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- SUBROUTINE SSPCO(AP,N,KPVT,RCOND,Z)
- INTEGER N,KPVT(1)
- REAL AP(1),Z(1)
- REAL RCOND
- C
- C SSPCO FACTORS A REAL SYMMETRIC MATRIX STORED IN PACKED
- C FORM BY ELIMINATION WITH SYMMETRIC PIVOTING AND ESTIMATES
- C THE CONDITION OF THE MATRIX.
- C
- C IF RCOND IS NOT NEEDED, SSPFA IS SLIGHTLY FASTER.
- C TO SOLVE A*X = B , FOLLOW SSPCO BY SSPSL.
- C TO COMPUTE INVERSE(A)*C , FOLLOW SSPCO BY SSPSL.
- C TO COMPUTE INVERSE(A) , FOLLOW SSPCO BY SSPDI.
- C TO COMPUTE DETERMINANT(A) , FOLLOW SSPCO BY SSPDI.
- C TO COMPUTE INERTIA(A), FOLLOW SSPCO BY SSPDI.
- C
- C ON ENTRY
- C
- C AP REAL (N*(N+1)/2)
- C THE PACKED FORM OF A SYMMETRIC MATRIX A . THE
- C COLUMNS OF THE UPPER TRIANGLE ARE STORED SEQUENTIALLY
- C IN A ONE-DIMENSIONAL ARRAY OF LENGTH N*(N+1)/2 .
- C SEE COMMENTS BELOW FOR DETAILS.
- C
- C N INTEGER
- C THE ORDER OF THE MATRIX A .
- C
- C OUTPUT
- C
- C AP A BLOCK DIAGONAL MATRIX AND THE MULTIPLIERS WHICH
- C WERE USED TO OBTAIN IT STORED IN PACKED FORM.
- C THE FACTORIZATION CAN BE WRITTEN A = U*D*TRANS(U)
- C WHERE U IS A PRODUCT OF PERMUTATION AND UNIT
- C UPPER TRIANGULAR MATRICES , TRANS(U) IS THE
- C TRANSPOSE OF U , AND D IS BLOCK DIAGONAL
- C WITH 1 BY 1 AND 2 BY 2 BLOCKS.
- C
- C KPVT INTEGER(N)
- C AN INTEGER VECTOR OF PIVOT INDICES.
- 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.
- C
- C Z REAL(N)
- C A WORK VECTOR WHOSE CONTENTS ARE USUALLY UNIMPORTANT.
- C IF A IS CLOSE TO A SINGULAR MATRIX, THEN Z IS
- C AN APPROXIMATE NULL VECTOR IN THE SENSE THAT
- C NORM(A*Z) = RCOND*NORM(A)*NORM(Z) .
- C
- C PACKED STORAGE
- C
- C THE FOLLOWING PROGRAM SEGMENT WILL PACK THE UPPER
- C TRIANGLE OF A SYMMETRIC MATRIX.
- C
- C K = 0
- C DO 20 J = 1, N
- C DO 10 I = 1, J
- C K = K + 1
- C AP(K) = A(I,J)
- C 10 CONTINUE
- C 20 CONTINUE
- 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 SSPFA
- C BLAS SAXPY,SDOT,SSCAL,SASUM
- C FORTRAN ABS,AMAX1,IABS,SIGN
- C
- C INTERNAL VARIABLES
- C
- REAL AK,AKM1,BK,BKM1,SDOT,DENOM,EK,T
- REAL ANORM,S,SASUM,YNORM
- INTEGER I,IJ,IK,IKM1,IKP1,INFO,J,JM1,J1
- INTEGER K,KK,KM1K,KM1KM1,KP,KPS,KS
- C
- C
- C FIND NORM OF A USING ONLY UPPER HALF
- C
- J1 = 1
- DO 30 J = 1, N
- Z(J) = SASUM(J,AP(J1),1)
- IJ = J1
- J1 = J1 + J
- JM1 = J - 1
- IF (JM1 .LT. 1) GO TO 20
- DO 10 I = 1, JM1
- Z(I) = Z(I) + ABS(AP(IJ))
- IJ = IJ + 1
- 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 SSPFA(AP,N,KPVT,INFO)
- 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 U*D*W = E .
- C THE VECTORS ARE FREQUENTLY RESCALED TO AVOID OVERFLOW.
- C
- C SOLVE U*D*W = E
- C
- EK = 1.0E0
- DO 50 J = 1, N
- Z(J) = 0.0E0
- 50 CONTINUE
- K = N
- IK = (N*(N - 1))/2
- 60 IF (K .EQ. 0) GO TO 120
- KK = IK + K
- IKM1 = IK - (K - 1)
- KS = 1
- IF (KPVT(K) .LT. 0) KS = 2
- KP = IABS(KPVT(K))
- KPS = K + 1 - KS
- IF (KP .EQ. KPS) GO TO 70
- T = Z(KPS)
- Z(KPS) = Z(KP)
- Z(KP) = T
- 70 CONTINUE
- IF (Z(K) .NE. 0.0E0) EK = SIGN(EK,Z(K))
- Z(K) = Z(K) + EK
- CALL SAXPY(K-KS,Z(K),AP(IK+1),1,Z(1),1)
- IF (KS .EQ. 1) GO TO 80
- IF (Z(K-1) .NE. 0.0E0) EK = SIGN(EK,Z(K-1))
- Z(K-1) = Z(K-1) + EK
- CALL SAXPY(K-KS,Z(K-1),AP(IKM1+1),1,Z(1),1)
- 80 CONTINUE
- IF (KS .EQ. 2) GO TO 100
- IF (ABS(Z(K)) .LE. ABS(AP(KK))) GO TO 90
- S = ABS(AP(KK))/ABS(Z(K))
- CALL SSCAL(N,S,Z,1)
- EK = S*EK
- 90 CONTINUE
- IF (AP(KK) .NE. 0.0E0) Z(K) = Z(K)/AP(KK)
- IF (AP(KK) .EQ. 0.0E0) Z(K) = 1.0E0
- GO TO 110
- 100 CONTINUE
- KM1K = IK + K - 1
- KM1KM1 = IKM1 + K - 1
- AK = AP(KK)/AP(KM1K)
- AKM1 = AP(KM1KM1)/AP(KM1K)
- BK = Z(K)/AP(KM1K)
- BKM1 = Z(K-1)/AP(KM1K)
- DENOM = AK*AKM1 - 1.0E0
- Z(K) = (AKM1*BK - BKM1)/DENOM
- Z(K-1) = (AK*BKM1 - BK)/DENOM
- 110 CONTINUE
- K = K - KS
- IK = IK - K
- IF (KS .EQ. 2) IK = IK - (K + 1)
- GO TO 60
- 120 CONTINUE
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- C
- C SOLVE TRANS(U)*Y = W
- C
- K = 1
- IK = 0
- 130 IF (K .GT. N) GO TO 160
- KS = 1
- IF (KPVT(K) .LT. 0) KS = 2
- IF (K .EQ. 1) GO TO 150
- Z(K) = Z(K) + SDOT(K-1,AP(IK+1),1,Z(1),1)
- IKP1 = IK + K
- IF (KS .EQ. 2)
- * Z(K+1) = Z(K+1) + SDOT(K-1,AP(IKP1+1),1,Z(1),1)
- KP = IABS(KPVT(K))
- IF (KP .EQ. K) GO TO 140
- T = Z(K)
- Z(K) = Z(KP)
- Z(KP) = T
- 140 CONTINUE
- 150 CONTINUE
- IK = IK + K
- IF (KS .EQ. 2) IK = IK + (K + 1)
- K = K + KS
- GO TO 130
- 160 CONTINUE
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- C
- YNORM = 1.0E0
- C
- C SOLVE U*D*V = Y
- C
- K = N
- IK = N*(N - 1)/2
- 170 IF (K .EQ. 0) GO TO 230
- KK = IK + K
- IKM1 = IK - (K - 1)
- KS = 1
- IF (KPVT(K) .LT. 0) KS = 2
- IF (K .EQ. KS) GO TO 190
- KP = IABS(KPVT(K))
- KPS = K + 1 - KS
- IF (KP .EQ. KPS) GO TO 180
- T = Z(KPS)
- Z(KPS) = Z(KP)
- Z(KP) = T
- 180 CONTINUE
- CALL SAXPY(K-KS,Z(K),AP(IK+1),1,Z(1),1)
- IF (KS .EQ. 2) CALL SAXPY(K-KS,Z(K-1),AP(IKM1+1),1,Z(1),1)
- 190 CONTINUE
- IF (KS .EQ. 2) GO TO 210
- IF (ABS(Z(K)) .LE. ABS(AP(KK))) GO TO 200
- S = ABS(AP(KK))/ABS(Z(K))
- CALL SSCAL(N,S,Z,1)
- YNORM = S*YNORM
- 200 CONTINUE
- IF (AP(KK) .NE. 0.0E0) Z(K) = Z(K)/AP(KK)
- IF (AP(KK) .EQ. 0.0E0) Z(K) = 1.0E0
- GO TO 220
- 210 CONTINUE
- KM1K = IK + K - 1
- KM1KM1 = IKM1 + K - 1
- AK = AP(KK)/AP(KM1K)
- AKM1 = AP(KM1KM1)/AP(KM1K)
- BK = Z(K)/AP(KM1K)
- BKM1 = Z(K-1)/AP(KM1K)
- DENOM = AK*AKM1 - 1.0E0
- Z(K) = (AKM1*BK - BKM1)/DENOM
- Z(K-1) = (AK*BKM1 - BK)/DENOM
- 220 CONTINUE
- K = K - KS
- IK = IK - K
- IF (KS .EQ. 2) IK = IK - (K + 1)
- GO TO 170
- 230 CONTINUE
- S = 1.0E0/SASUM(N,Z,1)
- CALL SSCAL(N,S,Z,1)
- YNORM = S*YNORM
- C
- C SOLVE TRANS(U)*Z = V
- C
- K = 1
- IK = 0
- 240 IF (K .GT. N) GO TO 270
- KS = 1
- IF (KPVT(K) .LT. 0) KS = 2
- IF (K .EQ. 1) GO TO 260
- Z(K) = Z(K) + SDOT(K-1,AP(IK+1),1,Z(1),1)
- IKP1 = IK + K
- IF (KS .EQ. 2)
- * Z(K+1) = Z(K+1) + SDOT(K-1,AP(IKP1+1),1,Z(1),1)
- KP = IABS(KPVT(K))
- IF (KP .EQ. K) GO TO 250
- T = Z(K)
- Z(K) = Z(KP)
- Z(KP) = T
- 250 CONTINUE
- 260 CONTINUE
- IK = IK + K
- IF (KS .EQ. 2) IK = IK + (K + 1)
- K = K + KS
- GO TO 240
- 270 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
- RETURN
- END