Narrowing down answer to NPR puzzle with Linux commands

Intro
This is for CentOS and RedHat Linux.

Narrow things down
$ egrep ′^[a-z]{6}$′ /usr/share/dict/linux.words |sed ′s/.//′|s
ort|uniq -c|sort -k1 -d -r > 6-ltr-last-5

Mind the line break in the display of this command – you have to join things back together.

This is a great string of commands to study if you want to unleash the power of the linux shell. Yuo have a matching operator, egrep, a simple regular expression, a substitution command, sed, a sort command, sort, a unique sort command, uniq, and a sort ordered by number and displayed in reverse order. I issue commands like this frequently against log files and can do much more import work than solving an NPR puzzle.

6-ltr-last-5 starts like this:

     14 itter
     14 ingle
     14 atter
     14 agged
     13 etter
     13 ester
     13 aster
     13 apper
     13 apped
     13 agger
...

It has 772 lines with 4 or greater occurrences – too many to process by hand.

Edit this file and only keep the top part of the file up until the last of the 4 occurrences.

Now go back and match these words against the dictionary.

$ cat 6-ltr-last-5 |awk ′{print $2}′|while read line;do egrep ′
^[a-z]′$line$ /usr/share/dict/linux.words >> 6-ltr-combos;echo " ">>6-ltr-combos; done

6-ltr-combos starts like this:

bitter
fitter
gitter
hitter
jitter
kitter
litter
nitter
pitter
ritter
sitter
titter
witter
zitter
 
bingle
cingle
dingle
gingle
hingle
jingle
...

Small program to process that file
OK, to work with that file we just created based on the logic of the problem statement, I created this custom perl script which I call 6-5.pl:

#!/usr/bin/perl
$DEBUG = 0;
$consonants = 'bcdfghjlmnpqrstvwxyz';
$oldplace = -1;
$pot = 0;
while(<STDIN>){
  if (/^\s/) {
    print "pot,start word = $pot, $startword\n" if $pot > 3;
# reset some  values
    $oldplace = -1;
    $pot = 0;
    $startword = $_;
  }
  chomp;
# get at first character
  ($char) = /^(\w)/;
# turn character into position number with this
  $place = index $consonants,$char;
  print "word,place: $_,$place\n" if $DEBUG;
  if ($place != $oldplace + 1) {
# clear things out
    print "pot,start word = $pot, $startword\n" if $pot > 3;
    $pot  = 1;
    $startword = $_;
  } else {
    $pot++;
  }
  print "pot: $pot\n" if $DEBUG;
  $oldplace = $place;
}

I really wish I knew Python – I bet it would be an even shorter script in that language. But this gets the job done. It’s warts and all as I have done enough debugging to get it to return mostly reasonable output, but it’s still not quite right. It’s good enough…

Run it:

$ ./6-5.pl < 6-ltr-combos

pot,start word = 4, fitter
pot,start word = 7, gingle
pot,start word = 4, latter
pot,start word = 8, dagged
pot,start word = 5, fetter
pot,start word = 5, jester
pot,start word = 6, dagger
...

The biggest problem is my dictionary contains too many uncommon words, but at least that guarantees that the answer will indeed be present. And it is. In fact I found three sets of what I consider common words. One set are very ordinary words so i guess that is the intended answer. I can’t give away everything right now – you’ll have to do some work! I’ll post the answers after Sunday.

References and related
A similar approach to a previous puzzle is here.