/// <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. var dataA = new DiffData(arrayA); // The B-Version of the data (modified data) to be compared. var dataB = new DiffData(arrayB); Lcs(dataA, 0, dataA.Length, dataB, 0, dataB.Length); return CreateDiffs(dataA, dataB); }
} // 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 } } // LCS()
/// <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 var downVector = new int[2 * max + 2]; // vector for the (u,v) to (N,M) search var 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."); }
/// <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 } }
/// <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) { var 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); }
/// <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. var h = new Hashtable(textA.Length + textB.Length); // The A-Version of the data (original data) to be compared. var dataA = new DiffData(DiffCodes(textA, h, trimSpace, ignoreSpace, ignoreCase)); // The B-Version of the data (modified data) to be compared. var dataB = new DiffData(DiffCodes(textB, h, trimSpace, ignoreSpace, ignoreCase)); Lcs(dataA, 0, dataA.Length, dataB, 0, dataB.Length); return CreateDiffs(dataA, dataB); }
} // 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 var downVector = new int[2 * max + 2]; // vector for the (u,v) to (N,M) search var 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