android-kvm / linux / 988f01683c7f2bf9f8fe2bae1cf4010fcd1baaf5 / . / tools / perf / util / levenshtein.c

// SPDX-License-Identifier: GPL-2.0 | |

#include "levenshtein.h" | |

#include <errno.h> | |

#include <stdlib.h> | |

#include <string.h> | |

/* | |

* This function implements the Damerau-Levenshtein algorithm to | |

* calculate a distance between strings. | |

* | |

* Basically, it says how many letters need to be swapped, substituted, | |

* deleted from, or added to string1, at least, to get string2. | |

* | |

* The idea is to build a distance matrix for the substrings of both | |

* strings. To avoid a large space complexity, only the last three rows | |

* are kept in memory (if swaps had the same or higher cost as one deletion | |

* plus one insertion, only two rows would be needed). | |

* | |

* At any stage, "i + 1" denotes the length of the current substring of | |

* string1 that the distance is calculated for. | |

* | |

* row2 holds the current row, row1 the previous row (i.e. for the substring | |

* of string1 of length "i"), and row0 the row before that. | |

* | |

* In other words, at the start of the big loop, row2[j + 1] contains the | |

* Damerau-Levenshtein distance between the substring of string1 of length | |

* "i" and the substring of string2 of length "j + 1". | |

* | |

* All the big loop does is determine the partial minimum-cost paths. | |

* | |

* It does so by calculating the costs of the path ending in characters | |

* i (in string1) and j (in string2), respectively, given that the last | |

* operation is a substitution, a swap, a deletion, or an insertion. | |

* | |

* This implementation allows the costs to be weighted: | |

* | |

* - w (as in "sWap") | |

* - s (as in "Substitution") | |

* - a (for insertion, AKA "Add") | |

* - d (as in "Deletion") | |

* | |

* Note that this algorithm calculates a distance _iff_ d == a. | |

*/ | |

int levenshtein(const char *string1, const char *string2, | |

int w, int s, int a, int d) | |

{ | |

int len1 = strlen(string1), len2 = strlen(string2); | |

int *row0 = malloc(sizeof(int) * (len2 + 1)); | |

int *row1 = malloc(sizeof(int) * (len2 + 1)); | |

int *row2 = malloc(sizeof(int) * (len2 + 1)); | |

int i, j; | |

for (j = 0; j <= len2; j++) | |

row1[j] = j * a; | |

for (i = 0; i < len1; i++) { | |

int *dummy; | |

row2[0] = (i + 1) * d; | |

for (j = 0; j < len2; j++) { | |

/* substitution */ | |

row2[j + 1] = row1[j] + s * (string1[i] != string2[j]); | |

/* swap */ | |

if (i > 0 && j > 0 && string1[i - 1] == string2[j] && | |

string1[i] == string2[j - 1] && | |

row2[j + 1] > row0[j - 1] + w) | |

row2[j + 1] = row0[j - 1] + w; | |

/* deletion */ | |

if (row2[j + 1] > row1[j + 1] + d) | |

row2[j + 1] = row1[j + 1] + d; | |

/* insertion */ | |

if (row2[j + 1] > row2[j] + a) | |

row2[j + 1] = row2[j] + a; | |

} | |

dummy = row0; | |

row0 = row1; | |

row1 = row2; | |

row2 = dummy; | |

} | |

i = row1[len2]; | |

free(row0); | |

free(row1); | |

free(row2); | |

return i; | |

} |