in reply to Challenge: 8 Letters, Most Words
Some theory:
For a minute I thought this can be trivially solved by counting the normalization of all words (<=8) from the dictionary in a hash... e.g.
DB<211> join "",sort split //,"electron" => "ceelnort" DB<212> join "",sort split //,"elector" => "ceelort"
As next step successively the count from all smaller words had to be added to covering words, e.g striking the "n" from "ceelnort" leads to "ceelort", so $count{ceelnort}+=$count{ceelort}
But than I realized that the best covering word from the dictionary is not necessarily the best solution.
take this counterexample for 3 out of 4 letters, the number showing the coverage-count
1 a 1 b 3 a b 2 a c 2 b c 4 a b d
so the word (a,b,d) is the maximum with a count 4, but the set (a,b,c) would cover 5 words!!!
(yes this also works with repeated letters)
IMHO this problem belongs to the family of Maximum coverage problem and Set_cover_problem, so finding a guarantied best solution shouldn't be trivial w/o brute force.
OTOH adapting the sort order of letters might already lead to very good solutions...
Cheers Rolf
( addicted to the Perl Programming Language)
update
Maybe you can use the above count hash to solve the dual problem:
"which of the n-8 letters cover the minimum of words" (n including repetition)
E.g. "d" is in only one out of 6 words with 4 letters => the remaining 3 letters cover 5 words.
"c" is only in 2 remaining words => (a,b) cover a maximum of 3 words and so on.
Not sure if this leads to the guarantied best solution, sounds to easy... =)