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- SUBROUTINE SPBCO(ABD,LDA,N,M,RCOND,Z,INFO)
- INTEGER LDA,N,M,INFO
- REAL ABD(LDA,1),Z(1)
- REAL RCOND
- C
- C SPBCO FACTORS A REAL SYMMETRIC POSITIVE DEFINITE
- C MATRIX STORED IN BAND FORM AND ESTIMATES THE CONDITION OF THE
- C MATRIX.
- C
- C IF RCOND IS NOT NEEDED, SPBFA IS SLIGHTLY FASTER.
- C TO SOLVE A*X = B , FOLLOW SPBCO BY SPBSL.
- C TO COMPUTE INVERSE(A)*C , FOLLOW SPBCO BY SPBSL.
- C TO COMPUTE DETERMINANT(A) , FOLLOW SPBCO BY SPBDI.
- C
- C ON ENTRY
- C
- C ABD REAL(LDA, N)
- C THE MATRIX TO BE FACTORED. THE COLUMNS OF THE UPPER
- C TRIANGLE ARE STORED IN THE COLUMNS OF ABD AND THE
- C DIAGONALS OF THE UPPER TRIANGLE ARE STORED IN THE
- C ROWS OF ABD . SEE THE COMMENTS BELOW FOR DETAILS.
- C
- C LDA INTEGER
- C THE LEADING DIMENSION OF THE ARRAY ABD .
- C LDA MUST BE .GE. M + 1 .
- C
- C N INTEGER
- C THE ORDER OF THE MATRIX A .
- C
- C M INTEGER
- C THE NUMBER OF DIAGONALS ABOVE THE MAIN DIAGONAL.
- C 0 .LE. M .LT. N .
- C
- C ON RETURN
- C
- C ABD AN UPPER TRIANGULAR MATRIX R , STORED IN BAND
- C FORM, SO THAT A = TRANS(R)*R .
- C IF INFO .NE. 0 , THE FACTORIZATION IS NOT COMPLETE.
- C
- C RCOND REAL
- C AN ESTIMATE OF THE RECIPROCAL CONDITION OF A .
- C FOR THE SYSTEM A*X = B , RELATIVE PERTURBATIONS
- C IN A AND B OF SIZE EPSILON MAY CAUSE
- C RELATIVE PERTURBATIONS IN X OF SIZE EPSILON/RCOND .
- C IF RCOND IS SO SMALL THAT THE LOGICAL EXPRESSION
- C 1.0 + RCOND .EQ. 1.0
- C IS TRUE, THEN A MAY BE SINGULAR TO WORKING
- C PRECISION. IN PARTICULAR, RCOND IS ZERO IF
- C EXACT SINGULARITY IS DETECTED OR THE ESTIMATE
- C UNDERFLOWS. IF INFO .NE. 0 , RCOND IS UNCHANGED.
- C
- C Z REAL(N)
- C A WORK VECTOR WHOSE CONTENTS ARE USUALLY UNIMPORTANT.
- C IF A IS SINGULAR TO WORKING PRECISION, THEN Z IS
- C AN APPROXIMATE NULL VECTOR IN THE SENSE THAT
- C NORM(A*Z) = RCOND*NORM(A)*NORM(Z) .
- C IF INFO .NE. 0 , Z IS UNCHANGED.
- 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 BAND STORAGE
- C
- C IF A IS A SYMMETRIC POSITIVE DEFINITE BAND MATRIX,
- C THE FOLLOWING PROGRAM SEGMENT WILL SET UP THE INPUT.
- C
- C M = (BAND WIDTH ABOVE DIAGONAL)
- C DO 20 J = 1, N
- C I1 = MAX0(1, J-M)
- C DO 10 I = I1, J
- C K = I-J+M+1
- C ABD(K,J) = A(I,J)
- C 10 CONTINUE
- C 20 CONTINUE
- C
- C THIS USES M + 1 ROWS OF A , EXCEPT FOR THE M BY M
- C UPPER LEFT TRIANGLE, WHICH IS IGNORED.
- C
- C EXAMPLE.. IF THE ORIGINAL MATRIX IS
- C
- C 11 12 13 0 0 0
- C 12 22 23 24 0 0
- C 13 23 33 34 35 0
- C 0 24 34 44 45 46
- C 0 0 35 45 55 56
- C 0 0 0 46 56 66
- C
- C THEN N = 6 , M = 2 AND ABD SHOULD CONTAIN
- C
- C * * 13 24 35 46
- C * 12 23 34 45 56
- C 11 22 33 44 55 66
- 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 LINPACK SPBFA
- C BLAS SAXPY,SDOT,SSCAL,SASUM
- C FORTRAN ABS,AMAX1,MAX0,MIN0,REAL,SIGN
- C
- C INTERNAL VARIABLES
- C
- REAL SDOT,EK,T,WK,WKM
- REAL ANORM,S,SASUM,SM,YNORM
- INTEGER I,J,J2,K,KB,KP1,L,LA,LB,LM,MU
- C
- C
- C FIND NORM OF A
- C
- DO 30 J = 1, N
- L = MIN0(J,M+1)
- MU = MAX0(M+2-J,1)
- Z(J) = SASUM(L,ABD(MU,J),1)
- K = J - L
- IF (M .LT. MU) GO TO 20
- DO 10 I = MU, M
- K = K + 1
- Z(K) = Z(K) + ABS(ABD(I,J))
- 10 CONTINUE
- 20 CONTINUE
- 30 CONTINUE
- ANORM = 0.0E0
- DO 40 J = 1, N
- ANORM = AMAX1(ANORM,Z(J))
- 40 CONTINUE
- C
- C FACTOR
- C
- CALL SPBFA(ABD,LDA,N,M,INFO)
- IF (INFO .NE. 0) GO TO 180
- C
- C RCOND = 1/(NORM(A)*(ESTIMATE OF NORM(INVERSE(A)))) .
- C ESTIMATE = NORM(Z)/NORM(Y) WHERE A*Z = Y AND A*Y = E .
- C THE COMPONENTS OF E ARE CHOSEN TO CAUSE MAXIMUM LOCAL
- C GROWTH IN THE ELEMENTS OF W WHERE TRANS(R)*W = E .
- C THE VECTORS ARE FREQUENTLY RESCALED TO AVOID OVERFLOW.
- C
- C SOLVE TRANS(R)*W = E
- C
- EK = 1.0E0
- DO 50 J = 1, N
- Z(J) = 0.0E0
- 50 CONTINUE
- DO 110 K = 1, N
- IF (Z(K) .NE. 0.0E0) EK = SIGN(EK,-Z(K))
- IF (ABS(EK-Z(K)) .LE. ABD(M+1,K)) GO TO 60
- S = ABD(M+1,K)/ABS(EK-Z(K))
- CALL SSCAL(N,S,Z,1)
- EK = S*EK
- 60 CONTINUE
- WK = EK - Z(K)
- WKM = -EK - Z(K)
- S = ABS(WK)
- SM = ABS(WKM)
- WK = WK/ABD(M+1,K)
- WKM = WKM/ABD(M+1,K)
- KP1 = K + 1
- J2 = MIN0(K+M,N)
- I = M + 1
- IF (KP1 .GT. J2) GO TO 100
- DO 70 J = KP1, J2
- I = I - 1
- SM = SM + ABS(Z(J)+WKM*ABD(I,J))
- Z(J) = Z(J) + WK*ABD(I,J)
- S = S + ABS(Z(J))
- 70 CONTINUE
- IF (S .GE. SM) GO TO 90
- T = WKM - WK
- WK = WKM
- I = M + 1
- DO 80 J = KP1, J2
- I = I - 1
- Z(J) = Z(J) + T*ABD(I,J)
- 80 CONTINUE
- 90 CONTINUE
- 100 CONTINUE
- Z(K) = WK
- 110 CONTINUE
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- C
- C SOLVE R*Y = W
- C
- DO 130 KB = 1, N
- K = N + 1 - KB
- IF (ABS(Z(K)) .LE. ABD(M+1,K)) GO TO 120
- S = ABD(M+1,K)/ABS(Z(K))
- CALL SSCAL(N,S,Z,1)
- 120 CONTINUE
- Z(K) = Z(K)/ABD(M+1,K)
- LM = MIN0(K-1,M)
- LA = M + 1 - LM
- LB = K - LM
- T = -Z(K)
- CALL SAXPY(LM,T,ABD(LA,K),1,Z(LB),1)
- 130 CONTINUE
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- C
- YNORM = 1.0E0
- C
- C SOLVE TRANS(R)*V = Y
- C
- DO 150 K = 1, N
- LM = MIN0(K-1,M)
- LA = M + 1 - LM
- LB = K - LM
- Z(K) = Z(K) - SDOT(LM,ABD(LA,K),1,Z(LB),1)
- IF (ABS(Z(K)) .LE. ABD(M+1,K)) GO TO 140
- S = ABD(M+1,K)/ABS(Z(K))
- CALL SSCAL(N,S,Z,1)
- YNORM = S*YNORM
- 140 CONTINUE
- Z(K) = Z(K)/ABD(M+1,K)
- 150 CONTINUE
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- YNORM = S*YNORM
- C
- C SOLVE R*Z = W
- C
- DO 170 KB = 1, N
- K = N + 1 - KB
- IF (ABS(Z(K)) .LE. ABD(M+1,K)) GO TO 160
- S = ABD(M+1,K)/ABS(Z(K))
- CALL SSCAL(N,S,Z,1)
- YNORM = S*YNORM
- 160 CONTINUE
- Z(K) = Z(K)/ABD(M+1,K)
- LM = MIN0(K-1,M)
- LA = M + 1 - LM
- LB = K - LM
- T = -Z(K)
- CALL SAXPY(LM,T,ABD(LA,K),1,Z(LB),1)
- 170 CONTINUE
- C MAKE ZNORM = 1.0
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- YNORM = S*YNORM
- C
- IF (ANORM .NE. 0.0E0) RCOND = YNORM/ANORM
- IF (ANORM .EQ. 0.0E0) RCOND = 0.0E0
- 180 CONTINUE
- RETURN
- END