Beispiel #1
0
        /// <summary>Solves a set of equation systems of type <c>A * X = B</c>.</summary>
        /// <param name="value">Right hand side matrix with as many rows as <c>A</c> and any number of columns.</param>
        /// <returns>Matrix <c>X</c> so that <c>L * L' * X = B</c>.</returns>
        /// <exception cref="T:System.ArgumentException">Matrix dimensions do not match.</exception>
        /// <exception cref="T:System.InvalidOperationException">Matrix is not symmetrix and positive definite.</exception>
        public Matrix Solve(Matrix value)
        {
            if (value == null)
            {
                throw new ArgumentNullException("value");
            }

            if (value.Rows != L.Rows)
            {
                throw new ArgumentException("Matrix dimensions do not match.");
            }

            if (!this.symmetric)
            {
                throw new InvalidOperationException("Matrix is not symmetric.");
            }

            if (!this.positiveDefinite)
            {
                throw new InvalidOperationException("Matrix is not positive definite.");
            }

            int dimension = L.Rows;
            int count     = value.Columns;

            Matrix B = (Matrix)value.Clone();

            double[][] l = L.Array;

            // Solve L*Y = B;
            for (int k = 0; k < L.Rows; k++)
            {
                for (int i = k + 1; i < dimension; i++)
                {
                    for (int j = 0; j < count; j++)
                    {
                        B[i, j] -= B[k, j] * l[i][k];
                    }
                }

                for (int j = 0; j < count; j++)
                {
                    B[k, j] /= l[k][k];
                }
            }

            // Solve L'*X = Y;
            for (int k = dimension - 1; k >= 0; k--)
            {
                for (int j = 0; j < count; j++)
                {
                    B[k, j] /= l[k][k];
                }

                for (int i = 0; i < k; i++)
                {
                    for (int j = 0; j < count; j++)
                    {
                        B[i, j] -= B[k, j] * l[k][i];
                    }
                }
            }

            return(B);
        }
Beispiel #2
0
        /// <summary>Construct singular value decomposition.</summary>
        public SingularValueDecomposition(Matrix value)
        {
            if (value == null)
            {
                throw new ArgumentNullException("value");
            }

            Matrix copy = (Matrix)value.Clone();

            double[][] a = copy.Array;
            m = value.Rows;
            n = value.Columns;
            int nu = Math.Min(m, n);

            s = new double [Math.Min(m + 1, n)];
            U = new Matrix(m, nu);
            V = new Matrix(n, n);
            double[][] u     = U.Array;
            double[][] v     = V.Array;
            double[]   e     = new double [n];
            double[]   work  = new double [m];
            bool       wantu = true;
            bool       wantv = true;

            // Reduce A to bidiagonal form, storing the diagonal elements in s and the super-diagonal elements in e.
            int nct = Math.Min(m - 1, n);
            int nrt = Math.Max(0, Math.Min(n - 2, m));

            for (int k = 0; k < Math.Max(nct, nrt); k++)
            {
                if (k < nct)
                {
                    // Compute the transformation for the k-th column and place the k-th diagonal in s[k].
                    // Compute 2-norm of k-th column without under/overflow.
                    s[k] = 0;
                    for (int i = k; i < m; i++)
                    {
                        s[k] = Hypotenuse(s[k], a[i][k]);
                    }

                    if (s[k] != 0.0)
                    {
                        if (a[k][k] < 0.0)
                        {
                            s[k] = -s[k];
                        }

                        for (int i = k; i < m; i++)
                        {
                            a[i][k] /= s[k];
                        }

                        a[k][k] += 1.0;
                    }

                    s[k] = -s[k];
                }

                for (int j = k + 1; j < n; j++)
                {
                    if ((k < nct) & (s[k] != 0.0))
                    {
                        // Apply the transformation.
                        double t = 0;
                        for (int i = k; i < m; i++)
                        {
                            t += a[i][k] * a[i][j];
                        }
                        t = -t / a[k][k];
                        for (int i = k; i < m; i++)
                        {
                            a[i][j] += t * a[i][k];
                        }
                    }

                    // Place the k-th row of A into e for the subsequent calculation of the row transformation.
                    e[j] = a[k][j];
                }

                if (wantu & (k < nct))
                {
                    // Place the transformation in U for subsequent back
                    // multiplication.
                    for (int i = k; i < m; i++)
                    {
                        u[i][k] = a[i][k];
                    }
                }

                if (k < nrt)
                {
                    // Compute the k-th row transformation and place the k-th super-diagonal in e[k].
                    // Compute 2-norm without under/overflow.
                    e[k] = 0;
                    for (int i = k + 1; i < n; i++)
                    {
                        e[k] = Hypotenuse(e[k], e[i]);
                    }

                    if (e[k] != 0.0)
                    {
                        if (e[k + 1] < 0.0)
                        {
                            e[k] = -e[k];
                        }

                        for (int i = k + 1; i < n; i++)
                        {
                            e[i] /= e[k];
                        }

                        e[k + 1] += 1.0;
                    }

                    e[k] = -e[k];
                    if ((k + 1 < m) & (e[k] != 0.0))
                    {
                        // Apply the transformation.
                        for (int i = k + 1; i < m; i++)
                        {
                            work[i] = 0.0;
                        }

                        for (int j = k + 1; j < n; j++)
                        {
                            for (int i = k + 1; i < m; i++)
                            {
                                work[i] += e[j] * a[i][j];
                            }
                        }

                        for (int j = k + 1; j < n; j++)
                        {
                            double t = -e[j] / e[k + 1];
                            for (int i = k + 1; i < m; i++)
                            {
                                a[i][j] += t * work[i];
                            }
                        }
                    }

                    if (wantv)
                    {
                        // Place the transformation in V for subsequent back multiplication.
                        for (int i = k + 1; i < n; i++)
                        {
                            v[i][k] = e[i];
                        }
                    }
                }
            }

            // Set up the final bidiagonal matrix or order p.
            int p = Math.Min(n, m + 1);

            if (nct < n)
            {
                s[nct] = a[nct][nct];
            }
            if (m < p)
            {
                s[p - 1] = 0.0;
            }
            if (nrt + 1 < p)
            {
                e[nrt] = a[nrt][p - 1];
            }
            e[p - 1] = 0.0;

            // If required, generate U.
            if (wantu)
            {
                for (int j = nct; j < nu; j++)
                {
                    for (int i = 0; i < m; i++)
                    {
                        u[i][j] = 0.0;
                    }
                    u[j][j] = 1.0;
                }

                for (int k = nct - 1; k >= 0; k--)
                {
                    if (s[k] != 0.0)
                    {
                        for (int j = k + 1; j < nu; j++)
                        {
                            double t = 0;
                            for (int i = k; i < m; i++)
                            {
                                t += u[i][k] * u[i][j];
                            }

                            t = -t / u[k][k];
                            for (int i = k; i < m; i++)
                            {
                                u[i][j] += t * u[i][k];
                            }
                        }

                        for (int i = k; i < m; i++)
                        {
                            u[i][k] = -u[i][k];
                        }

                        u[k][k] = 1.0 + u[k][k];
                        for (int i = 0; i < k - 1; i++)
                        {
                            u[i][k] = 0.0;
                        }
                    }
                    else
                    {
                        for (int i = 0; i < m; i++)
                        {
                            u[i][k] = 0.0;
                        }
                        u[k][k] = 1.0;
                    }
                }
            }

            // If required, generate V.
            if (wantv)
            {
                for (int k = n - 1; k >= 0; k--)
                {
                    if ((k < nrt) & (e[k] != 0.0))
                    {
                        for (int j = k + 1; j < nu; j++)
                        {
                            double t = 0;
                            for (int i = k + 1; i < n; i++)
                            {
                                t += v[i][k] * v[i][j];
                            }

                            t = -t / v[k + 1][k];
                            for (int i = k + 1; i < n; i++)
                            {
                                v[i][j] += t * v[i][k];
                            }
                        }
                    }

                    for (int i = 0; i < n; i++)
                    {
                        v[i][k] = 0.0;
                    }
                    v[k][k] = 1.0;
                }
            }

            // Main iteration loop for the singular values.
            int    pp   = p - 1;
            int    iter = 0;
            double eps  = Math.Pow(2.0, -52.0);

            while (p > 0)
            {
                int k, kase;

                // Here is where a test for too many iterations would go.
                // This section of the program inspects for
                // negligible elements in the s and e arrays.  On
                // completion the variables kase and k are set as follows.
                // kase = 1     if s(p) and e[k-1] are negligible and k<p
                // kase = 2     if s(k) is negligible and k<p
                // kase = 3     if e[k-1] is negligible, k<p, and s(k), ..., s(p) are not negligible (qr step).
                // kase = 4     if e(p-1) is negligible (convergence).
                for (k = p - 2; k >= -1; k--)
                {
                    if (k == -1)
                    {
                        break;
                    }

                    if (Math.Abs(e[k]) <= eps * (Math.Abs(s[k]) + Math.Abs(s[k + 1])))
                    {
                        e[k] = 0.0;
                        break;
                    }
                }

                if (k == p - 2)
                {
                    kase = 4;
                }
                else
                {
                    int ks;
                    for (ks = p - 1; ks >= k; ks--)
                    {
                        if (ks == k)
                        {
                            break;
                        }

                        double t = (ks != p ? Math.Abs(e[ks]) : 0.0) + (ks != k + 1 ? Math.Abs(e[ks - 1]) : 0.0);
                        if (Math.Abs(s[ks]) <= eps * t)
                        {
                            s[ks] = 0.0;
                            break;
                        }
                    }

                    if (ks == k)
                    {
                        kase = 3;
                    }
                    else if (ks == p - 1)
                    {
                        kase = 1;
                    }
                    else
                    {
                        kase = 2;
                        k    = ks;
                    }
                }

                k++;

                // Perform the task indicated by kase.
                switch (kase)
                {
                // Deflate negligible s(p).
                case 1:
                {
                    double f = e[p - 2];
                    e[p - 2] = 0.0;
                    for (int j = p - 2; j >= k; j--)
                    {
                        double t  = Hypotenuse(s[j], f);
                        double cs = s[j] / t;
                        double sn = f / t;
                        s[j] = t;
                        if (j != k)
                        {
                            f        = -sn * e[j - 1];
                            e[j - 1] = cs * e[j - 1];
                        }

                        if (wantv)
                        {
                            for (int i = 0; i < n; i++)
                            {
                                t           = cs * v[i][j] + sn * v[i][p - 1];
                                v[i][p - 1] = -sn * v[i][j] + cs * v[i][p - 1];
                                v[i][j]     = t;
                            }
                        }
                    }
                }
                break;

                // Split at negligible s(k).
                case 2:
                {
                    double f = e[k - 1];
                    e[k - 1] = 0.0;
                    for (int j = k; j < p; j++)
                    {
                        double t  = Hypotenuse(s[j], f);
                        double cs = s[j] / t;
                        double sn = f / t;
                        s[j] = t;
                        f    = -sn * e[j];
                        e[j] = cs * e[j];
                        if (wantu)
                        {
                            for (int i = 0; i < m; i++)
                            {
                                t           = cs * u[i][j] + sn * u[i][k - 1];
                                u[i][k - 1] = -sn * u[i][j] + cs * u[i][k - 1];
                                u[i][j]     = t;
                            }
                        }
                    }
                }
                break;

                // Perform one qr step.
                case 3:
                {
                    // Calculate the shift.
                    double scale = Math.Max(Math.Max(Math.Max(Math.Max(Math.Abs(s[p - 1]), Math.Abs(s[p - 2])), Math.Abs(e[p - 2])), Math.Abs(s[k])), Math.Abs(e[k]));
                    double sp    = s[p - 1] / scale;
                    double spm1  = s[p - 2] / scale;
                    double epm1  = e[p - 2] / scale;
                    double sk    = s[k] / scale;
                    double ek    = e[k] / scale;
                    double b     = ((spm1 + sp) * (spm1 - sp) + epm1 * epm1) / 2.0;
                    double c     = (sp * epm1) * (sp * epm1);
                    double shift = 0.0;
                    if ((b != 0.0) | (c != 0.0))
                    {
                        shift = Math.Sqrt(b * b + c);
                        if (b < 0.0)
                        {
                            shift = -shift;
                        }
                        shift = c / (b + shift);
                    }

                    double f = (sk + sp) * (sk - sp) + shift;
                    double g = sk * ek;

                    // Chase zeros.
                    for (int j = k; j < p - 1; j++)
                    {
                        double t  = Hypotenuse(f, g);
                        double cs = f / t;
                        double sn = g / t;
                        if (j != k)
                        {
                            e[j - 1] = t;
                        }
                        f        = cs * s[j] + sn * e[j];
                        e[j]     = cs * e[j] - sn * s[j];
                        g        = sn * s[j + 1];
                        s[j + 1] = cs * s[j + 1];
                        if (wantv)
                        {
                            for (int i = 0; i < n; i++)
                            {
                                t           = cs * v[i][j] + sn * v[i][j + 1];
                                v[i][j + 1] = -sn * v[i][j] + cs * v[i][j + 1];
                                v[i][j]     = t;
                            }
                        }

                        t        = Hypotenuse(f, g);
                        cs       = f / t;
                        sn       = g / t;
                        s[j]     = t;
                        f        = cs * e[j] + sn * s[j + 1];
                        s[j + 1] = -sn * e[j] + cs * s[j + 1];
                        g        = sn * e[j + 1];
                        e[j + 1] = cs * e[j + 1];
                        if (wantu && (j < m - 1))
                        {
                            for (int i = 0; i < m; i++)
                            {
                                t           = cs * u[i][j] + sn * u[i][j + 1];
                                u[i][j + 1] = -sn * u[i][j] + cs * u[i][j + 1];
                                u[i][j]     = t;
                            }
                        }
                    }

                    e[p - 2] = f;
                    iter     = iter + 1;
                }
                break;

                // Convergence.
                case 4:
                {
                    // Make the singular values positive.
                    if (s[k] <= 0.0)
                    {
                        s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
                        if (wantv)
                        {
                            for (int i = 0; i <= pp; i++)
                            {
                                v[i][k] = -v[i][k];
                            }
                        }
                    }

                    // Order the singular values.
                    while (k < pp)
                    {
                        if (s[k] >= s[k + 1])
                        {
                            break;
                        }

                        double t = s[k];
                        s[k]     = s[k + 1];
                        s[k + 1] = t;
                        if (wantv && (k < n - 1))
                        {
                            for (int i = 0; i < n; i++)
                            {
                                t           = v[i][k + 1];
                                v[i][k + 1] = v[i][k];
                                v[i][k]     = t;
                            }
                        }

                        if (wantu && (k < m - 1))
                        {
                            for (int i = 0; i < m; i++)
                            {
                                t           = u[i][k + 1];
                                u[i][k + 1] = u[i][k];
                                u[i][k]     = t;
                            }
                        }

                        k++;
                    }

                    iter = 0;
                    p--;
                }
                break;
                }
            }
        }
Beispiel #3
0
        /// <summary>Least squares solution of <c>A * X = B</c></summary>
        /// <param name="value">Right-hand-side matrix with as many rows as <c>A</c> and any number of columns.</param>
        /// <returns>A matrix that minimized the two norm of <c>Q * R * X - B</c>.</returns>
        /// <exception cref="T:System.ArgumentException">Matrix row dimensions must be the same.</exception>
        /// <exception cref="T:System.InvalidOperationException">Matrix is rank deficient.</exception>
        public Matrix Solve(Matrix value)
        {
            if (value == null)
            {
                throw new ArgumentNullException("value");
            }

            if (value.Rows != QR.Rows)
            {
                throw new ArgumentException("Matrix row dimensions must agree.");
            }

            if (!this.FullRank)
            {
                throw new InvalidOperationException("Matrix is rank deficient.");
            }

            // Copy right hand side
            int    count = value.Columns;
            Matrix X     = value.Clone();
            int    m     = QR.Rows;
            int    n     = QR.Columns;

            double[][] qr = QR.Array;

            // Compute Y = transpose(Q)*B
            for (int k = 0; k < n; k++)
            {
                for (int j = 0; j < count; j++)
                {
                    double s = 0.0;

                    for (int i = k; i < m; i++)
                    {
                        s += qr[i][k] * X[i, j];
                    }

                    s = -s / qr[k][k];

                    for (int i = k; i < m; i++)
                    {
                        X[i, j] += s * qr[i][k];
                    }
                }
            }

            // Solve R*X = Y;
            for (int k = n - 1; k >= 0; k--)
            {
                for (int j = 0; j < count; j++)
                {
                    X[k, j] /= Rdiag[k];
                }

                for (int i = 0; i < k; i++)
                {
                    for (int j = 0; j < count; j++)
                    {
                        X[i, j] -= X[k, j] * qr[i][k];
                    }
                }
            }

            return(X.Submatrix(0, n - 1, 0, count - 1));
        }
        /// <summary>Construct a LU decomposition.</summary>
        public LuDecomposition(Matrix value)
        {
            if (value == null)
            {
                throw new ArgumentNullException("value");
            }

            this.LU = (Matrix)value.Clone();
            double[][] lu      = LU.Array;
            int        rows    = value.Rows;
            int        columns = value.Columns;

            pivotVector = new int[rows];
            for (int i = 0; i < rows; i++)
            {
                pivotVector[i] = i;
            }

            pivotSign = 1;
            double[] LUrowi;
            double[] LUcolj = new double[rows];

            // Outer loop.
            for (int j = 0; j < columns; j++)
            {
                // Make a copy of the j-th column to localize references.
                for (int i = 0; i < rows; i++)
                {
                    LUcolj[i] = lu[i][j];
                }

                // Apply previous transformations.
                for (int i = 0; i < rows; i++)
                {
                    LUrowi = lu[i];

                    // Most of the time is spent in the following dot product.
                    int    kmax = Math.Min(i, j);
                    double s    = 0.0;
                    for (int k = 0; k < kmax; k++)
                    {
                        s += LUrowi[k] * LUcolj[k];
                    }
                    LUrowi[j] = LUcolj[i] -= s;
                }

                // Find pivot and exchange if necessary.
                int p = j;
                for (int i = j + 1; i < rows; i++)
                {
                    if (Math.Abs(LUcolj[i]) > Math.Abs(LUcolj[p]))
                    {
                        p = i;
                    }
                }

                if (p != j)
                {
                    for (int k = 0; k < columns; k++)
                    {
                        double t = lu[p][k];
                        lu[p][k] = lu[j][k];
                        lu[j][k] = t;
                    }

                    int v = pivotVector[p];
                    pivotVector[p] = pivotVector[j];
                    pivotVector[j] = v;

                    pivotSign = -pivotSign;
                }

                // Compute multipliers.

                if (j < rows & lu[j][j] != 0.0)
                {
                    for (int i = j + 1; i < rows; i++)
                    {
                        lu[i][j] /= lu[j][j];
                    }
                }
            }
        }