From: Daniel Finnie <danfinnie@...>
Date: 2007-01-08T09:32:25+09:00
Subject: Re: [QUIZ] Word Blender (#108)

Here is a version of the regex that works around that:
regex = 
/^([a-z])(?!\1)([a-z])(?!\1|\2)([a-z])(?!\1|\2|\3)([a-z])(?!\1|\2|\3|\4)([a-z])(?!\1|\2|\3|\4|\5)([a-z])$/

Some examples:
 >> regex.match 'abcdef'
=> #<MatchData:0xb7b46a18>
 >> regex.match 'abcdefg'
=> nil
 >> regex.match 'abcddf'
=> nil
 >> regex.match 'abbdtf'
=> nil
 >> regex.match 'Abbdtf'
=> nil
 >> regex.match 'awesom'
=> #<MatchData:0xb7b3a5b0>
 >> regex.match 'hollow'
=> nil

Fedor Labounko wrote:
> On 1/7/07, Daniel Finnie <danfinnie@optonline.net> wrote:
> 
>> # Find words that use the same letters
>> selectedWords = dict.scan(/^[#{baseWord}]{3,6}$/)
> 
> 
> I was really impressed when I first saw this. It doesn't quite work if you
> want to exclude reusing the same letter more than once
> ("hhh".scan(/^[hello]{3,6}$/) => ["hhh"]) but it comes so close to 
> something
> I've only ever thought about implementing as a recursive method.
> Unfortunately I don't know much about this but now I wonder if it's 
> possible
> to find all partial permutations of a word with a regexp.
>