/// <summary>
        /// Given we want to find x such as A * x = b
        ///   1. Decompose A = L * U, where L is the lower triangular and U is the upper triangular
        ///   2. We have L * U * x = b, which can be rewritten as L * y = b, where U * x = y
        ///   3. Solve y for L * y = b using forward substitution
        ///   4. Solve x for U * x = y using backward substitution
        /// </summary>
        /// <param name="A"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        public static IVector Solve(IMatrix A, IVector b)
        {
            IMatrix L;

            Cholesky.Factorize(A, out L);

            IMatrix U = L.Transpose(); // upper triangular matrix

            IVector y = ForwardSubstitution.Solve(L, b);

            IVector x = BackwardSubstitution.Solve(U, y);

            return(x);
        }
Пример #2
0
        /// <summary>
        /// Solve x for Ax = b, where A is a invertible matrix
        ///
        /// The method works as follows:
        ///  1. QR factorization A = Q * R
        ///  2. Let: c = Q.transpose * b, we have R * x = c, where R is a triangular matrix if A is n x n
        ///  3. Solve x using backward substitution
        /// </summary>
        /// <param name="A">An m x n matrix with linearly independent columns, where m >= n</param>
        /// <param name="b">An m x 1 column vector </param>
        /// <returns>An n x 1 column vector, x, such that Ax = b when m = n and Ax ~ b when m > n </returns>
        public static IVector Solve(IMatrix A, IVector b)
        {
            // Q is a m x m matrix, R is a m x n matrix
            // Q = [Q1 Q2] Q1 is a m x n matrix, Q2 is a m x (m-n) matrix
            // R = [R1; 0] R1 is a n x n upper triangular matrix matrix
            // A = Q * R = Q1 * R1
            IMatrix Q, R;

            QR.Factorize(A, out Q, out R);
            IVector c = Q.Transpose().Multiply(b);

            IVector x = BackwardSubstitution.Solve(R, c);

            return(x);
        }
        /// <summary>
        /// This is used for data fitting / regression
        /// A is a m x n matrix, where m >= n
        /// b is a m x 1 column vector
        /// The method solves for x, which is a n x 1 column vectors such that A * x is closest to b
        ///
        /// The method works as follows:
        ///   1. Let C = A.transpose * A, we have A.transpose * A * x = C * x = A.transpose * b
        ///   2. Decompose C : C = L * L.transpose = L * U, we have L * U * x = A.transpose * b
        ///   3. Let z = U * x, we have L * z = A.transpose * b
        ///   4. Solve z using forward substitution
        ///   5. Solve x from U * x = z using backward substitution
        /// </summary>
        /// <param name="A"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        public static IVector SolveLeastSquare(IMatrix A, IVector b)
        {
            IMatrix At = A.Transpose();
            IMatrix C  = At.Multiply(A); //C is a n x n matrix
            IVector d  = At.Multiply(b);

            IMatrix L;

            Cholesky.Factorize(C, out L);
            IMatrix U = L.Transpose();

            IVector z = ForwardSubstitution.Solve(L, d);
            IVector x = BackwardSubstitution.Solve(U, z);

            return(x);
        }
Пример #4
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        /// <summary>
        /// This is used for data fitting / regression
        /// A is a m x n matrix, where m >= n
        /// b is a m x 1 column vector
        /// The method solves for x, which is a n x 1 column vectors such that A * x is closest to b
        ///
        /// The method works as follows:
        ///   1. Let C = A.transpose * A, we have A.transpose * A * x = C * x = A.transpose * b
        ///   2. Decompose C : C = Q * R, we have Q * R * x = A.transpose * b
        ///   3. Multiply both side by Q.transpose = Q.inverse, we have Q.transpose * Q * R * x = Q.transpose * A.transpose * b
        ///   4. Since Q.tranpose * Q = I, we have R * x = Q.transpose * A.transpose * b
        ///   5. Solve x from R * x = Q.transpose * A.transpose * b using backward substitution
        /// </summary>
        /// <param name="A"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        public static IVector SolveLeastSquare(IMatrix A, IVector b)
        {
            IMatrix At = A.Transpose();
            IMatrix C  = At.Multiply(A); //C is a n x n matrix

            IMatrix Q, R;

            QR.Factorize(C, out Q, out R);

            IMatrix Qt = Q.Transpose();

            IVector d = Qt.Multiply(At).Multiply(b);

            IVector x = BackwardSubstitution.Solve(R, d);

            return(x);
        }