Finding what words a set of letters can create? - sql

Question

I am trying to write some SQL that will accept a set of letters and return all of the possible words it can make. My first thought was to create a basic three table database like so:
Words -- contains 200k words in real life
------
1 | act
2 | cat
Letters -- contains the whole alphabet in real life
--------
1 | a
3 | c
20 | t
WordLetters --First column is the WordId and the second column is the LetterId
------------
1 | 1
1 | 3
1 | 20
2 | 3
2 | 1
2 | 20
But I'm a bit stuck on how I would write a query that returns words that have an entry in WordLetters for every letter passed in. It also needs to account for words that have two of the same letter. I started with this query, but it obviously does not work:
SELECT DISTINCT w.Word
FROM Words w
INNER JOIN WordLetters wl
ON wl.LetterId = 20 AND wl.LetterId = 3 AND wl.LetterId = 1
How would I write a query to return only words that contain all of the letters passed in and accounting for duplicate letters?
Other info:
My Word table contains close to 200,000 words which is why I am trying to do this on the database side rather than in code. I am using the enable1 word list if anyone cares.

Answer 1

Ignoring, for the moment, the SQL part of the problem, the algorithm I'd use is fairly simple: start by taking each word in your dictionary, and producing a version of it with the letters in sorted order, along with a pointer back to the original version of that word.
This would give a table with entries like:
sorted_text word_id
act 123 /* we'll assume `act` was word number 123 in the original list */
act 321 /* we'll assume 'cat' was word number 321 in the original list */
Then when we receive an input (say, "tac") we sort it's letters, look it up in our table of sorted letters joined to the table of the original words, and that gives us a list of the words that can be created from that input.
If I were doing this, I'd have the tables for that in a SQL database, but probably use something else to pre-process the word list into the sorted form. Likewise, I'd probably leave sorting the letters of the user's input to whatever I was using to create the front-end, so SQL would be left to do what it's good at: relational database management.

Answer 2

If you use the solution you provide, you'll need to add an order column to the WordLetters table. Without that, there's no guarantee that you'll retrieve the rows that you retrieve are in the same order you inserted them.
However, I think I have a better solution. Based on your question, it appears that you want to find all words with the same component letters, independent of order or number of occurrences. This means that you have a limited number of possibilities. If you translate each letter of the alphabet into a different power of two, you can create a unique value for each combination of letters (aka a bitmask). You can then simply add together the values for each letter found in a word. This will make matching the words trivial, as all words with the same letters will map to the same value. Here's an example:
WITH letters
AS (SELECT Cast('a' AS VARCHAR) AS Letter,
1 AS LetterValue,
1 AS LetterNumber
UNION ALL
SELECT Cast(Char(97 + LetterNumber) AS VARCHAR),
Power(2, LetterNumber),
LetterNumber + 1
FROM letters
WHERE LetterNumber < 26),
words
AS (SELECT 1 AS wordid, 'act' AS word
UNION ALL SELECT 2, 'cat'
UNION ALL SELECT 3, 'tom'
UNION ALL SELECT 4, 'moot'
UNION ALL SELECT 5, 'mote')
SELECT wordid,
word,
Sum(distinct LetterValue) as WordValue
FROM letters
JOIN words
ON word LIKE '%' + letter + '%'
GROUP BY wordid, word
As you'll see if you run this query, "act" and "cat" have the same WordValue, as do "tom" and "moot", despite the difference in number of characters.
What makes this better than your solution? You don't have to build a lot of non-words to weed them out. This will constitute a massive savings of both storage and processing needed to perform the task.

Answer 3

There is a solution to this in SQL. It involves using a trick to count the number of times that each letter appears in a word. The following expression counts the number of times that 'a' appears:
select len(word) - len(replace(word, 'a', ''))
The idea is to count the total of all the letters in the word and see if that matches the overall length:
select w.word, (LEN(w.word) - SUM(LettersInWord))
from
(
select w.word, (LEN(w.word) - LEN(replace(w.word, wl.letter))) as LettersInWord
from word w
cross join wordletters wl
) wls
having (LEN(w.word) = SUM(LettersInWord))
This particular solution allows multiple occurrences of a letter. I'm not sure if this was desired in the original question or not. If we want up to a certain number of occurrences, then we might do the following:
select w.word, (LEN(w.word) - SUM(LettersInWord))
from
(
select w.word,
(case when (LEN(w.word) - LEN(replace(w.word, wl.letter))) <= maxcount
then (LEN(w.word) - LEN(replace(w.word, wl.letter)))
else maxcount end) as LettersInWord
from word w
cross join
(
select letter, count(*) as maxcount
from wordletters wl
group by letter
) wl
) wls
having (LEN(w.word) = SUM(LettersInWord))
If you want an exact match to the letters, then the case statement should use " = maxcount" instead of " <= maxcount".
In my experience, I have actually seen decent performance with small cross joins. This might actually work server-side. There are two big advantages to doing this work on the server. First, it takes advantage of the parallelism on the box. Second, a much smaller set of data needs to be transfered across the network.