five pairs back.) 
#!/usr/bin/perl
use strict;

# NPR puzzler helper, 11/23/2003
# Find a pair of words, [X]rn and [X]m, which are antonyms.
# We can pull all such pairs from /usr/share/dict/words.

open(FH,"/usr/share/dict/words");

my @big_arr;
# grep through it the slow way
while (<FH>) {
    if (/([a-zA-Z]*rn$)/) {
        push @big_arr, $1;
    }
}

#print @big_arr;
close FH;

# there's got to be a better way to do this.
my @solutions;
#print "|all_words| = ".$#all_words."\n";
my $curr_word;
foreach (@big_arr) {
    $curr_word = substr($_,0,-2)."m";
    my $cmd = 'grep ^'.$curr_word.'$ /usr/share/dict/words';
    my $result = `$cmd`;
    if ($result) { 
        push @solutions, "$_ / $curr_word\n";
    }
}
close FH;

print @solutions;

[download]
Makes me cringe just to look at it. Yes, I first find every -rn word in the dictionary, then check each and every one of them by grepping (externally, no less!) through the whole dictionary again. Brain-damaged, to say the least. But it worked, and I got the puzzler. I came back to it a little while later, to make it at least seem smarter. This time, I decided to first pluck out all of the -rn words, then walk through the dictionary a second time, taking advantage of the fact that @rn_words is already sorted---for instance, there's no point in seeing if 'spurn' has a counterpart in the dictionary if we're already in the t's. 
#!/usr/bin/perl
use strict;

# NPR puzzler helper, 11/23/2003
# rewritten 12/15/2003 around 4 am.
# Find a pair of words, [X]rn and [X]m, which are antonyms.
# We can pull all such pairs from /usr/share/dict/words.

open(FH,"/usr/share/dict/words");

my (@rn_words, @solutions);

@rn_words = grep /([a-zA-Z]*rn$)/, <FH>;
chomp(@rn_words);

seek FH,0,0;

my $ptr = 0;
my $this = substr($rn_words[$ptr],0,-2).'m';
LOOP: while (<FH>) {
    chomp;
    next LOOP if ($_ lt $this);
    if ($_ eq $this)
        { push @solutions, "$rn_words[$ptr] / $_\n"; }
    $ptr++;
    $this = substr($rn_words[$ptr],0,-2).'m';
    last unless defined $rn_words[$ptr];
}
close FH;

print @solutions;

[download]
I'm pretty sure that two is the mininum number of times the dictionary can be traversed. I considered trying to remember the last -m word seen, and waiting for the corresponding -rn word to see if it existed... but this didn't really work out. I wonder if there's a good way to benchmark the efficiency of these approaches, since both run quickly on my system, though the second feels a little faster. More importantly, I'm not ashamed of the second one. How quickly we improve.

In reply to The evolution of a solution to the 11/23 NPR puzzler. by grendelkhan

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