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/* 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));

}