|There's more than one way to do things|
Re: Finding longest palindrome from a stringby Aristotle (Chancellor)
|on Aug 13, 2004 at 14:30 UTC||Need Help??|
The XOR dance, with a rotating mirror copy.
If you don't understand what's going on, run this for a visual demonstration:
Update: changed 0 .. POSIX::ceil( length( $str ) / 2 ) to 0 .. length( $str ) - 1. It was a vestige from an earlier trail of thought that was no longer valid.
Update: since I was asked how this works, I'm adding an explanation here.
It is pretty simple: a XOR b = 0 when a = b. If you XOR two strings with each other, you will get a NULL at all locations with identical characters. Now obviously, if you XOR a string with a mirror copy of itself and get all NULLs, then it's a palindrome, because all characters in the rotated copy were identical with all characters of the original.
That's the gist of it. The particular problem given for this thread is complicated by the fact that we have to look for embedded palindromes, and rotating the string unfortunately displaces the rotated copy of an embedded palindrome. To find all embedded palindromes, the mirror copy must be XORed against the original string at each offset. The code does this by rotating the copy n times for a string of length n.
There is one nasty trap left. If the string consists of two adjacent palindromes, such as abbabbafef. Mirroring that yields fefabbabba. If you rotate this three times to the left, the mirror copy becomes abbabbafef and XORing them yields a string of all NULLs, which would indicate that the palindrome is abbabbafef. Oops. The problem is that we forgot to keep track of where inside the mirror copy its original start and end used to be. Palindromes obviously cannot run across that location. That is what the substr $mask, $rotate_count, 0, "\1"; is about: a non-NULL is added to break a string of NULLs running across that location. Of course, now we have to account for that extra character in offsets in the mask.
And that's it. The bulk of the work happens in a single XOR and a pattern match, and other auxiliary tasks are done using very few builtins. That's where it gets its speed. The bulk of the code is merely simple math.
Makeshifts last the longest.