示例#1
0
    public static void r8po_sl(double[] a, int lda, int n, ref double[] b, int aIndex = 0, int bIndex = 0)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    R8PO_SL solves a linear system factored by R8PO_FA.
    //
    //  Discussion:
    //
    //    A division by zero will occur if the input factor contains
    //    a zero on the diagonal.  Technically this indicates
    //    singularity but it is usually caused by improper subroutine
    //    arguments.  It will not occur if the subroutines are called
    //    correctly and INFO == 0.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 May 2005
    //
    //  Author:
    //
    //    FORTRAN77 original version by Dongarra, Moler, Bunch, Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Dongarra, Moler, Bunch and Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input, double A[LDA*N], the output from R8PO_FA.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input/output, double B[N].  On input, the right hand side.
    //    On output, the solution.
    //
    {
        int    k;
        double t;

        //
        //  Solve R' * Y = B.
        //
        for (k = 1; k <= n; k++)
        {
            t = BLAS1D.ddot(k - 1, a, 1, b, 1, aIndex + 0 + (k - 1) * lda, bIndex);
            b[bIndex + k - 1] = (b[bIndex + k - 1] - t) / a[aIndex + k - 1 + (k - 1) * lda];
        }

        //
        //  Solve R * X = Y.
        //
        for (k = n; 1 <= k; k--)
        {
            b[bIndex + k - 1] /= a[aIndex + k - 1 + (k - 1) * lda];
            t = -b[bIndex + k - 1];
            BLAS1D.daxpy(k - 1, t, a, 1, ref b, 1, aIndex + 0 + (k - 1) * lda, bIndex);
        }
    }
示例#2
0
    public static int r8po_fa(ref double[] a, int lda, int n)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    R8PO_FA factors a real symmetric positive definite matrix.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 May 2005
    //
    //  Author:
    //
    //    FORTRAN77 original version by Dongarra, Moler, Bunch, Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Dongarra, Moler, Bunch and Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double A[LDA*N].  On input, the symmetric matrix
    //    to be  factored.  Only the diagonal and upper triangle are used.
    //    On output, an upper triangular matrix R so that A = R'*R
    //    where R' is the transpose.  The strict lower triangle is unaltered.
    //    If INFO /= 0, the factorization is not complete.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, int R8PO_FA, error flag.
    //    0, for normal return.
    //    K, signals an error condition.  The leading minor of order K is not
    //    positive definite.
    //
    {
        int info;
        int j;

        for (j = 1; j <= n; j++)
        {
            double s = 0.0;

            int k;
            for (k = 1; k <= j - 1; k++)
            {
                double t = a[k - 1 + (j - 1) * lda] -
                           BLAS1D.ddot(k - 1, a, 1, a, 1, +0 + (k - 1) * lda, +0 + (j - 1) * lda);
                t /= a[k - 1 + (k - 1) * lda];
                a[k - 1 + (j - 1) * lda] = t;
                s += t * t;
            }

            s = a[j - 1 + (j - 1) * lda] - s;

            switch (s)
            {
            case <= 0.0:
                info = j;
                return(info);

            default:
                a[j - 1 + (j - 1) * lda] = Math.Sqrt(s);
                break;
            }
        }

        info = 0;

        return(info);
    }
示例#3
0
    public static double dpoco(ref double[] a, int lda, int n, ref double[] z)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPOCO factors a real symmetric positive definite matrix and estimates its condition.
    //
    //  Discussion:
    //
    //    If RCOND is not needed, DPOFA is slightly faster.
    //
    //    To solve A*X = B, follow DPOCO by DPOSL.
    //
    //    To compute inverse(A)*C, follow DPOCO by DPOSL.
    //
    //    To compute determinant(A), follow DPOCO by DPODI.
    //
    //    To compute inverse(A), follow DPOCO by DPODI.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    06 June 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double A[LDA*N].  On input, the symmetric
    //    matrix to be factored.  Only the diagonal and upper triangle are used.
    //    On output, an upper triangular matrix R so that A = R'*R where R'
    //    is the transpose.  The strict lower triangle is unaltered.
    //    If INFO /= 0, the factorization is not complete.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, double Z[N], a work vector whose contents are usually
    //    unimportant.  If A is close to a singular matrix, then Z is an
    //    approximate null vector in the sense that
    //      norm(A*Z) = RCOND * norm(A) * norm(Z).
    //    If INFO /= 0, Z is unchanged.
    //
    //    Output, double DPOCO, an estimate of the reciprocal
    //    condition of A.  For the system A*X = B, relative perturbations in
    //    A and B of size EPSILON may cause relative perturbations in X of
    //    size EPSILON/RCOND.  If RCOND is so small that the logical expression
    //      1.0D+00 + RCOND == 1.0D+00
    //    is true, then A may be singular to working precision.  In particular,
    //    RCOND is zero if exact singularity is detected or the estimate underflows.
    //
    {
        int    i;
        int    j;
        int    k;
        double rcond;
        double s;
        double t;

        //
        //  Find norm of A using only upper half.
        //
        for (j = 1; j <= n; j++)
        {
            z[j - 1] = BLAS1D.dasum(j, a, 1, index: +0 + (j - 1) * lda);
            for (i = 1; i <= j - 1; i++)
            {
                z[i - 1] += Math.Abs(a[i - 1 + (j - 1) * lda]);
            }
        }

        double anorm = 0.0;

        for (i = 1; i <= n; i++)
        {
            anorm = Math.Max(anorm, z[i - 1]);
        }

        //
        //  Factor.
        //
        int info = DPOFA.dpofa(ref a, lda, n);

        if (info != 0)
        {
            rcond = 0.0;
            return(rcond);
        }

        //
        //  RCOND = 1/(norm(A)*(estimate of norm(inverse(A)))).
        //
        //  Estimate = norm(Z)/norm(Y) where A*Z = Y and A*Y = E.
        //
        //  The components of E are chosen to cause maximum local
        //  growth in the elements of W where R'*W = E.
        //
        //  The vectors are frequently rescaled to avoid overflow.
        //
        //  Solve R' * W = E.
        //
        double ek = 1.0;

        for (i = 1; i <= n; i++)
        {
            z[i - 1] = 0.0;
        }

        for (k = 1; k <= n; k++)
        {
            if (z[k - 1] != 0.0)
            {
                ek *= typeMethods.r8_sign(-z[k - 1]);
            }

            if (a[k - 1 + (k - 1) * lda] < Math.Abs(ek - z[k - 1]))
            {
                s = a[k - 1 + (k - 1) * lda] / Math.Abs(ek - z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ek = s * ek;
            }

            double wk  = ek - z[k - 1];
            double wkm = -ek - z[k - 1];
            s = Math.Abs(wk);
            double sm = Math.Abs(wkm);
            wk  /= a[k - 1 + (k - 1) * lda];
            wkm /= a[k - 1 + (k - 1) * lda];

            if (k + 1 <= n)
            {
                for (j = k + 1; j <= n; j++)
                {
                    sm       += Math.Abs(z[j - 1] + wkm * a[k - 1 + (j - 1) * lda]);
                    z[j - 1] += wk * a[k - 1 + (j - 1) * lda];
                    s        += Math.Abs(z[j - 1]);
                }

                if (s < sm)
                {
                    t  = wkm - wk;
                    wk = wkm;
                    for (j = k + 1; j <= n; j++)
                    {
                        z[j - 1] += t * a[k - 1 + (j - 1) * lda];
                    }
                }
            }

            z[k - 1] = wk;
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        //
        //  Solve R * Y = W.
        //
        for (k = n; 1 <= k; k--)
        {
            if (a[k - 1 + (k - 1) * lda] < Math.Abs(z[k - 1]))
            {
                s = a[k - 1 + (k - 1) * lda] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }
            }

            z[k - 1] /= a[k - 1 + (k - 1) * lda];
            t         = -z[k - 1];
            BLAS1D.daxpy(k - 1, t, a, 1, ref z, 1, xIndex: +0 + (k - 1) * lda);
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        double ynorm = 1.0;

        //
        //  Solve R' * V = Y.
        //
        for (k = 1; k <= n; k++)
        {
            z[k - 1] -= BLAS1D.ddot(k - 1, a, 1, z, 1, xIndex: +0 + (k - 1) * lda);

            if (a[k - 1 + (k - 1) * lda] < Math.Abs(z[k - 1]))
            {
                s = a[k - 1 + (k - 1) * lda] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            z[k - 1] /= a[k - 1 + (k - 1) * lda];
        }

        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;
        //
        //  Solve R * Z = V.
        //
        for (k = n; 1 <= k; k--)
        {
            if (a[k - 1 + (k - 1) * lda] < Math.Abs(z[k - 1]))
            {
                s = a[k - 1 + (k - 1) * lda] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            z[k - 1] /= a[k - 1 + (k - 1) * lda];
            t         = -z[k - 1];
            BLAS1D.daxpy(k - 1, t, a, 1, ref z, 1, xIndex: +0 + (k - 1) * lda);
        }

        //
        //  Make ZNORM = 1.0.
        //
        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;

        if (anorm != 0.0)
        {
            rcond = ynorm / anorm;
        }
        else
        {
            rcond = 0.0;
        }

        return(rcond);
    }
示例#4
0
    public static double dgeco(ref double[] a, int lda, int n, ref int[] ipvt, ref double[] z)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGECO factors a real matrix and estimates its condition number.
    //
    //  Discussion:
    //
    //    If RCOND is not needed, DGEFA is slightly faster.
    //
    //    To solve A * X = B, follow DGECO by DGESL.
    //
    //    To compute inverse ( A ) * C, follow DGECO by DGESL.
    //
    //    To compute determinant ( A ), follow DGECO by DGEDI.
    //
    //    To compute inverse ( A ), follow DGECO by DGEDI.
    //
    //    For the system A * X = B, relative perturbations in A and B
    //    of size EPSILON may cause relative perturbations in X of size
    //    EPSILON/RCOND.
    //
    //    If RCOND is so small that the logical expression
    //      1.0D+00 + RCOND == 1.0D+00
    //    is true, then A may be singular to working precision.  In particular,
    //    RCOND is zero if exact singularity is detected or the estimate
    //    underflows.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    25 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double A[LDA*N].  On input, a matrix to be
    //    factored.  On output, the LU factorization of the matrix.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix A.
    //
    //    Output, int IPVT[N], the pivot indices.
    //
    //    Output, double Z[N], a work vector whose contents are usually
    //    unimportant.  If A is close to a singular matrix, then Z is an
    //    approximate null vector in the sense that
    //      norm ( A * Z ) = RCOND * norm ( A ) * norm ( Z ).
    //
    //    Output, double DGECO, the value of RCOND, an estimate
    //    of the reciprocal condition number of A.
    //
    {
        int    i;
        int    j;
        int    k;
        int    l;
        double rcond;
        double s;
        double t;
        //
        //  Compute the L1 norm of A.
        //
        double anorm = 0.0;

        for (j = 1; j <= n; j++)
        {
            anorm = Math.Max(anorm, BLAS1D.dasum(n, a, 1, index:  +0 + (j - 1) * lda));
        }

        //
        //  Compute the LU factorization.
        //
        DGEFA.dgefa(ref a, lda, n, ref ipvt);
        //
        //  RCOND = 1 / ( norm(A) * (estimate of norm(inverse(A))) )
        //
        //  estimate of norm(inverse(A)) = norm(Z) / norm(Y)
        //
        //  where
        //    A * Z = Y
        //  and
        //    A' * Y = E
        //
        //  The components of E are chosen to cause maximum local growth in the
        //  elements of W, where U'*W = E.  The vectors are frequently rescaled
        //  to avoid overflow.
        //
        //  Solve U' * W = E.
        //
        double ek = 1.0;

        for (i = 1; i <= n; i++)
        {
            z[i - 1] = 0.0;
        }

        for (k = 1; k <= n; k++)
        {
            if (z[k - 1] != 0.0)
            {
                ek *= typeMethods.r8_sign(-z[k - 1]);
            }

            if (Math.Abs(a[k - 1 + (k - 1) * lda]) < Math.Abs(ek - z[k - 1]))
            {
                s = Math.Abs(a[k - 1 + (k - 1) * lda]) / Math.Abs(ek - z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ek = s * ek;
            }

            double wk  = ek - z[k - 1];
            double wkm = -ek - z[k - 1];
            s = Math.Abs(wk);
            double sm = Math.Abs(wkm);

            if (a[k - 1 + (k - 1) * lda] != 0.0)
            {
                wk  /= a[k - 1 + (k - 1) * lda];
                wkm /= a[k - 1 + (k - 1) * lda];
            }
            else
            {
                wk  = 1.0;
                wkm = 1.0;
            }

            if (k + 1 <= n)
            {
                for (j = k + 1; j <= n; j++)
                {
                    sm       += Math.Abs(z[j - 1] + wkm * a[k - 1 + (j - 1) * lda]);
                    z[j - 1] += wk * a[k - 1 + (j - 1) * lda];
                    s        += Math.Abs(z[j - 1]);
                }

                if (s < sm)
                {
                    t  = wkm - wk;
                    wk = wkm;
                    for (i = k + 1; i <= n; i++)
                    {
                        z[i - 1] += t * a[k - 1 + (i - 1) * lda];
                    }
                }
            }

            z[k - 1] = wk;
        }

        t = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= t;
        }

        //
        //  Solve L' * Y = W
        //
        for (k = n; 1 <= k; k--)
        {
            z[k - 1] += BLAS1D.ddot(n - k, a, 1, z, 1, xIndex:  +k + (k - 1) * lda, yIndex: +k);

            switch (Math.Abs(z[k - 1]))
            {
            case > 1.0:
            {
                t = Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] /= t;
                }

                break;
            }
            }

            l = ipvt[k - 1];

            t        = z[l - 1];
            z[l - 1] = z[k - 1];
            z[k - 1] = t;
        }

        t = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= t;
        }

        double ynorm = 1.0;

        //
        //  Solve L * V = Y.
        //
        for (k = 1; k <= n; k++)
        {
            l = ipvt[k - 1];

            t        = z[l - 1];
            z[l - 1] = z[k - 1];
            z[k - 1] = t;

            for (i = k + 1; i <= n; i++)
            {
                z[i - 1] += t * a[i - 1 + (k - 1) * lda];
            }

            switch (Math.Abs(z[k - 1]))
            {
            case > 1.0:
            {
                ynorm /= Math.Abs(z[k - 1]);
                t      = Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] /= t;
                }

                break;
            }
            }
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        ynorm /= s;
        //
        //  Solve U * Z = V.
        //
        for (k = n; 1 <= k; k--)
        {
            if (Math.Abs(a[k - 1 + (k - 1) * lda]) < Math.Abs(z[k - 1]))
            {
                s = Math.Abs(a[k - 1 + (k - 1) * lda]) / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            if (a[k - 1 + (k - 1) * lda] != 0.0)
            {
                z[k - 1] /= a[k - 1 + (k - 1) * lda];
            }
            else
            {
                z[k - 1] = 1.0;
            }

            for (i = 1; i <= k - 1; i++)
            {
                z[i - 1] -= z[k - 1] * a[i - 1 + (k - 1) * lda];
            }
        }

        //
        //  Normalize Z in the L1 norm.
        //
        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;

        if (anorm != 0.0)
        {
            rcond = ynorm / anorm;
        }
        else
        {
            rcond = 0.0;
        }

        return(rcond);
    }
示例#5
0
    public static int dchdd(ref double[] r, int ldr, int p, double[] x, ref double[] z, int ldz,
                            int nz, double[] y, ref double[] rho, ref double[] c, ref double[] s)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DCHDD downdates an augmented Cholesky decomposition.
    //
    //  Discussion:
    //
    //    DCHDD can also downdate the triangular factor of an augmented QR
    //    decomposition.
    //
    //    Specifically, given an upper triangular matrix R of order P, a
    //    row vector X, a column vector Z, and a scalar Y, DCHDD
    //    determines an orthogonal matrix U and a scalar ZETA such that
    //
    //          (R   Z )     (RR  ZZ)
    //      U * (      )  =  (      ),
    //          (0 ZETA)     ( X   Y)
    //
    //    where RR is upper triangular.
    //
    //    If R and Z have been obtained from the factorization of a least squares
    //    problem, then RR and ZZ are the factors corresponding to the problem
    //    with the observation (X,Y) removed.  In this case, if RHO
    //    is the norm of the residual vector, then the norm of
    //    the residual vector of the downdated problem is
    //    sqrt ( RHO * RHO - ZETA * ZETA ). DCHDD will simultaneously downdate
    //    several triplets (Z, Y, RHO) along with R.
    //
    //    For a less terse description of what DCHDD does and how
    //    it may be applied, see the LINPACK guide.
    //
    //    The matrix U is determined as the product U(1)*...*U(P)
    //    where U(I) is a rotation in the (P+1,I)-plane of the form
    //
    //      ( C(I)      -S(I)    )
    //      (                    ).
    //      ( S(I)       C(I)    )
    //
    //    The rotations are chosen so that C(I) is real.
    //
    //    The user is warned that a given downdating problem may be impossible
    //    to accomplish or may produce inaccurate results.  For example, this
    //    can happen if X is near a vector whose removal will reduce the
    //    rank of R.  Beware.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 June 2009
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double R[LDR*P], the upper triangular matrix that
    //    is to be  downdated.  The part of R below the diagonal is not referenced.
    //
    //    Input, int LDR, the leading dimension of the array R.
    //    LDR must be at least P.
    //
    //    Input, int P, the order of the matrix R.
    //
    //    Input, double X[P], the row vector that is to be removed from R.
    //
    //    Input/output, double Z[LDZ*NZ], an array of NZ P-vectors
    //    which are to be downdated along with R.
    //
    //    Input, int LDZ, the leading dimension of the array Z.
    //    LDZ must be at least P.
    //
    //    Input, int NZ, the number of vectors to be downdated.
    //    NZ may be zero, in which case Z, Y, and RHO are not referenced.
    //
    //    Input, double Y[NZ], the scalars for the downdating of
    //    the vectors Z.
    //
    //    Input/output, double RHO[NZ], the norms of the residual vectors.
    //    On output these have been changed along with R and Z.
    //
    //    Output, double C[P], S[P], the cosines and sines of the
    //    transforming rotations.
    //
    //    Output, int DCHDD, return flag.
    //     0, the entire downdating was successful.
    //    -1, if R could not be downdated.  In this case, all quantities
    //        are left unaltered.
    //     1, if some RHO could not be downdated.  The offending RHO's are
    //        set to -1.
    //
    {
        int i;
        int ii;
        int j;
        //
        //  Solve R' * A = X, placing the result in the array S.
        //
        int info = 0;

        s[0] = x[0] / r[0 + 0 * ldr];

        for (j = 2; j <= p; j++)
        {
            s[j - 1]  = x[j - 1] - BLAS1D.ddot(j - 1, r, 1, s, 1, xIndex: +0 + (j - 1) * ldr);
            s[j - 1] /= r[j - 1 + (j - 1) * ldr];
        }

        double norm = BLAS1D.dnrm2(p, s, 1);

        switch (norm)
        {
        case >= 1.0:
            info = -1;
            return(info);
        }

        double alpha = Math.Sqrt(1.0 - norm * norm);

        //
        //  Determine the transformations.
        //
        for (ii = 1; ii <= p; ii++)
        {
            i = p - ii + 1;
            double scale = alpha + Math.Abs(s[i - 1]);
            double a     = alpha / scale;
            double b     = s[i - 1] / scale;
            norm     = Math.Sqrt(a * a + b * b);
            c[i - 1] = a / norm;
            s[i - 1] = b / norm;
            alpha    = scale * norm;
        }

        //
        //  Apply the transformations to R.
        //
        for (j = 1; j <= p; j++)
        {
            double xx = 0.0;
            for (ii = 1; ii <= j; ii++)
            {
                i = j - ii + 1;
                double t = c[i - 1] * xx + s[i - 1] * r[i - 1 + (j - 1) * ldr];
                r[i - 1 + (j - 1) * ldr] = c[i - 1] * r[i - 1 + (j - 1) * ldr] - s[i - 1] * xx;
                xx = t;
            }
        }

        //
        //  If required, downdate Z and RHO.
        //
        for (j = 1; j <= nz; j++)
        {
            double zeta = y[j - 1];
            for (i = 1; i <= p; i++)
            {
                z[i - 1 + (j - 1) * ldz] = (z[i - 1 + (j - 1) * ldz] - s[i - 1] * zeta) / c[i - 1];
                zeta = c[i - 1] * zeta - s[i - 1] * z[i - 1 + (j - 1) * ldz];
            }

            double azeta = Math.Abs(zeta);

            if (rho[j - 1] < azeta)
            {
                info       = 1;
                rho[j - 1] = -1.0;
            }
            else
            {
                rho[j - 1] *= Math.Sqrt(1.0 - Math.Pow(azeta / rho[j - 1], 2));
            }
        }

        return(info);
    }
示例#6
0
    private static void ddot_test()

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DDOT_TEST demonstrates DDOT.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    15 May 2006
    //
    //  Author:
    //
    //    John Burkardt
    //
    {
        const int N   = 5;
        const int LDA = 10;
        const int LDB = 7;
        const int LDC = 6;

        double[] a = new double[LDA * LDA];
        double[] b = new double[LDB * LDB];
        double[] c = new double[LDC * LDC];
        double[] x = new double[N];
        double[] y = new double[N];

        Console.WriteLine("");
        Console.WriteLine("DDOT_TEST");
        Console.WriteLine("  DDOT computes the dot product of vectors.");
        Console.WriteLine("");

        for (int i = 0; i < N; i++)
        {
            x[i] = i + 1;
        }

        for (int i = 0; i < N; i++)
        {
            y[i] = -(double)(i + 1);
        }

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                a[i + j * LDA] = i + 1 + j + 1;
            }
        }

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                b[i + j * LDB] = i + 1 - (j + 1);
            }
        }

        double sum1 = BLAS1D.ddot(N, x, 1, y, 1);

        Console.WriteLine("");
        Console.WriteLine("  Dot product of X and Y is " + sum1 + "");
        //
        //  To multiply a ROW of a matrix A times a vector X, we need to
        //  specify the increment between successive entries of the row of A:
        //
        sum1 = BLAS1D.ddot(N, a, LDA, x, 1, 1);

        Console.WriteLine("");
        Console.WriteLine("  Product of row 2 of A and X is " + sum1 + "");
        //
        //  Product of a column of A and a vector is simpler:
        //
        sum1 = BLAS1D.ddot(N, a, 1, x, 1, +0 + 1 * LDA);

        Console.WriteLine("");
        Console.WriteLine("  Product of column 2 of A and X is " + sum1 + "");
        //
        //  Here's how matrix multiplication, c = a*b, could be done
        //  with DDOT:
        //
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                c[i + j * LDC] = BLAS1D.ddot(N, a, LDA, b, 1, +i, +0 + j * LDB);
            }
        }

        Console.WriteLine("");
        Console.WriteLine("  Matrix product computed with DDOT:");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            string cout = "";
            for (int j = 0; j < N; j++)
            {
                cout += "  " + c[i + j * LDC].ToString(CultureInfo.InvariantCulture).PadLeft(14);
            }

            Console.WriteLine(cout);
        }
    }
示例#7
0
    public static void dgesl(double[] a, int lda, int n, int[] ipvt, ref double[] b, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGESL solves a real general linear system A * X = B.
    //
    //  Discussion:
    //
    //    DGESL can solve either of the systems A * X = B or A' * X = B.
    //
    //    The system matrix must have been factored by DGECO or DGEFA.
    //
    //    A division by zero will occur if the input factor contains a
    //    zero on the diagonal.  Technically this indicates singularity
    //    but it is often caused by improper arguments or improper
    //    setting of LDA.  It will not occur if the subroutines are
    //    called correctly and if DGECO has set 0.0 < RCOND
    //    or DGEFA has set INFO == 0.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    16 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input, double A[LDA*N], the output from DGECO or DGEFA.
    //
    //    Input, int LDA, the leading dimension of A.
    //
    //    Input, int N, the order of the matrix A.
    //
    //    Input, int IPVT[N], the pivot vector from DGECO or DGEFA.
    //
    //    Input/output, double B[N].
    //    On input, the right hand side vector.
    //    On output, the solution vector.
    //
    //    Input, int JOB.
    //    0, solve A * X = B;
    //    nonzero, solve A' * X = B.
    //
    {
        int    k;
        int    l;
        double t;

        switch (job)
        {
        //
        //  Solve A * X = B.
        //
        case 0:
        {
            for (k = 1; k <= n - 1; k++)
            {
                l = ipvt[k - 1];
                t = b[l - 1];

                if (l != k)
                {
                    b[l - 1] = b[k - 1];
                    b[k - 1] = t;
                }

                BLAS1D.daxpy(n - k, t, a, 1, ref b, 1, +k + (k - 1) * lda, +k);
            }

            for (k = n; 1 <= k; k--)
            {
                b[k - 1] /= a[k - 1 + (k - 1) * lda];
                t         = -b[k - 1];
                BLAS1D.daxpy(k - 1, t, a, 1, ref b, 1, +0 + (k - 1) * lda);
            }

            break;
        }

        //
        default:
        {
            for (k = 1; k <= n; k++)
            {
                t        = BLAS1D.ddot(k - 1, a, 1, b, 1, +0 + (k - 1) * lda);
                b[k - 1] = (b[k - 1] - t) / a[k - 1 + (k - 1) * lda];
            }

            for (k = n - 1; 1 <= k; k--)
            {
                b[k - 1] += BLAS1D.ddot(n - k, a, 1, b, 1, +k + (k - 1) * lda, +k);
                l         = ipvt[k - 1];

                if (l == k)
                {
                    continue;
                }

                t        = b[l - 1];
                b[l - 1] = b[k - 1];
                b[k - 1] = t;
            }

            break;
        }
        }
    }
示例#8
0
    public static double dgbco(ref double[] abd, int lda, int n, int ml, int mu, ref int[] ipvt,
                               double[] z)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGBCO factors a real band matrix and estimates its condition.
    //
    //  Discussion:
    //
    //    If RCOND is not needed, DGBFA is slightly faster.
    //
    //    To solve A*X = B, follow DGBCO by DGBSL.
    //
    //    To compute inverse(A)*C, follow DGBCO by DGBSL.
    //
    //    To compute determinant(A), follow DGBCO by DGBDI.
    //
    //  Example:
    //
    //    If the original matrix is
    //
    //      11 12 13  0  0  0
    //      21 22 23 24  0  0
    //       0 32 33 34 35  0
    //       0  0 43 44 45 46
    //       0  0  0 54 55 56
    //       0  0  0  0 65 66
    //
    //    then for proper band storage,
    //
    //      N = 6, ML = 1, MU = 2, 5 <= LDA and ABD should contain
    //
    //       *  *  *  +  +  +      * = not used
    //       *  * 13 24 35 46      + = used for pivoting
    //       * 12 23 34 45 56
    //      11 22 33 44 55 66
    //      21 32 43 54 65  *
    //
    //  Band storage:
    //
    //    If A is a band matrix, the following program segment
    //    will set up the input.
    //
    //      ml = (band width below the diagonal)
    //      mu = (band width above the diagonal)
    //      m = ml + mu + 1
    //
    //      do j = 1, n
    //        i1 = max ( 1, j-mu )
    //        i2 = min ( n, j+ml )
    //        do i = i1, i2
    //          k = i - j + m
    //          abd(k,j) = a(i,j)
    //        }
    //      }
    //
    //    This uses rows ML+1 through 2*ML+MU+1 of ABD.  In addition, the first
    //    ML rows in ABD are used for elements generated during the
    //    triangularization.  The total number of rows needed in ABD is
    //    2*ML+MU+1.  The ML+MU by ML+MU upper left triangle and the ML by ML
    //    lower right triangle are not referenced.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    07 June 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double ABD[LDA*N].  On input, the matrix in band
    //    storage.  The columns of the matrix are stored in the columns of ABD and
    //    the diagonals of the matrix are stored in rows ML+1 through 2*ML+MU+1
    //    of ABD.  On output, an upper triangular matrix in band storage and
    //    the multipliers which were used to obtain it.  The factorization can
    //    be written A = L*U where L is a product of permutation and unit lower
    //    triangular matrices and U is upper triangular.
    //
    //    Input, int LDA, the leading dimension of the array ABD.
    //    2*ML + MU + 1 <= LDA is required.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int ML, MU, the number of diagonals below and above the
    //    main diagonal.  0 <= ML < N, 0 <= MU < N.
    //
    //    Output, int IPVT[N], the pivot indices.
    //
    //    Workspace, double Z[N], a work vector whose contents are
    //    usually unimportant.  If A is close to a singular matrix, then Z is an
    //    approximate null vector in the sense that
    //      norm(A*Z) = RCOND * norm(A) * norm(Z).
    //
    //    Output, double DGBCO, an estimate of the reciprocal condition number RCOND
    //    of A.  For the system A*X = B, relative perturbations in A and B of size
    //    EPSILON may cause relative perturbations in X of size EPSILON/RCOND.
    //    If RCOND is so small that the logical expression
    //      1.0 + RCOND == 1.0D+00
    //    is true, then A may be singular to working precision.  In particular,
    //    RCOND is zero if exact singularity is detected or the estimate underflows.
    //
    {
        int    i;
        int    j;
        int    k;
        int    lm;
        double rcond;
        double s;
        double t;
        //
        //  Compute the 1-norm of A.
        //
        double anorm = 0.0;
        int    l     = ml + 1;
        int    is_   = l + mu;

        for (j = 1; j <= n; j++)
        {
            anorm = Math.Max(anorm, BLAS1D.dasum(l, abd, 1, index: +is_ - 1 + (j - 1) * lda));
            if (ml + 1 < is_)
            {
                is_ -= 1;
            }

            if (j <= mu)
            {
                l += 1;
            }

            if (n - ml <= j)
            {
                l -= 1;
            }
        }

        //
        //  Factor.
        //
        DGBFA.dgbfa(ref abd, lda, n, ml, mu, ref ipvt);
        //
        //  RCOND = 1/(norm(A)*(estimate of norm(inverse(A)))).
        //
        //  Estimate = norm(Z)/norm(Y) where  a*z = y  and A'*Y = E.
        //
        //  A' is the transpose of A.  The components of E are
        //  chosen to cause maximum local growth in the elements of W where
        //  U'*W = E.  The vectors are frequently rescaled to avoid
        //  overflow.
        //
        //  Solve U' * W = E.
        //
        double ek = 1.0;

        for (i = 1; i <= n; i++)
        {
            z[i - 1] = 0.0;
        }

        int m  = ml + mu + 1;
        int ju = 0;

        for (k = 1; k <= n; k++)
        {
            if (z[k - 1] != 0.0)
            {
                ek *= typeMethods.r8_sign(-z[k - 1]);
            }

            if (Math.Abs(abd[m - 1 + (k - 1) * lda]) < Math.Abs(ek - z[k - 1]))
            {
                s = Math.Abs(abd[m - 1 + (k - 1) * lda]) / Math.Abs(ek - z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ek = s * ek;
            }

            double wk  = ek - z[k - 1];
            double wkm = -ek - z[k - 1];
            s = Math.Abs(wk);
            double sm = Math.Abs(wkm);

            if (abd[m - 1 + (k - 1) * lda] != 0.0)
            {
                wk  /= abd[m - 1 + (k - 1) * lda];
                wkm /= abd[m - 1 + (k - 1) * lda];
            }
            else
            {
                wk  = 1.0;
                wkm = 1.0;
            }

            ju = Math.Min(Math.Max(ju, mu + ipvt[k - 1]), n);
            int mm = m;

            if (k + 1 <= ju)
            {
                for (j = k + 1; j <= ju; j++)
                {
                    mm       -= 1;
                    sm       += Math.Abs(z[j - 1] + wkm * abd[mm - 1 + (j - 1) * lda]);
                    z[j - 1] += wk * abd[mm - 1 + (j - 1) * lda];
                    s        += Math.Abs(z[j - 1]);
                }

                if (s < sm)
                {
                    t  = wkm - wk;
                    wk = wkm;
                    mm = m;
                    for (j = k + 1; j <= ju; ju++)
                    {
                        mm       -= 1;
                        z[j - 1] += t * abd[mm - 1 + (j - 1) * lda];
                    }
                }
            }

            z[k - 1] = wk;
        }

        s = BLAS1D.dasum(n, z, 1);

        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        //
        //  Solve L' * Y = W.
        //
        for (k = n; 1 <= k; k--)
        {
            lm = Math.Min(ml, n - k);

            if (k < m)
            {
                z[k - 1] += BLAS1D.ddot(lm, abd, 1, z, 1, xIndex: +m + (k - 1) * lda, yIndex: +k);
            }

            switch (Math.Abs(z[k - 1]))
            {
            case > 1.0:
            {
                s = 1.0 / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                break;
            }
            }

            l        = ipvt[k - 1];
            t        = z[l - 1];
            z[l - 1] = z[k - 1];
            z[k - 1] = t;
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        double ynorm = 1.0;

        //
        //  Solve L * V = Y.
        //
        for (k = 1; k <= n; k++)
        {
            l        = ipvt[k - 1];
            t        = z[l - 1];
            z[l - 1] = z[k - 1];
            z[k - 1] = t;
            lm       = Math.Min(ml, n - k);

            if (k < n)
            {
                BLAS1D.daxpy(lm, t, abd, 1, ref z, 1, xIndex: +m + (k - 1) * lda, yIndex: +k);
            }

            switch (Math.Abs(z[k - 1]))
            {
            case > 1.0:
            {
                s = 1.0 / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
                break;
            }
            }
        }

        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;
        //
        //  Solve U * Z = W.
        //
        for (k = n; 1 <= k; k--)
        {
            if (Math.Abs(abd[m - 1 + (k - 1) * lda]) < Math.Abs(z[k - 1]))
            {
                s = Math.Abs(abd[m - 1 + (k - 1) * lda]) / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            if (abd[m - 1 + (k - 1) * lda] != 0.0)
            {
                z[k - 1] /= abd[m - 1 + (k - 1) * lda];
            }
            else
            {
                z[k - 1] = 1.0;
            }

            lm = Math.Min(k, m) - 1;
            int la = m - lm;
            int lz = k - lm;
            t = -z[k - 1];
            BLAS1D.daxpy(lm, t, abd, 1, ref z, 1, xIndex:  +la - 1 + (k - 1) * lda, yIndex:  +lz - 1);
        }

        //
        //  Make ZNORM = 1.0.
        //
        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;

        if (anorm != 0.0)
        {
            rcond = ynorm / anorm;
        }
        else
        {
            rcond = 0.0;
        }

        return(rcond);
    }
示例#9
0
    public static void dspsl(double[] ap, int n, int[] kpvt, ref double[] b)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DSPSL solves the real symmetric system factored by DSPFA.
    //
    //  Discussion:
    //
    //    To compute inverse(A) * C where C is a matrix with P columns:
    //
    //      call dspfa ( ap, n, kpvt, info )
    //
    //      if ( info /= 0 ) go to ...
    //
    //      do j = 1, p
    //        call dspsl ( ap, n, kpvt, c(1,j) )
    //      end do
    //
    //    A division by zero may occur if DSPCO has set RCOND == 0.0D+00
    //    or DSPFA has set INFO /= 0.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    25 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input, double AP[(N*(N+1))/2], the output from DSPFA.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int KPVT[N], the pivot vector from DSPFA.
    //
    //    Input/output, double B[N].  On input, the right hand side.
    //    On output, the solution.
    //
    {
        int    kp;
        double temp;
        //
        //  Loop backward applying the transformations and D inverse to B.
        //
        int k  = n;
        int ik = n * (n - 1) / 2;

        while (0 < k)
        {
            int kk = ik + k;

            switch (kpvt[k - 1])
            {
            case >= 0:
            {
                //
                //  1 x 1 pivot block.
                //
                if (k != 1)
                {
                    kp = kpvt[k - 1];
                    //
                    //  Interchange.
                    //
                    if (kp != k)
                    {
                        temp      = b[k - 1];
                        b[k - 1]  = b[kp - 1];
                        b[kp - 1] = temp;
                    }

                    //
                    //  Apply the transformation.
                    //
                    BLAS1D.daxpy(k - 1, b[k - 1], ap, 1, ref b, 1, xIndex: +ik);
                }

                //
                //  Apply D inverse.
                //
                b[k - 1] /= ap[kk - 1];
                k        -= 1;
                ik       -= k;
                break;
            }

            default:
            {
                //
                //  2 x 2 pivot block.
                //
                int ikm1 = ik - (k - 1);

                if (k != 2)
                {
                    kp = Math.Abs(kpvt[k - 1]);
                    //
                    //  Interchange.
                    //
                    if (kp != k - 1)
                    {
                        temp      = b[k - 2];
                        b[k - 2]  = b[kp - 1];
                        b[kp - 1] = temp;
                    }

                    //
                    //  Apply the transformation.
                    //
                    BLAS1D.daxpy(k - 2, b[k - 1], ap, 1, ref b, 1, xIndex: +ik);
                    BLAS1D.daxpy(k - 2, b[k - 2], ap, 1, ref b, 1, xIndex: +ikm1);
                }

                //
                //  Apply D inverse.
                //
                int km1k = ik + k - 1;
                kk = ik + k;
                double ak     = ap[kk - 1] / ap[km1k - 1];
                int    km1km1 = ikm1 + k - 1;
                double akm1   = ap[km1km1 - 1] / ap[km1k - 1];
                double bk     = b[k - 1] / ap[km1k - 1];
                double bkm1   = b[k - 2] / ap[km1k - 1];
                double denom  = ak * akm1 - 1.0;
                b[k - 1] = (akm1 * bk - bkm1) / denom;
                b[k - 2] = (ak * bkm1 - bk) / denom;
                k       -= 2;
                ik       = ik - (k + 1) - k;
                break;
            }
            }
        }

        //
        //  Loop forward applying the transformations.
        //
        k  = 1;
        ik = 0;

        while (k <= n)
        {
            switch (kpvt[k - 1])
            {
            case >= 0:
            {
                //
                //  1 x 1 pivot block.
                //
                if (k != 1)
                {
                    //
                    //  Apply the transformation.
                    //
                    b[k - 1] += BLAS1D.ddot(k - 1, ap, 1, b, 1, xIndex: +ik);
                    kp        = kpvt[k - 1];
                    //
                    //  Interchange.
                    //
                    if (kp != k)
                    {
                        temp      = b[k - 1];
                        b[k - 1]  = b[kp - 1];
                        b[kp - 1] = temp;
                    }
                }

                ik += k;
                k  += 1;
                break;
            }

            default:
            {
                //
                //  2 x 2 pivot block.
                //
                if (k != 1)
                {
                    //
                    //  Apply the transformation.
                    //
                    b[k - 1] += BLAS1D.ddot(k - 1, ap, 1, b, 1, xIndex: +ik);
                    int ikp1 = ik + k;
                    b[k] += BLAS1D.ddot(k - 1, ap, 1, b, 1, xIndex: +ikp1);
                    kp    = Math.Abs(kpvt[k - 1]);
                    //
                    //  Interchange.
                    //
                    if (kp != k)
                    {
                        temp      = b[k - 1];
                        b[k - 1]  = b[kp - 1];
                        b[kp - 1] = temp;
                    }
                }

                ik = ik + k + k + 1;
                k += 2;
                break;
            }
            }
        }
    }
示例#10
0
    public static double dppco(ref double[] ap, int n, ref double[] z)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPPCO factors a real symmetric positive definite matrix in packed form.
    //
    //  Discussion:
    //
    //    DPPCO also estimates the condition of the matrix.
    //
    //    If RCOND is not needed, DPPFA is slightly faster.
    //
    //    To solve A*X = B, follow DPPCO by DPPSL.
    //
    //    To compute inverse(A)*C, follow DPPCO by DPPSL.
    //
    //    To compute determinant(A), follow DPPCO by DPPDI.
    //
    //    To compute inverse(A), follow DPPCO by DPPDI.
    //
    //  Packed storage:
    //
    //    The following program segment will pack the upper triangle of
    //    a symmetric matrix.
    //
    //      k = 0
    //      do j = 1, n
    //        do i = 1, j
    //          k = k + 1
    //          ap[k-1] = a(i,j)
    //        }
    //      }
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    07 June 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double AP[N*(N+1)/2].  On input, the packed
    //    form of a symmetric matrix A.  The columns of the upper triangle are
    //    stored sequentially in a one-dimensional array.  On output, an upper
    //    triangular matrix R, stored in packed form, so that A = R'*R.
    //    If INFO /= 0, the factorization is not complete.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, double Z[N], a work vector whose contents are usually
    //    unimportant.  If A is singular to working precision, then Z is an
    //    approximate null vector in the sense that
    //      norm(A*Z) = RCOND * norm(A) * norm(Z).
    //    If INFO /= 0, Z is unchanged.
    //
    //    Output, double DPPCO, an estimate of the reciprocal condition number RCOND
    //    of A.  For the system A*X = B, relative perturbations in A and B of size
    //    EPSILON may cause relative perturbations in X of size EPSILON/RCOND.
    //    If RCOND is so small that the logical expression
    //      1.0 + RCOND == 1.0D+00
    //    is true, then A may be singular to working precision.  In particular,
    //    RCOND is zero if exact singularity is detected or the estimate underflows.
    //
    {
        int    i;
        int    j;
        int    k;
        double rcond;
        double s;
        double t;
        //
        //  Find the norm of A.
        //
        int j1 = 1;

        for (j = 1; j <= n; j++)
        {
            z[j - 1] = BLAS1D.dasum(j, ap, 1, index: +j1 - 1);
            int ij = j1;
            j1 += j;
            for (i = 1; i <= j - 1; i++)
            {
                z[i - 1] += Math.Abs(ap[ij - 1]);
                ij       += 1;
            }
        }

        double anorm = 0.0;

        for (i = 1; i <= n; i++)
        {
            anorm = Math.Max(anorm, z[i - 1]);
        }

        //
        //  Factor.
        //
        int info = DPPFA.dppfa(ref ap, n);

        if (info != 0)
        {
            rcond = 0.0;
            return(rcond);
        }

        //
        //  RCOND = 1/(norm(A)*(estimate of norm(inverse(A)))).
        //
        //  Estimate = norm(Z)/norm(Y) where A * Z = Y and A * Y = E.
        //
        //  The components of E are chosen to cause maximum local
        //  growth in the elements of W where R'*W = E.
        //
        //  The vectors are frequently rescaled to avoid overflow.
        //
        //  Solve R' * W = E.
        //
        double ek = 1.0;

        for (i = 1; i <= n; i++)
        {
            z[i - 1] = 0.0;
        }

        int kk = 0;

        for (k = 1; k <= n; k++)
        {
            kk += k;

            if (z[k - 1] != 0.0)
            {
                ek *= typeMethods.r8_sign(-z[k - 1]);
            }

            if (ap[kk - 1] < Math.Abs(ek - z[k - 1]))
            {
                s = ap[kk - 1] / Math.Abs(ek - z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ek = s * ek;
            }

            double wk  = ek - z[k - 1];
            double wkm = -ek - z[k - 1];
            s = Math.Abs(wk);
            double sm = Math.Abs(wkm);
            wk  /= ap[kk - 1];
            wkm /= ap[kk - 1];
            int kj = kk + k;

            if (k + 1 <= n)
            {
                for (j = k + 1; j <= n; j++)
                {
                    sm       += Math.Abs(z[j - 1] + wkm * ap[kj - 1]);
                    z[j - 1] += wk * ap[kj - 1];
                    s        += Math.Abs(z[j - 1]);
                    kj       += j;
                }

                if (s < sm)
                {
                    t  = wkm - wk;
                    wk = wkm;
                    kj = kk + k;

                    for (j = k + 1; j <= n; j++)
                    {
                        z[j - 1] += t * ap[kj - 1];
                        kj       += j;
                    }
                }
            }

            z[k - 1] = wk;
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        //
        //  Solve R * Y = W.
        //
        for (k = n; 1 <= k; k--)
        {
            if (ap[kk - 1] < Math.Abs(z[k - 1]))
            {
                s = ap[kk - 1] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }
            }

            z[k - 1] /= ap[kk - 1];
            kk       -= k;
            t         = -z[k - 1];
            BLAS1D.daxpy(k - 1, t, ap, 1, ref z, 1, xIndex: +kk);
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        double ynorm = 1.0;

        //
        //  Solve R' * V = Y.
        //
        for (k = 1; k <= n; k++)
        {
            z[k - 1] -= BLAS1D.ddot(k - 1, ap, 1, z, 1, xIndex: +kk);
            kk       += k;

            if (ap[kk - 1] < Math.Abs(z[k - 1]))
            {
                s = ap[kk - 1] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            z[k - 1] /= ap[kk - 1];
        }

        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;
        //
        //  Solve R * Z = V.
        //
        for (k = n; 1 <= k; k--)
        {
            if (ap[kk - 1] < Math.Abs(z[k - 1]))
            {
                s = ap[kk - 1] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            z[k - 1] /= ap[kk - 1];
            kk       -= k;
            t         = -z[k - 1];
            BLAS1D.daxpy(k - 1, t, ap, 1, ref z, 1, xIndex: +kk);
        }

        //
        //  Make ZNORM = 1.0.
        //
        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;

        if (anorm != 0.0)
        {
            rcond = ynorm / anorm;
        }
        else
        {
            rcond = 0.0;
        }

        return(rcond);
    }
示例#11
0
    public static int dtrsl(double[] t, int ldt, int n, ref double[] b, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DTRSL solves triangular linear systems.
    //
    //  Discussion:
    //
    //    DTRSL can solve T * X = B or T' * X = B where T is a triangular
    //    matrix of order N.
    //
    //    Here T' denotes the transpose of the matrix T.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    19 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input, double T[LDT*N], the matrix of the system.  The zero
    //    elements of the matrix are not referenced, and the corresponding
    //    elements of the array can be used to store other information.
    //
    //    Input, int LDT, the leading dimension of the array T.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input/output, double B[N].  On input, the right hand side.
    //    On output, the solution.
    //
    //    Input, int JOB, specifies what kind of system is to be solved:
    //    00, solve T * X = B, T lower triangular,
    //    01, solve T * X = B, T upper triangular,
    //    10, solve T'* X = B, T lower triangular,
    //    11, solve T'* X = B, T upper triangular.
    //
    //    Output, int DTRSL, singularity indicator.
    //    0, the system is nonsingular.
    //    nonzero, the index of the first zero diagonal element of T.
    //
    {
        int    info;
        int    j;
        int    jj;
        double temp;

        //
        //  Check for zero diagonal elements.
        //
        for (j = 1; j <= n; j++)
        {
            switch (t[j - 1 + (j - 1) * ldt])
            {
            case 0.0:
                info = j;
                return(info);
            }
        }

        info = 0;
        int kase = (job % 10) switch
        {
            //
            //  Determine the task and go to it.
            //
            0 => 1,
            _ => 2
        };

        if (job % 100 / 10 != 0)
        {
            kase += 2;
        }

        switch (kase)
        {
        //
        //  Solve T * X = B for T lower triangular.
        //
        case 1:
        {
            b[0] /= t[0 + 0 * ldt];
            for (j = 2; j <= n; j++)
            {
                temp = -b[j - 2];
                BLAS1D.daxpy(n - j + 1, temp, t, 1, ref b, 1, xIndex: +(j - 1) + (j - 2) * ldt, yIndex: +j - 1);
                b[j - 1] /= t[j - 1 + (j - 1) * ldt];
            }

            break;
        }

        //
        //  Solve T * X = B for T upper triangular.
        //
        case 2:
        {
            b[n - 1] /= t[n - 1 + (n - 1) * ldt];
            for (jj = 2; jj <= n; jj++)
            {
                j    = n - jj + 1;
                temp = -b[j];
                BLAS1D.daxpy(j, temp, t, 1, ref b, 1, xIndex: +0 + j * ldt);
                b[j - 1] /= t[j - 1 + (j - 1) * ldt];
            }

            break;
        }

        //
        //  Solve T' * X = B for T lower triangular.
        //
        case 3:
        {
            b[n - 1] /= t[n - 1 + (n - 1) * ldt];
            for (jj = 2; jj <= n; jj++)
            {
                j         = n - jj + 1;
                b[j - 1] -= BLAS1D.ddot(jj - 1, t, 1, b, 1, xIndex: +j + (j - 1) * ldt, yIndex: +j);
                b[j - 1] /= t[j - 1 + (j - 1) * ldt];
            }

            break;
        }

        //
        //  Solve T' * X = B for T upper triangular.
        //
        case 4:
        {
            b[0] /= t[0 + 0 * ldt];
            for (j = 2; j <= n; j++)
            {
                b[j - 1] -= BLAS1D.ddot(j - 1, t, 1, b, 1, xIndex: +0 + (j - 1) * ldt);
                b[j - 1] /= t[j - 1 + (j - 1) * ldt];
            }

            break;
        }
        }

        return(info);
    }
}
示例#12
0
    public static void dsidi(ref double[] a, int lda, int n, int[] kpvt, ref double[] det,
                             ref int[] inert, double[] work, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DSIDI computes the determinant, inertia and inverse of a real symmetric matrix.
    //
    //  Discussion:
    //
    //    DSIDI uses the factors from DSIFA.
    //
    //    A division by zero may occur if the inverse is requested
    //    and DSICO has set RCOND == 0.0D+00 or DSIFA has set INFO /= 0.
    //
    //    Variables not requested by JOB are not used.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    25 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double A(LDA,N).  On input, the output from DSIFA.
    //    On output, the upper triangle of the inverse of the original matrix,
    //    if requested.  The strict lower triangle is never referenced.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int KPVT[N], the pivot vector from DSIFA.
    //
    //    Output, double DET[2], the determinant of the original matrix,
    //    if requested.
    //      determinant = DET[0] * 10.0**DET[1]
    //    with 1.0D+00 <= abs ( DET[0] ) < 10.0D+00 or DET[0] = 0.0.
    //
    //    Output, int INERT(3), the inertia of the original matrix,
    //    if requested.
    //    INERT(1) = number of positive eigenvalues.
    //    INERT(2) = number of negative eigenvalues.
    //    INERT(3) = number of zero eigenvalues.
    //
    //    Workspace, double WORK[N].
    //
    //    Input, int JOB, specifies the tasks.
    //    JOB has the decimal expansion ABC where
    //    If C /= 0, the inverse is computed,
    //    If B /= 0, the determinant is computed,
    //    If A /= 0, the inertia is computed.
    //    For example, JOB = 111 gives all three.
    //
    {
        double d;
        int    k;
        double t;

        bool doinv = job % 10 != 0;
        bool dodet = job % 100 / 10 != 0;
        bool doert = job % 1000 / 100 != 0;

        if (dodet || doert)
        {
            switch (doert)
            {
            case true:
                inert[0] = 0;
                inert[1] = 0;
                inert[2] = 0;
                break;
            }

            switch (dodet)
            {
            case true:
                det[0] = 1.0;
                det[1] = 0.0;
                break;
            }

            t = 0.0;

            for (k = 1; k <= n; k++)
            {
                d = a[k - 1 + (k - 1) * lda];
                switch (kpvt[k - 1])
                {
                //
                //  2 by 2 block.
                //
                //  use det (d  s)  =  (d/t * c - t) * t,  t = abs ( s )
                //          (s  c)
                //  to avoid underflow/overflow troubles.
                //
                //  Take two passes through scaling.  Use T for flag.
                //
                case <= 0 when t == 0.0:
                    t = Math.Abs(a[k - 1 + k * lda]);
                    d = d / t * a[k + k * lda] - t;
                    break;

                case <= 0:
                    d = t;
                    t = 0.0;
                    break;
                }

                switch (doert)
                {
                case true:
                    switch (d)
                    {
                    case > 0.0:
                        inert[0] += 1;
                        break;

                    case < 0.0:
                        inert[1] += 1;
                        break;

                    case 0.0:
                        inert[2] += 1;
                        break;
                    }

                    break;
                }

                switch (dodet)
                {
                case true:
                {
                    det[0] *= d;

                    if (det[0] != 0.0)
                    {
                        while (Math.Abs(det[0]) < 1.0)
                        {
                            det[0] *= 10.0;
                            det[1] -= 1.0;
                        }

                        while (10.0 <= Math.Abs(det[0]))
                        {
                            det[0] /= 10.0;
                            det[1] += 1.0;
                        }
                    }

                    break;
                }
                }
            }
        }

        switch (doinv)
        {
        //
        //  Compute inverse(A).
        //
        case true:
        {
            k = 1;

            while (k <= n)
            {
                int j;
                int kstep;
                switch (kpvt[k - 1])
                {
                case >= 0:
                {
                    //
                    //  1 by 1.
                    //
                    a[k - 1 + (k - 1) * lda] = 1.0 / a[k - 1 + (k - 1) * lda];

                    switch (k)
                    {
                    case >= 2:
                    {
                        BLAS1D.dcopy(k - 1, a, 1, ref work, 1, xIndex: +0 + (k - 1) * lda);

                        for (j = 1; j <= k - 1; j++)
                        {
                            a[j - 1 + (k - 1) * lda] = BLAS1D.ddot(j, a, 1, work, 1, xIndex: +0 + (j - 1) * lda);
                            BLAS1D.daxpy(j - 1, work[j - 1], a, 1, ref a, 1, xIndex: +0 + (j - 1) * lda,
                                         yIndex: +0 + (k - 1) * lda);
                        }

                        a[k - 1 + (k - 1) * lda] += BLAS1D.ddot(k - 1, work, 1, a, 1, yIndex: +0 + (k - 1) * lda);
                        break;
                    }
                    }

                    kstep = 1;
                    break;
                }

                //
                default:
                {
                    t = Math.Abs(a[k - 1 + k * lda]);
                    double ak    = a[k - 1 + (k - 1) * lda] / t;
                    double akp1  = a[k + k * lda] / t;
                    double akkp1 = a[k - 1 + k * lda] / t;
                    d = t * (ak * akp1 - 1.0);
                    a[k - 1 + (k - 1) * lda] = akp1 / d;
                    a[k + k * lda]           = ak / d;
                    a[k - 1 + k * lda]       = -akkp1 / d;

                    switch (k)
                    {
                    case >= 2:
                    {
                        BLAS1D.dcopy(k - 1, a, 1, ref work, 1, xIndex: +0 + k * lda);

                        for (j = 1; j <= k - 1; j++)
                        {
                            a[j - 1 + k * lda] = BLAS1D.ddot(j, a, 1, work, 1, xIndex: +0 + (j - 1) * lda);
                            BLAS1D.daxpy(j - 1, work[j - 1], a, 1, ref a, 1, xIndex: +0 + (j - 1) * lda,
                                         yIndex: +0 + k * lda);
                        }

                        a[k + k * lda]     += BLAS1D.ddot(k - 1, work, 1, a, 1, yIndex: +0 + k * lda);
                        a[k - 1 + k * lda] += BLAS1D.ddot(k - 1, a, 1, a, 1, xIndex: +0 + (k - 1) * lda,
                                                          yIndex: +0 + k * lda);
                        BLAS1D.dcopy(k - 1, a, 1, ref work, 1, xIndex: +0 + (k - 1) * lda);

                        for (j = 1; j <= k - 1; j++)
                        {
                            a[j - 1 + (k - 1) * lda] = BLAS1D.ddot(j, a, 1, work, 1, xIndex: +0 + (j - 1) * lda);
                            BLAS1D.daxpy(j - 1, work[j - 1], a, 1, ref a, 1, xIndex: +0 + (j - 1) * lda,
                                         yIndex: +0 + (k - 1) * lda);
                        }

                        a[k - 1 + (k - 1) * lda] += BLAS1D.ddot(k - 1, work, 1, a, 1, yIndex: +0 + (k - 1) * lda);
                        break;
                    }
                    }

                    kstep = 2;
                    break;
                }
                }

                //
                //  Swap.
                //
                int ks = Math.Abs(kpvt[k - 1]);

                if (ks != k)
                {
                    BLAS1D.dswap(ks, ref a, 1, ref a, 1, xIndex: +0 + (ks - 1) * lda, yIndex: +0 + (k - 1) * lda);

                    int    jb;
                    double temp;
                    for (jb = ks; jb <= k; jb++)
                    {
                        j    = k + ks - jb;
                        temp = a[j - 1 + (k - 1) * lda];
                        a[j - 1 + (k - 1) * lda]  = a[ks - 1 + (j - 1) * lda];
                        a[ks - 1 + (j - 1) * lda] = temp;
                    }

                    if (kstep != 1)
                    {
                        temp = a[ks - 1 + k * lda];
                        a[ks - 1 + k * lda] = a[k - 1 + k * lda];
                        a[k - 1 + k * lda]  = temp;
                    }
                }

                k += kstep;
            }

            break;
        }
        }
    }
示例#13
0
    public static int dpbfa(ref double[] abd, int lda, int n, int m)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPBFA factors a symmetric positive definite matrix stored in band form.
    //
    //  Discussion:
    //
    //    DPBFA is usually called by DPBCO, but it can be called
    //    directly with a saving in time if RCOND is not needed.
    //
    //    If A is a symmetric positive definite band matrix,
    //    the following program segment will set up the input.
    //
    //      m = (band width above diagonal)
    //      do j = 1, n
    //        i1 = max ( 1, j-m )
    //        do i = i1, j
    //          k = i-j+m+1
    //          abd(k,j) = a(i,j)
    //        end do
    //      end do
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double ABD[LDA*N].  On input, the matrix to be
    //    factored.  The columns of the upper triangle are stored in the columns
    //    of ABD and the diagonals of the upper triangle are stored in the
    //    rows of ABD.  On output, an upper triangular matrix R, stored in band
    //    form, so that A = R' * R.
    //
    //    Input, int LDA, the leading dimension of the array ABD.
    //    M+1 <= LDA is required.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int M, the number of diagonals above the main diagonal.
    //
    //    Output, int DPBFA, error indicator.
    //    0, for normal return.
    //    K, if the leading minor of order K is not positive definite.
    //
    {
        int info;
        int j;

        for (j = 1; j <= n; j++)
        {
            double s  = 0.0;
            int    ik = m + 1;
            int    jk = Math.Max(j - m, 1);
            int    mu = Math.Max(m + 2 - j, 1);

            int k;
            for (k = mu; k <= m; k++)
            {
                double t = abd[k - 1 + (j - 1) * lda]
                           - BLAS1D.ddot(k - mu, abd, 1, abd, 1, xIndex:  +ik - 1 + (jk - 1) * lda, yIndex: +mu - 1 + (j - 1) * lda);
                t /= abd[m + (jk - 1) * lda];
                abd[k - 1 + (j - 1) * lda] = t;
                s  += t * t;
                ik -= 1;
                jk += 1;
            }

            s = abd[m + (j - 1) * lda] - s;

            switch (s)
            {
            case <= 0.0:
                info = j;
                return(info);

            default:
                abd[m + (j - 1) * lda] = Math.Sqrt(s);
                break;
            }
        }

        info = 0;

        return(info);
    }
示例#14
0
    public static void dspdi(ref double[] ap, int n, int[] kpvt, ref double[] det, ref int[] inert,
                             double[] work, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DSPDI computes the determinant, inertia and inverse of a real symmetric matrix.
    //
    //  Discussion:
    //
    //    DSPDI uses the factors from DSPFA, where the matrix is stored in
    //    packed form.
    //
    //    A division by zero will occur if the inverse is requested
    //    and DSPCO has set RCOND == 0.0D+00 or DSPFA has set INFO /= 0.
    //
    //    Variables not requested by JOB are not used.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    25 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double AP[(N*(N+1))/2].  On input, the output from
    //    DSPFA.  On output, the upper triangle of the inverse of the original
    //    matrix, stored in packed form, if requested.  The columns of the upper
    //    triangle are stored sequentially in a one-dimensional array.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int KPVT[N], the pivot vector from DSPFA.
    //
    //    Output, double DET[2], the determinant of the original matrix,
    //    if requested.
    //      determinant = DET[0] * 10.0**DET[1]
    //    with 1.0D+00 <= abs ( DET[0] ) < 10.0D+00 or DET[0] = 0.0.
    //
    //    Output, int INERT[3], the inertia of the original matrix, if requested.
    //    INERT(1) = number of positive eigenvalues.
    //    INERT(2) = number of negative eigenvalues.
    //    INERT(3) = number of zero eigenvalues.
    //
    //    Workspace, double WORK[N].
    //
    //    Input, int JOB, has the decimal expansion ABC where:
    //      if A /= 0, the inertia is computed,
    //      if B /= 0, the determinant is computed,
    //      if C /= 0, the inverse is computed.
    //    For example, JOB = 111  gives all three.
    //
    {
        double d;
        int    ik;
        int    ikp1;
        int    k;
        int    kk;
        int    kkp1;
        double t;

        bool doinv = job % 10 != 0;
        bool dodet = job % 100 / 10 != 0;
        bool doert = job % 1000 / 100 != 0;

        if (dodet || doert)
        {
            switch (doert)
            {
            case true:
                inert[0] = 0;
                inert[1] = 0;
                inert[2] = 0;
                break;
            }

            switch (dodet)
            {
            case true:
                det[0] = 1.0;
                det[1] = 0.0;
                break;
            }

            t  = 0.0;
            ik = 0;

            for (k = 1; k <= n; k++)
            {
                kk = ik + k;
                d  = ap[kk - 1];
                switch (kpvt[k - 1])
                {
                //
                //  2 by 2 block
                //  use det (d  s)  =  (d/t * c - t) * t,  t = abs ( s )
                //          (s  c)
                //  to avoid underflow/overflow troubles.
                //
                //  Take two passes through scaling.  Use T for flag.
                //
                case <= 0 when t == 0.0:
                    ikp1 = ik + k;
                    kkp1 = ikp1 + k;
                    t    = Math.Abs(ap[kkp1 - 1]);
                    d    = d / t * ap[kkp1] - t;
                    break;

                case <= 0:
                    d = t;
                    t = 0.0;
                    break;
                }

                switch (doert)
                {
                case true:
                    switch (d)
                    {
                    case > 0.0:
                        inert[0] += 1;
                        break;

                    case < 0.0:
                        inert[1] += 1;
                        break;

                    case 0.0:
                        inert[2] += 1;
                        break;
                    }

                    break;
                }

                switch (dodet)
                {
                case true:
                {
                    det[0] *= d;

                    if (det[0] != 0.0)
                    {
                        while (Math.Abs(det[0]) < 1.0)
                        {
                            det[0] *= 10.0;
                            det[1] -= 1.0;
                        }

                        while (10.0 <= Math.Abs(det[0]))
                        {
                            det[0] /= 10.0;
                            det[1] += 1.0;
                        }
                    }

                    break;
                }
                }

                ik += k;
            }
        }

        switch (doinv)
        {
        //
        //  Compute inverse(A).
        //
        case true:
        {
            k  = 1;
            ik = 0;

            while (k <= n)
            {
                int km1 = k - 1;
                kk   = ik + k;
                ikp1 = ik + k;
                kkp1 = ikp1 + k;

                int kstep;
                int jk;
                int j;
                int ij;
                switch (kpvt[k - 1])
                {
                case >= 0:
                {
                    //
                    //  1 by 1.
                    //
                    ap[kk - 1] = 1.0 / ap[kk - 1];

                    switch (k)
                    {
                    case >= 2:
                    {
                        BLAS1D.dcopy(k - 1, ap, 1, ref work, 1, xIndex: +ik);
                        ij = 0;

                        for (j = 1; j <= k - 1; j++)
                        {
                            jk         = ik + j;
                            ap[jk - 1] = BLAS1D.ddot(j, ap, 1, work, 1, xIndex: +ij);
                            BLAS1D.daxpy(j - 1, work[j - 1], ap, 1, ref ap, 1, xIndex: +ij, yIndex: +ik);
                            ij += j;
                        }

                        ap[kk - 1] += BLAS1D.ddot(k - 1, work, 1, ap, 1, yIndex: +ik);
                        break;
                    }
                    }

                    kstep = 1;
                    break;
                }

                default:
                {
                    //
                    //  2 by 2.
                    //
                    t = Math.Abs(ap[kkp1 - 1]);
                    double ak    = ap[kk - 1] / t;
                    double akp1  = ap[kkp1] / t;
                    double akkp1 = ap[kkp1 - 1] / t;
                    d            = t * (ak * akp1 - 1.0);
                    ap[kk - 1]   = akp1 / d;
                    ap[kkp1]     = ak / d;
                    ap[kkp1 - 1] = -akkp1 / d;

                    switch (km1)
                    {
                    case >= 1:
                    {
                        BLAS1D.dcopy(km1, ap, 1, ref work, 1, xIndex: +ikp1);
                        ij = 0;

                        for (j = 1; j <= km1; j++)
                        {
                            int jkp1 = ikp1 + j;
                            ap[jkp1 - 1] = BLAS1D.ddot(j, ap, 1, work, 1, xIndex: +ij);
                            BLAS1D.daxpy(j - 1, work[j - 1], ap, 1, ref ap, 1, xIndex: +ij, yIndex: +ikp1);
                            ij += j;
                        }

                        ap[kkp1]     += BLAS1D.ddot(km1, work, 1, ap, 1, yIndex: +ikp1);
                        ap[kkp1 - 1] += BLAS1D.ddot(km1, ap, 1, ap, 1, xIndex: +ik, yIndex: +ikp1);
                        BLAS1D.dcopy(km1, ap, 1, ref work, 1, xIndex: +ik);
                        ij = 0;

                        for (j = 1; j <= km1; j++)
                        {
                            jk         = ik + j;
                            ap[jk - 1] = BLAS1D.ddot(j, ap, 1, work, 1, xIndex: +ij);
                            BLAS1D.daxpy(j - 1, work[j - 1], ap, 1, ref ap, 1, xIndex: +ij, yIndex: +ik);
                            ij += j;
                        }

                        ap[kk - 1] += BLAS1D.ddot(km1, work, 1, ap, 1, yIndex: +ik);
                        break;
                    }
                    }

                    kstep = 2;
                    break;
                }
                }

                //
                //  Swap.
                //
                int ks = Math.Abs(kpvt[k - 1]);

                if (ks != k)
                {
                    int iks = ks * (ks - 1) / 2;
                    BLAS1D.dswap(ks, ref ap, 1, ref ap, 1, xIndex: +iks, yIndex: +ik);
                    int ksj = ik + ks;

                    double temp;
                    int    jb;
                    for (jb = ks; jb <= k; jb++)
                    {
                        j           = k + ks - jb;
                        jk          = ik + j;
                        temp        = ap[jk - 1];
                        ap[jk - 1]  = ap[ksj - 1];
                        ap[ksj - 1] = temp;
                        ksj        -= j - 1;
                    }

                    if (kstep != 1)
                    {
                        int kskp1 = ikp1 + ks;
                        temp          = ap[kskp1 - 1];
                        ap[kskp1 - 1] = ap[kkp1 - 1];
                        ap[kkp1 - 1]  = temp;
                    }
                }

                ik += k;
                ik  = kstep switch
                {
                    2 => ik + k + 1,
                    _ => ik
                };

                k += kstep;
            }

            break;
        }
        }
    }
}
示例#15
0
    public static void dposl(double[] a, int lda, int n, ref double[] b)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPOSL solves a linear system factored by DPOCO or DPOFA.
    //
    //  Discussion:
    //
    //    To compute inverse(A) * C where C is a matrix with P columns:
    //
    //      call dpoco ( a, lda, n, rcond, z, info )
    //
    //      if ( rcond is not too small .and. info == 0 ) then
    //        do j = 1, p
    //          call dposl ( a, lda, n, c(1,j) )
    //        end do
    //      end if
    //
    //    A division by zero will occur if the input factor contains
    //    a zero on the diagonal.  Technically this indicates
    //    singularity but it is usually caused by improper subroutine
    //    arguments.  It will not occur if the subroutines are called
    //    correctly and INFO == 0.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input, double A[LDA*N], the output from DPOCO or DPOFA.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input/output, double B[N].  On input, the right hand side.
    //    On output, the solution.
    //
    {
        int    k;
        double t;

        //
        //  Solve R' * Y = B.
        //
        for (k = 1; k <= n; k++)
        {
            t        = BLAS1D.ddot(k - 1, a, 1, b, 1, xIndex: +0 + (k - 1) * lda);
            b[k - 1] = (b[k - 1] - t) / a[k - 1 + (k - 1) * lda];
        }

        //
        //  Solve R * X = Y.
        //
        for (k = n; 1 <= k; k--)
        {
            b[k - 1] /= a[k - 1 + (k - 1) * lda];
            t         = -b[k - 1];
            BLAS1D.daxpy(k - 1, t, a, 1, ref b, 1, xIndex: +0 + (k - 1) * lda);
        }
    }
示例#16
0
    public static void dgbsl(double[] abd, int lda, int n, int ml, int mu, int[] ipvt,
                             ref double[] b, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGBSL solves a real banded system factored by DGBCO or DGBFA.
    //
    //  Discussion:
    //
    //    DGBSL can solve either A * X = B  or  A' * X = B.
    //
    //    A division by zero will occur if the input factor contains a
    //    zero on the diagonal.  Technically this indicates singularity
    //    but it is often caused by improper arguments or improper
    //    setting of LDA.  It will not occur if the subroutines are
    //    called correctly and if DGBCO has set 0.0 < RCOND
    //    or DGBFA has set INFO == 0.
    //
    //    To compute inverse(A) * C  where C is a matrix with P columns:
    //
    //      call dgbco ( abd, lda, n, ml, mu, ipvt, rcond, z )
    //
    //      if ( rcond is too small ) then
    //        exit
    //      end if
    //
    //      do j = 1, p
    //        call dgbsl ( abd, lda, n, ml, mu, ipvt, c(1,j), 0 )
    //      end do
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input, double ABD[LDA*N], the output from DGBCO or DGBFA.
    //
    //    Input, integer LDA, the leading dimension of the array ABD.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int ML, MU, the number of diagonals below and above the
    //    main diagonal.  0 <= ML < N, 0 <= MU < N.
    //
    //    Input, int IPVT[N], the pivot vector from DGBCO or DGBFA.
    //
    //    Input/output, double B[N].  On input, the right hand side.
    //    On output, the solution.
    //
    //    Input, int JOB, job choice.
    //    0, solve A*X=B.
    //    nonzero, solve A'*X=B.
    //
    {
        int    k;
        int    l;
        int    la;
        int    lb;
        int    lm;
        double t;

        int m = mu + ml + 1;

        switch (job)
        {
        //
        //  JOB = 0, Solve A * x = b.
        //
        //  First solve L * y = b.
        //
        case 0:
        {
            switch (ml)
            {
            case > 0:
            {
                for (k = 1; k <= n - 1; k++)
                {
                    lm = Math.Min(ml, n - k);
                    l  = ipvt[k - 1];
                    t  = b[l - 1];
                    if (l != k)
                    {
                        b[l - 1] = b[k - 1];
                        b[k - 1] = t;
                    }

                    BLAS1D.daxpy(lm, t, abd, 1, ref b, 1, +m + (k - 1) * lda, k);
                }

                break;
            }
            }

            //
            //  Now solve U * x = y.
            //
            for (k = n; 1 <= k; k--)
            {
                b[k - 1] /= abd[m - 1 + (k - 1) * lda];
                lm        = Math.Min(k, m) - 1;
                la        = m - lm;
                lb        = k - lm;
                t         = -b[k - 1];
                BLAS1D.daxpy(lm, t, abd, 1, ref b, 1, +la - 1 + (k - 1) * lda, +lb - 1);
            }

            break;
        }

        //
        default:
        {
            for (k = 1; k <= n; k++)
            {
                lm       = Math.Min(k, m) - 1;
                la       = m - lm;
                lb       = k - lm;
                t        = BLAS1D.ddot(lm, abd, 1, b, 1, +la - 1 + (k - 1) * lda, +lb - 1);
                b[k - 1] = (b[k - 1] - t) / abd[m - 1 + (k - 1) * lda];
            }

            switch (ml)
            {
            //
            //  Now solve L' * x = y.
            //
            case > 0:
            {
                for (k = n - 1; 1 <= k; k--)
                {
                    lm        = Math.Min(ml, n - k);
                    b[k - 1] += BLAS1D.ddot(lm, abd, 1, b, 1, +m + (k - 1) * lda, +k);
                    l         = ipvt[k - 1];
                    if (l == k)
                    {
                        continue;
                    }

                    t        = b[l - 1];
                    b[l - 1] = b[k - 1];
                    b[k - 1] = t;
                }

                break;
            }
            }

            break;
        }
        }
    }
示例#17
0
    public static int dppfa(ref double[] ap, int n)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPPFA factors a real symmetric positive definite matrix in packed form.
    //
    //  Discussion:
    //
    //    DPPFA is usually called by DPPCO, but it can be called
    //    directly with a saving in time if RCOND is not needed.
    //
    //  Packed storage:
    //
    //    The following program segment will pack the upper
    //    triangle of a symmetric matrix.
    //
    //      k = 0
    //      do j = 1, n
    //        do i = 1, j
    //          k = k + 1
    //          ap(k) = a(i,j)
    //        end do
    //      end do
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    24 May 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double AP[N*(N+1)/2].  On input, the packed
    //    form of a symmetric matrix A.  The columns of the upper triangle are
    //    stored sequentially in a one-dimensional array.  On output, an upper
    //    triangular matrix R, stored in packed form, so that A = R'*R.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, int DPPFA, error flag.
    //    0, for normal return.
    //    K, if the leading minor of order K is not positive definite.
    //
    {
        int j;

        int info = 0;
        int jj   = 0;

        for (j = 1; j <= n; j++)
        {
            double s  = 0.0;
            int    kj = jj;
            int    kk = 0;

            int k;
            for (k = 1; k <= j - 1; k++)
            {
                kj += 1;
                double t = ap[kj - 1] - BLAS1D.ddot(k - 1, ap, 1, ap, 1, xIndex: +kk, yIndex: +jj);
                kk        += k;
                t         /= ap[kk - 1];
                ap[kj - 1] = t;
                s         += t * t;
            }

            jj += j;
            s   = ap[jj - 1] - s;

            switch (s)
            {
            case <= 0.0:
                info = j;
                return(info);

            default:
                ap[jj - 1] = Math.Sqrt(s);
                break;
            }
        }

        return(info);
    }
示例#18
0
    public static void dpbsl(double[] abd, int lda, int n, int m, ref double[] b)

        //****************************************************************************80
        //
        //  Purpose:
        //
        //    DPBSL solves a real SPD band system factored by DPBCO or DPBFA.
        //
        //  Discussion:
        //
        //    The matrix is assumed to be a symmetric positive definite (SPD)
        //    band matrix.
        //
        //    To compute inverse(A) * C  where C is a matrix with P columns:
        //
        //      call dpbco ( abd, lda, n, rcond, z, info )
        //
        //      if ( rcond is too small .or. info /= 0) go to ...
        //
        //      do j = 1, p
        //        call dpbsl ( abd, lda, n, c(1,j) )
        //      end do
        //
        //    A division by zero will occur if the input factor contains
        //    a zero on the diagonal.  Technically this indicates
        //    singularity but it is usually caused by improper subroutine
        //    arguments.  It will not occur if the subroutines are called
        //    correctly and INFO == 0.
        //
        //  Licensing:
        //
        //    This code is distributed under the GNU LGPL license. 
        //
        //  Modified:
        //
        //    23 May 2005
        //
        //  Author:
        //
        //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch, 
        //    Pete Stewart.
        //    C++ version by John Burkardt.
        //
        //  Reference:
        //
        //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
        //    LINPACK User's Guide,
        //    SIAM, (Society for Industrial and Applied Mathematics),
        //    3600 University City Science Center,
        //    Philadelphia, PA, 19104-2688.
        //    ISBN 0-89871-172-X
        //
        //  Parameters:
        //
        //    Input, double ABD[LDA*N], the output from DPBCO or DPBFA.
        //
        //    Input, int LDA, the leading dimension of the array ABD.
        //
        //    Input, int N, the order of the matrix.
        //
        //    Input, int M, the number of diagonals above the main diagonal.
        //
        //    Input/output, double B[N].  On input, the right hand side.
        //    On output, the solution.
        //
    {
        int k;
        int la;
        int lb;
        int lm;
        double t;
        //
        //  Solve R'*Y = B.
        //
        for (k = 1; k <= n; k++)
        {
            lm = Math.Min(k - 1, m);
            la = m + 1 - lm;
            lb = k - lm;
            t = BLAS1D.ddot(lm, abd, 1, b, 1, xIndex: +la - 1 + (k - 1) * lda, yIndex: +lb - 1);
            b[k - 1] = (b[k - 1] - t) / abd[m + (k - 1) * lda];
        }

        //
        //  Solve R*X = Y.
        //
        for (k = n; 1 <= k; k--)
        {
            lm = Math.Min(k - 1, m);
            la = m + 1 - lm;
            lb = k - lm;
            b[k - 1] /= abd[m + (k - 1) * lda];
            t = -b[k - 1];
            BLAS1D.daxpy(lm, t, abd, 1, ref b, 1, xIndex: +la - 1 + (k - 1) * lda, yIndex: +lb - 1);
        }
    }
示例#19
0
    public static double dpbco(ref double[] abd, int lda, int n, int m, ref double[] z)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPBCO factors a real symmetric positive definite banded matrix.
    //
    //  Discussion:
    //
    //    DPBCO also estimates the condition of the matrix.
    //
    //    If RCOND is not needed, DPBFA is slightly faster.
    //
    //    To solve A*X = B, follow DPBCO by DPBSL.
    //
    //    To compute inverse(A)*C, follow DPBCO by DPBSL.
    //
    //    To compute determinant(A), follow DPBCO by DPBDI.
    //
    //  Band storage:
    //
    //    If A is a symmetric positive definite band matrix, the following
    //    program segment will set up the input.
    //
    //      m = (band width above diagonal)
    //      do j = 1, n
    //        i1 = max (1, j-m)
    //        do i = i1, j
    //          k = i-j+m+1
    //          abd(k,j) = a(i,j)
    //        }
    //      }
    //
    //    This uses M + 1 rows of A, except for the M by M upper left triangle,
    //    which is ignored.
    //
    //    For example, if the original matrix is
    //
    //      11 12 13  0  0  0
    //      12 22 23 24  0  0
    //      13 23 33 34 35  0
    //       0 24 34 44 45 46
    //       0  0 35 45 55 56
    //       0  0  0 46 56 66
    //
    //    then N = 6, M = 2  and ABD should contain
    //
    //       *  * 13 24 35 46
    //       * 12 23 34 45 56
    //      11 22 33 44 55 66
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    07 June 2005
    //
    //  Author:
    //
    //    Original FORTRAN77 version by Jack Dongarra, Cleve Moler, Jim Bunch,
    //    Pete Stewart.
    //    C++ version by John Burkardt.
    //
    //  Reference:
    //
    //    Jack Dongarra, Cleve Moler, Jim Bunch and Pete Stewart,
    //    LINPACK User's Guide,
    //    SIAM, (Society for Industrial and Applied Mathematics),
    //    3600 University City Science Center,
    //    Philadelphia, PA, 19104-2688.
    //    ISBN 0-89871-172-X
    //
    //  Parameters:
    //
    //    Input/output, double ABD[LDA*N].  On input, the matrix to be
    //    factored.  The columns of the upper triangle are stored in the columns
    //    of ABD and the diagonals of the upper triangle are stored in the rows
    //    of ABD.  On output, an upper triangular matrix R, stored in band form,
    //    so that A = R'*R.  If INFO /= 0, the factorization is not complete.
    //
    //    Input, int LDA, the leading dimension of the array ABD.
    //    M+1 <= LDA is required.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Input, int M, the number of diagonals above the main diagonal.
    //
    //    Output, double Z[N], a work vector whose contents are usually
    //    unimportant.  If A is singular to working precision, then Z is an
    //    approximate null vector in the sense that
    //      norm(A*Z) = RCOND * norm(A) * norm(Z).
    //    If INFO /= 0, Z is unchanged.
    //
    //    Output, double DPBCO, an estimate of the reciprocal condition number
    //    RCOND.  For the system A*X = B, relative perturbations in A and B of size
    //    EPSILON may cause relative perturbations in X of size EPSILON/RCOND.
    //    If RCOND is so small that the logical expression
    //      1.0 + RCOND == 1.0D+00
    //    is true, then A may be singular to working precision.  In particular,
    //    RCOND is zero if exact singularity is detected or the estimate underflows.
    //
    {
        int    i;
        int    j;
        int    k;
        int    la;
        int    lb;
        int    lm;
        double rcond;
        double s;
        double t;

        //
        //  Find the norm of A.
        //
        for (j = 1; j <= n; j++)
        {
            int l  = Math.Min(j, m + 1);
            int mu = Math.Max(m + 2 - j, 1);
            z[j - 1] = BLAS1D.dasum(l, abd, 1, index: +mu - 1 + (j - 1) * lda);
            k        = j - l;
            for (i = mu; i <= m; i++)
            {
                k        += 1;
                z[k - 1] += Math.Abs(abd[i - 1 + (j - 1) * lda]);
            }
        }

        double anorm = 0.0;

        for (i = 1; i <= n; i++)
        {
            anorm = Math.Max(anorm, z[i - 1]);
        }

        //
        //  Factor.
        //
        int info = DPBFA.dpbfa(ref abd, lda, n, m);

        if (info != 0)
        {
            rcond = 0.0;
            return(rcond);
        }

        //
        //  RCOND = 1/(norm(A)*(estimate of norm(inverse(A)))).
        //
        //  Estimate = norm(Z)/norm(Y) where A*Z = Y and A*Y = E.
        //
        //  The components of E are chosen to cause maximum local
        //  growth in the elements of W where R'*W = E.
        //
        //  The vectors are frequently rescaled to avoid overflow.
        //
        //  Solve R' * W = E.
        //
        double ek = 1.0;

        for (i = 1; i <= n; i++)
        {
            z[i - 1] = 0.0;
        }

        for (k = 1; k <= n; k++)
        {
            if (z[k - 1] != 0.0)
            {
                ek *= typeMethods.r8_sign(-z[k - 1]);
            }

            if (abd[m + (k - 1) * lda] < Math.Abs(ek - z[k - 1]))
            {
                s = abd[m + (k - 1) * lda] / Math.Abs(ek - z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ek = s * ek;
            }

            double wk  = ek - z[k - 1];
            double wkm = -ek - z[k - 1];
            s = Math.Abs(wk);
            double sm = Math.Abs(wkm);
            wk  /= abd[m + (k - 1) * lda];
            wkm /= abd[m + (k - 1) * lda];
            int j2 = Math.Min(k + m, n);
            i = m + 1;

            if (k + 1 <= j2)
            {
                for (j = k + 1; j <= j2; j++)
                {
                    i        -= 1;
                    sm       += Math.Abs(z[j - 1] + wkm * abd[i - 1 + (j - 1) * lda]);
                    z[j - 1] += wk * abd[i - 1 + (j - 1) * lda];
                    s        += Math.Abs(z[j - 1]);
                }

                if (s < sm)
                {
                    t  = wkm - wk;
                    wk = wkm;
                    i  = m + 1;

                    for (j = k + 1; j <= j2; j++)
                    {
                        i        -= 1;
                        z[j - 1] += t * abd[i - 1 + (j - 1) * lda];
                    }
                }
            }

            z[k - 1] = wk;
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        //
        //  Solve R * Y = W.
        //
        for (k = n; 1 <= k; k--)
        {
            if (abd[m + (k - 1) * lda] < Math.Abs(z[k - 1]))
            {
                s = abd[m + (k - 1) * lda] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }
            }

            z[k - 1] /= abd[m + (k - 1) * lda];
            lm        = Math.Min(k - 1, m);
            la        = m + 1 - lm;
            lb        = k - lm;
            t         = -z[k - 1];
            BLAS1D.daxpy(lm, t, abd, 1, ref z, 1, xIndex: +la - 1 + (k - 1) * lda, yIndex: +lb - 1);
        }

        s = BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] /= s;
        }

        double ynorm = 1.0;

        //
        //  Solve R' * V = Y.
        //
        for (k = 1; k <= n; k++)
        {
            lm = Math.Min(k - 1, m);
            la = m + 1 - lm;
            lb = k - lm;

            z[k - 1] -= BLAS1D.ddot(lm, abd, 1, z, 1, xIndex:  +la - 1 + (k - 1) * lda, yIndex:  +lb - 1);

            if (abd[m + (k - 1) * lda] < Math.Abs(z[k - 1]))
            {
                s = abd[m + (k - 1) * lda] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            z[k - 1] /= abd[m + (k - 1) * lda];
        }

        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;
        //
        //  Solve R * Z = W.
        //
        for (k = n; 1 <= k; k--)
        {
            if (abd[m + (k - 1) * lda] < Math.Abs(z[k - 1]))
            {
                s = abd[m + (k - 1) * lda] / Math.Abs(z[k - 1]);
                for (i = 1; i <= n; i++)
                {
                    z[i - 1] = s * z[i - 1];
                }

                ynorm = s * ynorm;
            }

            z[k - 1] /= abd[m + (k - 1) * lda];
            lm        = Math.Min(k - 1, m);
            la        = m + 1 - lm;
            lb        = k - lm;
            t         = -z[k - 1];
            BLAS1D.daxpy(lm, t, abd, 1, ref z, 1, xIndex:  +la - 1 + (k - 1) * lda, yIndex:  +lb - 1);
        }

        //
        //  Make ZNORM = 1.0.
        //
        s = 1.0 / BLAS1D.dasum(n, z, 1);
        for (i = 1; i <= n; i++)
        {
            z[i - 1] = s * z[i - 1];
        }

        ynorm = s * ynorm;

        if (anorm != 0.0)
        {
            rcond = ynorm / anorm;
        }
        else
        {
            rcond = 0.0;
        }

        return(rcond);
    }