git-svn-id: svn://10.65.10.50/branches/R_10_00@22947 c028cbd2-c16b-5b4b-a496-9718f37d4682
		
			
				
	
	
		
			610 lines
		
	
	
		
			18 KiB
		
	
	
	
		
			C++
		
	
	
	
	
	
			
		
		
	
	
			610 lines
		
	
	
		
			18 KiB
		
	
	
	
		
			C++
		
	
	
	
	
	
/* Functions to make fuzzy comparisons between strings
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   Copyright (C) 1988-1989, 1992-1993, 1995, 2001-2003 Free Software 
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Foundation, Inc.
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   This program is free software; you can redistribute it and/or modify
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   it under the terms of the GNU General Public License as published by
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   the Free Software Foundation; either version 2 of the License, or (at
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   your option) any later version.
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   This program is distributed in the hope that it will be useful, but
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   WITHOUT ANY WARRANTY; without even the implied warranty of
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   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
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   General Public License for more details.
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   You should have received a copy of the GNU General Public License
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   along with this program; if not, write to the Free Software
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   Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
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   Derived from GNU diff 2.7, analyze.c et al.
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   The basic algorithm is described in:
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   "An O(ND) Difference Algorithm and its Variations", Eugene Myers,
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   Algorithmica Vol. 1 No. 2, 1986, pp. 251-266;
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   see especially section 4.2, which describes the variation used below.
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   The basic algorithm was independently discovered as described in:
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   "Algorithms for Approximate String Matching", E. Ukkonen,
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   Information and Control Vol. 64, 1985, pp. 100-118.
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   Modified to work on strings rather than files
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   by Peter Miller <pmiller@xxxxxxxxxxx>, October 1995 */
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//#include "config.h"
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/* Specification.  */
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#include "fstrcmp.h"
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#include <string.h>
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#include <limits.h>
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#include <stdlib.h>
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/*
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 * Data on one input string being compared.
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 */
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struct string_data
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{
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  /* The string to be compared. */
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  const char *data;
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  /* The length of the string to be compared. */
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  int data_length;
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  /* The number of characters inserted or deleted. */
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  int edit_count;
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};
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static struct string_data string[2];
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#ifdef MINUS_H_FLAG
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/* This corresponds to the diff -H flag.  With this heuristic, for
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   strings with a constant small density of changes, the algorithm is
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   linear in the strings size.  This is unlikely in typical uses of
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   fstrcmp, and so is usually compiled out.  Besides, there is no
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   interface to set it true.  */
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static int heuristic;
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#endif
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/* Vector, indexed by diagonal, containing 1 + the X coordinate of the
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   point furthest along the given diagonal in the forward search of the
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   edit matrix.  */
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static int *fdiag;
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/* Vector, indexed by diagonal, containing the X coordinate of the point
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   furthest along the given diagonal in the backward search of the edit
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   matrix.  */
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static int *bdiag;
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/* Edit scripts longer than this are too expensive to compute.  */
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static int too_expensive;
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/* Snakes bigger than this are considered `big'.  */
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#define SNAKE_LIMIT     20
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struct partition
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{
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  /* Midpoints of this partition.  */
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  int xmid, ymid;
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  /* Nonzero if low half will be analyzed minimally.  */
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  int lo_minimal;
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  /* Likewise for high half.  */
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  int hi_minimal;
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};
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/* NAME
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        diag - find diagonal path
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   SYNOPSIS
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        int diag(int xoff, int xlim, int yoff, int ylim, int minimal,
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                struct partition *part);
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   DESCRIPTION
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        Find the midpoint of the shortest edit script for a specified
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        portion of the two strings.
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        Scan from the beginnings of the strings, and simultaneously from
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        the ends, doing a breadth-first search through the space of
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        edit-sequence.  When the two searches meet, we have found the
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        midpoint of the shortest edit sequence.
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        If MINIMAL is nonzero, find the minimal edit script regardless
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        of expense.  Otherwise, if the search is too expensive, use
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        heuristics to stop the search and report a suboptimal answer.
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   RETURNS
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        Set PART->(XMID,YMID) to the midpoint (XMID,YMID).  The diagonal
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        number XMID - YMID equals the number of inserted characters
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        minus the number of deleted characters (counting only characters
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        before the midpoint).  Return the approximate edit cost; this is
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        the total number of characters inserted or deleted (counting
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        only characters before the midpoint), unless a heuristic is used
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        to terminate the search prematurely.
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        Set PART->LEFT_MINIMAL to nonzero iff the minimal edit script
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        for the left half of the partition is known; similarly for
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        PART->RIGHT_MINIMAL.
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   CAVEAT
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        This function assumes that the first characters of the specified
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        portions of the two strings do not match, and likewise that the
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        last characters do not match.  The caller must trim matching
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        characters from the beginning and end of the portions it is
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        going to specify.
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        If we return the "wrong" partitions, the worst this can do is
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        cause suboptimal diff output.  It cannot cause incorrect diff
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        output.  */
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static int
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diag (int xoff, int xlim, int yoff, int ylim, int minimal,
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      struct partition *part)
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{
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  int *const fd = fdiag;        /* Give the compiler a chance. */
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  int *const bd = bdiag;        /* Additional help for the compiler. */
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  const char *const xv = string[0].data;        /* Still more help for the 
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compiler. */
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  const char *const yv = string[1].data;        /* And more and more . . . */
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  const int dmin = xoff - ylim; /* Minimum valid diagonal. */
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  const int dmax = xlim - yoff; /* Maximum valid diagonal. */
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  const int fmid = xoff - yoff; /* Center diagonal of top-down search. */
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  const int bmid = xlim - ylim; /* Center diagonal of bottom-up search. */
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  int fmin = fmid;
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  int fmax = fmid;              /* Limits of top-down search. */
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  int bmin = bmid;
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  int bmax = bmid;              /* Limits of bottom-up search. */
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  int c;                        /* Cost. */
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  int odd = (fmid - bmid) & 1;
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  /*
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         * True if southeast corner is on an odd diagonal with respect
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         * to the northwest.
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         */
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  fd[fmid] = xoff;
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  bd[bmid] = xlim;
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  for (c = 1;; ++c)
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    {
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      int d;                    /* Active diagonal. */
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      int big_snake;
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      big_snake = 0;
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      /* Extend the top-down search by an edit step in each diagonal. */
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      if (fmin > dmin)
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        fd[--fmin - 1] = -1;
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      else
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        ++fmin;
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      if (fmax < dmax)
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        fd[++fmax + 1] = -1;
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      else
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        --fmax;
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      for (d = fmax; d >= fmin; d -= 2)
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        {
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          int x;
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          int y;
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          int oldx;
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          int tlo;
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          int thi;
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          tlo = fd[d - 1],
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            thi = fd[d + 1];
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          if (tlo >= thi)
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            x = tlo + 1;
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          else
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            x = thi;
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          oldx = x;
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          y = x - d;
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          while (x < xlim && y < ylim && xv[x] == yv[y])
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            {
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              ++x;
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              ++y;
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            }
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          if (x - oldx > SNAKE_LIMIT)
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            big_snake = 1;
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          fd[d] = x;
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          if (odd && bmin <= d && d <= bmax && bd[d] <= x)
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            {
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              part->xmid = x;
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              part->ymid = y;
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              part->lo_minimal = part->hi_minimal = 1;
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              return 2 * c - 1;
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            }
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        }
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      /* Similarly extend the bottom-up search.  */
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      if (bmin > dmin)
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        bd[--bmin - 1] = INT_MAX;
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      else
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        ++bmin;
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      if (bmax < dmax)
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        bd[++bmax + 1] = INT_MAX;
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      else
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        --bmax;
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      for (d = bmax; d >= bmin; d -= 2)
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        {
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          int x;
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          int y;
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          int oldx;
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          int tlo;
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          int thi;
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          tlo = bd[d - 1],
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            thi = bd[d + 1];
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          if (tlo < thi)
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            x = tlo;
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          else
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            x = thi - 1;
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          oldx = x;
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          y = x - d;
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          while (x > xoff && y > yoff && xv[x - 1] == yv[y - 1])
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            {
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              --x;
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              --y;
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            }
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          if (oldx - x > SNAKE_LIMIT)
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            big_snake = 1;
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          bd[d] = x;
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          if (!odd && fmin <= d && d <= fmax && x <= fd[d])
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            {
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              part->xmid = x;
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              part->ymid = y;
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              part->lo_minimal = part->hi_minimal = 1;
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              return 2 * c;
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            }
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        }
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      if (minimal)
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        continue;
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#ifdef MINUS_H_FLAG
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      /* Heuristic: check occasionally for a diagonal that has made lots
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         of progress compared with the edit distance.  If we have any
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         such, find the one that has made the most progress and return
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         it as if it had succeeded.
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         With this heuristic, for strings with a constant small density
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         of changes, the algorithm is linear in the strings size.  */
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      if (c > 200 && big_snake && heuristic)
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        {
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          int best;
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          best = 0;
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          for (d = fmax; d >= fmin; d -= 2)
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            {
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              int dd;
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              int x;
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              int y;
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              int v;
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              dd = d - fmid;
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              x = fd[d];
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              y = x - d;
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              v = (x - xoff) * 2 - dd;
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              if (v > 12 * (c + (dd < 0 ? -dd : dd)))
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                {
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                  if
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                    (
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                      v > best
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                      &&
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                      xoff + SNAKE_LIMIT <= x
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                      &&
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                      x < xlim
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                      &&
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                      yoff + SNAKE_LIMIT <= y
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                      &&
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                      y < ylim
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                    )
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                    {
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                      /* We have a good enough best diagonal; now insist
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                         that it end with a significant snake.  */
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                      int k;
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                      for (k = 1; xv[x - k] == yv[y - k]; k++)
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                        {
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                          if (k == SNAKE_LIMIT)
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                            {
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                              best = v;
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                              part->xmid = x;
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                              part->ymid = y;
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                              break;
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                            }
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                        }
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                    }
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                }
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            }
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          if (best > 0)
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            {
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              part->lo_minimal = 1;
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              part->hi_minimal = 0;
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              return 2 * c - 1;
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            }
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          best = 0;
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          for (d = bmax; d >= bmin; d -= 2)
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            {
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              int dd;
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              int x;
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              int y;
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              int v;
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              dd = d - bmid;
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              x = bd[d];
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              y = x - d;
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              v = (xlim - x) * 2 + dd;
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              if (v > 12 * (c + (dd < 0 ? -dd : dd)))
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                {
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                  if (v > best && xoff < x && x <= xlim - SNAKE_LIMIT &&
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                      yoff < y && y <= ylim - SNAKE_LIMIT)
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                    {
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                      /* We have a good enough best diagonal; now insist
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                         that it end with a significant snake.  */
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                      int k;
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                      for (k = 0; xv[x + k] == yv[y + k]; k++)
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                        {
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                          if (k == SNAKE_LIMIT - 1)
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                            {
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                              best = v;
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                              part->xmid = x;
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                              part->ymid = y;
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                              break;
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                            }
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                        }
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                    }
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                }
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            }
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          if (best > 0)
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            {
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              part->lo_minimal = 0;
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              part->hi_minimal = 1;
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              return 2 * c - 1;
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            }
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        }
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#endif /* MINUS_H_FLAG */
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      /* Heuristic: if we've gone well beyond the call of duty, give up
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         and report halfway between our best results so far.  */
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      if (c >= too_expensive)
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        {
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          int fxybest;
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          int fxbest;
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          int bxybest;
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          int bxbest;
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          /* Pacify `gcc -Wall'. */
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          fxbest = 0;
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          bxbest = 0;
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          /* Find forward diagonal that maximizes X + Y.  */
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          fxybest = -1;
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          for (d = fmax; d >= fmin; d -= 2)
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            {
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              int x;
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              int y;
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              x = fd[d] < xlim ? fd[d] : xlim;
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              y = x - d;
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              if (ylim < y)
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                {
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                  x = ylim + d;
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                  y = ylim;
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                }
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              if (fxybest < x + y)
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                {
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                  fxybest = x + y;
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                  fxbest = x;
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                }
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            }
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          /* Find backward diagonal that minimizes X + Y.  */
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          bxybest = INT_MAX;
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          for (d = bmax; d >= bmin; d -= 2)
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            {
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              int x;
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              int y;
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              x = xoff > bd[d] ? xoff : bd[d];
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              y = x - d;
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              if (y < yoff)
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                {
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                  x = yoff + d;
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                  y = yoff;
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                }
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              if (x + y < bxybest)
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                {
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                  bxybest = x + y;
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                  bxbest = x;
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                }
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            }
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          /* Use the better of the two diagonals.  */
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          if ((xlim + ylim) - bxybest < fxybest - (xoff + yoff))
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            {
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              part->xmid = fxbest;
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              part->ymid = fxybest - fxbest;
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              part->lo_minimal = 1;
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              part->hi_minimal = 0;
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            }
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          else
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            {
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              part->xmid = bxbest;
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              part->ymid = bxybest - bxbest;
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              part->lo_minimal = 0;
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              part->hi_minimal = 1;
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            }
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          return 2 * c - 1;
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        }
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    }
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}
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/* NAME
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        compareseq - find edit sequence
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   SYNOPSIS
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        void compareseq(int xoff, int xlim, int yoff, int ylim, int minimal);
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   DESCRIPTION
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        Compare in detail contiguous subsequences of the two strings
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        which are known, as a whole, to match each other.
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        The subsequence of string 0 is [XOFF, XLIM) and likewise for
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        string 1.
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        Note that XLIM, YLIM are exclusive bounds.  All character
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        numbers are origin-0.
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        If MINIMAL is nonzero, find a minimal difference no matter how
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        expensive it is.  */
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static void
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compareseq (int xoff, int xlim, int yoff, int ylim, int minimal)
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{
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  const char *const xv = string[0].data;        /* Help the compiler.  */
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  const char *const yv = string[1].data;
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 | 
						|
  /* 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)
 | 
						|
{
 | 
						|
  static int *fdiag_buf = NULL;
 | 
						|
  static size_t fdiag_max = 0;
 | 
						|
  
 | 
						|
  int i;
 | 
						|
  size_t fdiag_len;
 | 
						|
 | 
						|
  /* 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 = (int*)realloc (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));
 | 
						|
}
 |