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SUBROUTINE DPTRFS( N, NRHS, D, E, DF, EF, B, LDB, X, LDX, FERR,
$ BERR, WORK, INFO )
*
* -- LAPACK routine (version 3.2) --
* -- LAPACK is a software package provided by Univ. of Tennessee, --
* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
* November 2006
*
* .. Scalar Arguments ..
INTEGER INFO, LDB, LDX, N, NRHS
* ..
* .. Array Arguments ..
DOUBLE PRECISION B( LDB, * ), BERR( * ), D( * ), DF( * ),
$ E( * ), EF( * ), FERR( * ), WORK( * ),
$ X( LDX, * )
* ..
*
* Purpose
* =======
*
* DPTRFS improves the computed solution to a system of linear
* equations when the coefficient matrix is symmetric positive definite
* and tridiagonal, and provides error bounds and backward error
* estimates for the solution.
*
* Arguments
* =========
*
* N (input) INTEGER
* The order of the matrix A. N >= 0.
*
* NRHS (input) INTEGER
* The number of right hand sides, i.e., the number of columns
* of the matrix B. NRHS >= 0.
*
* D (input) DOUBLE PRECISION array, dimension (N)
* The n diagonal elements of the tridiagonal matrix A.
*
* E (input) DOUBLE PRECISION array, dimension (N-1)
* The (n-1) subdiagonal elements of the tridiagonal matrix A.
*
* DF (input) DOUBLE PRECISION array, dimension (N)
* The n diagonal elements of the diagonal matrix D from the
* factorization computed by DPTTRF.
*
* EF (input) DOUBLE PRECISION array, dimension (N-1)
* The (n-1) subdiagonal elements of the unit bidiagonal factor
* L from the factorization computed by DPTTRF.
*
* B (input) DOUBLE PRECISION array, dimension (LDB,NRHS)
* The right hand side matrix B.
*
* LDB (input) INTEGER
* The leading dimension of the array B. LDB >= max(1,N).
*
* X (input/output) DOUBLE PRECISION array, dimension (LDX,NRHS)
* On entry, the solution matrix X, as computed by DPTTRS.
* On exit, the improved solution matrix X.
*
* LDX (input) INTEGER
* The leading dimension of the array X. LDX >= max(1,N).
*
* FERR (output) DOUBLE PRECISION array, dimension (NRHS)
* The forward error bound for each solution vector
* X(j) (the j-th column of the solution matrix X).
* If XTRUE is the true solution corresponding to X(j), FERR(j)
* is an estimated upper bound for the magnitude of the largest
* element in (X(j) - XTRUE) divided by the magnitude of the
* largest element in X(j).
*
* BERR (output) DOUBLE PRECISION array, dimension (NRHS)
* The componentwise relative backward error of each solution
* vector X(j) (i.e., the smallest relative change in
* any element of A or B that makes X(j) an exact solution).
*
* WORK (workspace) DOUBLE PRECISION array, dimension (2*N)
*
* INFO (output) INTEGER
* = 0: successful exit
* < 0: if INFO = -i, the i-th argument had an illegal value
*
* Internal Parameters
* ===================
*
* ITMAX is the maximum number of steps of iterative refinement.
*
* =====================================================================
*
* .. Parameters ..
INTEGER ITMAX
PARAMETER ( ITMAX = 5 )
DOUBLE PRECISION ZERO
PARAMETER ( ZERO = 0.0D+0 )
DOUBLE PRECISION ONE
PARAMETER ( ONE = 1.0D+0 )
DOUBLE PRECISION TWO
PARAMETER ( TWO = 2.0D+0 )
DOUBLE PRECISION THREE
PARAMETER ( THREE = 3.0D+0 )
* ..
* .. Local Scalars ..
INTEGER COUNT, I, IX, J, NZ
DOUBLE PRECISION BI, CX, DX, EPS, EX, LSTRES, S, SAFE1, SAFE2,
$ SAFMIN
* ..
* .. External Subroutines ..
EXTERNAL DAXPY, DPTTRS, XERBLA
* ..
* .. Intrinsic Functions ..
INTRINSIC ABS, MAX
* ..
* .. External Functions ..
INTEGER IDAMAX
DOUBLE PRECISION DLAMCH
EXTERNAL IDAMAX, DLAMCH
* ..
* .. Executable Statements ..
*
* Test the input parameters.
*
INFO = 0
IF( N.LT.0 ) THEN
INFO = -1
ELSE IF( NRHS.LT.0 ) THEN
INFO = -2
ELSE IF( LDB.LT.MAX( 1, N ) ) THEN
INFO = -8
ELSE IF( LDX.LT.MAX( 1, N ) ) THEN
INFO = -10
END IF
IF( INFO.NE.0 ) THEN
CALL XERBLA( 'DPTRFS', -INFO )
RETURN
END IF
*
* Quick return if possible
*
IF( N.EQ.0 .OR. NRHS.EQ.0 ) THEN
DO 10 J = 1, NRHS
FERR( J ) = ZERO
BERR( J ) = ZERO
10 CONTINUE
RETURN
END IF
*
* NZ = maximum number of nonzero elements in each row of A, plus 1
*
NZ = 4
EPS = DLAMCH( 'Epsilon' )
SAFMIN = DLAMCH( 'Safe minimum' )
SAFE1 = NZ*SAFMIN
SAFE2 = SAFE1 / EPS
*
* Do for each right hand side
*
DO 90 J = 1, NRHS
*
COUNT = 1
LSTRES = THREE
20 CONTINUE
*
* Loop until stopping criterion is satisfied.
*
* Compute residual R = B - A * X. Also compute
* abs(A)*abs(x) + abs(b) for use in the backward error bound.
*
IF( N.EQ.1 ) THEN
BI = B( 1, J )
DX = D( 1 )*X( 1, J )
WORK( N+1 ) = BI - DX
WORK( 1 ) = ABS( BI ) + ABS( DX )
ELSE
BI = B( 1, J )
DX = D( 1 )*X( 1, J )
EX = E( 1 )*X( 2, J )
WORK( N+1 ) = BI - DX - EX
WORK( 1 ) = ABS( BI ) + ABS( DX ) + ABS( EX )
DO 30 I = 2, N - 1
BI = B( I, J )
CX = E( I-1 )*X( I-1, J )
DX = D( I )*X( I, J )
EX = E( I )*X( I+1, J )
WORK( N+I ) = BI - CX - DX - EX
WORK( I ) = ABS( BI ) + ABS( CX ) + ABS( DX ) + ABS( EX )
30 CONTINUE
BI = B( N, J )
CX = E( N-1 )*X( N-1, J )
DX = D( N )*X( N, J )
WORK( N+N ) = BI - CX - DX
WORK( N ) = ABS( BI ) + ABS( CX ) + ABS( DX )
END IF
*
* Compute componentwise relative backward error from formula
*
* max(i) ( abs(R(i)) / ( abs(A)*abs(X) + abs(B) )(i) )
*
* where abs(Z) is the componentwise absolute value of the matrix
* or vector Z. If the i-th component of the denominator is less
* than SAFE2, then SAFE1 is added to the i-th components of the
* numerator and denominator before dividing.
*
S = ZERO
DO 40 I = 1, N
IF( WORK( I ).GT.SAFE2 ) THEN
S = MAX( S, ABS( WORK( N+I ) ) / WORK( I ) )
ELSE
S = MAX( S, ( ABS( WORK( N+I ) )+SAFE1 ) /
$ ( WORK( I )+SAFE1 ) )
END IF
40 CONTINUE
BERR( J ) = S
*
* Test stopping criterion. Continue iterating if
* 1) The residual BERR(J) is larger than machine epsilon, and
* 2) BERR(J) decreased by at least a factor of 2 during the
* last iteration, and
* 3) At most ITMAX iterations tried.
*
IF( BERR( J ).GT.EPS .AND. TWO*BERR( J ).LE.LSTRES .AND.
$ COUNT.LE.ITMAX ) THEN
*
* Update solution and try again.
*
CALL DPTTRS( N, 1, DF, EF, WORK( N+1 ), N, INFO )
CALL DAXPY( N, ONE, WORK( N+1 ), 1, X( 1, J ), 1 )
LSTRES = BERR( J )
COUNT = COUNT + 1
GO TO 20
END IF
*
* Bound error from formula
*
* norm(X - XTRUE) / norm(X) .le. FERR =
* norm( abs(inv(A))*
* ( abs(R) + NZ*EPS*( abs(A)*abs(X)+abs(B) ))) / norm(X)
*
* where
* norm(Z) is the magnitude of the largest component of Z
* inv(A) is the inverse of A
* abs(Z) is the componentwise absolute value of the matrix or
* vector Z
* NZ is the maximum number of nonzeros in any row of A, plus 1
* EPS is machine epsilon
*
* The i-th component of abs(R)+NZ*EPS*(abs(A)*abs(X)+abs(B))
* is incremented by SAFE1 if the i-th component of
* abs(A)*abs(X) + abs(B) is less than SAFE2.
*
DO 50 I = 1, N
IF( WORK( I ).GT.SAFE2 ) THEN
WORK( I ) = ABS( WORK( N+I ) ) + NZ*EPS*WORK( I )
ELSE
WORK( I ) = ABS( WORK( N+I ) ) + NZ*EPS*WORK( I ) + SAFE1
END IF
50 CONTINUE
IX = IDAMAX( N, WORK, 1 )
FERR( J ) = WORK( IX )
*
* Estimate the norm of inv(A).
*
* Solve M(A) * x = e, where M(A) = (m(i,j)) is given by
*
* m(i,j) = abs(A(i,j)), i = j,
* m(i,j) = -abs(A(i,j)), i .ne. j,
*
* and e = [ 1, 1, ..., 1 ]**T. Note M(A) = M(L)*D*M(L)**T.
*
* Solve M(L) * x = e.
*
WORK( 1 ) = ONE
DO 60 I = 2, N
WORK( I ) = ONE + WORK( I-1 )*ABS( EF( I-1 ) )
60 CONTINUE
*
* Solve D * M(L)**T * x = b.
*
WORK( N ) = WORK( N ) / DF( N )
DO 70 I = N - 1, 1, -1
WORK( I ) = WORK( I ) / DF( I ) + WORK( I+1 )*ABS( EF( I ) )
70 CONTINUE
*
* Compute norm(inv(A)) = max(x(i)), 1<=i<=n.
*
IX = IDAMAX( N, WORK, 1 )
FERR( J ) = FERR( J )*ABS( WORK( IX ) )
*
* Normalize error.
*
LSTRES = ZERO
DO 80 I = 1, N
LSTRES = MAX( LSTRES, ABS( X( I, J ) ) )
80 CONTINUE
IF( LSTRES.NE.ZERO )
$ FERR( J ) = FERR( J ) / LSTRES
*
90 CONTINUE
*
RETURN
*
* End of DPTRFS
*
END
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