} // Optimize
/// <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);
int MAX = DataA.Length + DataB.Length + 1;
/// 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];
LCS(DataA, 0, DataA.Length, DataB, 0, DataB.Length, DownVector, UpVector);
return(CreateDiffs(DataA, DataB));
} // Diff
} // 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>
/// <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>
private static void LCS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB, int[] DownVector, int[] UpVector)
{
// 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, 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
}
} // LCS()
} // 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>
/// <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>
/// <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,
int[] DownVector, int[] UpVector)
{
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;
// 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