Greplin issued a programming challenge recently, where the first question involved finding the longest palindrome substring. It was then posted as a challenge to the Programming Praxis blog, and I thought I would contribute a solution in Factor.
First, some vocabularies that we will be using:
USING: fry kernel locals make math.ranges sequences unicode.case unicode.categories ;
As part of the Factor language tutorial, the first program many people write in Factor is a word for detecting if something is a palindrome. The implementation of palindrome?
(extended to support case-insensitive comparisons using the unicode vocabulary) looks like this:
: normalize ( str -- str' ) [ Letter? ] filter >lower ; : palindrome? ( str -- ? ) normalize dup reverse = ;
The "obvious" (but not that fast) way to solve the problem is to examine every possible substring, adding to a list if it is a palindrome. The list of palindrome substrings can then be used to answer the question. This is how we'll implement it.
I thought it would be useful to split the problem into two steps. First, we need a way to enumerate all possible substrings (not including the "empty" substring), applying a quotation to each in turn.
:: each-subseq ( ... seq quot: ( ... x -- ... ) -- ... ) seq length [0,b] [ :> from from seq length (a,b] [ :> to from to seq subseq quot call( x -- ) ] each ] each ;
You can try this out in the listener, to see how it works:
( scratchpad ) "abc" [ . ] each-subseq "a" "ab" "abc" "b" "bc" "c"
Once we have that, it's pretty easy to build a word to look for palindrome substrings:
: palindromes ( str -- seq ) [ [ dup palindrome? [ , ] [ drop ] if ] each-subseq ] { } make ;
We can use the longest
word that I implemented for my anagrams post to find the longest palindrome substring:
Using this on the 1169-character string from the original challenge, we find 52 unique palindromes of 2 or more characters. The longest palindrome substring I found was a 7-character sequence.: longest ( seq -- subseq ) dup 0 [ length max ] reduce '[ length _ = ] filter ;
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