示例#1
0
        /**
         * <p>
         * Performs a matrix inversion operation on the specified matrix and stores the results
         * in the same matrix.<br>
         * <br>
         * a = a<sup>-1</sup>
         * </p>
         *
         * <p>
         * If the algorithm could not invert the matrix then false is returned.  If it returns true
         * that just means the algorithm finished.  The results could still be bad
         * because the matrix is singular or nearly singular.
         * </p>
         *
         * @param A The matrix that is to be inverted.  Results are stored here.  Modified.
         * @return true if it could invert the matrix false if it could not.
         */
        public static bool invert(CMatrixRMaj A)
        {
            LinearSolverDense <CMatrixRMaj> solver = LinearSolverFactory_CDRM.lu(A.numRows);

            if (solver.setA(A))
            {
                solver.invert(A);
            }
            else
            {
                return(false);
            }
            return(true);
        }
示例#2
0
        /**
         * <p>
         * Performs a matrix inversion operation that does not modify the original
         * and stores the results in another matrix.  The two matrices must have the
         * same dimension.<br>
         * <br>
         * b = a<sup>-1</sup>
         * </p>
         *
         * <p>
         * If the algorithm could not invert the matrix then false is returned.  If it returns true
         * that just means the algorithm finished.  The results could still be bad
         * because the matrix is singular or nearly singular.
         * </p>
         *
         * <p>
         * For medium to large matrices there might be a slight performance boost to using
         * {@link LinearSolverFactory_CDRM} instead.
         * </p>
         *
         * @param input The matrix that is to be inverted. Not modified.
         * @param output Where the inverse matrix is stored.  Modified.
         * @return true if it could invert the matrix false if it could not.
         */
        public static bool invert(CMatrixRMaj input, CMatrixRMaj output)
        {
            LinearSolverDense <CMatrixRMaj> solver = LinearSolverFactory_CDRM.lu(input.numRows);

            if (solver.modifiesA())
            {
                input = (CMatrixRMaj)input.copy();
            }

            if (!solver.setA(input))
            {
                return(false);
            }
            solver.invert(output);
            return(true);
        }
示例#3
0
        /**
         * <p>
         * Solves for x in the following equation:<br>
         * <br>
         * A*x = b
         * </p>
         *
         * <p>
         * If the system could not be solved then false is returned.  If it returns true
         * that just means the algorithm finished operating, but the results could still be bad
         * because 'A' is singular or nearly singular.
         * </p>
         *
         * <p>
         * If repeat calls to solve are being made then one should consider using {@link LinearSolverFactory_CDRM}
         * instead.
         * </p>
         *
         * <p>
         * It is ok for 'b' and 'x' to be the same matrix.
         * </p>
         *
         * @param a A matrix that is m by n. Not modified.
         * @param b A matrix that is n by k. Not modified.
         * @param x A matrix that is m by k. Modified.
         *
         * @return true if it could invert the matrix false if it could not.
         */
        public static bool solve(CMatrixRMaj a, CMatrixRMaj b, CMatrixRMaj x)
        {
            LinearSolverDense <CMatrixRMaj> solver;

            if (a.numCols == a.numRows)
            {
                solver = LinearSolverFactory_CDRM.lu(a.numRows);
            }
            else
            {
                solver = LinearSolverFactory_CDRM.qr(a.numRows, a.numCols);
            }

            // make sure the inputs 'a' and 'b' are not modified
            solver = new LinearSolverSafe <CMatrixRMaj>(solver);

            if (!solver.setA(a))
            {
                return(false);
            }

            solver.solve(b, x);
            return(true);
        }