3 ms·
I gave only a quick read, and I'm completely unfamiliar with Scrabble... but: why not use a Markov chain? I would start looking at the two step case, using only
by shef 16y ago
I gave only a quick read, and I'm completely unfamiliar with Scrabble... but: why not use a Markov chain? I would start looking at the two step case, using only the most probable letter. In this case you have to remember a 26x26 matrix, if this doesn't work you can extend it to the top-n letters.
- deleted 16y ago[deleted]
- danger 16y agoI'm not sure I understand. Let me try to recap: you can store a transition matrix for P(letter | previous letter) using 26x26=676 real numbers. By adding 26 more parameters saying how often a word begins with each letter, you can get P(word) = P(first letter) * P(second letter | first letter) * P(third letter | second letter). The Danny complexity of this is pretty high, though. I'd go so far as to say it's harder to memorize 702 arbitrary real numbers than it is to memorize 972 three letter words. But I like a variation on this, which is to find a low rank representation of the transition matrix, P(next letter | letter) = A x B, where A is a 26 x k matrix, and B is a k x 26 matrix. Setting k to be 2 or 3 and rounding to say 2 or 3 decimal places might be manageable to memorize.