From: Daniel Finnie <danfinnie@...>
Date: 2007-01-08T09:41:10+09:00
Subject: Re: [QUIZ] Word Blender (#108)

Oops, that doesn't match {3,6} just {6}.

 >> 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]))?$/
=> [a-z]1[a-z]12[a-z]:123[a-z]:1234[a-z]:12345[a-z]
 >> regex.match 'hhh'
=> nil
 >> regex.match 'hal'
=> #<MatchData:0xb7cfb7d0>
 >> regex.match 'sos'
=> nil
 >> regex.match 'sauce'
=> #<MatchData:0xb7c12bac>
 >> regex.match 'hatoff'
=> nil
 >>

That's a better one.

Daniel Finnie wrote:
> 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.
>>
> 
>