} // 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.");
}
