home
***
CD-ROM
|
disk
|
FTP
|
other
***
search
/
Simtel MSDOS 1992 September
/
Simtel20_Sept92.cdr
/
msdos
/
fortran
/
linpklib.arc
/
SSIDI.FOR
< prev
next >
Wrap
Text File
|
1984-01-07
|
7KB
|
213 lines
SUBROUTINE SSIDI(A,LDA,N,KPVT,DET,INERT,WORK,JOB)
INTEGER LDA,N,JOB
REAL A(LDA,1),WORK(1)
REAL DET(2)
INTEGER KPVT(1),INERT(3)
C
C SSIDI COMPUTES THE DETERMINANT, INERTIA AND INVERSE
C OF A REAL SYMMETRIC MATRIX USING THE FACTORS FROM SSIFA.
C
C ON ENTRY
C
C A REAL(LDA,N)
C THE OUTPUT FROM SSIFA.
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 KPVT INTEGER(N)
C THE PIVOT VECTOR FROM SSIFA.
C
C WORK REAL(N)
C WORK VECTOR. CONTENTS DESTROYED.
C
C JOB INTEGER
C JOB HAS THE DECIMAL EXPANSION ABC WHERE
C IF C .NE. 0, THE INVERSE IS COMPUTED,
C IF B .NE. 0, THE DETERMINANT IS COMPUTED,
C IF A .NE. 0, THE INERTIA IS COMPUTED.
C
C FOR EXAMPLE, JOB = 111 GIVES ALL THREE.
C
C ON RETURN
C
C VARIABLES NOT REQUESTED BY JOB ARE NOT USED.
C
C A CONTAINS THE UPPER TRIANGLE OF THE INVERSE OF
C THE ORIGINAL MATRIX. THE STRICT LOWER TRIANGLE
C IS NEVER REFERENCED.
C
C DET REAL(2)
C DETERMINANT OF ORIGINAL MATRIX.
C DETERMINANT = DET(1) * 10.0**DET(2)
C WITH 1.0 .LE. ABS(DET(1)) .LT. 10.0
C OR DET(1) = 0.0.
C
C INERT INTEGER(3)
C THE INERTIA OF THE ORIGINAL MATRIX.
C INERT(1) = NUMBER OF POSITIVE EIGENVALUES.
C INERT(2) = NUMBER OF NEGATIVE EIGENVALUES.
C INERT(3) = NUMBER OF ZERO EIGENVALUES.
C
C ERROR CONDITION
C
C A DIVISION BY ZERO MAY OCCUR IF THE INVERSE IS REQUESTED
C AND SSICO HAS SET RCOND .EQ. 0.0
C OR SSIFA HAS SET INFO .NE. 0 .
C
C LINPACK. THIS VERSION DATED 08/14/78 .
C JAMES BUNCH, UNIV. CALIF. SAN DIEGO, ARGONNE NAT. LAB
C
C SUBROUTINES AND FUNCTIONS
C
C BLAS SAXPY,SCOPY,SDOT,SSWAP
C FORTRAN ABS,IABS,MOD
C
C INTERNAL VARIABLES.
C
REAL AKKP1,SDOT,TEMP
REAL TEN,D,T,AK,AKP1
INTEGER J,JB,K,KM1,KS,KSTEP
LOGICAL NOINV,NODET,NOERT
C
NOINV = MOD(JOB,10) .EQ. 0
NODET = MOD(JOB,100)/10 .EQ. 0
NOERT = MOD(JOB,1000)/100 .EQ. 0
C
IF (NODET .AND. NOERT) GO TO 140
IF (NOERT) GO TO 10
INERT(1) = 0
INERT(2) = 0
INERT(3) = 0
10 CONTINUE
IF (NODET) GO TO 20
DET(1) = 1.0E0
DET(2) = 0.0E0
TEN = 10.0E0
20 CONTINUE
T = 0.0E0
DO 130 K = 1, N
D = A(K,K)
C
C CHECK IF 1 BY 1
C
IF (KPVT(K) .GT. 0) GO TO 50
C
C 2 BY 2 BLOCK
C USE DET (D S) = (D/T * C - T) * T , T = ABS(S)
C (S C)
C TO AVOID UNDERFLOW/OVERFLOW TROUBLES.
C TAKE TWO PASSES THROUGH SCALING. USE T FOR FLAG.
C
IF (T .NE. 0.0E0) GO TO 30
T = ABS(A(K,K+1))
D = (D/T)*A(K+1,K+1) - T
GO TO 40
30 CONTINUE
D = T
T = 0.0E0
40 CONTINUE
50 CONTINUE
C
IF (NOERT) GO TO 60
IF (D .GT. 0.0E0) INERT(1) = INERT(1) + 1
IF (D .LT. 0.0E0) INERT(2) = INERT(2) + 1
IF (D .EQ. 0.0E0) INERT(3) = INERT(3) + 1
60 CONTINUE
C
IF (NODET) GO TO 120
DET(1) = D*DET(1)
IF (DET(1) .EQ. 0.0E0) GO TO 110
70 IF (ABS(DET(1)) .GE. 1.0E0) GO TO 80
DET(1) = TEN*DET(1)
DET(2) = DET(2) - 1.0E0
GO TO 70
80 CONTINUE
90 IF (ABS(DET(1)) .LT. TEN) GO TO 100
DET(1) = DET(1)/TEN
DET(2) = DET(2) + 1.0E0
GO TO 90
100 CONTINUE
110 CONTINUE
120 CONTINUE
130 CONTINUE
140 CONTINUE
C
C COMPUTE INVERSE(A)
C
IF (NOINV) GO TO 270
K = 1
150 IF (K .GT. N) GO TO 260
KM1 = K - 1
IF (KPVT(K) .LT. 0) GO TO 180
C
C 1 BY 1
C
A(K,K) = 1.0E0/A(K,K)
IF (KM1 .LT. 1) GO TO 170
CALL SCOPY(KM1,A(1,K),1,WORK,1)
DO 160 J = 1, KM1
A(J,K) = SDOT(J,A(1,J),1,WORK,1)
CALL SAXPY(J-1,WORK(J),A(1,J),1,A(1,K),1)
160 CONTINUE
A(K,K) = A(K,K) + SDOT(KM1,WORK,1,A(1,K),1)
170 CONTINUE
KSTEP = 1
GO TO 220
180 CONTINUE
C
C 2 BY 2
C
T = ABS(A(K,K+1))
AK = A(K,K)/T
AKP1 = A(K+1,K+1)/T
AKKP1 = A(K,K+1)/T
D = T*(AK*AKP1 - 1.0E0)
A(K,K) = AKP1/D
A(K+1,K+1) = AK/D
A(K,K+1) = -AKKP1/D
IF (KM1 .LT. 1) GO TO 210
CALL SCOPY(KM1,A(1,K+1),1,WORK,1)
DO 190 J = 1, KM1
A(J,K+1) = SDOT(J,A(1,J),1,WORK,1)
CALL SAXPY(J-1,WORK(J),A(1,J),1,A(1,K+1),1)
190 CONTINUE
A(K+1,K+1) = A(K+1,K+1) + SDOT(KM1,WORK,1,A(1,K+1),1)
A(K,K+1) = A(K,K+1) + SDOT(KM1,A(1,K),1,A(1,K+1),1)
CALL SCOPY(KM1,A(1,K),1,WORK,1)
DO 200 J = 1, KM1
A(J,K) = SDOT(J,A(1,J),1,WORK,1)
CALL SAXPY(J-1,WORK(J),A(1,J),1,A(1,K),1)
200 CONTINUE
A(K,K) = A(K,K) + SDOT(KM1,WORK,1,A(1,K),1)
210 CONTINUE
KSTEP = 2
220 CONTINUE
C
C SWAP
C
KS = IABS(KPVT(K))
IF (KS .EQ. K) GO TO 250
CALL SSWAP(KS,A(1,KS),1,A(1,K),1)
DO 230 JB = KS, K
J = K + KS - JB
TEMP = A(J,K)
A(J,K) = A(KS,J)
A(KS,J) = TEMP
230 CONTINUE
IF (KSTEP .EQ. 1) GO TO 240
TEMP = A(KS,K+1)
A(KS,K+1) = A(K,K+1)
A(K,K+1) = TEMP
240 CONTINUE
250 CONTINUE
K = K + KSTEP
GO TO 150
260 CONTINUE
270 CONTINUE
RETURN
END