sptrfs.f.html

来自「famous linear algebra library (LAPACK) p」· HTML 代码 · 共 326 行 · 第 1/2 页

</span><span class="comment">*</span><span class="comment">     NZ = maximum number of nonzero elements in each row of A, plus 1
</span><span class="comment">*</span><span class="comment">
</span>      NZ = 4
      EPS = <a name="SLAMCH.148"></a><a href="slamch.f.html#SLAMCH.1">SLAMCH</a>( <span class="string">'Epsilon'</span> )
      SAFMIN = <a name="SLAMCH.149"></a><a href="slamch.f.html#SLAMCH.1">SLAMCH</a>( <span class="string">'Safe minimum'</span> )
      SAFE1 = NZ*SAFMIN
      SAFE2 = SAFE1 / EPS
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">     Do for each right hand side
</span><span class="comment">*</span><span class="comment">
</span>      DO 90 J = 1, NRHS
<span class="comment">*</span><span class="comment">
</span>         COUNT = 1
         LSTRES = THREE
   20    CONTINUE
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Loop until stopping criterion is satisfied.
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Compute residual R = B - A * X.  Also compute
</span><span class="comment">*</span><span class="comment">        abs(A)*abs(x) + abs(b) for use in the backward error bound.
</span><span class="comment">*</span><span class="comment">
</span>         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
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Compute componentwise relative backward error from formula
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        max(i) ( abs(R(i)) / ( abs(A)*abs(X) + abs(B) )(i) )
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        where abs(Z) is the componentwise absolute value of the matrix
</span><span class="comment">*</span><span class="comment">        or vector Z.  If the i-th component of the denominator is less
</span><span class="comment">*</span><span class="comment">        than SAFE2, then SAFE1 is added to the i-th components of the
</span><span class="comment">*</span><span class="comment">        numerator and denominator before dividing.
</span><span class="comment">*</span><span class="comment">
</span>         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
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Test stopping criterion. Continue iterating if
</span><span class="comment">*</span><span class="comment">           1) The residual BERR(J) is larger than machine epsilon, and
</span><span class="comment">*</span><span class="comment">           2) BERR(J) decreased by at least a factor of 2 during the
</span><span class="comment">*</span><span class="comment">              last iteration, and
</span><span class="comment">*</span><span class="comment">           3) At most ITMAX iterations tried.
</span><span class="comment">*</span><span class="comment">
</span>         IF( BERR( J ).GT.EPS .AND. TWO*BERR( J ).LE.LSTRES .AND.
     $       COUNT.LE.ITMAX ) THEN
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">           Update solution and try again.
</span><span class="comment">*</span><span class="comment">
</span>            CALL <a name="SPTTRS.223"></a><a href="spttrs.f.html#SPTTRS.1">SPTTRS</a>( 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
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Bound error from formula
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        norm(X - XTRUE) / norm(X) .le. FERR =
</span><span class="comment">*</span><span class="comment">        norm( abs(inv(A))*
</span><span class="comment">*</span><span class="comment">           ( abs(R) + NZ*EPS*( abs(A)*abs(X)+abs(B) ))) / norm(X)
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        where
</span><span class="comment">*</span><span class="comment">          norm(Z) is the magnitude of the largest component of Z
</span><span class="comment">*</span><span class="comment">          inv(A) is the inverse of A
</span><span class="comment">*</span><span class="comment">          abs(Z) is the componentwise absolute value of the matrix or
</span><span class="comment">*</span><span class="comment">             vector Z
</span><span class="comment">*</span><span class="comment">          NZ is the maximum number of nonzeros in any row of A, plus 1
</span><span class="comment">*</span><span class="comment">          EPS is machine epsilon
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        The i-th component of abs(R)+NZ*EPS*(abs(A)*abs(X)+abs(B))
</span><span class="comment">*</span><span class="comment">        is incremented by SAFE1 if the i-th component of
</span><span class="comment">*</span><span class="comment">        abs(A)*abs(X) + abs(B) is less than SAFE2.
</span><span class="comment">*</span><span class="comment">
</span>         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 )
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Estimate the norm of inv(A).
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Solve M(A) * x = e, where M(A) = (m(i,j)) is given by
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">           m(i,j) =  abs(A(i,j)), i = j,
</span><span class="comment">*</span><span class="comment">           m(i,j) = -abs(A(i,j)), i .ne. j,
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        and e = [ 1, 1, ..., 1 ]'.  Note M(A) = M(L)*D*M(L)'.
</span><span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Solve M(L) * x = e.
</span><span class="comment">*</span><span class="comment">
</span>         WORK( 1 ) = ONE
         DO 60 I = 2, N
            WORK( I ) = ONE + WORK( I-1 )*ABS( EF( I-1 ) )
   60    CONTINUE
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Solve D * M(L)' * x = b.
</span><span class="comment">*</span><span class="comment">
</span>         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
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Compute norm(inv(A)) = max(x(i)), 1&lt;=i&lt;=n.
</span><span class="comment">*</span><span class="comment">
</span>         IX = ISAMAX( N, WORK, 1 )
         FERR( J ) = FERR( J )*ABS( WORK( IX ) )
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">        Normalize error.
</span><span class="comment">*</span><span class="comment">
</span>         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
<span class="comment">*</span><span class="comment">
</span>   90 CONTINUE
<span class="comment">*</span><span class="comment">
</span>      RETURN
<span class="comment">*</span><span class="comment">
</span><span class="comment">*</span><span class="comment">     End of <a name="SPTRFS.299"></a><a href="sptrfs.f.html#SPTRFS.1">SPTRFS</a>
</span><span class="comment">*</span><span class="comment">
</span>      END

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