*DECK SPOCO SUBROUTINE SPOCO (A, LDA, N, RCOND, Z, INFO) C***BEGIN PROLOGUE SPOCO C***PURPOSE Factor a real symmetric positive definite matrix C and estimate the condition number of the matrix. C***LIBRARY SLATEC (LINPACK) C***CATEGORY D2B1B C***TYPE SINGLE PRECISION (SPOCO-S, DPOCO-D, CPOCO-C) C***KEYWORDS CONDITION NUMBER, LINEAR ALGEBRA, LINPACK, C MATRIX FACTORIZATION, POSITIVE DEFINITE C***AUTHOR Moler, C. B., (U. of New Mexico) C***DESCRIPTION C C SPOCO factors a real symmetric positive definite matrix C and estimates the condition of the matrix. C C If RCOND is not needed, SPOFA is slightly faster. C To solve A*X = B , follow SPOCO by SPOSL. C To compute INVERSE(A)*C , follow SPOCO by SPOSL. C To compute DETERMINANT(A) , follow SPOCO by SPODI. C To compute INVERSE(A) , follow SPOCO by SPODI. C C On Entry C C A REAL(LDA, N) C the symmetric 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 = TRANS(R)*R C where TRANS(R) is the transpose. C The strict lower triangle is unaltered. 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 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 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***REFERENCES J. J. Dongarra, J. R. Bunch, C. B. Moler, and G. W. C Stewart, LINPACK Users' Guide, SIAM, 1979. C***ROUTINES CALLED SASUM, SAXPY, SDOT, SPOFA, SSCAL C***REVISION HISTORY (YYMMDD) C 780814 DATE WRITTEN C 890531 Changed all specific intrinsics to generic. (WRB) C 890831 Modified array declarations. (WRB) C 890831 REVISION DATE from Version 3.2 C 891214 Prologue converted to Version 4.0 format. (BAB) C 900326 Removed duplicate information from DESCRIPTION section. C (WRB) C 920501 Reformatted the REFERENCES section. (WRB) C***END PROLOGUE SPOCO INTEGER LDA,N,INFO REAL A(LDA,*),Z(*) REAL RCOND C REAL SDOT,EK,T,WK,WKM REAL ANORM,S,SASUM,SM,YNORM INTEGER I,J,JM1,K,KB,KP1 C C FIND NORM OF A USING ONLY UPPER HALF C C***FIRST EXECUTABLE STATEMENT SPOCO DO 30 J = 1, N Z(J) = SASUM(J,A(1,J),1) JM1 = J - 1 IF (JM1 .LT. 1) GO TO 20 DO 10 I = 1, JM1 Z(I) = Z(I) + ABS(A(I,J)) 10 CONTINUE 20 CONTINUE 30 CONTINUE ANORM = 0.0E0 DO 40 J = 1, N ANORM = MAX(ANORM,Z(J)) 40 CONTINUE C C FACTOR C CALL SPOFA(A,LDA,N,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. A(K,K)) GO TO 60 S = A(K,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/A(K,K) WKM = WKM/A(K,K) KP1 = K + 1 IF (KP1 .GT. N) GO TO 100 DO 70 J = KP1, N SM = SM + ABS(Z(J)+WKM*A(K,J)) Z(J) = Z(J) + WK*A(K,J) S = S + ABS(Z(J)) 70 CONTINUE IF (S .GE. SM) GO TO 90 T = WKM - WK WK = WKM DO 80 J = KP1, N Z(J) = Z(J) + T*A(K,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. A(K,K)) GO TO 120 S = A(K,K)/ABS(Z(K)) CALL SSCAL(N,S,Z,1) 120 CONTINUE Z(K) = Z(K)/A(K,K) T = -Z(K) CALL SAXPY(K-1,T,A(1,K),1,Z(1),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 Z(K) = Z(K) - SDOT(K-1,A(1,K),1,Z(1),1) IF (ABS(Z(K)) .LE. A(K,K)) GO TO 140 S = A(K,K)/ABS(Z(K)) CALL SSCAL(N,S,Z,1) YNORM = S*YNORM 140 CONTINUE Z(K) = Z(K)/A(K,K) 150 CONTINUE S = 1.0E0/SASUM(N,Z,1) CALL SSCAL(N,S,Z,1) YNORM = S*YNORM C C SOLVE R*Z = V C DO 170 KB = 1, N K = N + 1 - KB IF (ABS(Z(K)) .LE. A(K,K)) GO TO 160 S = A(K,K)/ABS(Z(K)) CALL SSCAL(N,S,Z,1) YNORM = S*YNORM 160 CONTINUE Z(K) = Z(K)/A(K,K) T = -Z(K) CALL SAXPY(K-1,T,A(1,K),1,Z(1),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