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, 7 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 *
#	User	Rev	Content
1	root	1.1	/* 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	root	1.2	const UV *data;
57	root	1.1
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	root	1.2	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	root	1.1	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	root	1.2	const UV const xv = string[0].data; / Help the compiler. */
492			const UV *const yv = string[1].data;
493	root	1.1
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	root	1.2	double fstrcmp(const ChaR s1, int l1, const UV s2, int l2, double);
567	root	1.1
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	root	1.2	fstrcmp (const UV *string1, int length1,
581			const UV *string2, int length2,
582	root	1.1	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			}