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187 lines
5.7 KiB
Fortran
187 lines
5.7 KiB
Fortran
C
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C Assumption: the triangular matrix has the diagonal element in the
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C "right" place, i.e. the last in its row for Lower and the first
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C for Upper.
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C
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SUBROUTINE ZCOOSV (UPLO,TRANS,DIAG,N,AS,IA,JA,INFOA,B,X,IERROR)
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COMPLEX*16 ZERO
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PARAMETER (ZERO=(0.0D0,0.0D0))
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LOGICAL DEBUG
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PARAMETER (DEBUG=.FALSE.)
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INTEGER N,IERROR
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CHARACTER DIAG, TRANS, UPLO
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COMPLEX*16 AS(*), B(*), X(*)
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INTEGER IA(*), JA(*),INFOA(*)
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COMPLEX*16 ACC
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INTEGER I, J, K, NNZ, II, JJ
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LOGICAL LOW, TRA, UNI
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if (debug) write(*,*) 'ZCOOSV ',n
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if (debug) write(*,*) 'ZCOOSV ',n,nnz,diag,trans,uplo
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UNI = (DIAG.EQ.'U')
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TRA = (TRANS.EQ.'T')
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LOW = (UPLO.EQ.'L')
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NNZ = INFOA(1)
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if (debug) write(*,*) 'ZCOOSV ',n,nnz,uni,tra,low,ia(1),ja(1)
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IERROR = 0
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if (debug) write(*,*) 'ZCOOSV ierror ',ierror
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IF ( .NOT. TRA) THEN
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if (debug) write(*,*) 'ZCOOSV NOT TRA'
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IF (LOW) THEN
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if (debug) write(*,*) 'ZCOOSV LOW'
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IF ( .NOT. UNI) THEN
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if (debug) write(*,*) 'ZCOOSV NOT UNI'
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I = 1
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J = I
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DO WHILE (I.LE.NNZ)
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DO WHILE ((J.LE.NNZ).AND.(IA(J).EQ.IA(I)))
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J = J+1
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ENDDO
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ACC = ZERO
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IR = IA(I)
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DO K = I, J-2
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JC = JA(K)
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ACC = ACC + AS(K)*X(JC)
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ENDDO
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X(IR) = (B(IR)-ACC)/AS(J-1)
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I = J
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ENDDO
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ELSE IF (UNI) THEN
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C
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C Bug warning: if UNI, some rows might be empty
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C
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I = 1
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if (debug) write(*,*) 'ZCOOSV UNILOW ',
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+ i,n,nnz,uni,tra,low
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DO II = 1, N
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if (debug) write(*,*) 'Loop1 COOSV',i,j,ii,x(ii),b(ii)
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DO WHILE ((I.LE.NNZ).AND.(IA(I).LT.II))
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I = I + 1
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c$$$ if (debug) write(*,*) 'Loop2 COOSV',i,ia(i),ii
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ENDDO
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ACC = ZERO
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IF ((I.LE.NNZ).AND.(IA(I).EQ.II)) THEN
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J = I + 1
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DO WHILE ((J.LE.NNZ).AND.(IA(J).EQ.IA(I)))
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J = J+1
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ENDDO
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DO K = I, J-1
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JC = JA(K)
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ACC = ACC + AS(K)*X(JC)
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ENDDO
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ELSE
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J = I
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ENDIF
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X(II) = (B(II)-ACC)
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if (debug) write(*,*) 'Loop COOSV',i,j,ii,x(ii),b(ii)
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I = J
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ENDDO
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END IF
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ELSE IF ( .NOT. LOW) THEN
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if (debug) write(*,*) 'ZCOOSV NOT LOW'
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IF ( .NOT. UNI) THEN
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if (debug) write(*,*) 'ZCOOSV NOT UNI'
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I = NNZ
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J = NNZ
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DO WHILE (I.GT.0)
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DO WHILE ((J.GT.0).AND.(IA(J).EQ.IA(I)))
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J = J-1
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ENDDO
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ACC = ZERO
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IR = IA(I)
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DO K = I, J+2,-1
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JC = JA(K)
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ACC = ACC + AS(K)*X(JC)
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ENDDO
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X(IR) = (B(IR)-ACC)/AS(J+1)
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I = J
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ENDDO
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ELSE IF (UNI) THEN
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if (debug) write(*,*) 'ZCOOSV UNI'
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I = NNZ
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DO II = N,1,-1
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DO WHILE ((I.GT.0).AND.(IA(I).GT.II))
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I = I -1
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ENDDO
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ACC = ZERO
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IF ((I.GT.0).AND.(IA(I).EQ.II)) THEN
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J = I - 1
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DO WHILE ((J.GT.0).AND.(IA(J).EQ.IA(I)))
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J = J-1
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ENDDO
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DO K = I, J+1, -1
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JC = JA(K)
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ACC = ACC + AS(K)*X(JC)
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ENDDO
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ELSE
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J = I
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ENDIF
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X(II) = (B(II)-ACC)
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if (debug) write(*,*) 'Loop COOSV',i,j,ii,x(ii),b(ii)
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I = J
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ENDDO
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END IF
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END IF
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ELSE IF (TRA) THEN
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IERROR = 3010
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return
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CCCCCCCCCCCCCCCC
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C
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C TBF
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C
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CCCCCCCCCCCCCCCC
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DO 180 I = 1, N
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X(I) = B(I)
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180 CONTINUE
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IF (LOW) THEN
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IF ( .NOT. UNI) THEN
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DO 220 I = N, 1, -1
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X(I) = X(I)/AS(IA(I+1)-1)
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ACC = X(I)
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DO 200 J = IA(I), IA(I+1) - 2
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K = JA(J)
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X(K) = X(K) - AS(J)*ACC
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200 CONTINUE
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220 CONTINUE
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ELSE IF (UNI) THEN
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DO 260 I = N, 1, -1
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ACC = X(I)
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DO 240 J = IA(I), IA(I+1) - 1
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K = JA(J)
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X(K) = X(K) - AS(J)*ACC
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240 CONTINUE
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260 CONTINUE
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END IF
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ELSE IF ( .NOT. LOW) THEN
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IF ( .NOT. UNI) THEN
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DO 300 I = 1, N
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X(I) = X(I)/AS(IA(I))
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ACC = X(I)
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DO 280 J = IA(I) + 1, IA(I+1) - 1
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K = JA(J)
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X(K) = X(K) - AS(J)*ACC
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280 CONTINUE
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300 CONTINUE
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ELSE IF (UNI) THEN
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DO 340 I = 1, N
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ACC = X(I)
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DO 320 J = IA(I), IA(I+1) - 1
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K = JA(J)
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X(K) = X(K) - AS(J)*ACC
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320 CONTINUE
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340 CONTINUE
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END IF
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END IF
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END IF
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RETURN
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END
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