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[netbsd-mini2440.git] / gnu / dist / gettext / gettext-tools / lib / fstrcmp.c

/* Functions to make fuzzy comparisons between strings
   Copyright (C) 1988-1989, 1992-1993, 1995, 2001-2003 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.

   Unless the 'minimal' flag is set, this code uses the TOO_EXPENSIVE
   heuristic, by Paul Eggert, to limit the cost to O(N**1.5 log N)
   at the price of producing suboptimal output for large inputs with
   many differences.

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

#ifdef HAVE_CONFIG_H
# include "config.h"
#endif

/* Specification.  */
#include "fstrcmp.h"

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

#include "xalloc.h"


/*
 * 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];


#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 (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 (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;

  /* 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 *, const char *);

   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, const char *string2)
{
  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 = strlen (string1);
  string[1].data = string2;
  string[1].data_length = strlen (string2);

  /* 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 = xrealloc (fdiag_buf, fdiag_max * (2 * sizeof (int)));
    }
  fdiag = fdiag_buf + string[1].data_length + 1;
  bdiag = fdiag + fdiag_len;

  /* 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));
}