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