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SUBROUTINE SPTRFS( N, NRHS, D, E, DF, EF, B, LDB, X, LDX, FERR,
$ BERR, WORK, INFO ) * * -- LAPACK routine (version 3.3.1) -- * -- LAPACK is a software package provided by Univ. of Tennessee, -- * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..-- * -- April 2011 -- * * .. Scalar Arguments .. INTEGER INFO, LDB, LDX, N, NRHS * .. * .. Array Arguments .. REAL B( LDB, * ), BERR( * ), D( * ), DF( * ), $ E( * ), EF( * ), FERR( * ), WORK( * ), $ X( LDX, * ) * .. * * Purpose * ======= * * SPTRFS 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) REAL array, dimension (N) * The n diagonal elements of the tridiagonal matrix A. * * E (input) REAL array, dimension (N-1) * The (n-1) subdiagonal elements of the tridiagonal matrix A. * * DF (input) REAL array, dimension (N) * The n diagonal elements of the diagonal matrix D from the * factorization computed by SPTTRF. * * EF (input) REAL array, dimension (N-1) * The (n-1) subdiagonal elements of the unit bidiagonal factor * L from the factorization computed by SPTTRF. * * B (input) REAL 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) REAL array, dimension (LDX,NRHS) * On entry, the solution matrix X, as computed by SPTTRS. * On exit, the improved solution matrix X. * * LDX (input) INTEGER * The leading dimension of the array X. LDX >= max(1,N). * * FERR (output) REAL 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) REAL 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) REAL 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 ) REAL ZERO PARAMETER ( ZERO = 0.0E+0 ) REAL ONE PARAMETER ( ONE = 1.0E+0 ) REAL TWO PARAMETER ( TWO = 2.0E+0 ) REAL THREE PARAMETER ( THREE = 3.0E+0 ) * .. * .. Local Scalars .. INTEGER COUNT, I, IX, J, NZ REAL BI, CX, DX, EPS, EX, LSTRES, S, SAFE1, SAFE2, $ SAFMIN * .. * .. External Subroutines .. EXTERNAL SAXPY, SPTTRS, XERBLA * .. * .. Intrinsic Functions .. INTRINSIC ABS, MAX * .. * .. External Functions .. INTEGER ISAMAX REAL SLAMCH EXTERNAL ISAMAX, SLAMCH * .. * .. 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( 'SPTRFS', -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 = SLAMCH( 'Epsilon' ) SAFMIN = SLAMCH( '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 SPTTRS( N, 1, DF, EF, WORK( N+1 ), N, INFO ) CALL SAXPY( 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 = ISAMAX( 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 = ISAMAX( 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 SPTRFS * END |