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"

/* moved here from fstrcmp.h to avoid problems on lose32 machines */
#define PARAMS(proto) proto
typedef UV CHAR;

/*
 * Data on one input string being compared.
 */
struct string_data
{
  /* The string to be compared. */
  const CHAR *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 CHAR *const xv = string[0].data;        /* Still more help for the compiler. */
  const CHAR *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 CHAR *const xv = string[0].data;        /* Help the compiler.  */
  const CHAR *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 CHAR *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 CHAR *string1, int length1,
         const CHAR *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:	1.1
Committed:	Sat Jun 25 09:55:53 2005 UTC (18 years, 11 months ago) by root
Content type:	text/plain
Branch:	MAIN
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	/* moved here from fstrcmp.h to avoid problems on lose32 machines */
49	#define PARAMS(proto) proto
50	typedef UV CHAR;
51
52	/*
53	* Data on one input string being compared.
54	*/
55	struct string_data
56	{
57	/* The string to be compared. */
58	const CHAR *data;
59
60	/* The length of the string to be compared. */
61	int data_length;
62
63	/* The number of characters inserted or deleted. */
64	int edit_count;
65	};
66
67	static struct string_data string[2];
68
69	static int max_edits; /* compareseq stops when edits > max_edits */
70
71	#ifdef MINUS_H_FLAG
72
73	/* This corresponds to the diff -H flag. With this heuristic, for
74	strings with a constant small density of changes, the algorithm is
75	linear in the strings size. This is unlikely in typical uses of
76	fstrcmp, and so is usually compiled out. Besides, there is no
77	interface to set it true. */
78	static int heuristic;
79
80	#endif
81
82
83	/* Vector, indexed by diagonal, containing 1 + the X coordinate of the
84	point furthest along the given diagonal in the forward search of the
85	edit matrix. */
86	static int *fdiag;
87
88	/* Vector, indexed by diagonal, containing the X coordinate of the point
89	furthest along the given diagonal in the backward search of the edit
90	matrix. */
91	static int *bdiag;
92
93	/* Edit scripts longer than this are too expensive to compute. */
94	static int too_expensive;
95
96	/* Snakes bigger than this are considered `big'. */
97	#define SNAKE_LIMIT 20
98
99	struct partition
100	{
101	/* Midpoints of this partition. */
102	int xmid, ymid;
103
104	/* Nonzero if low half will be analyzed minimally. */
105	int lo_minimal;
106
107	/* Likewise for high half. */
108	int hi_minimal;
109	};
110
111
112	/* NAME
113	diag - find diagonal path
114
115	SYNOPSIS
116	int diag(int xoff, int xlim, int yoff, int ylim, int minimal,
117	struct partition *part);
118
119	DESCRIPTION
120	Find the midpoint of the shortest edit script for a specified
121	portion of the two strings.
122
123	Scan from the beginnings of the strings, and simultaneously from
124	the ends, doing a breadth-first search through the space of
125	edit-sequence. When the two searches meet, we have found the
126	midpoint of the shortest edit sequence.
127
128	If MINIMAL is nonzero, find the minimal edit script regardless
129	of expense. Otherwise, if the search is too expensive, use
130	heuristics to stop the search and report a suboptimal answer.
131
132	RETURNS
133	Set PART->(XMID,YMID) to the midpoint (XMID,YMID). The diagonal
134	number XMID - YMID equals the number of inserted characters
135	minus the number of deleted characters (counting only characters
136	before the midpoint). Return the approximate edit cost; this is
137	the total number of characters inserted or deleted (counting
138	only characters before the midpoint), unless a heuristic is used
139	to terminate the search prematurely.
140
141	Set PART->LEFT_MINIMAL to nonzero iff the minimal edit script
142	for the left half of the partition is known; similarly for
143	PART->RIGHT_MINIMAL.
144
145	CAVEAT
146	This function assumes that the first characters of the specified
147	portions of the two strings do not match, and likewise that the
148	last characters do not match. The caller must trim matching
149	characters from the beginning and end of the portions it is
150	going to specify.
151
152	If we return the "wrong" partitions, the worst this can do is
153	cause suboptimal diff output. It cannot cause incorrect diff
154	output. */
155
156	static int diag PARAMS ((int, int, int, int, int, struct partition *));
157
158	static int
159	diag (xoff, xlim, yoff, ylim, minimal, part)
160	int xoff;
161	int xlim;
162	int yoff;
163	int ylim;
164	int minimal;
165	struct partition *part;
166	{
167	int const fd = fdiag; / Give the compiler a chance. */
168	int const bd = bdiag; / Additional help for the compiler. */
169	const CHAR const xv = string[0].data; / Still more help for the compiler. */
170	const CHAR const yv = string[1].data; / And more and more . . . */
171	const int dmin = xoff - ylim; /* Minimum valid diagonal. */
172	const int dmax = xlim - yoff; /* Maximum valid diagonal. */
173	const int fmid = xoff - yoff; /* Center diagonal of top-down search. */
174	const int bmid = xlim - ylim; /* Center diagonal of bottom-up search. */
175	int fmin = fmid;
176	int fmax = fmid; /* Limits of top-down search. */
177	int bmin = bmid;
178	int bmax = bmid; /* Limits of bottom-up search. */
179	int c; /* Cost. */
180	int odd = (fmid - bmid) & 1;
181
182	/*
183	* True if southeast corner is on an odd diagonal with respect
184	* to the northwest.
185	*/
186	fd[fmid] = xoff;
187	bd[bmid] = xlim;
188	for (c = 1;; ++c)
189	{
190	int d; /* Active diagonal. */
191	int big_snake;
192
193	big_snake = 0;
194	/* Extend the top-down search by an edit step in each diagonal. */
195	if (fmin > dmin)
196	fd[--fmin - 1] = -1;
197	else
198	++fmin;
199	if (fmax < dmax)
200	fd[++fmax + 1] = -1;
201	else
202	--fmax;
203	for (d = fmax; d >= fmin; d -= 2)
204	{
205	int x;
206	int y;
207	int oldx;
208	int tlo;
209	int thi;
210
211	tlo = fd[d - 1],
212	thi = fd[d + 1];
213
214	if (tlo >= thi)
215	x = tlo + 1;
216	else
217	x = thi;
218	oldx = x;
219	y = x - d;
220	while (x < xlim && y < ylim && xv[x] == yv[y])
221	{
222	++x;
223	++y;
224	}
225	if (x - oldx > SNAKE_LIMIT)
226	big_snake = 1;
227	fd[d] = x;
228	if (odd && bmin <= d && d <= bmax && bd[d] <= x)
229	{
230	part->xmid = x;
231	part->ymid = y;
232	part->lo_minimal = part->hi_minimal = 1;
233	return 2 * c - 1;
234	}
235	}
236	/* Similarly extend the bottom-up search. */
237	if (bmin > dmin)
238	bd[--bmin - 1] = INT_MAX;
239	else
240	++bmin;
241	if (bmax < dmax)
242	bd[++bmax + 1] = INT_MAX;
243	else
244	--bmax;
245	for (d = bmax; d >= bmin; d -= 2)
246	{
247	int x;
248	int y;
249	int oldx;
250	int tlo;
251	int thi;
252
253	tlo = bd[d - 1],
254	thi = bd[d + 1];
255	if (tlo < thi)
256	x = tlo;
257	else
258	x = thi - 1;
259	oldx = x;
260	y = x - d;
261	while (x > xoff && y > yoff && xv[x - 1] == yv[y - 1])
262	{
263	--x;
264	--y;
265	}
266	if (oldx - x > SNAKE_LIMIT)
267	big_snake = 1;
268	bd[d] = x;
269	if (!odd && fmin <= d && d <= fmax && x <= fd[d])
270	{
271	part->xmid = x;
272	part->ymid = y;
273	part->lo_minimal = part->hi_minimal = 1;
274	return 2 * c;
275	}
276	}
277
278	if (minimal)
279	continue;
280
281	#ifdef MINUS_H_FLAG
282	/* Heuristic: check occasionally for a diagonal that has made lots
283	of progress compared with the edit distance. If we have any
284	such, find the one that has made the most progress and return
285	it as if it had succeeded.
286
287	With this heuristic, for strings with a constant small density
288	of changes, the algorithm is linear in the strings size. */
289	if (c > 200 && big_snake && heuristic)
290	{
291	int best;
292
293	best = 0;
294	for (d = fmax; d >= fmin; d -= 2)
295	{
296	int dd;
297	int x;
298	int y;
299	int v;
300
301	dd = d - fmid;
302	x = fd[d];
303	y = x - d;
304	v = (x - xoff) * 2 - dd;
305
306	if (v > 12 * (c + (dd < 0 ? -dd : dd)))
307	{
308	if
309	(
310	v > best
311	&&
312	xoff + SNAKE_LIMIT <= x
313	&&
314	x < xlim
315	&&
316	yoff + SNAKE_LIMIT <= y
317	&&
318	y < ylim
319	)
320	{
321	/* We have a good enough best diagonal; now insist
322	that it end with a significant snake. */
323	int k;
324
325	for (k = 1; xv[x - k] == yv[y - k]; k++)
326	{
327	if (k == SNAKE_LIMIT)
328	{
329	best = v;
330	part->xmid = x;
331	part->ymid = y;
332	break;
333	}
334	}
335	}
336	}
337	}
338	if (best > 0)
339	{
340	part->lo_minimal = 1;
341	part->hi_minimal = 0;
342	return 2 * c - 1;
343	}
344	best = 0;
345	for (d = bmax; d >= bmin; d -= 2)
346	{
347	int dd;
348	int x;
349	int y;
350	int v;
351
352	dd = d - bmid;
353	x = bd[d];
354	y = x - d;
355	v = (xlim - x) * 2 + dd;
356
357	if (v > 12 * (c + (dd < 0 ? -dd : dd)))
358	{
359	if (v > best && xoff < x && x <= xlim - SNAKE_LIMIT &&
360	yoff < y && y <= ylim - SNAKE_LIMIT)
361	{
362	/* We have a good enough best diagonal; now insist
363	that it end with a significant snake. */
364	int k;
365
366	for (k = 0; xv[x + k] == yv[y + k]; k++)
367	{
368	if (k == SNAKE_LIMIT - 1)
369	{
370	best = v;
371	part->xmid = x;
372	part->ymid = y;
373	break;
374	}
375	}
376	}
377	}
378	}
379	if (best > 0)
380	{
381	part->lo_minimal = 0;
382	part->hi_minimal = 1;
383	return 2 * c - 1;
384	}
385	}
386	#endif /* MINUS_H_FLAG */
387
388	/* Heuristic: if we've gone well beyond the call of duty, give up
389	and report halfway between our best results so far. */
390	if (c >= too_expensive)
391	{
392	int fxybest;
393	int fxbest;
394	int bxybest;
395	int bxbest;
396
397	/* Pacify `gcc -Wall'. */
398	fxbest = 0;
399	bxbest = 0;
400
401	/* Find forward diagonal that maximizes X + Y. */
402	fxybest = -1;
403	for (d = fmax; d >= fmin; d -= 2)
404	{
405	int x;
406	int y;
407
408	x = fd[d] < xlim ? fd[d] : xlim;
409	y = x - d;
410
411	if (ylim < y)
412	{
413	x = ylim + d;
414	y = ylim;
415	}
416	if (fxybest < x + y)
417	{
418	fxybest = x + y;
419	fxbest = x;
420	}
421	}
422	/* Find backward diagonal that minimizes X + Y. */
423	bxybest = INT_MAX;
424	for (d = bmax; d >= bmin; d -= 2)
425	{
426	int x;
427	int y;
428
429	x = xoff > bd[d] ? xoff : bd[d];
430	y = x - d;
431
432	if (y < yoff)
433	{
434	x = yoff + d;
435	y = yoff;
436	}
437	if (x + y < bxybest)
438	{
439	bxybest = x + y;
440	bxbest = x;
441	}
442	}
443	/* Use the better of the two diagonals. */
444	if ((xlim + ylim) - bxybest < fxybest - (xoff + yoff))
445	{
446	part->xmid = fxbest;
447	part->ymid = fxybest - fxbest;
448	part->lo_minimal = 1;
449	part->hi_minimal = 0;
450	}
451	else
452	{
453	part->xmid = bxbest;
454	part->ymid = bxybest - bxbest;
455	part->lo_minimal = 0;
456	part->hi_minimal = 1;
457	}
458	return 2 * c - 1;
459	}
460	}
461	}
462
463
464	/* NAME
465	compareseq - find edit sequence
466
467	SYNOPSIS
468	void compareseq(int xoff, int xlim, int yoff, int ylim, int minimal);
469
470	DESCRIPTION
471	Compare in detail contiguous subsequences of the two strings
472	which are known, as a whole, to match each other.
473
474	The subsequence of string 0 is [XOFF, XLIM) and likewise for
475	string 1.
476
477	Note that XLIM, YLIM are exclusive bounds. All character
478	numbers are origin-0.
479
480	If MINIMAL is nonzero, find a minimal difference no matter how
481	expensive it is. */
482
483	static void compareseq PARAMS ((int, int, int, int, int));
484
485	static void
486	compareseq (xoff, xlim, yoff, ylim, minimal)
487	int xoff;
488	int xlim;
489	int yoff;
490	int ylim;
491	int minimal;
492	{
493	const CHAR const xv = string[0].data; / Help the compiler. */
494	const CHAR *const yv = string[1].data;
495
496	if (string[1].edit_count + string[0].edit_count > max_edits)
497	return;
498
499	/* Slide down the bottom initial diagonal. */
500	while (xoff < xlim && yoff < ylim && xv[xoff] == yv[yoff])
501	{
502	++xoff;
503	++yoff;
504	}
505
506	/* Slide up the top initial diagonal. */
507	while (xlim > xoff && ylim > yoff && xv[xlim - 1] == yv[ylim - 1])
508	{
509	--xlim;
510	--ylim;
511	}
512
513	/* Handle simple cases. */
514	if (xoff == xlim)
515	{
516	while (yoff < ylim)
517	{
518	++string[1].edit_count;
519	++yoff;
520	}
521	}
522	else if (yoff == ylim)
523	{
524	while (xoff < xlim)
525	{
526	++string[0].edit_count;
527	++xoff;
528	}
529	}
530	else
531	{
532	int c;
533	struct partition part;
534
535	/* Find a point of correspondence in the middle of the strings. */
536	c = diag (xoff, xlim, yoff, ylim, minimal, &part);
537	if (c == 1)
538	{
539	#if 0
540	/* This should be impossible, because it implies that one of
541	the two subsequences is empty, and that case was handled
542	above without calling `diag'. Let's verify that this is
543	true. */
544	abort ();
545	#else
546	/* The two subsequences differ by a single insert or delete;
547	record it and we are done. */
548	if (part.xmid - part.ymid < xoff - yoff)
549	++string[1].edit_count;
550	else
551	++string[0].edit_count;
552	#endif
553	}
554	else
555	{
556	/* Use the partitions to split this problem into subproblems. */
557	compareseq (xoff, part.xmid, yoff, part.ymid, part.lo_minimal);
558	compareseq (part.xmid, xlim, part.ymid, ylim, part.hi_minimal);
559	}
560	}
561	}
562
563
564	/* NAME
565	fstrcmp - fuzzy string compare
566
567	SYNOPSIS
568	double fstrcmp(const CHAR s1, int l1, const CHAR s2, int l2, double);
569
570	DESCRIPTION
571	The fstrcmp function may be used to compare two string for
572	similarity. It is very useful in reducing "cascade" or
573	"secondary" errors in compilers or other situations where
574	symbol tables occur.
575
576	RETURNS
577	double; 0 if the strings are entirly dissimilar, 1 if the
578	strings are identical, and a number in between if they are
579	similar. */
580
581	double
582	fstrcmp (const CHAR *string1, int length1,
583	const CHAR *string2, int length2,
584	double minimum)
585	{
586	int i;
587
588	size_t fdiag_len;
589	static int *fdiag_buf;
590	static size_t fdiag_max;
591
592	/* set the info for each string. */
593	string[0].data = string1;
594	string[0].data_length = length1;
595	string[1].data = string2;
596	string[1].data_length = length2;
597
598	/* short-circuit obvious comparisons */
599	if (string[0].data_length == 0 && string[1].data_length == 0)
600	return 1.0;
601	if (string[0].data_length == 0 \|\| string[1].data_length == 0)
602	return 0.0;
603
604	/* Set TOO_EXPENSIVE to be approximate square root of input size,
605	bounded below by 256. */
606	too_expensive = 1;
607	for (i = string[0].data_length + string[1].data_length; i != 0; i >>= 2)
608	too_expensive <<= 1;
609	if (too_expensive < 256)
610	too_expensive = 256;
611
612	/* Because fstrcmp is typically called multiple times, while scanning
613	symbol tables, etc, attempt to minimize the number of memory
614	allocations performed. Thus, we use a static buffer for the
615	diagonal vectors, and never free them. */
616	fdiag_len = string[0].data_length + string[1].data_length + 3;
617	if (fdiag_len > fdiag_max)
618	{
619	fdiag_max = fdiag_len;
620	fdiag_buf = realloc (fdiag_buf, fdiag_max * (2 * sizeof (int)));
621	}
622	fdiag = fdiag_buf + string[1].data_length + 1;
623	bdiag = fdiag + fdiag_len;
624
625	max_edits = 1 + (string[0].data_length + string[1].data_length) * (1. - minimum);
626
627	/* Now do the main comparison algorithm */
628	string[0].edit_count = 0;
629	string[1].edit_count = 0;
630	compareseq (0, string[0].data_length, 0, string[1].data_length, 0);
631
632	/* The result is
633	((number of chars in common) / (average length of the strings)).
634	This is admittedly biased towards finding that the strings are
635	similar, however it does produce meaningful results. */
636	return ((double)
637	(string[0].data_length + string[1].data_length - string[1].edit_count - string[0].edit_count)
638	/ (string[0].data_length + string[1].data_length));
639
640	}