Re: Factor

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
fromseq length (a,b][
:>to
fromtoseq subseq quotcall(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:

: longest( seq -- subseq )
 dup 0 [ length max] reduce '[ length _= ] filter ;

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.