/*
- Naïve implementation of Damerau-Levenshtein distance
- (Does not work when there are neighbouring transpositions)! */
function DamerauLevenshtein($S1, $S2)
{
$L1 = strlen($S1);
$L2 = strlen($S2);
if ($L1==0 || $L2==0) {
// Trivial case: one string is 0-length
return max($L1, $L2);
}
else {
// The cost of substituting the last character
$substitutionCost = ($S1[$L1-1] != $S2[$L2-1])? 1 : 0;
// {H1,H2} are {L1,L2} with the last character chopped off
$H1 = substr($S1, 0, $L1-1);
$H2 = substr($S2, 0, $L2-1);
if ($L1>1 && $L2>1 && $S1[$L1-1]==$S2[$L2-2] && $S1[$L1-2]==$S2[$L2-1]) {
return min (
DamerauLevenshtein($H1, $S2) + 1,
DamerauLevenshtein($S1, $H2) + 1,
DamerauLevenshtein($H1, $H2) + $substitutionCost,
DamerauLevenshtein(substr($S1, 0, $L1-2), substr($S2, 0, $L2-2)) + 1
);
}
return min (
DamerauLevenshtein($H1, $S2) + 1,
DamerauLevenshtein($S1, $H2) + 1,
DamerauLevenshtein($H1, $H2) + $substitutionCost
);
}
}
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