sgtsl.f
119 lines
| 3.1 KiB
| text/x-fortran
|
FortranFixedLexer
| r1601 | SUBROUTINE SGTSL(N,C,D,E,B,INFO) | |||
| INTEGER N,INFO | ||||
| REAL C(1),D(1),E(1),B(1) | ||||
| C | ||||
| C SGTSL GIVEN A GENERAL TRIDIAGONAL MATRIX AND A RIGHT HAND | ||||
| C SIDE WILL FIND THE SOLUTION. | ||||
| C | ||||
| C ON ENTRY | ||||
| C | ||||
| C N INTEGER | ||||
| C IS THE ORDER OF THE TRIDIAGONAL MATRIX. | ||||
| C | ||||
| C C REAL(N) | ||||
| C IS THE SUBDIAGONAL OF THE TRIDIAGONAL MATRIX. | ||||
| C C(2) THROUGH C(N) SHOULD CONTAIN THE SUBDIAGONAL. | ||||
| C ON OUTPUT C IS DESTROYED. | ||||
| C | ||||
| C D REAL(N) | ||||
| C IS THE DIAGONAL OF THE TRIDIAGONAL MATRIX. | ||||
| C ON OUTPUT D IS DESTROYED. | ||||
| C | ||||
| C E REAL(N) | ||||
| C IS THE SUPERDIAGONAL OF THE TRIDIAGONAL MATRIX. | ||||
| C E(1) THROUGH E(N-1) SHOULD CONTAIN THE SUPERDIAGONAL. | ||||
| C ON OUTPUT E IS DESTROYED. | ||||
| C | ||||
| C B REAL(N) | ||||
| C IS THE RIGHT HAND SIDE VECTOR. | ||||
| C | ||||
| C ON RETURN | ||||
| C | ||||
| C B IS THE SOLUTION VECTOR. | ||||
| C | ||||
| C INFO INTEGER | ||||
| C = 0 NORMAL VALUE. | ||||
| C = K IF THE K-TH ELEMENT OF THE DIAGONAL BECOMES | ||||
| C EXACTLY ZERO. THE SUBROUTINE RETURNS WHEN | ||||
| C THIS IS DETECTED. | ||||
| C | ||||
| C LINPACK. THIS VERSION DATED 08/14/78 . | ||||
| C JACK DONGARRA, ARGONNE NATIONAL LABORATORY. | ||||
| C | ||||
| C NO EXTERNALS | ||||
| C FORTRAN ABS | ||||
| C | ||||
| C INTERNAL VARIABLES | ||||
| C | ||||
| INTEGER K,KB,KP1,NM1,NM2 | ||||
| REAL T | ||||
| C BEGIN BLOCK PERMITTING ...EXITS TO 100 | ||||
| C | ||||
| INFO = 0 | ||||
| C(1) = D(1) | ||||
| NM1 = N - 1 | ||||
| IF (NM1 .LT. 1) GO TO 40 | ||||
| D(1) = E(1) | ||||
| E(1) = 0.0E0 | ||||
| E(N) = 0.0E0 | ||||
| C | ||||
| DO 30 K = 1, NM1 | ||||
| KP1 = K + 1 | ||||
| C | ||||
| C FIND THE LARGEST OF THE TWO ROWS | ||||
| C | ||||
| IF (ABS(C(KP1)) .LT. ABS(C(K))) GO TO 10 | ||||
| C | ||||
| C INTERCHANGE ROW | ||||
| C | ||||
| T = C(KP1) | ||||
| C(KP1) = C(K) | ||||
| C(K) = T | ||||
| T = D(KP1) | ||||
| D(KP1) = D(K) | ||||
| D(K) = T | ||||
| T = E(KP1) | ||||
| E(KP1) = E(K) | ||||
| E(K) = T | ||||
| T = B(KP1) | ||||
| B(KP1) = B(K) | ||||
| B(K) = T | ||||
| 10 CONTINUE | ||||
| C | ||||
| C ZERO ELEMENTS | ||||
| C | ||||
| IF (C(K) .NE. 0.0E0) GO TO 20 | ||||
| INFO = K | ||||
| C ............EXIT | ||||
| GO TO 100 | ||||
| 20 CONTINUE | ||||
| T = -C(KP1)/C(K) | ||||
| C(KP1) = D(KP1) + T*D(K) | ||||
| D(KP1) = E(KP1) + T*E(K) | ||||
| E(KP1) = 0.0E0 | ||||
| B(KP1) = B(KP1) + T*B(K) | ||||
| 30 CONTINUE | ||||
| 40 CONTINUE | ||||
| IF (C(N) .NE. 0.0E0) GO TO 50 | ||||
| INFO = N | ||||
| GO TO 90 | ||||
| 50 CONTINUE | ||||
| C | ||||
| C BACK SOLVE | ||||
| C | ||||
| NM2 = N - 2 | ||||
| B(N) = B(N)/C(N) | ||||
| IF (N .EQ. 1) GO TO 80 | ||||
| B(NM1) = (B(NM1) - D(NM1)*B(N))/C(NM1) | ||||
| IF (NM2 .LT. 1) GO TO 70 | ||||
| DO 60 KB = 1, NM2 | ||||
| K = NM2 - KB + 1 | ||||
| B(K) = (B(K) - D(K)*B(K+1) - E(K)*B(K+2))/C(K) | ||||
| 60 CONTINUE | ||||
| 70 CONTINUE | ||||
| 80 CONTINUE | ||||
| 90 CONTINUE | ||||
| 100 CONTINUE | ||||
| C | ||||
| RETURN | ||||
| END | ||||
