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