' Quote from
'Visual Basic 2005 Cookbook Solutions for VB 2005 Programmers
'by Tim Patrick (Author), John Craig (Author)
'# Publisher: O'Reilly Media, Inc. (September 21, 2006)
'# Language: English
'# ISBN-10: 0596101775
'# ISBN-13: 978-0596101770
Public Class Tester
Public Shared Sub Main
Dim matrixA(,) As Double = { _
{4, 5, 6}, _
{7, 8, 9}, _
{3, 2, 1}}
Console.WriteLine(MatrixHelper.MakeDisplayable(matrixA))
End Sub
End Class
Module MatrixHelper
Public Function MakeDisplayable(ByVal sourceMatrix(,) As Double) As String
' ----- Prepare a multi-line string that shows the contents
' of a matrix, a 2D array.
Dim rows As Integer
Dim cols As Integer
Dim eachRow As Integer
Dim eachCol As Integer
Dim result As New System.Text.StringBuilder
' ----- Process all rows of the matrix, generating one
' output line per row.
rows = UBound(sourceMatrix, 1) + 1
cols = UBound(sourceMatrix, 2) + 1
For eachRow = 0 To rows - 1
' ----- Process each column of the matrix on a single
' row, separating values by commas.
If (eachRow > 0) Then result.AppendLine()
For eachCol = 0 To cols - 1
' ----- Add a single matrix element to the output.
If (eachCol > 0) Then result.Append(",")
result.Append(sourceMatrix(eachRow, eachCol).ToString)
Next eachCol
Next eachRow
' ----- Finished.
Return result.ToString
End Function
Public Function MakeDisplayable(ByVal sourceArray() As Double) As String
' ----- Present an array as multiple lines of output.
Dim result As New System.Text.StringBuilder
Dim scanValue As Double
For Each scanValue In sourceArray
result.AppendLine(scanValue.ToString)
Next scanValue
Return result.ToString
End Function
Public Function Inverse(ByVal sourceMatrix(,) As Double) As Double(,)
' ----- Build a new matrix that is the mathematical inverse
' of the supplied matrix. Multiplying a matrix and its
' inverse together will give the identity matrix.
Dim eachCol As Integer
Dim eachRow As Integer
Dim rowsAndCols As Integer
' ----- Determine the size of each dimension of the matrix.
' Only square matrices can be inverted.
If (UBound(sourceMatrix, 1) <> UBound(sourceMatrix, 2)) Then
Throw New Exception("Matrix must be square.")
End If
Dim rank As Integer = UBound(sourceMatrix, 1)
' ----- Clone a copy of the matrix (not just a new reference).
Dim workMatrix(,) As Double = _
CType(sourceMatrix.Clone, Double(,))
' ----- Variables used for backsolving.
Dim destMatrix(rank, rank) As Double
Dim rightHandSide(rank) As Double
Dim solutions(rank) As Double
Dim rowPivots(rank) As Integer
Dim colPivots(rank) As Integer
' ----- Use LU decomposition to form a triangular matrix.
workMatrix = FormLU(workMatrix, rowPivots, colPivots, rowsAndCols)
' ----- Backsolve the triangular matrix to get the inverted
' value for each position in the final matrix.
For eachCol = 0 To rank
rightHandSide(eachCol) = 1
BackSolve(workMatrix, rightHandSide, solutions, rowPivots, colPivots)
For eachRow = 0 To rank
destMatrix(eachRow, eachCol) = solutions(eachRow)
rightHandSide(eachRow) = 0
Next eachRow
Next eachCol
' ----- Return the inverted matrix result.
Return destMatrix
End Function
Public Function Determinant(ByVal sourceMatrix(,) As Double) As Double
' ----- Calculate the determinant of a matrix.
Dim result As Double
Dim pivots As Integer
Dim count As Integer
' ----- Only calculate the determinants of square matrices.
If (UBound(sourceMatrix, 1) <> UBound(sourceMatrix, 2)) Then
Throw New Exception( _
"Determinant only calculated for square matrices.")
End If
Dim rank As Integer = UBound(sourceMatrix, 1)
' ----- Make a copy of the matrix so we can work inside of it.
Dim workMatrix(rank, rank) As Double
Array.Copy(sourceMatrix, workMatrix, sourceMatrix.Length)
' ----- Use LU decomposition to form a triangular matrix.
Dim rowPivots(rank) As Integer
Dim colPivots(rank) As Integer
workMatrix = FormLU(workMatrix, rowPivots, colPivots, count)
' ----- Get the product at each of the pivot points.
result = 1
For pivots = 0 To rank
result *= workMatrix(rowPivots(pivots), colPivots(pivots))
Next pivots
' ----- Determine the sign of the result using LaPlace's formula.
result = (-1) ^ count * result
Return result
End Function
Public Function SimultEq(ByVal sourceEquations(,) As Double, _
ByVal sourceRHS() As Double) As Double()
' ----- Use matrices to solve simultaneous equations.
Dim rowsAndCols As Integer
' ----- The matrix must be square and the array size must match.
Dim rank As Integer = UBound(sourceEquations, 1)
If (UBound(sourceEquations, 2) <> rank) Or _
(UBound(sourceRHS, 1) <> rank) Then
Throw New Exception("Size problem for simultaneous equations.")
End If
' ----- Create some arrays for doing all of the work.
Dim coefficientMatrix(rank, rank) As Double
Dim rightHandSide(rank) As Double
Dim solutions(rank) As Double
Dim rowPivots(rank) As Integer
Dim colPivots(rank) As Integer
' ----- Make copies of the original matrices so we don't
' mess them up.
Array.Copy(sourceEquations, coefficientMatrix, sourceEquations.Length)
Array.Copy(sourceRHS, rightHandSide, sourceRHS.Length)
' ----- Use LU decomposition to form a triangular matrix.
coefficientMatrix = FormLU(coefficientMatrix, rowPivots, _
colPivots, rowsAndCols)
' ----- Find the unique solution for the upper-triangle.
BackSolve(coefficientMatrix, rightHandSide, solutions, _
rowPivots, colPivots)
' ----- Return the simultaneous equations result in an array.
Return solutions
End Function
Private Function FormLU(ByVal sourceMatrix(,) As Double, _
ByRef rowPivots() As Integer, ByRef colPivots() As Integer, _
ByRef rowsAndCols As Integer) As Double(,)
' ----- Perform an LU (lower and upper) decomposition of a matrix,
' a modified form of Gaussian elimination.
Dim eachRow As Integer
Dim eachCol As Integer
Dim pivot As Integer
Dim rowIndex As Integer
Dim colIndex As Integer
Dim bestRow As Integer
Dim bestCol As Integer
Dim rowToPivot As Integer
Dim colToPivot As Integer
Dim maxValue As Double
Dim testValue As Double
Dim oldMax As Double
Const Deps As Double = 0.0000000000000001
' ----- Determine the size of the array.
Dim rank As Integer = UBound(sourceMatrix, 1)
Dim destMatrix(rank, rank) As Double
Dim rowNorm(rank) As Double
ReDim rowPivots(rank)
ReDim colPivots(rank)
' ----- Make a copy of the array so we don't mess it up.
Array.Copy(sourceMatrix, destMatrix, sourceMatrix.Length)
' ----- Initialize row and column pivot arrays.
For eachRow = 0 To rank
rowPivots(eachRow) = eachRow
colPivots(eachRow) = eachRow
For eachCol = 0 To rank
rowNorm(eachRow) += Math.Abs(destMatrix(eachRow, eachCol))
Next eachCol
If (rowNorm(eachRow) = 0) Then
Throw New Exception("Cannot invert a singular matrix.")
End If
Next eachRow
' ----- Use Gauss-Jordan elimination on the matrix rows.
For pivot = 0 To rank - 1
maxValue = 0
For eachRow = pivot To rank
rowIndex = rowPivots(eachRow)
For eachCol = pivot To rank
colIndex = colPivots(eachCol)
testValue = Math.Abs(destMatrix(rowIndex, colIndex)) _
/ rowNorm(rowIndex)
If (testValue > maxValue) Then
maxValue = testValue
bestRow = eachRow
bestCol = eachCol
End If
Next eachCol
Next eachRow
' ----- Detect a singular, or very nearly singular, matrix.
If (maxValue = 0) Then
Throw New Exception("Singular matrix used for LU.")
ElseIf (pivot > 1) Then
If (maxValue < (Deps * oldMax)) Then
Throw New Exception("Non-invertible matrix used for LU.")
End If
End If
oldMax = maxValue
' ----- Swap row pivot values for the best row.
If (rowPivots(pivot) <> rowPivots(bestRow)) Then
rowsAndCols += 1
Swap(rowPivots(pivot), rowPivots(bestRow))
End If
' ----- Swap column pivot values for the best column.
If (colPivots(pivot) <> colPivots(bestCol)) Then
rowsAndCols += 1
Swap(colPivots(pivot), colPivots(bestCol))
End If
' ----- Work with the current pivot points.
rowToPivot = rowPivots(pivot)
colToPivot = colPivots(pivot)
' ----- Modify the remaining rows from the pivot points.
For eachRow = (pivot + 1) To rank
rowIndex = rowPivots(eachRow)
destMatrix(rowIndex, colToPivot) = _
-destMatrix(rowIndex, colToPivot) / _
destMatrix(rowToPivot, colToPivot)
For eachCol = (pivot + 1) To rank
colIndex = colPivots(eachCol)
destMatrix(rowIndex, colIndex) += _
destMatrix(rowIndex, colToPivot) * _
destMatrix(rowToPivot, colIndex)
Next eachCol
Next eachRow
Next pivot
' ----- Detect a non-invertible matrix.
If (destMatrix(rowPivots(rank), colPivots(rank)) = 0) Then
Throw New Exception("Non-invertible matrix used for LU.")
ElseIf (Math.Abs(destMatrix(rowPivots(rank), colPivots(rank))) / _
rowNorm(rowPivots(rank))) < (Deps * oldMax) Then
Throw New Exception("Non-invertible matrix used for LU.")
End If
' ----- Success. Return the LU triangular matrix.
Return destMatrix
End Function
Private Sub Swap(ByRef firstValue As Integer, ByRef secondValue As Integer)
' ----- Reverse the values of two reference integers.
Dim holdValue As Integer
holdValue = firstValue
firstValue = secondValue
secondValue = holdValue
End Sub
Private Sub BackSolve(ByVal sourceMatrix(,) As Double, _
ByVal rightHandSide() As Double, ByVal solutions() As Double, _
ByRef rowPivots() As Integer, ByRef colPivots() As Integer)
' ----- Solve an upper-right-triangle matrix.
Dim pivot As Integer
Dim rowToPivot As Integer
Dim colToPivot As Integer
Dim eachRow As Integer
Dim eachCol As Integer
Dim rank As Integer = UBound(sourceMatrix, 1)
' ----- Work through all pivot points. This section builds
' the "B" in the AX=B formula.
For pivot = 0 To (rank - 1)
colToPivot = colPivots(pivot)
For eachRow = (pivot + 1) To rank
rowToPivot = rowPivots(eachRow)
rightHandSide(rowToPivot) += _
sourceMatrix(rowToPivot, colToPivot) _
* rightHandSide(rowPivots(pivot))
Next eachRow
Next pivot
' ----- Now solve for each X using the general formula
' x(i) = (b(i) - summation(a(i,j)x(j)))/a(i,i)
For eachRow = rank To 0 Step -1
colToPivot = colPivots(eachRow)
rowToPivot = rowPivots(eachRow)
solutions(colToPivot) = rightHandSide(rowToPivot)
For eachCol = (eachRow + 1) To rank
solutions(colToPivot) -= _
sourceMatrix(rowToPivot, colPivots(eachCol)) _
* solutions(colPivots(eachCol))
Next eachCol
solutions(colToPivot) /= sourceMatrix(rowToPivot, colToPivot)
Next eachRow
End Sub
End Module
4,5,6
7,8,9
3,2,1