} // 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()
private static void LCS(ref DiffData DataA, int LowerA, int UpperA, ref DiffData DataB, int LowerB, int UpperB, int[] DownVector, int[] UpVector) { // Fast walkthrough equal lines at the start while ((LowerA < UpperA && LowerB < UpperB && DataA.data[LowerA] == DataB.data[LowerB])) { LowerA += 1; LowerB += 1; } // Fast walkthrough equal lines at the end while ((LowerA < UpperA && LowerB < UpperB && DataA.data[UpperA - 1] == DataB.data[UpperB - 1])) { UpperA -= 1; UpperB -= 1; } if ((LowerA == UpperA)) { // mark as inserted lines. while ((LowerB < UpperB)) { //LowerB += 1 DataB.modified[LowerB] = true; LowerB += 1; } } else if ((LowerB == UpperB)) { // mark as deleted lines. while ((LowerA < UpperA)) { //LowerA += 1 DataA.modified[LowerA] = true; LowerA += 1; } } else { // Find the middle snakea and length of an optimal path for A and B SMSRD smsrd = SMS(ref DataA, LowerA, UpperA, ref DataB, LowerB, UpperB, DownVector, UpVector); // Debug.Write(2, "MiddleSnakeData", String.Format("{0},{1}", smsrd.x, smsrd.y)); // The path is from LowerX to (x,y) and (x,y) to UpperX LCS(ref DataA, LowerA, smsrd.x, ref DataB, LowerB, smsrd.y, DownVector, UpVector); LCS(ref DataA, smsrd.x, UpperA, ref DataB, smsrd.y, UpperB, DownVector, UpVector); // 2002.09.20: no need for 2 points } }
// SMS /// <summary> /// This is the divide-and-conquer implementation of the longest 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> /// <param name="downVector">a vector for the (0,0) to (x,y) search. Passed as a parameter for speed reasons.</param> /// <param name="upVector">a vector for the (u,v) to (N,M) search. Passed as a parameter for speed reasons.</param> static void LCS <T> (DiffData <T> dataA, int lowerA, int upperA, DiffData <T> dataB, int lowerB, int upperB, int[] downVector, int[] upVector) { // Fast walkthrough equal lines at the start while (lowerA < upperA && lowerB < upperB && dataA.Data[lowerA].Equals(dataB.Data[lowerB])) { lowerA++; lowerB++; } // Fast walkthrough equal lines at the end while (lowerA < upperA && lowerB < upperB && dataA.Data[upperA - 1].Equals(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, downVector, upVector); // Debug.Write(2, "MiddleSnakeData", String.Format("{0},{1}", smsrd.x, smsrd.y)); // The path is from LowerX to (x,y) and (x,y) to UpperX LCS(dataA, lowerA, smsrd.x, dataB, lowerB, smsrd.y, downVector, upVector); LCS(dataA, smsrd.x, upperA, dataB, smsrd.y, upperB, downVector, upVector); // 2002.09.20: no need for 2 points } }
private static SMSRD SMS(ref DiffData DataA, int LowerA, int UpperA, ref DiffData DataB, int LowerB, int UpperB, int[] DownVector, int[] UpVector) { SMSRD ret = new SMSRD(); 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; // 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 = 0; int y = 0; 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 = x + 1; y = y + 1; } DownVector[DownOffset + k] = x; // overlap ? if ((oddDelta == true && (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); } } } // 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 = 0; int y = 0; 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 } } y = x - k; while (((x > LowerA) && (y > LowerB) && (DataA.data[x - 1] == DataB.data[y - 1]))) { x = x - 1; y = y - 1; // diagonal } UpVector[UpOffset + k] = x; // overlap ? if ((oddDelta == false && (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); } } } } throw new ApplicationException("the algorithm should never come here."); }