Sub Ex_Dqgmres_r()
Const N = 3, Nnz = N * N, Tol = 0.0000000001 '1.0e-10
Dim A(Nnz - 1) As Double, Ia(N) As Long, Ja(Nnz - 1) As Long
Dim B(N - 1) As Double, X(N - 1) As Double
Dim XX(N - 1) As Double, YY(N - 1) As Double
Dim Iter As Long, Res As Double, IRev As Long, Info As Long, I As Long
A(0) = 0.2: A(1) = -0.11: A(2) = -0.93: A(3) = -0.32: A(4) = 0.81: A(5) = 0.37: A(6) = -0.8: A(7) = -0.92: A(8) = -0.29
Ia(0) = 0: Ia(1) = 3: Ia(2) = 6: Ia(3) = 9
Ja(0) = 0: Ja(1) = 1: Ja(2) = 2: Ja(3) = 0: Ja(4) = 1: Ja(5) = 2: Ja(6) = 0: Ja(7) = 1: Ja(8) = 2
B(0) = -0.3727: B(1) = 0.4319: B(2) = -1.4247
IRev = 0
Do
Call
Dqgmres_r(N, B(), X(), Info, XX(), YY(), IRev, Iter, Res)
If IRev = 1 Then '- Matvec
Call
CsrDusmv("N", N, N, 1, A(), Ia(), Ja(), XX(), 0, YY())
ElseIf IRev = 3 Then '- Psolve
For I = 0 To N - 1
XX(I) = YY(I)
Next
ElseIf IRev = 10 Then '- Check convergence
If Res < Tol Then IRev = 11
End If
Loop While IRev <> 0
Debug.Print "X =", X(0), X(1), X(2)
Debug.Print "Iter = " + CStr(Iter) + ", Res = " + CStr(Res) + ", Info = " + CStr(Info)
End Sub
Sub CsrDusmv(Trans As String, M As Long, N As Long, Alpha As Double, Val() As Double, Rowptr() As Long, Colind() As Long, X() As Double, Beta As Double, Y() As Double, Optional Info As Long, Optional Base As Long=-1, Optional IncX As Long=1, Optional IncY As Long=1)
y <- αAx + βy または y <- αATx + βy (CSR)
Sub Dqgmres_r(N As Long, B() As Double, X() As Double, Info As Long, XX() As Double, YY() As Double, IRev As Long, Optional Iter As Long, Optional Res As Double, Optional M As Long=0, Optional MaxIter As Long=500)
疑似最小残差(DQGMRES)法による連立一次方程式 Ax = b の解 (リバースコミュニケーション版)
1.9.7