sbsWrite(p, " | ", 3);
    sbsWriteLineno(p, lnRight);
    p->iEnd = nRight - nSuffix;
    sbsWriteText(p, pRight, SBS_NEWLINE);
  }
}

/*
** Minimum of two values
*/
static int minInt(int a, int b){ return a<b ? a : b; }

/*
** Return the number between 0 and 100 that is smaller the closer pA and
** pB match.  Return 0 for a perfect match.  Return 100 if pA and pB are
** completely different.
**
** The current algorithm is as follows:
**
** (1) Remove leading and trailing whitespace.
** (2) Truncate both strings to at most 250 characters
** (3) Find the length of the longest common subsequence
** (4) Longer common subsequences yield lower scores.
*/
static int match_dline(DLine *pA, DLine *pB){
  const char *zA;            /* Left string */
  const char *zB;            /* right string */
  int nA;                    /* Bytes in zA[] */
  int nB;                    /* Bytes in zB[] */
  int avg;                   /* Average length of A and B */
  int i, j, k;               /* Loop counters */
  int best = 0;              /* Longest match found so far */
  int score;                 /* Final score.  0..100 */
  unsigned char c;           /* Character being examined */
  unsigned char aFirst[256]; /* aFirst[X] = index in zB[] of first char X */
  unsigned char aNext[252];  /* aNext[i] = index in zB[] of next zB[i] char */

  zA = pA->z;
  zB = pB->z;
  nA = pA->h & LENGTH_MASK;
  nB = pB->h & LENGTH_MASK;
  while( nA>0 && fossil_isspace(zA[0]) ){ nA--; zA++; }
  while( nA>0 && fossil_isspace(zA[nA-1]) ){ nA--; }
  while( nB>0 && fossil_isspace(zB[0]) ){ nB--; zB++; }
  while( nB>0 && fossil_isspace(zB[nB-1]) ){ nB--; }
  if( nA>250 ) nA = 250;
  if( nB>250 ) nB = 250;
  avg = (nA+nB)/2;
  if( avg==0 ) return 0;
  memset(aFirst, 0, sizeof(aFirst));
  memset(aNext, 0, nB);
  zA--; zB--;   /* Make both zA[] and zB[] 1-indexed */
  for(i=nB; i>0; i--){
    c = (unsigned char)zB[i];
    aNext[i] = aFirst[c];
    aFirst[c] = i;
  }
  best = 0;
  for(i=1; i<=nA; i++){
    c = (unsigned char)zA[i];
    for(j=aFirst[c]; j>0; j = aNext[j]){

      int limit = minInt(nA-i, nB-j)+1;
      for(k=1; k<limit && zA[k+i]==zB[k+j]; k++){}
      if( k>best ) best = k;
    }
  }
  score = (best>avg) ? 0 : (avg - best)*100/avg;

#if 0
  fprintf(stderr, "A: [%.*s]\nB: [%.*s]\nbest=%d avg=%d score=%d\n",
  nA, zA+1, nB, zB+1, best, avg, score);
#endif

  /* Return the result */
  return score;
}

/*
** There is a change block in which nLeft lines of text on the left are
** converted into nRight lines of text on the right.  This routine computes

  }
  *piSX = iSXb;
  *piEX = iSXb + mxLength;
  *piSY = iSYb;
  *piEY = iSYb + mxLength;
}






/*
** Compare two blocks of text on lines iS1 through iE1-1 of the aFrom[]
** file and lines iS2 through iE2-1 of the aTo[] file.  Locate a sequence
** of lines in these two blocks that are exactly the same.  Return
** the bounds of the matching sequence.
**
** If there are two or more possible answers of the same length, the