From: JB Eriksson <mrkode@...>
Date: 2005-12-08T16:32:05+09:00
Subject: Re: is this a good way to find anagrams?

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umm... what's the dictionary for?

also, I think there's an easier way to do this. with a bit of spacing and
stuff, I wrote up a 10 line solution. ;)
I also did the splitting string into array thing. Then there's two instance
methods for Array, named "sort" and "=3D=3D" respectively.

An anagram is when two words contain the exact same characters, right? so i=
f
the two splitted strings are the same when sorted, they should be anagrams.

On 12/8/05, travis laduke <wrong@socal.rr.com> wrote:
>
> it seems to me, with computers these days, this should finish
> instantly, not take like 20 seconds.
> also, please help me make my ruby more ruby-like. i'm new to ruby,
> not that i know any other language.
>
>
> ##these are here from when i was first testing
> secret_word =3D "spine"
> rack =3D "spine"
>
> secret_word =3D secret_word.split(//)
> rack =3D rack.split(//)
> test_rack =3D rack.to_s
>
> ##are there enough of the right letters in the rack to spell the word
> from the dictionary?
> ##i think i'm going to use funny spanish names for my methods instead
> of descriptive names.
> def rodolfo(secret_word, rack)
> secret_word.each do |x|
>     rack =3D rack.to_s
>     if rack.include? x
>         rack =3D rack.sub(x, '')
>         ##p rack, x
>     else
>         ##puts x, ' rack don\'t work'
>         break
>     end
> end
> end
>
>
> puts "reading dictionary"
> dict =3D IO.readlines('/usr/share/dict/words')
>
> while rack
>     puts "enter rack"
>     rack =3D gets.chomp.split(//)
>
>     dict.each do |secret_word|
>
>         if rodolfo(secret_word.chomp.split(//), rack)
>             puts secret_word.to_s
>         end
>     end
> end
>
>

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