SUBROUTINE CGEFA(A,LDA,N,IPVT,INFO) INTEGER LDA,N,IPVT(*),INFO COMPLEX A(LDA,*) C C CGEFA FACTORS A COMPLEX MATRIX BY GAUSSIAN ELIMINATION. C C CGEFA IS USUALLY CALLED BY CGECO, BUT IT CAN BE CALLED C DIRECTLY WITH A SAVING IN TIME IF RCOND IS NOT NEEDED. C (TIME FOR CGECO) = (1 + 9/N)*(TIME FOR CGEFA) . C C ON ENTRY C C A COMPLEX(LDA, N) C THE MATRIX TO BE FACTORED. C C LDA INTEGER C THE LEADING DIMENSION OF THE ARRAY A . C C N INTEGER C THE ORDER OF THE MATRIX A . C C ON RETURN C C A AN UPPER TRIANGULAR MATRIX AND THE MULTIPLIERS C WHICH WERE USED TO OBTAIN IT. C THE FACTORIZATION CAN BE WRITTEN A = L*U WHERE C L IS A PRODUCT OF PERMUTATION AND UNIT LOWER C TRIANGULAR MATRICES AND U IS UPPER TRIANGULAR. C C IPVT INTEGER(N) C AN INTEGER VECTOR OF PIVOT INDICES. C C INFO INTEGER C = 0 NORMAL VALUE. C = K IF U(K,K) .EQ. 0.0 . THIS IS NOT AN ERROR C CONDITION FOR THIS SUBROUTINE, BUT IT DOES C INDICATE THAT CGESL OR CGEDI WILL DIVIDE BY ZERO C IF CALLED. USE RCOND IN CGECO FOR A RELIABLE C INDICATION OF SINGULARITY. C C LINPACK. THIS VERSION DATED 08/14/78 . C CLEVE MOLER, UNIVERSITY OF NEW MEXICO, ARGONNE NATIONAL LAB. C C SUBROUTINES AND FUNCTIONS C C BLAS CAXPY,CSCAL,ICAMAX C FORTRAN ABS,AIMAG,REAL C C INTERNAL VARIABLES C COMPLEX T INTEGER ICAMAX,J,K,KP1,L,NM1 C COMPLEX ZDUM REAL CABS1 CABS1(ZDUM) = ABS(REAL(ZDUM)) + ABS(AIMAG(ZDUM)) C C GAUSSIAN ELIMINATION WITH PARTIAL PIVOTING C INFO = 0 NM1 = N - 1 IF (NM1 .LT. 1) GO TO 70 DO 60 K = 1, NM1 KP1 = K + 1 C C FIND L = PIVOT INDEX C L = ICAMAX(N-K+1,A(K,K),1) + K - 1 IPVT(K) = L C C ZERO PIVOT IMPLIES THIS COLUMN ALREADY TRIANGULARIZED C IF (CABS1(A(L,K)) .EQ. 0.0E0) GO TO 40 C C INTERCHANGE IF NECESSARY C IF (L .EQ. K) GO TO 10 T = A(L,K) A(L,K) = A(K,K) A(K,K) = T 10 CONTINUE C C COMPUTE MULTIPLIERS C T = -(1.0E0,0.0E0)/A(K,K) CALL CSCAL(N-K,T,A(K+1,K),1) C C ROW ELIMINATION WITH COLUMN INDEXING C DO 30 J = KP1, N T = A(L,J) IF (L .EQ. K) GO TO 20 A(L,J) = A(K,J) A(K,J) = T 20 CONTINUE CALL CAXPY(N-K,T,A(K+1,K),1,A(K+1,J),1) 30 CONTINUE GO TO 50 40 CONTINUE INFO = K 50 CONTINUE 60 CONTINUE 70 CONTINUE IPVT(N) = N IF (CABS1(A(N,N)) .EQ. 0.0E0) INFO = N RETURN END SUBROUTINE CPOFA(A,LDA,N,INFO) INTEGER LDA,N,INFO COMPLEX A(LDA,*) C C CPOFA FACTORS A COMPLEX HERMITIAN POSITIVE DEFINITE MATRIX. C C CPOFA IS USUALLY CALLED BY CPOCO, BUT IT CAN BE CALLED C DIRECTLY WITH A SAVING IN TIME IF RCOND IS NOT NEEDED. C (TIME FOR CPOCO) = (1 + 18/N)*(TIME FOR CPOFA) . C C ON ENTRY C C A COMPLEX(LDA, N) C THE HERMITIAN MATRIX TO BE FACTORED. ONLY THE C DIAGONAL AND UPPER TRIANGLE ARE USED. C C LDA INTEGER C THE LEADING DIMENSION OF THE ARRAY A . C C N INTEGER C THE ORDER OF THE MATRIX A . C C ON RETURN C C A AN UPPER TRIANGULAR MATRIX R SO THAT A = C CTRANS(R)*R WHERE CTRANS(R) IS THE CONJUGATE C TRANSPOSE. THE STRICT LOWER TRIANGLE IS UNALTERED. C IF INFO .NE. 0 , THE FACTORIZATION IS NOT COMPLETE. C C INFO INTEGER C = 0 FOR NORMAL RETURN. C = K SIGNALS AN ERROR CONDITION. THE LEADING MINOR C OF ORDER K IS NOT POSITIVE DEFINITE. C C LINPACK. THIS VERSION DATED 08/14/78 . C CLEVE MOLER, UNIVERSITY OF NEW MEXICO, ARGONNE NATIONAL LAB. C C SUBROUTINES AND FUNCTIONS C C BLAS CDOTC C FORTRAN AIMAG,CMPLX,CONJG,REAL,SQRT C C INTERNAL VARIABLES C COMPLEX CDOTC,T REAL S INTEGER J,JM1,K C BEGIN BLOCK WITH ...EXITS TO 40 C C DO 30 J = 1, N INFO = J S = 0.0E0 JM1 = J - 1 IF (JM1 .LT. 1) GO TO 20 DO 10 K = 1, JM1 T = A(K,J) - CDOTC(K-1,A(1,K),1,A(1,J),1) T = T/A(K,K) A(K,J) = T S = S + REAL(T*CONJG(T)) 10 CONTINUE 20 CONTINUE S = REAL(A(J,J)) - S C ......EXIT IF (S .LE. 0.0E0 .OR. AIMAG(A(J,J)) .NE. 0.0E0) GO TO 40 A(J,J) = CMPLX(SQRT(S),0.0E0) 30 CONTINUE INFO = 0 40 CONTINUE RETURN END SUBROUTINE CGTSL(N,C,D,E,B,INFO) INTEGER N,INFO COMPLEX C(*),D(*),E(*),B(*) C C CGTSL GIVEN A GENERAL TRIDIAGONAL MATRIX AND A RIGHT HAND C SIDE WILL FIND THE SOLUTION. C C ON ENTRY C C N INTEGER C IS THE ORDER OF THE TRIDIAGONAL MATRIX. C C C COMPLEX(N) C IS THE SUBDIAGONAL OF THE TRIDIAGONAL MATRIX. C C(2) THROUGH C(N) SHOULD CONTAIN THE SUBDIAGONAL. C ON OUTPUT C IS DESTROYED. C C D COMPLEX(N) C IS THE DIAGONAL OF THE TRIDIAGONAL MATRIX. C ON OUTPUT D IS DESTROYED. C C E COMPLEX(N) C IS THE SUPERDIAGONAL OF THE TRIDIAGONAL MATRIX. C E(1) THROUGH E(N-1) SHOULD CONTAIN THE SUPERDIAGONAL. C ON OUTPUT E IS DESTROYED. C C B COMPLEX(N) C IS THE RIGHT HAND SIDE VECTOR. C C ON RETURN C C B IS THE SOLUTION VECTOR. C C INFO INTEGER C = 0 NORMAL VALUE. C = K IF THE K-TH ELEMENT OF THE DIAGONAL BECOMES C EXACTLY ZERO. THE SUBROUTINE RETURNS WHEN C THIS IS DETECTED. C C LINPACK. THIS VERSION DATED 08/14/78 . C JACK DONGARRA, ARGONNE NATIONAL LABORATORY. C C NO EXTERNALS C FORTRAN ABS,AIMAG,REAL C C INTERNAL VARIABLES C INTEGER K,KB,KP1,NM1,NM2 COMPLEX T COMPLEX ZDUM REAL CABS1 CABS1(ZDUM) = ABS(REAL(ZDUM)) + ABS(AIMAG(ZDUM)) C BEGIN BLOCK PERMITTING ...EXITS TO 100 C INFO = 0 C(1) = D(1) NM1 = N - 1 IF (NM1 .LT. 1) GO TO 40 D(1) = E(1) E(1) = (0.0E0,0.0E0) E(N) = (0.0E0,0.0E0) C DO 30 K = 1, NM1 KP1 = K + 1 C C FIND THE LARGEST OF THE TWO ROWS C IF (CABS1(C(KP1)) .LT. CABS1(C(K))) GO TO 10 C C INTERCHANGE ROW C T = C(KP1) C(KP1) = C(K) C(K) = T T = D(KP1) D(KP1) = D(K) D(K) = T T = E(KP1) E(KP1) = E(K) E(K) = T T = B(KP1) B(KP1) = B(K) B(K) = T 10 CONTINUE C C ZERO ELEMENTS C IF (CABS1(C(K)) .NE. 0.0E0) GO TO 20 INFO = K C ............EXIT GO TO 100 20 CONTINUE T = -C(KP1)/C(K) C(KP1) = D(KP1) + T*D(K) D(KP1) = E(KP1) + T*E(K) E(KP1) = (0.0E0,0.0E0) B(KP1) = B(KP1) + T*B(K) 30 CONTINUE 40 CONTINUE IF (CABS1(C(N)) .NE. 0.0E0) GO TO 50 INFO = N GO TO 90 50 CONTINUE C C BACK SOLVE C NM2 = N - 2 B(N) = B(N)/C(N) IF (N .EQ. 1) GO TO 80 B(NM1) = (B(NM1) - D(NM1)*B(N))/C(NM1) IF (NM2 .LT. 1) GO TO 70 DO 60 KB = 1, NM2 K = NM2 - KB + 1 B(K) = (B(K) - D(K)*B(K+1) - E(K)*B(K+2))/C(K) 60 CONTINUE 70 CONTINUE 80 CONTINUE 90 CONTINUE 100 CONTINUE C RETURN END SUBROUTINE CPTSL(N,D,E,B) INTEGER N COMPLEX D(*),E(*),B(*) C C CPTSL GIVEN A POSITIVE DEFINITE TRIDIAGONAL MATRIX AND A RIGHT C HAND SIDE WILL FIND THE SOLUTION. C C ON ENTRY C C N INTEGER C IS THE ORDER OF THE TRIDIAGONAL MATRIX. C C D COMPLEX(N) C IS THE DIAGONAL OF THE TRIDIAGONAL MATRIX. C ON OUTPUT D IS DESTROYED. C C E COMPLEX(N) C IS THE OFFDIAGONAL OF THE TRIDIAGONAL MATRIX. C E(1) THROUGH E(N-1) SHOULD CONTAIN THE C OFFDIAGONAL. C C B COMPLEX(N) C IS THE RIGHT HAND SIDE VECTOR. C C ON RETURN C C B CONTAINS THE SOULTION. C C LINPACK. THIS VERSION DATED 08/14/78 . C JACK DONGARRA, ARGONNE NATIONAL LABORATORY. C C NO EXTERNALS C FORTRAN CONJG,MOD C C INTERNAL VARIABLES C INTEGER K,KBM1,KE,KF,KP1,NM1,NM1D2 COMPLEX T1,T2 C C CHECK FOR 1 X 1 CASE C IF (N .NE. 1) GO TO 10 B(1) = B(1)/D(1) GO TO 70 10 CONTINUE NM1 = N - 1 NM1D2 = NM1/2 IF (N .EQ. 2) GO TO 30 KBM1 = N - 1 C C ZERO TOP HALF OF SUBDIAGONAL AND BOTTOM HALF OF C SUPERDIAGONAL C DO 20 K = 1, NM1D2 T1 = CONJG(E(K))/D(K) D(K+1) = D(K+1) - T1*E(K) B(K+1) = B(K+1) - T1*B(K) T2 = E(KBM1)/D(KBM1+1) D(KBM1) = D(KBM1) - T2*CONJG(E(KBM1)) B(KBM1) = B(KBM1) - T2*B(KBM1+1) KBM1 = KBM1 - 1 20 CONTINUE 30 CONTINUE KP1 = NM1D2 + 1 C C CLEAN UP FOR POSSIBLE 2 X 2 BLOCK AT CENTER C IF (MOD(N,2) .NE. 0) GO TO 40 T1 = CONJG(E(KP1))/D(KP1) D(KP1+1) = D(KP1+1) - T1*E(KP1) B(KP1+1) = B(KP1+1) - T1*B(KP1) KP1 = KP1 + 1 40 CONTINUE C C BACK SOLVE STARTING AT THE CENTER, GOING TOWARDS THE TOP C AND BOTTOM C B(KP1) = B(KP1)/D(KP1) IF (N .EQ. 2) GO TO 60 K = KP1 - 1 KE = KP1 + NM1D2 - 1 DO 50 KF = KP1, KE B(K) = (B(K) - E(K)*B(K+1))/D(K) B(KF+1) = (B(KF+1) - CONJG(E(KF))*B(KF))/D(KF+1) K = K - 1 50 CONTINUE 60 CONTINUE IF (MOD(N,2) .EQ. 0) B(1) = (B(1) - E(1)*B(2))/D(1) 70 CONTINUE RETURN END