Speaking of which, this is a fun problem. So if you want to give a try, you should stop reading here and come back after you've solved it!
I found this problem to be surprisingly difficult. I kept coming up with ideas but inevitably they were over simplified and wouldn't work. I was visualizing it as a tree.
It's an interesting problem because it has three different cases to consider. It's normal for a problem to have two cases, like the first item in the list, and every other item. But at least with the way I approached it, this problem had three cases: the first node (which has no parent node), the nodes in the first level of the tree (which all have the same parent), and every level after that (where there are many different parents). This kept causing me to come up with ideas that only worked for the first level, or ideas that worked for one of those levels and not the others...
I knew that I needed a breadth first search but I was really struggling with how to implement it while also keeping track of the path to each node. Usually a breadth first search effectively just loops over all the nodes in each level, and then calls the next level recursively. But I need to know what the parent of each node is, and what that parent's parent is, etc up to the root. My solution to this was to represent each node as a tuple containing the list of it's full path from root to that node, and the word of that node. This is what I passed through the recursive calls, therefore I always knew the full path to each node. This was simple, but has the downside of wasting memory since it duplicates a significant portion of the parent list on each node.
Another interesting element of this solution is that I prune words out of the tree that I've already seen, like the two red nodes in the picture above. This means I don't have to worry about "infinite loops" (they're not really loops, but you get what I mean) in the graph, so I can search until I run out of nodes. And its safe, because I need to find the shortest ladder, so if I've already encountered a word earlier in the graph, that same word can't be part of the solution later in the graph.
The git repo with my solution is here: https://github.com/kberridge/word-ladder-kata-fs
And here's the code:
Here are some notes on some of the things I found interesting about this code:
- The array slicing code in the findChildren function was fun.
- And using the range operator in the generateCandidates ['a'..'z'] was fun too.
- The findValidUniqueChildren function is an excellant example of something I've been struggling with in F#. What parameters does this function take? What does it return? It's NOT easy to figure this out, is it? This is also the first time I've used function composition for real!
- Notice in the queuechildren method how I'm using the concatenate list operator: "let newSearchNodes = searchNodes @ childnodes"? The Programming F# book says if you find yourself using this it probably means you're doing something in a non-idiomatic way... I suppose I could have written buildNodes to append the nodes to a provided list, but that seemed awkward.
- The match of findLadderWorker is pretty typical, but demonstrates another little pattern I keep finding. For the second match, I created a function to call since it's longer than 1 line, so I had to make up a name. I went with "testnode" which I don't really like, but I had to name it something!
Here's Anthony Coble's Haskell solution. His is very different and shows a great example of a different way to approach the problem.
Ben Lee also solved it in Erlang. His is async!
If you solve it, in any language, send me the link to your solution and I'll add it to the post!