From: Michael Linfield <globyy3000@...>
Date: 2007-10-20T04:38:13+09:00
Subject: Re: combinations listing

Brian Adkins wrote:

>> Feel free to improve it if you see a flaw or a better way of doing it
> 
> Can your program solve this anagram (a very common word)?
> "aaabcehilllpty"
> 
> It might take a while. The word provides a hint (in two separate ways)
> for an improvement.
> 
> Using my code from another post will speed up the permutation
> generation by a factor of 3, but that is no match for an O(n!)
> algorithm. I think you'll need a new technique for longer words.
> 
> Also, since the anagram and the actual word must be the same length to
> match, you can partition the dictionary by word size. That and the
> hint above should get you a long way down the road.
> 
> Brian Adkins

Alright i see your point lol, so are you suggesting that the actual 
dictionaries be split up? in a sense of...

word = "foobar"
res = word.split('')

if res.size > 3
  #use dictionary 1
end

if res.size > 6
  #use dictionary 2
end

ect...

one way or another, the only speed issue here is generating the 
permutations, searching the dictionary is pretty quick from what ive 
seen.
The clue was alphabetically, sadly i didnt want to wait for my program 
to finish that lol. I'm kind of hazy as to what that clue might suggest, 
my interpretation is to possibly grep out all the words that are of the 
same length as the word entered.

word = gets.chomp
res = word.split('')
size = res.length
# so now size would equal 14 if you used the word 'alphabetically'

file = open('dict1.txt') {|p|  p.readlines}
dict = []
 #when statement that shoves all words = to 14 into the dict array
end
-- 
Posted via http://www.ruby-forum.com/.