Ejemplo n.º 1
0
        /// <summary>
        /// Find the difference in 2 arrays of integers.
        /// </summary>
        /// <param name="ArrayA">A-version of the numbers (usualy the old one)</param>
        /// <param name="ArrayB">B-version of the numbers (usualy the new one)</param>
        /// <returns>Returns a array of Items that describe the differences.</returns>
        public static Item[] DiffInt(int[] ArrayA, int[] ArrayB)
        {
            // The A-Version of the data (original data) to be compared.
              DiffData DataA = new DiffData(ArrayA);

              // The B-Version of the data (modified data) to be compared.
              DiffData DataB = new DiffData(ArrayB);

              LCS(DataA, 0, DataA.Length, DataB, 0, DataB.Length);
              return CreateDiffs(DataA, DataB);
        }
Ejemplo n.º 2
0
        /// <summary>Scan the tables of which lines are inserted and deleted,
        /// producing an edit script in forward order.  
        /// </summary>
        /// dynamic array
        private static Item[] CreateDiffs(DiffData DataA, DiffData DataB)
        {
            ArrayList a = new ArrayList();
              Item aItem;
              Item []result;

              int StartA, StartB;
              int LineA, LineB;

              LineA = 0;
              LineB = 0;
              while (LineA < DataA.Length || LineB < DataB.Length) {
            if ((LineA < DataA.Length) && (! DataA.modified[LineA])
              && (LineB < DataB.Length) && (! DataB.modified[LineB])) {
              // equal lines
              LineA++;
              LineB++;

            } else {
              // maybe deleted and/or inserted lines
              StartA = LineA;
              StartB = LineB;

              while (LineA < DataA.Length && (LineB >= DataB.Length || DataA.modified[LineA]))
            // while (LineA < DataA.Length && DataA.modified[LineA])
            LineA++;

              while (LineB < DataB.Length && (LineA >= DataA.Length || DataB.modified[LineB]))
            // while (LineB < DataB.Length && DataB.modified[LineB])
            LineB++;

              if ((StartA < LineA) || (StartB < LineB)) {
            // store a new difference-item
            aItem = new Item();
            aItem.StartA = StartA;
            aItem.StartB = StartB;
            aItem.deletedA = LineA - StartA;
            aItem.insertedB = LineB - StartB;
            a.Add(aItem);
              } // if
            } // if
              } // while

              result = new Item[a.Count];
              a.CopyTo(result);

              return (result);
        }
Ejemplo n.º 3
0
        /// <summary>
        /// Find the difference in 2 text documents, comparing by textlines.
        /// The algorithm itself is comparing 2 arrays of numbers so when comparing 2 text documents
        /// each line is converted into a (hash) number. This hash-value is computed by storing all
        /// textlines into a common hashtable so i can find dublicates in there, and generating a 
        /// new number each time a new textline is inserted.
        /// </summary>
        /// <param name="TextA">A-version of the text (usualy the old one)</param>
        /// <param name="TextB">B-version of the text (usualy the new one)</param>
        /// <param name="trimSpace">When set to true, all leading and trailing whitespace characters are stripped out before the comparation is done.</param>
        /// <param name="ignoreSpace">When set to true, all whitespace characters are converted to a single space character before the comparation is done.</param>
        /// <param name="ignoreCase">When set to true, all characters are converted to their lowercase equivivalence before the comparation is done.</param>
        /// <returns>Returns a array of Items that describe the differences.</returns>
        public static Item[] DiffText(string TextA, string TextB, bool trimSpace, bool ignoreSpace, bool ignoreCase, params char[] separators)
        {
            // prepare the input-text and convert to comparable numbers.
              Hashtable h = new Hashtable(TextA.Length + TextB.Length);

              // The A-Version of the data (original data) to be compared.
              DiffData DataA = new DiffData(DiffCodes(TextA, h, trimSpace, ignoreSpace, ignoreCase, separators));

              // The B-Version of the data (modified data) to be compared.
              DiffData DataB = new DiffData(DiffCodes(TextB, h, trimSpace, ignoreSpace, ignoreCase, separators));

              h = null; // free up hashtable memory (maybe)

              LCS(DataA, 0, DataA.Length, DataB, 0, DataB.Length);
              return CreateDiffs(DataA, DataB);
        }
Ejemplo n.º 4
0
    } // DiffCodes


    /// <summary>
    /// This is the algorithm to find the Shortest Middle Snake (SMS).
    /// </summary>
    /// <param name="DataA">sequence A</param>
    /// <param name="LowerA">lower bound of the actual range in DataA</param>
    /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
    /// <param name="DataB">sequence B</param>
    /// <param name="LowerB">lower bound of the actual range in DataB</param>
    /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
    /// <returns>a MiddleSnakeData record containing x,y and u,v</returns>
    private static SMSRD SMS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB) {
      SMSRD ret;
      int MAX = DataA.Length + DataB.Length + 1;

      int DownK = LowerA - LowerB; // the k-line to start the forward search
      int UpK = UpperA - UpperB; // the k-line to start the reverse search

      int Delta = (UpperA - LowerA) - (UpperB - LowerB);
      bool oddDelta = (Delta & 1) != 0;

      /// vector for the (0,0) to (x,y) search
      int[] DownVector = new int[2* MAX + 2];

      /// vector for the (u,v) to (N,M) search
      int[] UpVector = new int[2 * MAX + 2];
      
      // The vectors in the publication accepts negative indexes. the vectors implemented here are 0-based
      // and are access using a specific offset: UpOffset UpVector and DownOffset for DownVektor
      int DownOffset = MAX - DownK;
      int UpOffset = MAX - UpK;
	
      int  MaxD = ((UpperA - LowerA + UpperB - LowerB) / 2) + 1;
		
      // Debug.Write(2, "SMS", String.Format("Search the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

      // init vectors
      DownVector[DownOffset + DownK + 1] = LowerA;
      UpVector[UpOffset + UpK - 1] = UpperA;
			
      for (int D = 0; D <= MaxD; D++) {

        // Extend the forward path.
        for (int k = DownK - D; k <= DownK + D; k += 2) {
          // Debug.Write(0, "SMS", "extend forward path " + k.ToString());

          // find the only or better starting point
          int x, y;
          if (k == DownK - D) {
            x = DownVector[DownOffset + k+1]; // down
          } else {
            x = DownVector[DownOffset + k-1] + 1; // a step to the right
            if ((k < DownK + D) && (DownVector[DownOffset + k+1] >= x))
              x = DownVector[DownOffset + k+1]; // down
          }
          y = x - k;

          // find the end of the furthest reaching forward D-path in diagonal k.
          while ((x < UpperA) && (y < UpperB) && (DataA.data[x] == DataB.data[y])) {
            x++; y++;
          }
          DownVector[DownOffset + k] = x;

          // overlap ?
          if (oddDelta && (UpK-D < k) && (k < UpK+D)) {
            if (UpVector[UpOffset + k] <= DownVector[DownOffset + k]) {
              ret.x = DownVector[DownOffset + k];
              ret.y = DownVector[DownOffset + k] - k;
              // ret.u = UpVector[UpOffset + k];      // 2002.09.20: no need for 2 points 
              // ret.v = UpVector[UpOffset + k] - k;
              return (ret);
            } // if
          } // if

        } // for k
				
        // Extend the reverse path.
        for (int k = UpK - D; k <= UpK + D; k += 2) {
          // Debug.Write(0, "SMS", "extend reverse path " + k.ToString());

          // find the only or better starting point
          int x, y;
          if (k == UpK + D) {
            x = UpVector[UpOffset + k-1]; // up
          } else {
            x = UpVector[UpOffset + k+1] - 1; // left
            if ((k > UpK - D) && (UpVector[UpOffset + k-1] < x))
              x = UpVector[UpOffset + k-1]; // up
          } // if
          y = x - k;

          while ((x > LowerA) && (y > LowerB) && (DataA.data[x-1] == DataB.data[y-1])) {
            x--; y--; // diagonal
          }
          UpVector[UpOffset + k] = x;

          // overlap ?
          if (! oddDelta && (DownK-D <= k) && (k <= DownK+D)) {
            if (UpVector[UpOffset + k] <= DownVector[DownOffset + k]) {
              ret.x = DownVector[DownOffset + k];
              ret.y = DownVector[DownOffset + k] - k;
              // ret.u = UpVector[UpOffset + k];     // 2002.09.20: no need for 2 points 
              // ret.v = UpVector[UpOffset + k] - k;
              return (ret);
            } // if
          } // if

        } // for k

      } // for D

      throw new ApplicationException("the algorithm should never come here.");
    } // SMS
Ejemplo n.º 5
0
        /// <summary>
        /// This is the algorithm to find the Shortest Middle Snake (SMS).
        /// </summary>
        /// <param name="DataA">sequence A</param>
        /// <param name="LowerA">lower bound of the actual range in DataA</param>
        /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
        /// <param name="DataB">sequence B</param>
        /// <param name="LowerB">lower bound of the actual range in DataB</param>
        /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
        /// <returns>a MiddleSnakeData record containing x,y and u,v</returns>
        private static SMSRD SMS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB)
        {
            SMSRD ret;
              int MAX = DataA.Length + DataB.Length + 1;

              int DownK = LowerA - LowerB; // the k-line to start the forward search
              int UpK = UpperA - UpperB; // the k-line to start the reverse search

              int Delta = (UpperA - LowerA) - (UpperB - LowerB);
              bool oddDelta = (Delta & 1) != 0;

              /// vector for the (0,0) to (x,y) search
              int[] DownVector = new int[2* MAX + 2];

              /// vector for the (u,v) to (N,M) search
              int[] UpVector = new int[2 * MAX + 2];

              // The vectors in the publication accepts negative indexes. the vectors implemented here are 0-based
              // and are access using a specific offset: UpOffset UpVector and DownOffset for DownVektor
              int DownOffset = MAX - DownK;
              int UpOffset = MAX - UpK;

              int  MaxD = ((UpperA - LowerA + UpperB - LowerB) / 2) + 1;

              // Debug.Write(2, "SMS", String.Format("Search the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

              // init vectors
              DownVector[DownOffset + DownK + 1] = LowerA;
              UpVector[UpOffset + UpK - 1] = UpperA;

              for (int D = 0; D <= MaxD; D++) {

            // Extend the forward path.
            for (int k = DownK - D; k <= DownK + D; k += 2) {
              // Debug.Write(0, "SMS", "extend forward path " + k.ToString());

              // find the only or better starting point
              int x, y;
              if (k == DownK - D) {
            x = DownVector[DownOffset + k+1]; // down
              } else {
            x = DownVector[DownOffset + k-1] + 1; // a step to the right
            if ((k < DownK + D) && (DownVector[DownOffset + k+1] >= x))
              x = DownVector[DownOffset + k+1]; // down
              }
              y = x - k;

              // find the end of the furthest reaching forward D-path in diagonal k.
              while ((x < UpperA) && (y < UpperB) && (DataA.data[x] == DataB.data[y])) {
            x++; y++;
              }
              DownVector[DownOffset + k] = x;

              // overlap ?
              if (oddDelta && (UpK-D < k) && (k < UpK+D)) {
            if (UpVector[UpOffset + k] <= DownVector[DownOffset + k]) {
              ret.x = DownVector[DownOffset + k];
              ret.y = DownVector[DownOffset + k] - k;
              // ret.u = UpVector[UpOffset + k];      // 2002.09.20: no need for 2 points
              // ret.v = UpVector[UpOffset + k] - k;
              return (ret);
            } // if
              } // if

            } // for k

            // Extend the reverse path.
            for (int k = UpK - D; k <= UpK + D; k += 2) {
              // Debug.Write(0, "SMS", "extend reverse path " + k.ToString());

              // find the only or better starting point
              int x, y;
              if (k == UpK + D) {
            x = UpVector[UpOffset + k-1]; // up
              } else {
            x = UpVector[UpOffset + k+1] - 1; // left
            if ((k > UpK - D) && (UpVector[UpOffset + k-1] < x))
              x = UpVector[UpOffset + k-1]; // up
              } // if
              y = x - k;

              while ((x > LowerA) && (y > LowerB) && (DataA.data[x-1] == DataB.data[y-1])) {
            x--; y--; // diagonal
              }
              UpVector[UpOffset + k] = x;

              // overlap ?
              if (! oddDelta && (DownK-D <= k) && (k <= DownK+D)) {
            if (UpVector[UpOffset + k] <= DownVector[DownOffset + k]) {
              ret.x = DownVector[DownOffset + k];
              ret.y = DownVector[DownOffset + k] - k;
              // ret.u = UpVector[UpOffset + k];     // 2002.09.20: no need for 2 points
              // ret.v = UpVector[UpOffset + k] - k;
              return (ret);
            } // if
              } // if

            } // for k

              } // for D

              throw new ApplicationException("the algorithm should never come here.");
        }
Ejemplo n.º 6
0
        /// <summary>
        /// This is the divide-and-conquer implementation of the longes common-subsequence (LCS) 
        /// algorithm.
        /// The published algorithm passes recursively parts of the A and B sequences.
        /// To avoid copying these arrays the lower and upper bounds are passed while the sequences stay constant.
        /// </summary>
        /// <param name="DataA">sequence A</param>
        /// <param name="LowerA">lower bound of the actual range in DataA</param>
        /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
        /// <param name="DataB">sequence B</param>
        /// <param name="LowerB">lower bound of the actual range in DataB</param>
        /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
        private static void LCS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB)
        {
            // Debug.Write(2, "LCS", String.Format("Analyse the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

              // Fast walkthrough equal lines at the start
              while (LowerA < UpperA && LowerB < UpperB && DataA.data[LowerA] == DataB.data[LowerB]) {
            LowerA++; LowerB++;
              }

              // Fast walkthrough equal lines at the end
              while (LowerA < UpperA && LowerB < UpperB && DataA.data[UpperA-1] == DataB.data[UpperB-1]) {
            --UpperA; --UpperB;
              }

              if (LowerA == UpperA) {
            // mark as inserted lines.
            while (LowerB < UpperB)
              DataB.modified[LowerB++] = true;

              } else if (LowerB == UpperB) {
            // mark as deleted lines.
            while (LowerA < UpperA)
              DataA.modified[LowerA++] = true;

              } else {
            // Find the middle snakea and length of an optimal path for A and B
            SMSRD smsrd = SMS(DataA, LowerA, UpperA, DataB, LowerB, UpperB);
            // Debug.Write(2, "MiddleSnakeData", String.Format("{0},{1}", smsrd.x, smsrd.y));

            // The path is from LowerX to (x,y) and (x,y) ot UpperX
            LCS(DataA, LowerA, smsrd.x, DataB, LowerB, smsrd.y);
            LCS(DataA, smsrd.x, UpperA, DataB, smsrd.y, UpperB);  // 2002.09.20: no need for 2 points
              }
        }