String-Similarity/fstrcmp.c

/* Functions to make fuzzy comparisons between strings
   Copyright (C) 1988, 1989, 1992, 1993, 1995 Free Software Foundation, Inc.

   This program is free software; you can redistribute it and/or modify
   it under the terms of the GNU General Public License as published by
   the Free Software Foundation; either version 2 of the License, or (at
   your option) any later version.

   This program is distributed in the hope that it will be useful, but
   WITHOUT ANY WARRANTY; without even the implied warranty of
   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
   General Public License for more details.

   You should have received a copy of the GNU General Public License
   along with this program; if not, write to the Free Software
   Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.


   Derived from GNU diff 2.7, analyze.c et al.

   The basic algorithm is described in:
   "An O(ND) Difference Algorithm and its Variations", Eugene Myers,
   Algorithmica Vol. 1 No. 2, 1986, pp. 251-266;
   see especially section 4.2, which describes the variation used below.

   The basic algorithm was independently discovered as described in:
   "Algorithms for Approximate String Matching", E. Ukkonen,
   Information and Control Vol. 64, 1985, pp. 100-118.

   Modified to work on strings rather than files
   by Peter Miller <pmiller@agso.gov.au>, October 1995

   Modified to accept a "minimum similarity limit" to stop analyzing the
   string when the similarity drops below the given limit by Marc Lehmann
   <schmorp@schmorp.de>.

   Modified to work on unicode (actually 31 bit are allowed) by Marc Lehmann
   <schmorp@schmorp.de>.
*/

#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#include "fstrcmp.h"

#define PARAMS(proto) proto

/*
 * Data on one input string being compared.
 */
struct string_data
{
  /* The string to be compared. */
  const UV *data;

  /* The length of the string to be compared. */
  int data_length;

  /* The number of characters inserted or deleted. */
  int edit_count;
};

static struct string_data string[2];

static int max_edits; /* compareseq stops when edits > max_edits */

#ifdef MINUS_H_FLAG

/* This corresponds to the diff -H flag.  With this heuristic, for
   strings with a constant small density of changes, the algorithm is
   linear in the strings size.  This is unlikely in typical uses of
   fstrcmp, and so is usually compiled out.  Besides, there is no
   interface to set it true.  */
static int heuristic;

#endif


/* Vector, indexed by diagonal, containing 1 + the X coordinate of the
   point furthest along the given diagonal in the forward search of the
   edit matrix.  */
static int *fdiag;

/* Vector, indexed by diagonal, containing the X coordinate of the point
   furthest along the given diagonal in the backward search of the edit
   matrix.  */
static int *bdiag;

/* Edit scripts longer than this are too expensive to compute.  */
static int too_expensive;

/* Snakes bigger than this are considered `big'.  */
#define SNAKE_LIMIT     20

struct partition
{
  /* Midpoints of this partition.  */
  int xmid, ymid;

  /* Nonzero if low half will be analyzed minimally.  */
  int lo_minimal;

  /* Likewise for high half.  */
  int hi_minimal;
};


/* NAME
        diag - find diagonal path

   SYNOPSIS
        int diag(int xoff, int xlim, int yoff, int ylim, int minimal,
                struct partition *part);

   DESCRIPTION
        Find the midpoint of the shortest edit script for a specified
        portion of the two strings.

        Scan from the beginnings of the strings, and simultaneously from
        the ends, doing a breadth-first search through the space of
        edit-sequence.  When the two searches meet, we have found the
        midpoint of the shortest edit sequence.

        If MINIMAL is nonzero, find the minimal edit script regardless
        of expense.  Otherwise, if the search is too expensive, use
        heuristics to stop the search and report a suboptimal answer.

   RETURNS
        Set PART->(XMID,YMID) to the midpoint (XMID,YMID).  The diagonal
        number XMID - YMID equals the number of inserted characters
        minus the number of deleted characters (counting only characters
        before the midpoint).  Return the approximate edit cost; this is
        the total number of characters inserted or deleted (counting
        only characters before the midpoint), unless a heuristic is used
        to terminate the search prematurely.

        Set PART->LEFT_MINIMAL to nonzero iff the minimal edit script
        for the left half of the partition is known; similarly for
        PART->RIGHT_MINIMAL.

   CAVEAT
        This function assumes that the first characters of the specified
        portions of the two strings do not match, and likewise that the
        last characters do not match.  The caller must trim matching
        characters from the beginning and end of the portions it is
        going to specify.

        If we return the "wrong" partitions, the worst this can do is
        cause suboptimal diff output.  It cannot cause incorrect diff
        output.  */

static int diag PARAMS ((int, int, int, int, int, struct partition *));

static int
diag (xoff, xlim, yoff, ylim, minimal, part)
     int xoff;
     int xlim;
     int yoff;
     int ylim;
     int minimal;
     struct partition *part;
{
  int *const fd = fdiag;        /* Give the compiler a chance. */
  int *const bd = bdiag;        /* Additional help for the compiler. */
  const UV *const xv = string[0].data;  /* Still more help for the compiler. */
  const UV *const yv = string[1].data;  /* And more and more . . . */
  const int dmin = xoff - ylim; /* Minimum valid diagonal. */
  const int dmax = xlim - yoff; /* Maximum valid diagonal. */
  const int fmid = xoff - yoff; /* Center diagonal of top-down search. */
  const int bmid = xlim - ylim; /* Center diagonal of bottom-up search. */
  int fmin = fmid;
  int fmax = fmid;              /* Limits of top-down search. */
  int bmin = bmid;
  int bmax = bmid;              /* Limits of bottom-up search. */
  int c;                        /* Cost. */
  int odd = (fmid - bmid) & 1;

  /*
         * True if southeast corner is on an odd diagonal with respect
         * to the northwest.
         */
  fd[fmid] = xoff;
  bd[bmid] = xlim;
  for (c = 1;; ++c)
    {
      int d;                    /* Active diagonal. */
      int big_snake;

      big_snake = 0;
      /* Extend the top-down search by an edit step in each diagonal. */
      if (fmin > dmin)
        fd[--fmin - 1] = -1;
      else
        ++fmin;
      if (fmax < dmax)
        fd[++fmax + 1] = -1;
      else
        --fmax;
      for (d = fmax; d >= fmin; d -= 2)
        {
          int x;
          int y;
          int oldx;
          int tlo;
          int thi;

          tlo = fd[d - 1],
            thi = fd[d + 1];

          if (tlo >= thi)
            x = tlo + 1;
          else
            x = thi;
          oldx = x;
          y = x - d;
          while (x < xlim && y < ylim && xv[x] == yv[y])
            {
              ++x;
              ++y;
            }
          if (x - oldx > SNAKE_LIMIT)
            big_snake = 1;
          fd[d] = x;
          if (odd && bmin <= d && d <= bmax && bd[d] <= x)
            {
              part->xmid = x;
              part->ymid = y;
              part->lo_minimal = part->hi_minimal = 1;
              return 2 * c - 1;
            }
        }
      /* Similarly extend the bottom-up search.  */
      if (bmin > dmin)
        bd[--bmin - 1] = INT_MAX;
      else
        ++bmin;
      if (bmax < dmax)
        bd[++bmax + 1] = INT_MAX;
      else
        --bmax;
      for (d = bmax; d >= bmin; d -= 2)
        {
          int x;
          int y;
          int oldx;
          int tlo;
          int thi;

          tlo = bd[d - 1],
            thi = bd[d + 1];
          if (tlo < thi)
            x = tlo;
          else
            x = thi - 1;
          oldx = x;
          y = x - d;
          while (x > xoff && y > yoff && xv[x - 1] == yv[y - 1])
            {
              --x;
              --y;
            }
          if (oldx - x > SNAKE_LIMIT)
            big_snake = 1;
          bd[d] = x;
          if (!odd && fmin <= d && d <= fmax && x <= fd[d])
            {
              part->xmid = x;
              part->ymid = y;
              part->lo_minimal = part->hi_minimal = 1;
              return 2 * c;
            }
        }

      if (minimal)
        continue;

#ifdef MINUS_H_FLAG
      /* Heuristic: check occasionally for a diagonal that has made lots
         of progress compared with the edit distance.  If we have any
         such, find the one that has made the most progress and return
         it as if it had succeeded.

         With this heuristic, for strings with a constant small density
         of changes, the algorithm is linear in the strings size.  */
      if (c > 200 && big_snake && heuristic)
        {
          int best;

          best = 0;
          for (d = fmax; d >= fmin; d -= 2)
            {
              int dd;
              int x;
              int y;
              int v;

              dd = d - fmid;
              x = fd[d];
              y = x - d;
              v = (x - xoff) * 2 - dd;

              if (v > 12 * (c + (dd < 0 ? -dd : dd)))
                {
                  if
                    (
                      v > best
                      &&
                      xoff + SNAKE_LIMIT <= x
                      &&
                      x < xlim
                      &&
                      yoff + SNAKE_LIMIT <= y
                      &&
                      y < ylim
                    )
                    {
                      /* We have a good enough best diagonal; now insist
                         that it end with a significant snake.  */
                      int k;

                      for (k = 1; xv[x - k] == yv[y - k]; k++)
                        {
                          if (k == SNAKE_LIMIT)
                            {
                              best = v;
                              part->xmid = x;
                              part->ymid = y;
                              break;
                            }
                        }
                    }
                }
            }
          if (best > 0)
            {
              part->lo_minimal = 1;
              part->hi_minimal = 0;
              return 2 * c - 1;
            }
          best = 0;
          for (d = bmax; d >= bmin; d -= 2)
            {
              int dd;
              int x;
              int y;
              int v;

              dd = d - bmid;
              x = bd[d];
              y = x - d;
              v = (xlim - x) * 2 + dd;

              if (v > 12 * (c + (dd < 0 ? -dd : dd)))
                {
                  if (v > best && xoff < x && x <= xlim - SNAKE_LIMIT &&
                      yoff < y && y <= ylim - SNAKE_LIMIT)
                    {
                      /* We have a good enough best diagonal; now insist
                         that it end with a significant snake.  */
                      int k;

                      for (k = 0; xv[x + k] == yv[y + k]; k++)
                        {
                          if (k == SNAKE_LIMIT - 1)
                            {
                              best = v;
                              part->xmid = x;
                              part->ymid = y;
                              break;
                            }
                        }
                    }
                }
            }
          if (best > 0)
            {
              part->lo_minimal = 0;
              part->hi_minimal = 1;
              return 2 * c - 1;
            }
        }
#endif /* MINUS_H_FLAG */

      /* Heuristic: if we've gone well beyond the call of duty, give up
         and report halfway between our best results so far.  */
      if (c >= too_expensive)
        {
          int fxybest;
          int fxbest;
          int bxybest;
          int bxbest;

          /* Pacify `gcc -Wall'. */
          fxbest = 0;
          bxbest = 0;

          /* Find forward diagonal that maximizes X + Y.  */
          fxybest = -1;
          for (d = fmax; d >= fmin; d -= 2)
            {
              int x;
              int y;

              x = fd[d] < xlim ? fd[d] : xlim;
              y = x - d;

              if (ylim < y)
                {
                  x = ylim + d;
                  y = ylim;
                }
              if (fxybest < x + y)
                {
                  fxybest = x + y;
                  fxbest = x;
                }
            }
          /* Find backward diagonal that minimizes X + Y.  */
          bxybest = INT_MAX;
          for (d = bmax; d >= bmin; d -= 2)
            {
              int x;
              int y;

              x = xoff > bd[d] ? xoff : bd[d];
              y = x - d;

              if (y < yoff)
                {
                  x = yoff + d;
                  y = yoff;
                }
              if (x + y < bxybest)
                {
                  bxybest = x + y;
                  bxbest = x;
                }
            }
          /* Use the better of the two diagonals.  */
          if ((xlim + ylim) - bxybest < fxybest - (xoff + yoff))
            {
              part->xmid = fxbest;
              part->ymid = fxybest - fxbest;
              part->lo_minimal = 1;
              part->hi_minimal = 0;
            }
          else
            {
              part->xmid = bxbest;
              part->ymid = bxybest - bxbest;
              part->lo_minimal = 0;
              part->hi_minimal = 1;
            }
          return 2 * c - 1;
        }
    }
}


/* NAME
        compareseq - find edit sequence

   SYNOPSIS
        void compareseq(int xoff, int xlim, int yoff, int ylim, int minimal);

   DESCRIPTION
        Compare in detail contiguous subsequences of the two strings
        which are known, as a whole, to match each other.

        The subsequence of string 0 is [XOFF, XLIM) and likewise for
        string 1.

        Note that XLIM, YLIM are exclusive bounds.  All character
        numbers are origin-0.

        If MINIMAL is nonzero, find a minimal difference no matter how
        expensive it is.  */

static void compareseq PARAMS ((int, int, int, int, int));

static void
compareseq (xoff, xlim, yoff, ylim, minimal)
     int xoff;
     int xlim;
     int yoff;
     int ylim;
     int minimal;
{
  const UV *const xv = string[0].data;  /* Help the compiler.  */
  const UV *const yv = string[1].data;

  if (string[1].edit_count + string[0].edit_count > max_edits)
    return;

  /* Slide down the bottom initial diagonal. */
  while (xoff < xlim && yoff < ylim && xv[xoff] == yv[yoff])
    {
      ++xoff;
      ++yoff;
    }

  /* Slide up the top initial diagonal. */
  while (xlim > xoff && ylim > yoff && xv[xlim - 1] == yv[ylim - 1])
    {
      --xlim;
      --ylim;
    }

  /* Handle simple cases. */
  if (xoff == xlim)
    {
      while (yoff < ylim)
        {
          ++string[1].edit_count;
          ++yoff;
        }
    }
  else if (yoff == ylim)
    {
      while (xoff < xlim)
        {
          ++string[0].edit_count;
          ++xoff;
        }
    }
  else
    {
      int c;
      struct partition part;

      /* Find a point of correspondence in the middle of the strings.  */
      c = diag (xoff, xlim, yoff, ylim, minimal, &part);
      if (c == 1)
        {
#if 0
          /* This should be impossible, because it implies that one of
             the two subsequences is empty, and that case was handled
             above without calling `diag'.  Let's verify that this is
             true.  */
          abort ();
#else
          /* The two subsequences differ by a single insert or delete;
             record it and we are done.  */
          if (part.xmid - part.ymid < xoff - yoff)
            ++string[1].edit_count;
          else
            ++string[0].edit_count;
#endif
        }
      else
        {
          /* Use the partitions to split this problem into subproblems.  */
          compareseq (xoff, part.xmid, yoff, part.ymid, part.lo_minimal);
          compareseq (part.xmid, xlim, part.ymid, ylim, part.hi_minimal);
        }
    }
}


/* NAME
        fstrcmp - fuzzy string compare

   SYNOPSIS
        double fstrcmp(const ChaR *s1, int l1, const UV *s2, int l2, double);

   DESCRIPTION
        The fstrcmp function may be used to compare two string for
        similarity.  It is very useful in reducing "cascade" or
        "secondary" errors in compilers or other situations where
        symbol tables occur.

   RETURNS
        double; 0 if the strings are entirly dissimilar, 1 if the
        strings are identical, and a number in between if they are
        similar.  */

double
fstrcmp (const UV *string1, int length1,
         const UV *string2, int length2,
         double minimum)
{
  int i;

  size_t fdiag_len;
  static int *fdiag_buf;
  static size_t fdiag_max;

  /* set the info for each string.  */
  string[0].data = string1;
  string[0].data_length = length1;
  string[1].data = string2;
  string[1].data_length = length2;

  /* short-circuit obvious comparisons */
  if (string[0].data_length == 0 && string[1].data_length == 0)
    return 1.0;
  if (string[0].data_length == 0 || string[1].data_length == 0)
    return 0.0;

  /* Set TOO_EXPENSIVE to be approximate square root of input size,
     bounded below by 256.  */
  too_expensive = 1;
  for (i = string[0].data_length + string[1].data_length; i != 0; i >>= 2)
    too_expensive <<= 1;
  if (too_expensive < 256)
    too_expensive = 256;

  /* Because fstrcmp is typically called multiple times, while scanning
     symbol tables, etc, attempt to minimize the number of memory
     allocations performed.  Thus, we use a static buffer for the
     diagonal vectors, and never free them.  */
  fdiag_len = string[0].data_length + string[1].data_length + 3;
  if (fdiag_len > fdiag_max)
    {
      fdiag_max = fdiag_len;
      fdiag_buf = realloc (fdiag_buf, fdiag_max * (2 * sizeof (int)));
    }
  fdiag = fdiag_buf + string[1].data_length + 1;
  bdiag = fdiag + fdiag_len;

  max_edits = 1 + (string[0].data_length + string[1].data_length) * (1. - minimum);

  /* Now do the main comparison algorithm */
  string[0].edit_count = 0;
  string[1].edit_count = 0;
  compareseq (0, string[0].data_length, 0, string[1].data_length, 0);

  /* The result is
        ((number of chars in common) / (average length of the strings)).
     This is admittedly biased towards finding that the strings are
     similar, however it does produce meaningful results.  */
  return ((double)
             (string[0].data_length + string[1].data_length - string[1].edit_count - string[0].edit_count)
           / (string[0].data_length + string[1].data_length));

}
Revision:	1.2
Committed:	Tue Nov 4 15:31:38 2008 UTC (15 years, 6 months ago) by root
Content type:	text/plain
Branch:	MAIN
CVS Tags:	rel-1_04, HEAD
Changes since 1.1:	+8 -10 lines
Log Message:	* empty log message *
#	Content
1	/* Functions to make fuzzy comparisons between strings
2	Copyright (C) 1988, 1989, 1992, 1993, 1995 Free Software Foundation, Inc.
3
4	This program is free software; you can redistribute it and/or modify
5	it under the terms of the GNU General Public License as published by
6	the Free Software Foundation; either version 2 of the License, or (at
7	your option) any later version.
8
9	This program is distributed in the hope that it will be useful, but
10	WITHOUT ANY WARRANTY; without even the implied warranty of
11	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
12	General Public License for more details.
13
14	You should have received a copy of the GNU General Public License
15	along with this program; if not, write to the Free Software
16	Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
17
18
19	Derived from GNU diff 2.7, analyze.c et al.
20
21	The basic algorithm is described in:
22	"An O(ND) Difference Algorithm and its Variations", Eugene Myers,
23	Algorithmica Vol. 1 No. 2, 1986, pp. 251-266;
24	see especially section 4.2, which describes the variation used below.
25
26	The basic algorithm was independently discovered as described in:
27	"Algorithms for Approximate String Matching", E. Ukkonen,
28	Information and Control Vol. 64, 1985, pp. 100-118.
29
30	Modified to work on strings rather than files
31	by Peter Miller <pmiller@agso.gov.au>, October 1995
32
33	Modified to accept a "minimum similarity limit" to stop analyzing the
34	string when the similarity drops below the given limit by Marc Lehmann
35	<schmorp@schmorp.de>.
36
37	Modified to work on unicode (actually 31 bit are allowed) by Marc Lehmann
38	<schmorp@schmorp.de>.
39	*/
40
41	#include <string.h>
42	#include <stdio.h>
43	#include <stdlib.h>
44	#include <limits.h>
45
46	#include "fstrcmp.h"
47
48	#define PARAMS(proto) proto
49
50	/*
51	* Data on one input string being compared.
52	*/
53	struct string_data
54	{
55	/* The string to be compared. */
56	const UV *data;
57
58	/* The length of the string to be compared. */
59	int data_length;
60
61	/* The number of characters inserted or deleted. */
62	int edit_count;
63	};
64
65	static struct string_data string[2];
66
67	static int max_edits; /* compareseq stops when edits > max_edits */
68
69	#ifdef MINUS_H_FLAG
70
71	/* This corresponds to the diff -H flag. With this heuristic, for
72	strings with a constant small density of changes, the algorithm is
73	linear in the strings size. This is unlikely in typical uses of
74	fstrcmp, and so is usually compiled out. Besides, there is no
75	interface to set it true. */
76	static int heuristic;
77
78	#endif
79
80
81	/* Vector, indexed by diagonal, containing 1 + the X coordinate of the
82	point furthest along the given diagonal in the forward search of the
83	edit matrix. */
84	static int *fdiag;
85
86	/* Vector, indexed by diagonal, containing the X coordinate of the point
87	furthest along the given diagonal in the backward search of the edit
88	matrix. */
89	static int *bdiag;
90
91	/* Edit scripts longer than this are too expensive to compute. */
92	static int too_expensive;
93
94	/* Snakes bigger than this are considered `big'. */
95	#define SNAKE_LIMIT 20
96
97	struct partition
98	{
99	/* Midpoints of this partition. */
100	int xmid, ymid;
101
102	/* Nonzero if low half will be analyzed minimally. */
103	int lo_minimal;
104
105	/* Likewise for high half. */
106	int hi_minimal;
107	};
108
109
110	/* NAME
111	diag - find diagonal path
112
113	SYNOPSIS
114	int diag(int xoff, int xlim, int yoff, int ylim, int minimal,
115	struct partition *part);
116
117	DESCRIPTION
118	Find the midpoint of the shortest edit script for a specified
119	portion of the two strings.
120
121	Scan from the beginnings of the strings, and simultaneously from
122	the ends, doing a breadth-first search through the space of
123	edit-sequence. When the two searches meet, we have found the
124	midpoint of the shortest edit sequence.
125
126	If MINIMAL is nonzero, find the minimal edit script regardless
127	of expense. Otherwise, if the search is too expensive, use
128	heuristics to stop the search and report a suboptimal answer.
129
130	RETURNS
131	Set PART->(XMID,YMID) to the midpoint (XMID,YMID). The diagonal
132	number XMID - YMID equals the number of inserted characters
133	minus the number of deleted characters (counting only characters
134	before the midpoint). Return the approximate edit cost; this is
135	the total number of characters inserted or deleted (counting
136	only characters before the midpoint), unless a heuristic is used
137	to terminate the search prematurely.
138
139	Set PART->LEFT_MINIMAL to nonzero iff the minimal edit script
140	for the left half of the partition is known; similarly for
141	PART->RIGHT_MINIMAL.
142
143	CAVEAT
144	This function assumes that the first characters of the specified
145	portions of the two strings do not match, and likewise that the
146	last characters do not match. The caller must trim matching
147	characters from the beginning and end of the portions it is
148	going to specify.
149
150	If we return the "wrong" partitions, the worst this can do is
151	cause suboptimal diff output. It cannot cause incorrect diff
152	output. */
153
154	static int diag PARAMS ((int, int, int, int, int, struct partition *));
155
156	static int
157	diag (xoff, xlim, yoff, ylim, minimal, part)
158	int xoff;
159	int xlim;
160	int yoff;
161	int ylim;
162	int minimal;
163	struct partition *part;
164	{
165	int const fd = fdiag; / Give the compiler a chance. */
166	int const bd = bdiag; / Additional help for the compiler. */
167	const UV const xv = string[0].data; / Still more help for the compiler. */
168	const UV const yv = string[1].data; / And more and more . . . */
169	const int dmin = xoff - ylim; /* Minimum valid diagonal. */
170	const int dmax = xlim - yoff; /* Maximum valid diagonal. */
171	const int fmid = xoff - yoff; /* Center diagonal of top-down search. */
172	const int bmid = xlim - ylim; /* Center diagonal of bottom-up search. */
173	int fmin = fmid;
174	int fmax = fmid; /* Limits of top-down search. */
175	int bmin = bmid;
176	int bmax = bmid; /* Limits of bottom-up search. */
177	int c; /* Cost. */
178	int odd = (fmid - bmid) & 1;
179
180	/*
181	* True if southeast corner is on an odd diagonal with respect
182	* to the northwest.
183	*/
184	fd[fmid] = xoff;
185	bd[bmid] = xlim;
186	for (c = 1;; ++c)
187	{
188	int d; /* Active diagonal. */
189	int big_snake;
190
191	big_snake = 0;
192	/* Extend the top-down search by an edit step in each diagonal. */
193	if (fmin > dmin)
194	fd[--fmin - 1] = -1;
195	else
196	++fmin;
197	if (fmax < dmax)
198	fd[++fmax + 1] = -1;
199	else
200	--fmax;
201	for (d = fmax; d >= fmin; d -= 2)
202	{
203	int x;
204	int y;
205	int oldx;
206	int tlo;
207	int thi;
208
209	tlo = fd[d - 1],
210	thi = fd[d + 1];
211
212	if (tlo >= thi)
213	x = tlo + 1;
214	else
215	x = thi;
216	oldx = x;
217	y = x - d;
218	while (x < xlim && y < ylim && xv[x] == yv[y])
219	{
220	++x;
221	++y;
222	}
223	if (x - oldx > SNAKE_LIMIT)
224	big_snake = 1;
225	fd[d] = x;
226	if (odd && bmin <= d && d <= bmax && bd[d] <= x)
227	{
228	part->xmid = x;
229	part->ymid = y;
230	part->lo_minimal = part->hi_minimal = 1;
231	return 2 * c - 1;
232	}
233	}
234	/* Similarly extend the bottom-up search. */
235	if (bmin > dmin)
236	bd[--bmin - 1] = INT_MAX;
237	else
238	++bmin;
239	if (bmax < dmax)
240	bd[++bmax + 1] = INT_MAX;
241	else
242	--bmax;
243	for (d = bmax; d >= bmin; d -= 2)
244	{
245	int x;
246	int y;
247	int oldx;
248	int tlo;
249	int thi;
250
251	tlo = bd[d - 1],
252	thi = bd[d + 1];
253	if (tlo < thi)
254	x = tlo;
255	else
256	x = thi - 1;
257	oldx = x;
258	y = x - d;
259	while (x > xoff && y > yoff && xv[x - 1] == yv[y - 1])
260	{
261	--x;
262	--y;
263	}
264	if (oldx - x > SNAKE_LIMIT)
265	big_snake = 1;
266	bd[d] = x;
267	if (!odd && fmin <= d && d <= fmax && x <= fd[d])
268	{
269	part->xmid = x;
270	part->ymid = y;
271	part->lo_minimal = part->hi_minimal = 1;
272	return 2 * c;
273	}
274	}
275
276	if (minimal)
277	continue;
278
279	#ifdef MINUS_H_FLAG
280	/* Heuristic: check occasionally for a diagonal that has made lots
281	of progress compared with the edit distance. If we have any
282	such, find the one that has made the most progress and return
283	it as if it had succeeded.
284
285	With this heuristic, for strings with a constant small density
286	of changes, the algorithm is linear in the strings size. */
287	if (c > 200 && big_snake && heuristic)
288	{
289	int best;
290
291	best = 0;
292	for (d = fmax; d >= fmin; d -= 2)
293	{
294	int dd;
295	int x;
296	int y;
297	int v;
298
299	dd = d - fmid;
300	x = fd[d];
301	y = x - d;
302	v = (x - xoff) * 2 - dd;
303
304	if (v > 12 * (c + (dd < 0 ? -dd : dd)))
305	{
306	if
307	(
308	v > best
309	&&
310	xoff + SNAKE_LIMIT <= x
311	&&
312	x < xlim
313	&&
314	yoff + SNAKE_LIMIT <= y
315	&&
316	y < ylim
317	)
318	{
319	/* We have a good enough best diagonal; now insist
320	that it end with a significant snake. */
321	int k;
322
323	for (k = 1; xv[x - k] == yv[y - k]; k++)
324	{
325	if (k == SNAKE_LIMIT)
326	{
327	best = v;
328	part->xmid = x;
329	part->ymid = y;
330	break;
331	}
332	}
333	}
334	}
335	}
336	if (best > 0)
337	{
338	part->lo_minimal = 1;
339	part->hi_minimal = 0;
340	return 2 * c - 1;
341	}
342	best = 0;
343	for (d = bmax; d >= bmin; d -= 2)
344	{
345	int dd;
346	int x;
347	int y;
348	int v;
349
350	dd = d - bmid;
351	x = bd[d];
352	y = x - d;
353	v = (xlim - x) * 2 + dd;
354
355	if (v > 12 * (c + (dd < 0 ? -dd : dd)))
356	{
357	if (v > best && xoff < x && x <= xlim - SNAKE_LIMIT &&
358	yoff < y && y <= ylim - SNAKE_LIMIT)
359	{
360	/* We have a good enough best diagonal; now insist
361	that it end with a significant snake. */
362	int k;
363
364	for (k = 0; xv[x + k] == yv[y + k]; k++)
365	{
366	if (k == SNAKE_LIMIT - 1)
367	{
368	best = v;
369	part->xmid = x;
370	part->ymid = y;
371	break;
372	}
373	}
374	}
375	}
376	}
377	if (best > 0)
378	{
379	part->lo_minimal = 0;
380	part->hi_minimal = 1;
381	return 2 * c - 1;
382	}
383	}
384	#endif /* MINUS_H_FLAG */
385
386	/* Heuristic: if we've gone well beyond the call of duty, give up
387	and report halfway between our best results so far. */
388	if (c >= too_expensive)
389	{
390	int fxybest;
391	int fxbest;
392	int bxybest;
393	int bxbest;
394
395	/* Pacify `gcc -Wall'. */
396	fxbest = 0;
397	bxbest = 0;
398
399	/* Find forward diagonal that maximizes X + Y. */
400	fxybest = -1;
401	for (d = fmax; d >= fmin; d -= 2)
402	{
403	int x;
404	int y;
405
406	x = fd[d] < xlim ? fd[d] : xlim;
407	y = x - d;
408
409	if (ylim < y)
410	{
411	x = ylim + d;
412	y = ylim;
413	}
414	if (fxybest < x + y)
415	{
416	fxybest = x + y;
417	fxbest = x;
418	}
419	}
420	/* Find backward diagonal that minimizes X + Y. */
421	bxybest = INT_MAX;
422	for (d = bmax; d >= bmin; d -= 2)
423	{
424	int x;
425	int y;
426
427	x = xoff > bd[d] ? xoff : bd[d];
428	y = x - d;
429
430	if (y < yoff)
431	{
432	x = yoff + d;
433	y = yoff;
434	}
435	if (x + y < bxybest)
436	{
437	bxybest = x + y;
438	bxbest = x;
439	}
440	}
441	/* Use the better of the two diagonals. */
442	if ((xlim + ylim) - bxybest < fxybest - (xoff + yoff))
443	{
444	part->xmid = fxbest;
445	part->ymid = fxybest - fxbest;
446	part->lo_minimal = 1;
447	part->hi_minimal = 0;
448	}
449	else
450	{
451	part->xmid = bxbest;
452	part->ymid = bxybest - bxbest;
453	part->lo_minimal = 0;
454	part->hi_minimal = 1;
455	}
456	return 2 * c - 1;
457	}
458	}
459	}
460
461
462	/* NAME
463	compareseq - find edit sequence
464
465	SYNOPSIS
466	void compareseq(int xoff, int xlim, int yoff, int ylim, int minimal);
467
468	DESCRIPTION
469	Compare in detail contiguous subsequences of the two strings
470	which are known, as a whole, to match each other.
471
472	The subsequence of string 0 is [XOFF, XLIM) and likewise for
473	string 1.
474
475	Note that XLIM, YLIM are exclusive bounds. All character
476	numbers are origin-0.
477
478	If MINIMAL is nonzero, find a minimal difference no matter how
479	expensive it is. */
480
481	static void compareseq PARAMS ((int, int, int, int, int));
482
483	static void
484	compareseq (xoff, xlim, yoff, ylim, minimal)
485	int xoff;
486	int xlim;
487	int yoff;
488	int ylim;
489	int minimal;
490	{
491	const UV const xv = string[0].data; / Help the compiler. */
492	const UV *const yv = string[1].data;
493
494	if (string[1].edit_count + string[0].edit_count > max_edits)
495	return;
496
497	/* Slide down the bottom initial diagonal. */
498	while (xoff < xlim && yoff < ylim && xv[xoff] == yv[yoff])
499	{
500	++xoff;
501	++yoff;
502	}
503
504	/* Slide up the top initial diagonal. */
505	while (xlim > xoff && ylim > yoff && xv[xlim - 1] == yv[ylim - 1])
506	{
507	--xlim;
508	--ylim;
509	}
510
511	/* Handle simple cases. */
512	if (xoff == xlim)
513	{
514	while (yoff < ylim)
515	{
516	++string[1].edit_count;
517	++yoff;
518	}
519	}
520	else if (yoff == ylim)
521	{
522	while (xoff < xlim)
523	{
524	++string[0].edit_count;
525	++xoff;
526	}
527	}
528	else
529	{
530	int c;
531	struct partition part;
532
533	/* Find a point of correspondence in the middle of the strings. */
534	c = diag (xoff, xlim, yoff, ylim, minimal, &part);
535	if (c == 1)
536	{
537	#if 0
538	/* This should be impossible, because it implies that one of
539	the two subsequences is empty, and that case was handled
540	above without calling `diag'. Let's verify that this is
541	true. */
542	abort ();
543	#else
544	/* The two subsequences differ by a single insert or delete;
545	record it and we are done. */
546	if (part.xmid - part.ymid < xoff - yoff)
547	++string[1].edit_count;
548	else
549	++string[0].edit_count;
550	#endif
551	}
552	else
553	{
554	/* Use the partitions to split this problem into subproblems. */
555	compareseq (xoff, part.xmid, yoff, part.ymid, part.lo_minimal);
556	compareseq (part.xmid, xlim, part.ymid, ylim, part.hi_minimal);
557	}
558	}
559	}
560
561
562	/* NAME
563	fstrcmp - fuzzy string compare
564
565	SYNOPSIS
566	double fstrcmp(const ChaR s1, int l1, const UV s2, int l2, double);
567
568	DESCRIPTION
569	The fstrcmp function may be used to compare two string for
570	similarity. It is very useful in reducing "cascade" or
571	"secondary" errors in compilers or other situations where
572	symbol tables occur.
573
574	RETURNS
575	double; 0 if the strings are entirly dissimilar, 1 if the
576	strings are identical, and a number in between if they are
577	similar. */
578
579	double
580	fstrcmp (const UV *string1, int length1,
581	const UV *string2, int length2,
582	double minimum)
583	{
584	int i;
585
586	size_t fdiag_len;
587	static int *fdiag_buf;
588	static size_t fdiag_max;
589
590	/* set the info for each string. */
591	string[0].data = string1;
592	string[0].data_length = length1;
593	string[1].data = string2;
594	string[1].data_length = length2;
595
596	/* short-circuit obvious comparisons */
597	if (string[0].data_length == 0 && string[1].data_length == 0)
598	return 1.0;
599	if (string[0].data_length == 0 \|\| string[1].data_length == 0)
600	return 0.0;
601
602	/* Set TOO_EXPENSIVE to be approximate square root of input size,
603	bounded below by 256. */
604	too_expensive = 1;
605	for (i = string[0].data_length + string[1].data_length; i != 0; i >>= 2)
606	too_expensive <<= 1;
607	if (too_expensive < 256)
608	too_expensive = 256;
609
610	/* Because fstrcmp is typically called multiple times, while scanning
611	symbol tables, etc, attempt to minimize the number of memory
612	allocations performed. Thus, we use a static buffer for the
613	diagonal vectors, and never free them. */
614	fdiag_len = string[0].data_length + string[1].data_length + 3;
615	if (fdiag_len > fdiag_max)
616	{
617	fdiag_max = fdiag_len;
618	fdiag_buf = realloc (fdiag_buf, fdiag_max * (2 * sizeof (int)));
619	}
620	fdiag = fdiag_buf + string[1].data_length + 1;
621	bdiag = fdiag + fdiag_len;
622
623	max_edits = 1 + (string[0].data_length + string[1].data_length) * (1. - minimum);
624
625	/* Now do the main comparison algorithm */
626	string[0].edit_count = 0;
627	string[1].edit_count = 0;
628	compareseq (0, string[0].data_length, 0, string[1].data_length, 0);
629
630	/* The result is
631	((number of chars in common) / (average length of the strings)).
632	This is admittedly biased towards finding that the strings are
633	similar, however it does produce meaningful results. */
634	return ((double)
635	(string[0].data_length + string[1].data_length - string[1].edit_count - string[0].edit_count)
636	/ (string[0].data_length + string[1].data_length));
637
638	}