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levenshteinUDF.c
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/*
* @author Juan Miguel Cejuela, @jmcejuela
*
*
* INSTALLATION
* yum install MariaDB-devel
* gcc -o levenshteinUDF.so -shared levenshteinUDF.c `mysql_config --cflags` -fPIC
*
*
* Put the shared library as described in: http://dev.mysql.com/doc/refman/5.0/en/udf-compiling.html
*
* Afterwards in SQL:
*
* CREATE FUNCTION levenshteinUDF RETURNS INT SONAME 'levenshteinUDF.so';
*
*
* Some credit for simple levenshtein to: Joshua Drew, SpinWeb Net Designs
*
* Other contributors:
* - popthestack
* - sarbaudie
* - linuxjedi
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdarg.h>
#include <assert.h>
#define META_MALLOC(v,n,t) \
(v = (t*)malloc(((n)*sizeof(t))))
#define META_REALLOC(v,n,t) \
(v = (t*)realloc((v),((n)*sizeof(t))))
#define META_FREE(x) free((x))
/* STANDARD is defined, don't use any mysql functions */
#include <string.h>
typedef unsigned long long ulonglong;
typedef long long longlong;
#include <mysql.h>
#include <ctype.h>
/*
* See MySQL UDF documentation pages for details on their implementation.
*
*/
/**
* Levenshtein distance
*
* @param s string 1 to compare, length n
* @param t string 2 to compare, length m
* @result levenshtein distance between s and t
*
* @time O(nm), quadratic
* @space O(nm)
*/
my_bool levenshteinUDF_init(UDF_INIT *initid, UDF_ARGS *args, char *message);
void levenshteinUDF_deinit(UDF_INIT *initid);
longlong levenshteinUDF(UDF_INIT *initid, UDF_ARGS *args, char *is_null, char *error);
/**
* Levenshtein distance with threshold k (maximum allowed distance)
*
* @param s string 1 to compare, length n
* @param t string 2 to compare, length m
* @param k maximum threshold
* @result levenshtein distance between s and t or >k (not specified) if this is greater than k
*
* @time O(kl), linear; where l = min(n, m)
* @space O(k), constant
*/
my_bool levenshtein_k_init(UDF_INIT *initid, UDF_ARGS *args, char *message);
void levenshtein_k_deinit(UDF_INIT *initid);
longlong levenshtein_k(UDF_INIT *initid, UDF_ARGS *args, char *is_null, char *error);
/**
* Levenshtein ratio
*
* @param s string 1 to compare, length n
* @param t string 2 to compare, length m
* @result levenshtein ratio between s and t
*
* @time O(nm), quadratic
* @space O(nm)
*/
my_bool levenshteinratio_init(UDF_INIT *initid, UDF_ARGS *args, char *message);
void levenshteinratio_deinit(UDF_INIT *initid);
double levenshteinratio(UDF_INIT *initid, UDF_ARGS *args, char *is_null, char *error);
//(Expected) maximum number of digits to return
#define LEVENSHTEIN_MAX 3
my_bool levenshteinUDF_init(UDF_INIT *initid, UDF_ARGS *args, char *message) {
if ((args->arg_count != 2) ||
(args->arg_type[0] != STRING_RESULT || args->arg_type[1] != STRING_RESULT)) {
strcpy(message, "Function requires 2 arguments, (string, string)");
return 1;
}
//matrix for levenshtein calculations of size n+1 x m+1 (+1 for base values)
int *d = (int *) malloc(sizeof(int) * (args->lengths[0] + 1) * (args->lengths[1] + 1));
if (d == NULL) {
strcpy(message, "Failed to allocate memory");
return 1;
}
initid->ptr = (char*) d;
initid->max_length = LEVENSHTEIN_MAX;
initid->maybe_null = 0; //doesn't return null
return 0;
}
void levenshteinUDF_deinit(UDF_INIT *initid) {
if (initid->ptr != NULL) {
free(initid->ptr);
}
}
inline int minimum(int a, int b, int c) {
int min = a;
if (b < min)
min = b;
if (c < min)
min = c;
return min;
}
inline int maximum(int a, int b) {
int max = a;
if (b > max)
max = b;
return max;
}
longlong levenshteinUDF(UDF_INIT *initid, UDF_ARGS *args, char *is_null, char *error) {
const char *s = args->args[0];
const char *t = args->args[1];
int n = (s == NULL) ? 0 : args->lengths[0];
int m = (t == NULL) ? 0 : args->lengths[1];
if (0 == n)
return m;
if (0 == m)
return n;
int *d = (int*) initid->ptr;
/* Initialization */
n++; m++;
int i;
for (i = 0; i < n; i++)
d[i] = i;
int j;
for (j = 0; j < m; j++)
d[n * j] = j;
/* Recurrence */
int p, h; //indices for d matrix seen as a vector, see below
int im1 = 0; //i minus 1
for (i = 1; i < n; i++) {
p = i;
int jm1 = 0; //j minus 1
for (j = 1; j < m; j++) {
h = p; // d[h] = d[i,j-1], h = (j * n + i - n) = ((j - 1) * n + i)
p += n; // d[p] = d[i,j], p = (j * n + i)
if (s[im1] == t[jm1]) {
d[p] = d[h-1]; //no operation required, d[i-1,j-1]
}
else {
d[p] = minimum(d[p-1], //deletion, d[i-1, j]
d[h], //insertion, d[i, j-1]
d[h-1] //substitution, d[i-1,j-1]
) + 1; //can put +1 outside because the cost is the same
}
jm1 = j;
}
im1 = i;
}
return (longlong) d[p];
}
my_bool levenshteinratio_init(UDF_INIT *initid, UDF_ARGS *args, char *message) {
if ((args->arg_count != 2) ||
(args->arg_type[0] != STRING_RESULT || args->arg_type[1] != STRING_RESULT)) {
strcpy(message, "Function requires 2 arguments, (string, string)");
return 1;
}
//matrix for levenshtein calculations of size n+1 x m+1 (+1 for base values)
int *d = (int *) malloc(sizeof(int) * (args->lengths[0] + 1) * (args->lengths[1] + 1));
if (d == NULL) {
strcpy(message, "Failed to allocate memory");
return 1;
}
initid->ptr = (char*) d;
initid->max_length = LEVENSHTEIN_MAX;
initid->maybe_null = 0; //doesn't return null
return 0;
}
void levenshteinratio_deinit(UDF_INIT *initid) {
if (initid->ptr != NULL) {
free(initid->ptr);
}
}
double levenshteinratio(UDF_INIT *initid, UDF_ARGS *args, char *is_null, char *error) {
const char *s = args->args[0];
const char *t = args->args[1];
int n = (s == NULL) ? 0 : args->lengths[0];
int m = (t == NULL) ? 0 : args->lengths[1];
double dist = (double) levenshteinUDF(initid, args, is_null, error);
double maxlen = maximum(n, m);
if (maxlen == 0) return 0.0;
else return 1.0 - dist/maxlen;
}
my_bool levenshtein_k_init(UDF_INIT *initid, UDF_ARGS *args, char *message) {
if ((args->arg_count != 3) ||
(args->arg_type[0] != STRING_RESULT || args->arg_type[1] != STRING_RESULT || args->arg_type[2] != INT_RESULT)) {
strcpy(message, "Function requires 3 arguments, (string, string, int)");
return 1;
}
initid->max_length = LEVENSHTEIN_MAX;
initid->maybe_null = 0; //doesn't return null
return 0;
}
//Not necessary
//
/* void levenshtein_k_deinit(UDF_INIT *initid) { */
/* //nothing */
/* fflush(stderr); //for debugging */
/* } */
/*
* 1st observation: time O(kl)
* (see Algorithms on Strings, Trees, and Sequences, Dan Gusfield, pg. 263)
*
* When the levenshtein distance is limited to k the alignment path cannot be >k
* cells off the main diagonal, either from the left (insertions) or from the
* right (deletions) This means we don't have to fill the whole recurrence
* matrix; it suffices to fill a strip of 2k + 1 cells in a row. Also, r <= k,
* for r = m - n, is a necessary condition for there to be any solution.
*
*
* 2nd observation: exactly k + 1
*
* The number of _unpaired_ insertions/deletions (the absolute difference in the
* number of these operations) is as maximum k. When n == m (r == 0) any
* movement to the left, insertion, in the recurrence matrix is corresponded to
* a movement to the right, deletion, or viceversa. Analogously when m > n only
* a maximum of k insertions/deletions operations can be left unpaired. This
* makes the required strip size to be k + 1 cells.
*
* Call l = min(n, m). Place the string with length l on the y axis, and the
* other on the x axis. By doing this, the number of rows is equal or less than
* the number of columns and so the algorithm effectively runs in O(kl) Also, by
* this arrangement of s and t in the recurrence matrix, r indicates the forced
* number of unpaired deletions, i.e., since r>=0, there cannot be unpaired
* deletions. Observe that for there to be any insertion k must be equal or
* greater than 2.
*
* This arrangement leaves (k-r)/2 possible insertions (left) and (k-r)/2 + r
* possible deletions (right) --> (k-r)/2 + (k-r)/2 + r + 1 == k + 1
*
*
* 3rd observation: space O(k)
*
* As with the original levenshtein algorithm, only the current and last rows
* are needed (that is, when we don't do traceback because we need not the
* alignment path). By the previous observation, we just need 2 rows of k + 1
* cells. For this, considering the last row, the previous diagonal is actually
* in the same column (substitution) and the above cell is actually in the next
* column (deletion). Observe that by doing this, we're virtually creating a
* recurrence matrix of n * (k+1) being now the original diagonal in the middle
* column (-r) (matrix which could be used to do the traceback)
*
*/
longlong levenshtein_k(UDF_INIT *initid, UDF_ARGS *args, char *is_null, char *error) {
char *s = args->args[0];
char *t = args->args[1];
int n = (s == NULL) ? 0 : args->lengths[0];
int m = (t == NULL) ? 0 : args->lengths[1];
//order the strings so that the first always has the minimum length l
if (n > m) {
int aux = n;
n = m;
m = aux;
char *auxs = s;
s = t;
t = auxs;
}
const int k = *((int*) args->args[2]);
const int ignore = k + 1; //lev dist between s and t is at least greater than k
const int r = m - n;
if (0 == n)
return (m > k) ? ignore : m;
if (0 == m)
return (n > k) ? ignore : n;
if (r > k)
return ignore;
const int lsize = (((k > m) ? m : k) - r) / 2; //left space for insertions
const int rsize = lsize + r; //right space for deletions, rsize >= lsize (rsize == lsize iff r == 0)
const int stripsize = lsize + rsize + 1; // + 1 for the diagonal cell
const int stripsizem1 = stripsize - 1; //see later, not to repeat calculations
int d[2 * stripsize]; //Current and last rows
int currentrow;
int lastrow;
/* Initialization */
//currentrow = 0
int i;
for (i = lsize; i < stripsize; i++) //start from diagonal cell
d[i] = i - lsize;
/* Recurrence */
currentrow = stripsize;
lastrow = 0;
//j index for virtual recurrence matrix, jv index for rows
//bl & br = left & right bounds for j
int j, jv, bl, br;
int im1 = 0, jm1;
int a, b, c, min; //for minimum function, coded directly here for maximum speed
for (i = 1; i <= n; i++) {
//bl = max(i - lsize, 0), br = min(i + rsize, m)
bl = i - lsize;
if (bl < 0) {
jv = abs(bl); //no space for all allowed insertions
bl = 0;
}
else
jv = 0;
br = i + rsize;
if (br > m)
br = m;
jm1 = bl - 1;
for (j = bl; j <= br; j++) {
if (0 == j) //postponed part of initialization
d[currentrow + jv] = i;
else {
//By observation 3, the indices change for the lastrow (always +1)
if (s[im1] == t[jm1]) {
d[currentrow + jv] = d[lastrow + jv];
}
else {
//get the minimum of these 3 operations
a = (0 == jv) ? ignore : d[currentrow + jv - 1]; //deletion
b = (stripsizem1 == jv) ? ignore : d[lastrow + jv + 1]; //insertion
c = d[lastrow + jv]; //substitution
min = a;
if (b < min)
min = b;
if (c < min)
min = c;
d[currentrow + jv] = min + 1;
}
}
jv++;
jm1 = j;
}
//obsv: the cost of a following diagonal never decreases
if (d[currentrow + lsize + r] > k)
return ignore;
im1 = i;
//swap
currentrow = currentrow ^ stripsize;
lastrow = lastrow ^ stripsize;
}
//only here if levenhstein(s, t) <= k
return (longlong) d[lastrow + lsize + r]; //d[n, m]
}