sub comp {
    local $_ = $_[0] ^ $_[1];
    return s/([^\x00])\1//g==1 && !tr/\x00/1/c;
}
[download]

That doesn't return the index of where the transposition happens, does it?

Abigail

So here it is with bonus, just a little longer... And now, its just a little faster than yours. Except on large identical strings where it is 3 times faster because of the first line. Without it the regexp would perform very slow.

sub comp {
    return -1 if $_[0] eq $_[1];
    local $_ = $_[0] ^ $_[1];
    /^([\x00]*)([^\x00])\2[\x00]*$/;
    $1 ? length$1 : -1;
}
[download]



T I M T O W T D I

No, that was only a bonus ;)

T I M T O W T D I

Well, if you do an XOR of two strings (scalars that haven't been used as numbers), perl will do a bitwise XOR on the characters of the strings, instead of the bits on the numbers. Two characters that are the same will have all the bits the same, XORing those characters will give you the character of which all the bits are 0, aka "\x00". If two characters are not the same, there is at least one bitpositions where the bits are different - resulting in a 1 bit when XORing. Hence, the resulting character will be anything but "\x00".

Abigail

It works because the xor of two bits is zero if and only if the two bits are the same:

0 ^ 0 = 0
1 ^ 1 = 0
0 ^ 1 = 1
1 ^ 0 = 1

So the xor of two characters (which are a string of 8 bits) will be 0 (8 0-bits) if and only if the two characters are the same.

Using xor on strings xors the characters from the first string with the characters in the second string, so in the places where the characters are identical the resulting character will be zero, and in all places where they differ it will be nonzero.

Would it be possible for the product of $f ^ $s to contain a "1" and hence pollute your $_ string?

Yes, but no. It could certainly be possible that the XORred string contains a character "1". But this is ok, the tr/\x00/1/c turns anything that is not "\x00" into a "1" anyway - after the tr, the string will only contain "\x00" and "1" characters.

Abigail

Oh of course they are. Looks like I'm being even more stupid than usual

Sorry to have troubled you