Data on one input file being compared.
예제 #1
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.");
        }
예제 #2
0
		// LCS()


		/// <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);
		}
예제 #3
0
		// SMS


		/// <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 
			}
		}
예제 #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.");
		}
예제 #5
0
		// DiffText


		/// <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);
		}
예제 #6
0
		// DiffText


		/// <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)
		{
			// 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));

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

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

			LCS(DataA, 0, DataA.Length, DataB, 0, DataB.Length);
			return CreateDiffs(DataA, DataB);
		}