You can notice for example that the "S$ with no children" pattern is extremely common, and turn the trie into: ^root^ The trie does a great job of sharing nodes at the beginning but a lousy job of sharing them at the end. Notice how there is a lot of repetition because in English, lots of words end the same, just as lots of words begin the same. You can optimize the memory usage by building a DAWG out of the trie. It can be time-consuming to build the trie out of the dictionary, so you might want to do so once and then figure out some way to store the trie on disk in a form that is amenable to rebuilding it quickly from disk. This simple version of the algorithm will be plenty fast enough to generate all the possible words on a rack.ĭon't forget that of course you only have to compute the trie/dawg once if the dictionary is not changing over time.
#WORD FINDER SCRABBLE PROFESSIONAL#
This is the basic algorithm that professional Scrabble AI programs use, though of course they also have to deal with things like board position, rack management and so on, which complicate the algorithms somewhat. It should be fast enough that doing it 26 times is cheap (or doing it 26 x 26 times if you have two blanks.) If you have PO but no T, you don't have to investigate POTSHERD or POTATO or POTASH or POTLATCH or POTABLE all those expensive and fruitless searches go away very quickly.Īdapting the system to deal with "wild" tiles is pretty straightforward if you have OPS?, then just run the search algorithm 26 times, on OPSA, OPSB, OPSC. You see how this data structure makes it very efficient? Once you have determined that you do not have the letters on the rack to make the beginning of a word, you don't have to investigate any dictionary words that start with that beginning. But we never had to investigate OPTS, POTS, STOP or STOPS because we didn't have a T. We've tried every possible word in the dictionary and discovered that OP, OPS and SOP all match. We don't go down the ST branch because we don't have a T. Eventually we go down the SOP branch and find an end-of-word on SOP, so "SOP" matches this rack.
Backtrack to the root.Īnd again, you see how this goes. Go down the PO branch and match S - that fails. Now the problem is to match OS against the P branch. Now backtrack up to the root.Ĭan you go down the P branch? Yes. Does it have an end-of-word marker? Yes! So OPS matches also. Now you have the empty rack and you have to match it against the OPS branch. Now the problem is matching "S" against the OP branch. Does it have an end-of-word marker? Yes, so OP is a match. So now the problem is matching "PS" against the O branch. ^root^Īnd you have the rack "OPS" - what do you do?įirst you say "can I go down the O branch?" Yes, you can. Suppose you have the dictionary "OP, OPS, OPT, OPTS, POT, POTS, SOP, SOPS, STOP, STOPS" From that you build this trie: (Nodes with a $ are those that are marked as "word can end here". Once you have a trie/dawg it becomes very inexpensive to test every word in the dictionary against a given rack, because you can "prune out" whole huge branches of the dictionary that the rack cannot possibly match. What you do instead is you build a trie data structure out of the dictionary (or, if you're really buff, you build a dawg - a directed acyclic word graph - which is a sort of compressed trie.) A relational database table is not a suitable data structure for solving this problem as efficiently as you need to.
If the user inputs “TEST” the result can’t contain words with more than 1 E and S and words with more than 2 T, a result with “TESTER” in it would be bad. Users can input any number of letters and also use wildcards (representing any letter). How do I get the right words from a MySQL database containing a dictionary from user letter input? The application will be used to help users to find possible words to play while playing games like Scrabble. I’m currently building a mobile web application using C#, MySQL, HTML5 and Javascript. I’ve got a problem and it seems some before me have had similar problems but I haven’t been able to find a working solution for me.
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