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