/// <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
