changed: update version strings for beta4

[xbmc.git] / xbmc / utils / fstrcmp.cpp

/* 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
  <pcg@goof.com>.

*/

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

#include "fstrcmp.h"

#ifndef __GNUC__
#pragma warning( disable : 4244 )
#endif

/*
 * 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 (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 (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 *, const char *, 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, const char *string2, 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 = (int)strlen (string1);
  string[1].data = string2;
  string[1].data_length = (int)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 = (int*)realloc (fdiag_buf, fdiag_max * (2 * sizeof (int)));
  }
  fdiag = fdiag_buf + string[1].data_length + 1;
  bdiag = fdiag + fdiag_len;

  max_edits = (int) (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));

}