Exemplo n.º 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);
        }
    }
Exemplo n.º 2
0
    public static int dtrdi(ref double[] t, int ldt, int n, ref double[] det, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DTRDI computes the determinant and inverse of a real triangular matrix.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    23 March 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 T[LDT*N].
    //    On input, T contains the triangular matrix.  The zero elements of the
    //    matrix are not referenced, and the corresponding elements of the array
    //    can be used to store other information.
    //    On output, T contains the inverse matrix, if it was requested.
    //
    //    Input, int LDT, the leading dimension of T.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, double DET[2], the determinant of the matrix, if
    //    requested.  The determinant = DET[0] * 10.0**DET[1], with
    //    1.0 <= abs ( DET[0] ) < 10.0, or DET[0] == 0.
    //
    //    Input, int JOB, specifies the shape of T, and the task.
    //    010, inverse of lower triangular matrix.
    //    011, inverse of upper triangular matrix.
    //    100, determinant only.
    //    110, determinant and inverse of lower triangular.
    //    111, determinant and inverse of upper triangular.
    //
    //    Output, int DTRDI.
    //    If the inverse was requested, then
    //    0, if the system was nonsingular;
    //    nonzero, if the system was singular.
    //
    {
        int    j;
        int    k;
        double temp;
        //
        //  Determinant.
        //
        int info = 0;

        if (job / 100 != 0)
        {
            det[0] = 1.0;
            det[1] = 0.0;

            int i;
            for (i = 1; i <= n; i++)
            {
                det[0] *= t[i - 1 + (i - 1) * ldt];

                if (det[0] == 0.0)
                {
                    break;
                }

                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;
                }
            }
        }

        switch (job / 10 % 10)
        {
        case 0:
            return(info);
        }

        //
        //  Inverse of an upper triangular matrix.
        //
        if (job % 10 != 0)
        {
            info = 0;

            for (k = 1; k <= n; k++)
            {
                if (t[k - 1 + (k - 1) * ldt] == 0.0)
                {
                    info = k;
                    break;
                }

                t[k - 1 + (k - 1) * ldt] = 1.0 / t[k - 1 + (k - 1) * ldt];
                temp = -t[k - 1 + (k - 1) * ldt];
                BLAS1D.dscal(k - 1, temp, ref t, 1, index: +0 + (k - 1) * ldt);

                for (j = k + 1; j <= n; j++)
                {
                    temp = t[k - 1 + (j - 1) * ldt];
                    t[k - 1 + (j - 1) * ldt] = 0.0;
                    BLAS1D.daxpy(k, temp, t, 1, ref t, 1, xIndex: +0 + (k - 1) * ldt, yIndex: +0 + (j - 1) * ldt);
                }
            }
        }
        //
        //  Inverse of a lower triangular matrix.
        //
        else
        {
            info = 0;

            for (k = n; 1 <= k; k--)
            {
                if (t[k - 1 + (k - 1) * ldt] == 0.0)
                {
                    info = k;
                    break;
                }

                t[k - 1 + (k - 1) * ldt] = 1.0 / t[k - 1 + (k - 1) * ldt];
                temp = -t[k - 1 + (k - 1) * ldt];

                if (k != n)
                {
                    BLAS1D.dscal(n - k, temp, ref t, 1, index: +k + (k - 1) * ldt);
                }

                for (j = 1; j <= k - 1; j++)
                {
                    temp = t[k - 1 + (j - 1) * ldt];
                    t[k - 1 + (j - 1) * ldt] = 0.0;
                    BLAS1D.daxpy(n - k + 1, temp, t, 1, ref t, 1, xIndex: +k - 1 + (k - 1) * ldt,
                                 yIndex: +k - 1 + (j - 1) * ldt);
                }
            }
        }

        return(info);
    }
Exemplo n.º 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);
    }
Exemplo n.º 4
0
    public static int dgefa(ref double[] a, int lda, int n, ref int[] ipvt)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGEFA factors a real general matrix.
    //
    //  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/output, double A[LDA*N].
    //    On intput, the matrix to be factored.
    //    On output, an upper triangular matrix and the multipliers 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 A.
    //
    //    Input, int N, the order of the matrix A.
    //
    //    Output, int IPVT[N], the pivot indices.
    //
    //    Output, int DGEFA, singularity indicator.
    //    0, normal value.
    //    K, if U(K,K) == 0.  This is not an error condition for this subroutine,
    //    but it does indicate that DGESL or DGEDI will divide by zero if called.
    //    Use RCOND in DGECO for a reliable indication of singularity.
    //
    {
        int k;
        //
        //  Gaussian elimination with partial pivoting.
        //
        int info = 0;

        for (k = 1; k <= n - 1; k++)
        {
            //
            //  Find L = pivot index.
            //
            int l = BLAS1D.idamax(n - k + 1, a, 1, index:  +(k - 1) + (k - 1) * lda) + k - 1;
            ipvt[k - 1] = l;
            switch (a[l - 1 + (k - 1) * lda])
            {
            //
            //  Zero pivot implies this column already triangularized.
            //
            case 0.0:
                info = k;
                continue;
            }

            //
            //  Interchange if necessary.
            //
            double t;
            if (l != k)
            {
                t = a[l - 1 + (k - 1) * lda];
                a[l - 1 + (k - 1) * lda] = a[k - 1 + (k - 1) * lda];
                a[k - 1 + (k - 1) * lda] = t;
            }

            //
            //  Compute multipliers.
            //
            t = -1.0 / a[k - 1 + (k - 1) * lda];

            BLAS1D.dscal(n - k, t, ref a, 1, index: +k + (k - 1) * lda);
            //
            //  Row elimination with column indexing.
            //
            int j;
            for (j = k + 1; j <= n; j++)
            {
                t = a[l - 1 + (j - 1) * lda];
                if (l != k)
                {
                    a[l - 1 + (j - 1) * lda] = a[k - 1 + (j - 1) * lda];
                    a[k - 1 + (j - 1) * lda] = t;
                }

                BLAS1D.daxpy(n - k, t, a, 1, ref a, 1, xIndex:  +k + (k - 1) * lda, yIndex:  +k + (j - 1) * lda);
            }
        }

        ipvt[n - 1] = n;

        info = a[n - 1 + (n - 1) * lda] switch
        {
            0.0 => n,
            _ => info
        };

        return(info);
    }
}
Exemplo n.º 5
0
    public static int dsifa(ref double[] a, int lda, int n, ref int[] kpvt)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DSIFA factors a real symmetric matrix.
    //
    //  Discussion:
    //
    //    To solve A*X = B, follow DSIFA by DSISL.
    //
    //    To compute inverse(A)*C, follow DSIFA by DSISL.
    //
    //    To compute determinant(A), follow DSIFA by DSIDI.
    //
    //    To compute inertia(A), follow DSIFA by DSIDI.
    //
    //    To compute inverse(A), follow DSIFA by DSIDI.
    //
    //  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 symmetric matrix
    //    to be factored.  Only the diagonal and upper triangle are used.
    //    On output, a block diagonal matrix and the multipliers which
    //    were used to obtain it.  The factorization can be written A = U*D*U'
    //    where U is a product of permutation and unit upper triangular
    //    matrices, U' is the transpose of U, and D is block diagonal
    //    with 1 by 1 and 2 by 2 blocks.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, int KPVT[N], the pivot indices.
    //
    //    Output, integer DSIFA, error flag.
    //    0, normal value.
    //    K, if the K-th pivot block is singular.  This is not an error
    //    condition for this subroutine, but it does indicate that DSISL
    //    or DSIDI may divide by zero if called.
    //
    {
        //
        //  ALPHA is used in choosing pivot block size.
        //
        double alpha = (1.0 + Math.Sqrt(17.0)) / 8.0;

        int info = 0;
        //
        //  Main loop on K, which goes from N to 1.
        //
        int k = n;

        while (0 < k)
        {
            switch (k)
            {
            case 1:
            {
                kpvt[0] = 1;
                info    = a[0 + 0 * lda] switch
                {
                    0.0 => 1,
                    _ => info
                };

                return(info);
            }
            }

            //
            //  This section of code determines the kind of
            //  elimination to be performed.  When it is completed,
            //  KSTEP will be set to the size of the pivot block, and
            //  SWAP will be set to .true. if an interchange is required.
            //
            double absakk = Math.Abs(a[k - 1 + (k - 1) * lda]);
            //
            //  Determine the largest off-diagonal element in column K.
            //
            int    imax   = BLAS1D.idamax(k - 1, a, 1, index: +0 + (k - 1) * lda);
            double colmax = Math.Abs(a[imax - 1 + (k - 1) * lda]);

            int  j;
            int  kstep;
            bool swap;
            if (alpha * colmax <= absakk)
            {
                kstep = 1;
                swap  = false;
            }
            //
            //  Determine the largest off-diagonal element in row IMAX.
            //
            else
            {
                double rowmax = 0.0;
                int    imaxp1 = imax + 1;
                for (j = imaxp1; j <= k; j++)
                {
                    rowmax = Math.Max(rowmax, Math.Abs(a[imax - 1 + (j - 1) * lda]));
                }

                if (imax != 1)
                {
                    int jmax = BLAS1D.idamax(imax - 1, a, 1, index: +0 + (imax - 1) * lda);
                    rowmax = Math.Max(rowmax, Math.Abs(a[jmax - 1 + (imax - 1) * lda]));
                }

                if (alpha * rowmax <= Math.Abs(a[imax - 1 + (imax - 1) * lda]))
                {
                    kstep = 1;
                    swap  = true;
                }
                else if (alpha * colmax * (colmax / rowmax) <= absakk)
                {
                    kstep = 1;
                    swap  = false;
                }
                else
                {
                    kstep = 2;
                    swap  = imax != k - 1;
                }
            }

            switch (Math.Max(absakk, colmax))
            {
            //
            //  Column K is zero.
            //  Set INFO and iterate the loop.
            //
            case 0.0:
                kpvt[k - 1] = k;
                info        = k;
                break;

            //
            default:
            {
                int    jj;
                double mulk;
                double t;
                if (kstep != 2)
                {
                    switch (swap)
                    {
                    case true:
                    {
                        BLAS1D.dswap(imax, ref a, 1, ref a, 1, xIndex: +0 + (imax - 1) * lda,
                                     yIndex: +0 + (k - 1) * lda);

                        for (jj = imax; jj <= k; jj++)
                        {
                            j = k + imax - jj;
                            t = a[j - 1 + (k - 1) * lda];
                            a[j - 1 + (k - 1) * lda]    = a[imax - 1 + (j - 1) * lda];
                            a[imax - 1 + (j - 1) * lda] = t;
                        }

                        break;
                    }
                    }

                    //
                    //  Perform the elimination.
                    //
                    for (jj = 1; jj <= k - 1; jj++)
                    {
                        j    = k - jj;
                        mulk = -a[j - 1 + (k - 1) * lda] / a[k - 1 + (k - 1) * lda];
                        t    = mulk;
                        BLAS1D.daxpy(j, t, a, 1, ref a, 1, xIndex: +0 + (k - 1) * lda, yIndex: +0 + (j - 1) * lda);
                        a[j - 1 + (k - 1) * lda] = mulk;
                    }

                    kpvt[k - 1] = swap switch
                    {
                        //
                        //  Set the pivot array.
                        //
                        true => imax,
                        _ => k
                    };
                }
                //
                //  2 x 2 pivot block.
                //
                //  Perform an interchange.
                //
                else
                {
                    switch (swap)
                    {
                    case true:
                    {
                        BLAS1D.dswap(imax, ref a, 1, ref a, 1, xIndex: +0 + (imax - 1) * lda,
                                     yIndex: +0 + (k - 2) * lda);

                        for (jj = imax; jj <= k - 1; jj++)
                        {
                            j = k - 1 + imax - jj;
                            t = a[j - 1 + (k - 1) * lda];
                            a[j - 1 + (k - 1) * lda]    = a[imax - 1 + (j - 1) * lda];
                            a[imax - 1 + (j - 1) * lda] = t;
                        }

                        t = a[k - 2 + (k - 1) * lda];
                        a[k - 2 + (k - 1) * lda]    = a[imax - 1 + (k - 1) * lda];
                        a[imax - 1 + (k - 1) * lda] = t;
                        break;
                    }
                    }

                    //
                    //  Perform the elimination.
                    //
                    if (k - 2 != 0)
                    {
                        double ak    = a[k - 1 + (k - 1) * lda] / a[k - 2 + (k - 1) * lda];
                        double akm1  = a[k - 2 + (k - 2) * lda] / a[k - 2 + (k - 1) * lda];
                        double denom = 1.0 - ak * akm1;

                        for (jj = 1; jj <= k - 2; jj++)
                        {
                            j = k - 1 - jj;
                            double bk   = a[j - 1 + (k - 1) * lda] / a[k - 2 + (k - 1) * lda];
                            double bkm1 = a[j - 1 + (k - 2) * lda] / a[k - 2 + (k - 1) * lda];
                            mulk = (akm1 * bk - bkm1) / denom;
                            double mulkm1 = (ak * bkm1 - bk) / denom;
                            t = mulk;
                            BLAS1D.daxpy(j, t, a, 1, ref a, 1, xIndex: +0 + (k - 1) * lda, yIndex: +0 + (j - 1) * lda);
                            t = mulkm1;
                            BLAS1D.daxpy(j, t, a, 1, ref a, 1, xIndex: +0 + (k - 2) * lda, yIndex: +0 + (j - 1) * lda);
                            a[j - 1 + (k - 1) * lda] = mulk;
                            a[j - 1 + (k - 2) * lda] = mulkm1;
                        }
                    }

                    kpvt[k - 1] = swap switch
                    {
                        //
                        //  Set the pivot array.
                        //
                        true => - imax,
                        _ => 1 - k
                    };

                    kpvt[k - 2] = kpvt[k - 1];
                }

                break;
            }
            }

            k -= kstep;
        }

        return(info);
    }
}
Exemplo n.º 6
0
    public static void dppdi(ref double[] ap, int n, ref double[] det, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPPDI computes the determinant and inverse of a matrix factored by DPPCO or DPPFA.
    //
    //  Discussion:
    //
    //    A division by zero will occur if the input factor contains
    //    a zero on the diagonal and the inverse is requested.
    //    It will not occur if the subroutines are called correctly
    //    and if DPOCO or DPOFA has set INFO == 0.
    //
    //  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 output from
    //    DPPCO or DPPFA.  On output, the upper triangular half of the
    //    inverse, if requested.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, double DET[2], the determinant of the original matrix
    //    if requested.
    //      determinant = DET[0] * 10.0**DET[1]
    //    with  1.0D+00 <= DET[0] < 10.0D+00 or DET[0] == 0.0D+00.
    //
    //    Input, int JOB, job request.
    //    11, both determinant and inverse.
    //    01, inverse only.
    //    10, determinant only.
    //
    {
        //
        //  Compute the determinant.
        //
        if (job / 10 != 0)
        {
            det[0] = 1.0;
            det[1] = 0.0;
            const double s  = 10.0;
            int          ii = 0;

            int i;
            for (i = 1; i <= n; i++)
            {
                ii += i;

                det[0] = det[0] * ap[ii - 1] * ap[ii - 1];

                if (det[0] == 0.0)
                {
                    break;
                }

                while (det[0] < 1.0)
                {
                    det[0] *= s;
                    det[1] -= 1.0;
                }

                while (s <= det[0])
                {
                    det[0] /= s;
                    det[1] += 1.0;
                }
            }
        }

        //
        //  Compute inverse(R).
        //
        if (job % 10 == 0)
        {
            return;
        }

        int kk = 0;

        int    k;
        int    k1;
        int    kj;
        int    j1;
        double t;
        int    j;

        for (k = 1; k <= n; k++)
        {
            k1         = kk + 1;
            kk        += k;
            ap[kk - 1] = 1.0 / ap[kk - 1];
            t          = -ap[kk - 1];
            BLAS1D.dscal(k - 1, t, ref ap, 1, index: +k1 - 1);
            j1 = kk + 1;
            kj = kk + k;

            for (j = k + 1; j <= n; j++)
            {
                t          = ap[kj - 1];
                ap[kj - 1] = 0.0;
                BLAS1D.daxpy(k, t, ap, 1, ref ap, 1, xIndex: +k1 - 1, yIndex: +j1 - 1);
                j1 += j;
                kj += j;
            }
        }

        //
        //  Form inverse(R) * (inverse(R))'.
        //
        int jj = 0;

        for (j = 1; j <= n; j++)
        {
            j1  = jj + 1;
            jj += j;
            k1  = 1;
            kj  = j1;

            for (k = 1; k <= j - 1; k++)
            {
                t = ap[kj - 1];
                BLAS1D.daxpy(k, t, ap, 1, ref ap, 1, xIndex: +j1 - 1, yIndex: +k1 - 1);
                k1 += k;
                kj += 1;
            }

            t = ap[jj - 1];
            BLAS1D.dscal(j, t, ref ap, 1, index: +j1 - 1);
        }
    }
Exemplo n.º 7
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;
        }
        }
    }
Exemplo n.º 8
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;
            }
            }
        }
    }
Exemplo n.º 9
0
    public static int dchdc(ref double[] a, int lda, int p, double[] work, ref int[] ipvt, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DCHDC computes the Cholesky decomposition of a positive definite matrix.
    //
    //  Discussion:
    //
    //    A pivoting option allows the user to estimate the condition of a
    //    positive definite matrix or determine the rank of a positive
    //    semidefinite matrix.
    //
    //    For positive definite matrices, INFO = P is the normal return.
    //
    //    For pivoting with positive semidefinite matrices, INFO will
    //    in general be less than P.  However, INFO may be greater than
    //    the rank of A, since rounding error can cause an otherwise zero
    //    element to be positive.  Indefinite systems will always cause
    //    INFO to be less than P.
    //
    //  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 A[LDA*P].
    //    On input, A contains the matrix whose decomposition is to
    //    be computed.  Only the upper half of A need be stored.
    //    The lower part of the array a is not referenced.
    //    On output, A contains in its upper half the Cholesky factor
    //    of the input matrix, as it has been permuted by pivoting.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int P, the order of the matrix.
    //
    //    Input, double WORK[P] is a work array.
    //
    //    Input/output, int IPVT[P].
    //    On input, IPVT contains integers that control the selection
    //    of the pivot elements, if pivoting has been requested.
    //    Each diagonal element A(K,K) is placed in one of three classes
    //    according to the value of IPVT(K).
    //
    //      > 0, then X(K) is an initial element.
    //      = 0, then X(K) is a free element.
    //      < 0, then X(K) is a final element.
    //
    //    Before the decomposition is computed, initial elements are moved by
    //    symmetric row and column interchanges to the beginning of the array A
    //    and final elements to the end.  Both initial and final elements are
    //    frozen in place during the computation and only free elements are moved.
    //    At the K-th stage of the reduction, if A(K,K) is occupied by a free
    //    element, it is interchanged with the largest free element A(L,L) with
    //    K <= L.  IPVT is not referenced if JOB is 0.
    //
    //    On output, IPVT(J) contains the index of the diagonal element
    //    of A that was moved into the J-th position, if pivoting was requested.
    //
    //    Input, int JOB, initiates column pivoting.
    //    0, no pivoting is done.
    //    nonzero, pivoting is done.
    //
    //    Output, int DCHDC, contains the index of the last positive diagonal
    //    element of the Cholesky factor.
    //
    {
        int    j;
        int    k;
        double temp;

        int pl   = 1;
        int pu   = 0;
        int info = p;

        //
        //  Pivoting has been requested.
        //  Rearrange the the elements according to IPVT.
        //
        if (job != 0)
        {
            for (k = 1; k <= p; k++)
            {
                bool swapk = 0 < ipvt[k - 1];
                bool negk  = ipvt[k - 1] < 0;

                ipvt[k - 1] = negk switch
                {
                    true => - k,
                    _ => k
                };

                switch (swapk)
                {
                case true:
                {
                    if (k != pl)
                    {
                        BLAS1D.dswap(pl - 1, ref a, 1, ref a, 1, xIndex:  +0 + (k - 1) * lda, yIndex:  +0 + (pl - 1) * lda);

                        temp = a[k - 1 + (k - 1) * lda];
                        a[k - 1 + (k - 1) * lda]   = a[pl - 1 + (pl - 1) * lda];
                        a[pl - 1 + (pl - 1) * lda] = temp;

                        for (j = pl + 1; j <= p; j++)
                        {
                            if (j < k)
                            {
                                temp = a[pl - 1 + (j - 1) * lda];
                                a[pl - 1 + (j - 1) * lda] = a[j - 1 + (k - 1) * lda];
                                a[j - 1 + (k - 1) * lda]  = temp;
                            }
                            else if (k < j)
                            {
                                temp = a[k - 1 + (j - 1) * lda];
                                a[k - 1 + (j - 1) * lda]  = a[pl - 1 + (j - 1) * lda];
                                a[pl - 1 + (j - 1) * lda] = temp;
                            }
                        }

                        ipvt[k - 1]  = ipvt[pl - 1];
                        ipvt[pl - 1] = k;
                    }

                    pl += 1;
                    break;
                }
                }
            }

            pu = p;

            for (k = p; pl <= k; k--)
            {
                switch (ipvt[k - 1])
                {
                case < 0:
                {
                    ipvt[k - 1] = -ipvt[k - 1];

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

                        temp = a[k - 1 + (k - 1) * lda];
                        a[k - 1 + (k - 1) * lda]   = a[pu - 1 + (pu - 1) * lda];
                        a[pu - 1 + (pu - 1) * lda] = temp;

                        for (j = k + 1; j <= p; j++)
                        {
                            if (j < pu)
                            {
                                temp = a[k - 1 + (j - 1) * lda];
                                a[k - 1 + (j - 1) * lda]  = a[j - 1 + (pu - 1) * lda];
                                a[j - 1 + (pu - 1) * lda] = temp;
                            }
                            else if (pu < j)
                            {
                                temp = a[k - 1 + (j - 1) * lda];
                                a[k - 1 + (j - 1) * lda]  = a[pu - 1 + (j - 1) * lda];
                                a[pu - 1 + (j - 1) * lda] = temp;
                            }
                        }

                        (ipvt[k - 1], ipvt[pu - 1]) = (ipvt[pu - 1], ipvt[k - 1]);
                    }

                    pu -= 1;
                    break;
                }
                }
            }
        }

        for (k = 1; k <= p; k++)
        {
            //
            //  Reduction loop.
            //
            double maxdia = a[k - 1 + (k - 1) * lda];
            int    maxl   = k;
            //
            //  Determine the pivot element.
            //
            if (pl <= k && k < pu)
            {
                int l;
                for (l = k + 1; l <= pu; l++)
                {
                    if (!(maxdia < a[l - 1 + (l - 1) * lda]))
                    {
                        continue;
                    }

                    maxdia = a[l - 1 + (l - 1) * lda];
                    maxl   = l;
                }
            }

            switch (maxdia)
            {
            //
            //  Quit if the pivot element is not positive.
            //
            case <= 0.0:
                info = k - 1;
                return(info);
            }

            //
            //  Start the pivoting and update IPVT.
            //
            if (k != maxl)
            {
                BLAS1D.dswap(k - 1, ref a, 1, ref a, 1, xIndex:  +0 + (k - 1) * lda, yIndex:  +0 + (maxl - 1) * lda);
                a[maxl - 1 + (maxl - 1) * lda] = a[k - 1 + (k - 1) * lda];
                a[k - 1 + (k - 1) * lda]       = maxdia;
                (ipvt[maxl - 1], ipvt[k - 1])  = (ipvt[k - 1], ipvt[maxl - 1]);
            }

            //
            //  Reduction step.
            //  Pivoting is contained across the rows.
            //
            work[k - 1] = Math.Sqrt(a[k - 1 + (k - 1) * lda]);
            a[k - 1 + (k - 1) * lda] = work[k - 1];

            for (j = k + 1; j <= p; j++)
            {
                if (k != maxl)
                {
                    if (j < maxl)
                    {
                        temp = a[k - 1 + (j - 1) * lda];
                        a[k - 1 + (j - 1) * lda]    = a[j - 1 + (maxl - 1) * lda];
                        a[j - 1 + (maxl - 1) * lda] = temp;
                    }
                    else if (maxl < j)
                    {
                        temp = a[k - 1 + (j - 1) * lda];
                        a[k - 1 + (j - 1) * lda]    = a[maxl - 1 + (j - 1) * lda];
                        a[maxl - 1 + (j - 1) * lda] = temp;
                    }
                }

                a[k - 1 + (j - 1) * lda] /= work[k - 1];
                work[j - 1] = a[k - 1 + (j - 1) * lda];
                temp        = -a[k - 1 + (j - 1) * lda];
                BLAS1D.daxpy(j - k, temp, work, 1, ref a, 1, xIndex: +k, yIndex:  +k + (j - 1) * lda);
            }
        }

        return(info);
    }
}
Exemplo n.º 10
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);
    }
}
Exemplo n.º 11
0
    public static void dgedi(ref double[] a, int lda, int n, int[] ipvt, ref double[] det,
                             double[] work, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGEDI computes the determinant and inverse of a matrix factored by DGECO or DGEFA.
    //
    //  Discussion:
    //
    //    A division by zero will occur if the input factor contains
    //    a zero on the diagonal and the inverse is requested.
    //    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:
    //
    //    17 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 LU factor information,
    //    as output by DGECO or DGEFA.  On output, the inverse
    //    matrix if requested.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix A.
    //
    //    Input, int IPVT[N], the pivot vector from DGECO or DGEFA.
    //
    //    Workspace, double WORK[N].
    //
    //    Output, double DET[2], the determinant of original matrix if
    //    requested.  The determinant = DET[0] * pow ( 10.0, DET[1] )
    //    with  1.0 <= abs ( DET[0] ) < 10.0 or DET[0] == 0.0.
    //
    //    Input, int JOB, specifies what is to be computed.
    //    11, both determinant and inverse.
    //    01, inverse only.
    //    10, determinant only.
    //
    {
        int i;

        //
        //  Compute the determinant.
        //
        if (job / 10 != 0)
        {
            det[0] = 1.0;
            det[1] = 0.0;

            for (i = 1; i <= n; i++)
            {
                if (ipvt[i - 1] != i)
                {
                    det[0] = -det[0];
                }

                det[0] *= a[i - 1 + (i - 1) * lda];

                if (det[0] == 0.0)
                {
                    break;
                }

                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;
                }
            }
        }

        //
        //  Compute inverse(U).
        //
        if (job % 10 == 0)
        {
            return;
        }

        int    j;
        double t;
        int    k;

        for (k = 1; k <= n; k++)
        {
            a[k - 1 + (k - 1) * lda] = 1.0 / a[k - 1 + (k - 1) * lda];
            t = -a[k - 1 + (k - 1) * lda];
            BLAS1D.dscal(k - 1, t, ref a, 1, index:  +0 + (k - 1) * lda);

            for (j = k + 1; j <= n; j++)
            {
                t = a[k - 1 + (j - 1) * lda];
                a[k - 1 + (j - 1) * lda] = 0.0;
                BLAS1D.daxpy(k, t, a, 1, ref a, 1, xIndex:  +0 + (k - 1) * lda, yIndex:  +0 + (j - 1) * lda);
            }
        }

        //
        //  Form inverse(U) * inverse(L).
        //
        for (k = n - 1; 1 <= k; k--)
        {
            for (i = k + 1; i <= n; i++)
            {
                work[i - 1] = a[i - 1 + (k - 1) * lda];
                a[i - 1 + (k - 1) * lda] = 0.0;
            }

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

            int l = ipvt[k - 1];
            if (l != k)
            {
                BLAS1D.dswap(n, ref a, 1, ref a, 1, xIndex:  +0 + (k - 1) * lda, yIndex:  +0 + (l - 1) * lda);
            }
        }
    }
Exemplo n.º 12
0
    public static void dpodi(ref double[] a, int lda, int n, ref double[] det, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DPODI computes the determinant and inverse of a certain matrix.
    //
    //  Discussion:
    //
    //    The matrix is real symmetric positive definite.
    //    DPODI uses the factors computed by DPOCO, DPOFA or DQRDC.
    //
    //    A division by zero will occur if the input factor contains
    //    a zero on the diagonal and the inverse is requested.
    //    It will not occur if the subroutines are called correctly
    //    and if DPOCO or DPOFA has set 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/output, double A[LDA*N].  On input, the output A from
    //    DPOCO or DPOFA, or the output X from DQRDC.  On output, if DPOCO or
    //    DPOFA was used to factor A then DPODI produces the upper half of
    //    inverse(A).  If DQRDC was used to decompose X then DPODI produces
    //    the upper half of inverse(X'*X) where X' is the transpose.
    //    Elements of A below the diagonal are unchanged.  If the units digit
    //    of JOB is zero, A is unchanged.
    //
    //    Input, int LDA, the leading dimension of the array A.
    //
    //    Input, int N, the order of the matrix A.
    //
    //    Input, int JOB, specifies the task.
    //    11, both determinant and inverse.
    //    01, inverse only.
    //    10, determinant only.
    //
    //    Output, double DET[2], the determinant of A or of X'*X
    //    if requested.
    //      determinant = DET[0] * 10.0**DET[1]
    //    with 1.0D+00 <= DET[0] < 10.0D+00 or DET[0] == 0.0D+00.
    //
    {
        //
        //  Compute the determinant.
        //
        if (job / 10 != 0)
        {
            det[0] = 1.0;
            det[1] = 0.0;
            const double s = 10.0;

            int i;
            for (i = 1; i <= n; i++)
            {
                det[0] = det[0] * a[i - 1 + (i - 1) * lda] * a[i - 1 + (i - 1) * lda];

                if (det[0] == 0.0)
                {
                    break;
                }

                while (det[0] < 1.0)
                {
                    det[0] *= s;
                    det[1] -= 1.0;
                }

                while (s <= det[0])
                {
                    det[0] /= s;
                    det[1] += 1.0;
                }
            }
        }

        //
        //  Compute inverse(R).
        //
        if (job % 10 == 0)
        {
            return;
        }

        double t;
        int    k;
        int    j;

        for (k = 1; k <= n; k++)
        {
            a[k - 1 + (k - 1) * lda] = 1.0 / a[k - 1 + (k - 1) * lda];
            t = -a[k - 1 + (k - 1) * lda];
            BLAS1D.dscal(k - 1, t, ref a, 1, index: +0 + (k - 1) * lda);

            for (j = k + 1; j <= n; j++)
            {
                t = a[k - 1 + (j - 1) * lda];
                a[k - 1 + (j - 1) * lda] = 0.0;
                BLAS1D.daxpy(k, t, a, 1, ref a, 1, xIndex: +0 + (k - 1) * lda, yIndex: +0 + (j - 1) * lda);
            }
        }

        //
        //  Form inverse(R) * (inverse(R))'.
        //
        for (j = 1; j <= n; j++)
        {
            for (k = 1; k <= j - 1; k++)
            {
                t = a[k - 1 + (j - 1) * lda];
                BLAS1D.daxpy(k, t, a, 1, ref a, 1, xIndex:  +0 + (j - 1) * lda, yIndex:  +0 + (k - 1) * lda);
            }

            t = a[j - 1 + (j - 1) * lda];
            BLAS1D.dscal(j, t, ref a, 1, index:  +0 + (j - 1) * lda);
        }
    }
Exemplo n.º 13
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;
        }
        }
    }
Exemplo n.º 14
0
    public static double dtrco(double[] t, int ldt, int n, ref double[] z, int job)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DTRCO estimates the condition of a real triangular matrix.
    //
    //  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 T[LDT*N], the triangular matrix.  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.
    //
    //    Output, double Z[N] a work vector whose contents are usually
    //    unimportant.  If T 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).
    //
    //    Input, int JOB, indicates the shape of T:
    //    0, T is lower triangular.
    //    nonzero, T is upper triangular.
    //
    //    Output, double DTRCO, an estimate of the reciprocal condition RCOND
    //    of T.  For the system T*X = B, relative perturbations in T 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 T may be singular to working precision.  In particular,
    //    RCOND is zero if exact singularity is detected or the estimate underflows.
    //
    {
        int    i;
        int    i1;
        int    j;
        int    k;
        int    kk;
        double rcond;
        double s;
        double w;

        bool lower = job == 0;
        //
        //  Compute the 1-norm of T.
        //
        double tnorm = 0.0;

        for (j = 1; j <= n; j++)
        {
            int l;
            switch (lower)
            {
            case true:
                l  = n + 1 - j;
                i1 = j;
                break;

            default:
                l  = j;
                i1 = 1;
                break;
            }

            tnorm = Math.Max(tnorm, BLAS1D.dasum(l, t, 1, index: +i1 - 1 + (j - 1) * ldt));
        }

        //
        //  RCOND = 1/(norm(T)*(estimate of norm(inverse(T)))).
        //
        //  Estimate = norm(Z)/norm(Y) where T * Z = Y and T' * Y = E.
        //
        //  T' is the transpose of T.
        //
        //  The components of E are chosen to cause maximum local
        //  growth in the elements of Y.
        //
        //  The vectors are frequently rescaled to avoid overflow.
        //
        //  Solve T' * Y = E.
        //
        double ek = 1.0;

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

        for (kk = 1; kk <= n; kk++)
        {
            k = lower switch
            {
                true => n + 1 - kk,
                _ => kk
            };

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

            if (Math.Abs(t[k - 1 + (k - 1) * ldt]) < Math.Abs(ek - z[k - 1]))
            {
                s = Math.Abs(t[k - 1 + (k - 1) * ldt]) / 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 (t[k - 1 + (k - 1) * ldt] != 0.0)
            {
                wk  /= t[k - 1 + (k - 1) * ldt];
                wkm /= t[k - 1 + (k - 1) * ldt];
            }
            else
            {
                wk  = 1.0;
                wkm = 1.0;
            }

            if (kk != n)
            {
                int j2;
                int j1;
                switch (lower)
                {
                case true:
                    j1 = 1;
                    j2 = k - 1;
                    break;

                default:
                    j1 = k + 1;
                    j2 = n;
                    break;
                }

                for (j = j1; j <= j2; j++)
                {
                    sm       += Math.Abs(z[j - 1] + wkm * t[k - 1 + (j - 1) * ldt]);
                    z[j - 1] += wk * t[k - 1 + (j - 1) * ldt];
                    s        += Math.Abs(z[j - 1]);
                }

                if (s < sm)
                {
                    w  = wkm - wk;
                    wk = wkm;
                    for (j = j1; j <= j2; j++)
                    {
                        z[j - 1] += w * t[k - 1 + (j - 1) * ldt];
                    }
                }
            }

            z[k - 1] = wk;
        }

        double temp = BLAS1D.dasum(n, z, 1);

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

        double ynorm = 1.0;

        //
        //  Solve T * Z = Y.
        //
        for (kk = 1; kk <= n; kk++)
        {
            k = lower switch
            {
                true => kk,
                _ => n + 1 - kk
            };

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

                ynorm = s * ynorm;
            }

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

            i1 = lower switch
            {
                true => k + 1,
                _ => 1
            };

            if (kk >= n)
            {
                continue;
            }

            w = -z[k - 1];
            BLAS1D.daxpy(n - kk, w, t, 1, ref z, 1, xIndex: +i1 - 1 + (k - 1) * ldt, yIndex: +i1 - 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 (tnorm != 0.0)
        {
            rcond = ynorm / tnorm;
        }
        else
        {
            rcond = 0.0;
        }

        return(rcond);
    }
}
Exemplo n.º 15
0
    public static int dspfa(ref double[] ap, int n, ref int[] kpvt)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DSPFA factors a real symmetric matrix stored in packed form.
    //
    //  Discussion:
    //
    //    To solve A*X = B, follow DSPFA by DSPSL.
    //
    //    To compute inverse(A)*C, follow DSPFA by DSPSL.
    //
    //    To compute determinant(A), follow DSPFA by DSPDI.
    //
    //    To compute inertia(A), follow DSPFA by DSPDI.
    //
    //    To compute inverse(A), follow DSPFA by DSPDI.
    //
    //  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:
    //
    //    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 packed form of a
    //    symmetric matrix A.  The columns of the upper triangle are stored
    //    sequentially in a one-dimensional array.  On output, a block diagonal
    //    matrix and the multipliers which were used to obtain it stored in
    //    packed form.  The factorization can be written A = U*D*U' where U
    //    is a product of permutation and unit upper triangular matrices, U'
    //    is the transpose of U, and D is block diagonal with 1 by 1 and 2
    //    by 2 blocks.
    //
    //    Input, int N, the order of the matrix.
    //
    //    Output, int KPVT[N], the pivot indices.
    //
    //    Output, int DSPFA, error flag.
    //    0, normal value.
    //    K, if the K-th pivot block is singular.  This is not an error
    //    condition for this subroutine, but it does indicate that DSPSL or
    //    DSPDI may divide by zero if called.
    //
    {
        int im = 0;
        //
        //  ALPHA is used in choosing pivot block size.
        //
        double alpha = (1.0 + Math.Sqrt(17.0)) / 8.0;

        int info = 0;
        //
        //  Main loop on K, which goes from N to 1.
        //
        int k  = n;
        int ik = n * (n - 1) / 2;

        for (;;)
        {
            //
            //  Leave the loop if K = 0 or K = 1.
            //
            if (k == 0)
            {
                break;
            }

            if (k == 1)
            {
                kpvt[0] = 1;
                info    = ap[0] switch
                {
                    0.0 => 1,
                    _ => info
                };

                break;
            }

            //
            //  This section of code determines the kind of elimination to be performed.
            //  When it is completed, KSTEP will be set to the size of the pivot block,
            //  and SWAP will be set to .true. if an interchange is required.
            //
            int    km1    = k - 1;
            int    kk     = ik + k;
            double absakk = Math.Abs(ap[kk - 1]);
            //
            //  Determine the largest off-diagonal element in column K.
            //
            int    imax   = BLAS1D.idamax(k - 1, ap, 1, index: +ik);
            int    imk    = ik + imax;
            double colmax = Math.Abs(ap[imk - 1]);

            int  kstep;
            bool swap;
            int  j;
            int  imj;
            if (alpha * colmax <= absakk)
            {
                kstep = 1;
                swap  = false;
            }
            //
            //  Determine the largest off-diagonal element in row IMAX.
            //
            else
            {
                double rowmax = 0.0;
                int    imaxp1 = imax + 1;
                im  = imax * (imax - 1) / 2;
                imj = im + 2 * imax;

                for (j = imaxp1; j <= k; j++)
                {
                    rowmax = Math.Max(rowmax, Math.Abs(ap[imj - 1]));
                    imj   += j;
                }

                if (imax != 1)
                {
                    int jmax = BLAS1D.idamax(imax - 1, ap, 1, index: +im);
                    int jmim = jmax + im;
                    rowmax = Math.Max(rowmax, Math.Abs(ap[jmim - 1]));
                }

                int imim = imax + im;

                if (alpha * rowmax <= Math.Abs(ap[imim - 1]))
                {
                    kstep = 1;
                    swap  = true;
                }
                else if (alpha * colmax * (colmax / rowmax) <= absakk)
                {
                    kstep = 1;
                    swap  = false;
                }
                else
                {
                    kstep = 2;
                    swap  = imax != km1;
                }
            }

            switch (Math.Max(absakk, colmax))
            {
            //
            //  Column K is zero.  Set INFO and iterate the loop.
            //
            case 0.0:
                kpvt[k - 1] = k;
                info        = k;
                break;

            default:
            {
                double mulk;
                int    jk;
                int    jj;
                double t;
                int    ij;
                if (kstep != 2)
                {
                    switch (swap)
                    {
                    //
                    //  1 x 1 pivot block.
                    //
                    case true:
                    {
                        //
                        //  Perform an interchange.
                        //
                        BLAS1D.dswap(imax, ref ap, 1, ref ap, 1, xIndex: +im, yIndex: +ik);
                        imj = ik + imax;

                        for (jj = imax; jj <= k; jj++)
                        {
                            j           = k + imax - jj;
                            jk          = ik + j;
                            t           = ap[jk - 1];
                            ap[jk - 1]  = ap[imj - 1];
                            ap[imj - 1] = t;
                            imj        -= j - 1;
                        }

                        break;
                    }
                    }

                    //
                    //  Perform the elimination.
                    //
                    ij = ik - (k - 1);

                    for (jj = 1; jj <= km1; jj++)
                    {
                        j    = k - jj;
                        jk   = ik + j;
                        mulk = -ap[jk - 1] / ap[kk - 1];
                        t    = mulk;
                        BLAS1D.daxpy(j, t, ap, 1, ref ap, 1, xIndex: +ik, yIndex: +ij);
                        ap[jk - 1] = mulk;
                        ij        -= j - 1;
                    }

                    kpvt[k - 1] = swap switch
                    {
                        //
                        //  Set the pivot array.
                        //
                        true => imax,
                        _ => k
                    };
                }
                else
                {
                    //
                    //  2 x 2 pivot block.
                    //
                    int km1k = ik + k - 1;
                    int ikm1 = ik - (k - 1);
                    int jkm1;
                    switch (swap)
                    {
                    //
                    //  Perform an interchange.
                    //
                    case true:
                    {
                        BLAS1D.dswap(imax, ref ap, 1, ref ap, 1, xIndex: +im, yIndex: +ikm1);
                        imj = ikm1 + imax;

                        for (jj = imax; jj <= km1; jj++)
                        {
                            j            = km1 + imax - jj;
                            jkm1         = ikm1 + j;
                            t            = ap[jkm1 - 1];
                            ap[jkm1 - 1] = ap[imj - 1];
                            ap[imj - 1]  = t;
                            imj         -= j - 1;
                        }

                        t            = ap[km1k - 1];
                        ap[km1k - 1] = ap[imk - 1];
                        ap[imk - 1]  = t;
                        break;
                    }
                    }

                    //
                    //  Perform the elimination.
                    //
                    if (k - 2 != 0)
                    {
                        double ak     = ap[kk - 1] / ap[km1k - 1];
                        int    km1km1 = ikm1 + k - 1;
                        double akm1   = ap[km1km1 - 1] / ap[km1k - 1];
                        double denom  = 1.0 - ak * akm1;
                        ij = ik - (k - 1) - (k - 2);

                        for (jj = 1; jj <= k - 2; jj++)
                        {
                            j  = km1 - jj;
                            jk = ik + j;
                            double bk = ap[jk - 1] / ap[km1k - 1];
                            jkm1 = ikm1 + j;
                            double bkm1 = ap[jkm1 - 1] / ap[km1k - 1];
                            mulk = (akm1 * bk - bkm1) / denom;
                            double mulkm1 = (ak * bkm1 - bk) / denom;
                            t = mulk;
                            BLAS1D.daxpy(j, t, ap, 1, ref ap, 1, xIndex: +ik, yIndex: +ij);
                            t = mulkm1;
                            BLAS1D.daxpy(j, t, ap, 1, ref ap, 1, xIndex: +ikm1, yIndex: +ij);
                            ap[jk - 1]   = mulk;
                            ap[jkm1 - 1] = mulkm1;
                            ij          -= j - 1;
                        }
                    }

                    kpvt[k - 1] = swap switch
                    {
                        //
                        //  Set the pivot array.
                        //
                        true => - imax,
                        _ => 1 - k
                    };

                    kpvt[k - 2] = kpvt[k - 1];
                }

                break;
            }
            }

            ik -= k - 1;
            switch (kstep)
            {
            case 2:
                ik -= k - 2;
                break;
            }

            k -= kstep;
        }

        return(info);
    }
}
Exemplo n.º 16
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);
    }
Exemplo n.º 17
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;
        }
        }
    }
}
Exemplo n.º 18
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;
        }
        }
    }
Exemplo n.º 19
0
    public static int dgbfa(ref double[] abd, int lda, int n, int ml, int mu, ref int[] ipvt)

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DGBFA factors a real band matrix by elimination.
    //
    //  Discussion:
    //
    //    DGBFA is usually called by DGBCO, but it can be called
    //    directly with a saving in time if RCOND is not needed.
    //
    //  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 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.
    //
    //    Output, integer DGBFA, error flag.
    //    0, normal value.
    //    K, if U(K,K) == 0.0D+00.  This is not an error condition for this
    //      subroutine, but it does indicate that DGBSL will divide by zero if
    //      called.  Use RCOND in DGBCO for a reliable indication of singularity.
    //
    {
        int i;
        int jz;
        int k;

        int m    = ml + mu + 1;
        int info = 0;
        //
        //  Zero initial fill-in columns.
        //
        int j0 = mu + 2;
        int j1 = Math.Min(n, m) - 1;

        for (jz = j0; jz <= j1; jz++)
        {
            int i0 = m + 1 - jz;
            for (i = i0; i <= ml; i++)
            {
                abd[i - 1 + (jz - 1) * lda] = 0.0;
            }
        }

        jz = j1;
        int ju = 0;

        //
        //  Gaussian elimination with partial pivoting.
        //
        for (k = 1; k <= n - 1; k++)
        {
            //
            //  Zero out the next fill-in column.
            //
            jz += 1;
            if (jz <= n)
            {
                for (i = 1; i <= ml; i++)
                {
                    abd[i - 1 + (jz - 1) * lda] = 0.0;
                }
            }

            //
            //  Find L = pivot index.
            //
            int lm = Math.Min(ml, n - k);
            int l  = BLAS1D.idamax(lm + 1, abd, 1, +m - 1 + (k - 1) * lda) + m - 1;
            ipvt[k - 1] = l + k - m;
            switch (abd[l - 1 + (k - 1) * lda])
            {
            //
            //  Zero pivot implies this column already triangularized.
            //
            case 0.0:
                info = k;
                break;

            //
            default:
            {
                double t;
                if (l != m)
                {
                    t = abd[l - 1 + (k - 1) * lda];
                    abd[l - 1 + (k - 1) * lda] = abd[m - 1 + (k - 1) * lda];
                    abd[m - 1 + (k - 1) * lda] = t;
                }

                //
                //  Compute multipliers.
                //
                t = -1.0 / abd[m - 1 + (k - 1) * lda];
                BLAS1D.dscal(lm, t, ref abd, 1, +m + (k - 1) * lda);
                //
                //  Row elimination with column indexing.
                //
                ju = Math.Min(Math.Max(ju, mu + ipvt[k - 1]), n);
                int mm = m;

                int j;
                for (j = k + 1; j <= ju; j++)
                {
                    l  -= 1;
                    mm -= 1;
                    t   = abd[l - 1 + (j - 1) * lda];
                    if (l != mm)
                    {
                        abd[l - 1 + (j - 1) * lda]  = abd[mm - 1 + (j - 1) * lda];
                        abd[mm - 1 + (j - 1) * lda] = t;
                    }

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

                break;
            }
            }
        }

        ipvt[n - 1] = n;

        info = abd[m - 1 + (n - 1) * lda] switch
        {
            0.0 => n,
            _ => info
        };

        return(info);
    }
}
Exemplo n.º 20
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);
        }
    }
Exemplo n.º 21
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);
    }
Exemplo n.º 22
0
    private static void daxpy_test()

    //****************************************************************************80
    //
    //  Purpose:
    //
    //    DAXPY_TEST tests DAXPY.
    //
    //  Licensing:
    //
    //    This code is distributed under the GNU LGPL license.
    //
    //  Modified:
    //
    //    15 May 2006
    //
    //  Author:
    //
    //    John Burkardt
    //
    {
        const int N = 6;

        double[] x = new double[N];
        double[] y = new double[N];

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

        for (int i = 0; i < N; i++)
        {
            y[i] = 100 * (i + 1);
        }

        Console.WriteLine("");
        Console.WriteLine("DAXPY_TEST");
        Console.WriteLine("  DAXPY adds a multiple of vector X to vector Y.");
        Console.WriteLine("");
        Console.WriteLine("  X =");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            Console.WriteLine("  "
                              + (i + 1).ToString(CultureInfo.InvariantCulture).PadLeft(6) + "  "
                              + x[i].ToString(CultureInfo.InvariantCulture).PadLeft(14) + "");
        }

        Console.WriteLine("");
        Console.WriteLine("  Y =");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            Console.WriteLine("  "
                              + (i + 1).ToString(CultureInfo.InvariantCulture).PadLeft(6) + "  "
                              + y[i].ToString(CultureInfo.InvariantCulture).PadLeft(14) + "");
        }

        double da = 1.0;

        BLAS1D.daxpy(N, da, x, 1, ref y, 1);
        Console.WriteLine("");
        Console.WriteLine("  DAXPY ( N, " + da + ", X, 1, Y, 1 )");
        Console.WriteLine("");
        Console.WriteLine("  Y =");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            Console.WriteLine("  "
                              + (i + 1).ToString(CultureInfo.InvariantCulture).PadLeft(6) + "  "
                              + y[i].ToString(CultureInfo.InvariantCulture).PadLeft(14) + "");
        }

        for (int i = 0; i < N; i++)
        {
            y[i] = 100 * (i + 1);
        }

        da = -2.0;
        BLAS1D.daxpy(N, da, x, 1, ref y, 1);
        Console.WriteLine("");
        Console.WriteLine("  DAXPY ( N, " + da + ", X, 1, Y, 1 )");
        Console.WriteLine("");
        Console.WriteLine("  Y =");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            Console.WriteLine("  "
                              + (i + 1).ToString(CultureInfo.InvariantCulture).PadLeft(6) + "  "
                              + y[i].ToString(CultureInfo.InvariantCulture).PadLeft(14) + "");
        }

        for (int i = 0; i < N; i++)
        {
            y[i] = 100 * (i + 1);
        }

        da = 3.0;
        BLAS1D.daxpy(3, da, x, 2, ref y, 1);
        Console.WriteLine("");
        Console.WriteLine("  DAXPY ( 3, " + da + ", X, 2, Y, 1 )");
        Console.WriteLine("");
        Console.WriteLine("  Y =");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            Console.WriteLine("  "
                              + (i + 1).ToString(CultureInfo.InvariantCulture).PadLeft(6) + "  "
                              + y[i].ToString(CultureInfo.InvariantCulture).PadLeft(14) + "");
        }

        for (int i = 0; i < N; i++)
        {
            y[i] = 100 * (i + 1);
        }

        da = -4.0;
        BLAS1D.daxpy(3, da, x, 1, ref y, 2);
        Console.WriteLine("");
        Console.WriteLine("  DAXPY ( 3, " + da + ", X, 1, Y, 2 )");
        Console.WriteLine("");
        Console.WriteLine("  Y =");
        Console.WriteLine("");
        for (int i = 0; i < N; i++)
        {
            Console.WriteLine("  "
                              + (i + 1).ToString(CultureInfo.InvariantCulture).PadLeft(6) + "  "
                              + y[i].ToString(CultureInfo.InvariantCulture).PadLeft(14) + "");
        }
    }
Exemplo n.º 23
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);
        }
    }
Exemplo n.º 24
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);
    }