Write a program that solves word-chain puzzles.
cat cot cog dog
The objective of this kata is to write a program that accepts start and end words and, using words from the dictionary, builds a word chain between them. For added programming fun, return the shortest word chain that solves each puzzle. For example, my Powerbook running Ruby can turn "lead" into "gold" in four steps (lead, load, goad, gold), taking about 20 seconds. Turning "ruby" into "code" takes six steps (ruby, rubs, robs, rods, rode, code) and 90 seconds, while turning "code" into "ruby" (again in six steps) takes about an hour. Go figure…
Update: It turns out that my original algorithm was pretty dumb: a better approach greatly speeds up search and makes it symetrical. It now does all the above examples in between 0.5s and 1s.
We just did this Kata the other night at the Chicago Ruby meetup. I played with it afterward and came up with a (semi?) simple Hadoop MapReduce implementation that works in fully-distributed mode. Definitely not as fast as a simple Ruby implementation, but watching an entire cluster solve it in parallel quickly is pretty snazzy :-).
http://github.com/cchandler/codekata19-mapreduce
Posted by: Squanderingtime | April 18, 2010 at 12:41 AM
Any chance of getting the dictionary reposted, or even an indication of the number of words would be helpful.
I am using the 360k Moby word list. Is anyone else?
Posted by: jon | November 25, 2009 at 02:52 PM
I'd be interested to see how long it takes people to get from "the" to "end". That particular beast takes the longest for me, about 2.2s. Everything else is under .5s.
I'm not sure about all the particulars of how Matt did it (ternary search tree?!). I went ahead and made a customized object that would store its relative path, and retrieve all words that a) weren't in its path and b) were one-off in terms of letters.
cat-dog takes 49ms, ruby-code 13ms, lead-gold 33ms, the-end !2240ms!
Is everyone else's searches slow for the-end?
Posted by: Loren | October 19, 2009 at 03:59 PM
I took a look at this in C# - and so far have a nice solution that is very fast. However my first implementation will only complete if the new letter appears in the final word.
However this fits in with the 3 cases above:
cat-cot-cog-dog
ruby-rube-rude-rode-code
however at the moment if there was no rube it would need to find
a replacment for U B or Y that was either O D or E in the correct place.
So its great for finding the shortest path in under 0.5s however i might need to review it for longer chains. Any recommended word sets?
Posted by: Luke | August 05, 2009 at 09:45 AM
Your word list link seems to have died. Maybe you could put up another file or link to some generic word list like http://wordlist.sourceforge.net/
Posted by: Johannes Spielmann | March 25, 2009 at 07:38 AM
If your wordlist has 'rube'... Mine doesn't. :)
Posted by: Dave Thomas | May 14, 2007 at 03:38 PM
Hi Dave,
Isn't #('ruby' 'rube' 'rude' 'rode' 'code') the shortest path from ruby to code?
Posted by: Michael Davies | May 14, 2007 at 03:30 PM
I don't believe I'm incorrect: given a word of length four, it must always transform to another word of length four.
However, such transformations are not always possible.
Posted by: Dave Thomas | April 26, 2007 at 08:07 AM
Dave is incorrect. The word list provided on this site contains many words which cannot be transformed to all the other words of the same length in the list. e.g. ebb, Abba, Aaron, etc.
*spoiler alert*
The technique I used to solve this Kata was to do a breadth-first search of the tree of possible chains from a particular word, generating the tree as I went. This technique gives the shortest possible chain of words from one word to another (if it exists). In order to make the algorithm efficient I used a ternary search tree to store a dictionary of words of the given length. As each word was added to the tree of chains, it was removed from the dictionary, preventing it from being considered again. I haven't done any rigorous benchmarking but it runs in well under a second for any given two words (in C#).
i would be interested to hear of other successful approaches.
Posted by: Matt Howells | April 26, 2007 at 04:33 AM
Yes
Posted by: Dave Thomas | March 09, 2007 at 05:40 PM
So does the word conversion always happen between words of similar length?
Posted by: Amit | March 09, 2007 at 05:33 PM