예제 #1
0
 public void SquareRandomMatrixEigenvalues()
 {
     for (int d = 1; d <= 67; d = d + 11)
     {
         SquareMatrix       I      = TestUtilities.CreateSquareUnitMatrix(d);
         SquareMatrix       M      = CreateSquareRandomMatrix(d, d);
         double             tr     = M.Trace();
         DateTime           start  = DateTime.Now;
         ComplexEigensystem E      = M.Eigensystem();
         DateTime           finish = DateTime.Now;
         Console.WriteLine("d={0} t={1} ms", d, (finish - start).Milliseconds);
         Assert.IsTrue(E.Dimension == d);
         Complex[] es = new Complex[d];
         for (int i = 0; i < d; i++)
         {
             es[i] = E.Eigenvalue(i);
             Complex[] v = E.Eigenvector(i);
             Assert.IsTrue(TestUtilities.IsNearlyEigenpair(M, v, es[i]));
         }
         Assert.IsTrue(TestUtilities.IsSumNearlyEqual(es, tr));
     }
 }
예제 #2
0
        public void SquareMatrixStochasticEigensystem()
        {
            // this is a simplifed form of a Markov matrix that arose in the Monopoly problem
            // and failed to converge

            int n = 12;
            // originally failed for n=12 due to lack of 2x2 detection at top
            // now tested for n up to 40, but n=14 appears to still be a problem,
            // probably due to a six-times degenerate eigenvalue

            SquareMatrix R = new SquareMatrix(n);

            for (int c = 0; c < n; c++)
            {
                R[(c + 2) % n, c]  = 1.0 / 36.0;
                R[(c + 3) % n, c]  = 2.0 / 36.0;
                R[(c + 4) % n, c]  = 3.0 / 36.0;
                R[(c + 5) % n, c]  = 4.0 / 36.0;
                R[(c + 6) % n, c]  = 5.0 / 36.0;
                R[(c + 7) % n, c]  = 6.0 / 36.0;
                R[(c + 8) % n, c]  = 5.0 / 36.0;
                R[(c + 9) % n, c]  = 4.0 / 36.0;
                R[(c + 10) % n, c] = 3.0 / 36.0;
                R[(c + 11) % n, c] = 2.0 / 36.0;
                R[(c + 12) % n, c] = 1.0 / 36.0;
            }

            //Complex[] v = R.Eigenvalues();
            ComplexEigensystem E = R.Eigensystem();

            for (int i = 0; i < E.Dimension; i++)
            {
                Console.WriteLine(E.Eigenvalue(i));
                Assert.IsTrue(TestUtilities.IsNearlyEigenpair(R, E.Eigenvector(i), E.Eigenvalue(i)));
            }
        }
예제 #3
0
        private void GetParameters()
        {
            // A x^2 + B x y + C y^2 + D x + E y + F = 0
            double[] par       = new double[6];
            int      numPoints = pointList.Count;

            if (numPoints > 2)
            {
                Matrix       D1 = new Matrix(numPoints, 3);
                Matrix       D2 = new Matrix(numPoints, 3);
                SquareMatrix S1 = new SquareMatrix(3);
                SquareMatrix S2 = new SquareMatrix(3);
                SquareMatrix S3 = new SquareMatrix(3);
                SquareMatrix T  = new SquareMatrix(3);
                SquareMatrix M  = new SquareMatrix(3);
                SquareMatrix C1 = new SquareMatrix(3);
                Matrix       a1 = new Matrix(3, 1);
                Matrix       a2 = new Matrix(3, 1);
                Matrix       temp;

                C1[0, 0] = 0;
                C1[0, 1] = 0;
                C1[0, 2] = 0.5;
                C1[1, 0] = 0;
                C1[1, 1] = -1;
                C1[1, 2] = 0;
                C1[2, 0] = 0.5;
                C1[2, 1] = 0;
                C1[2, 2] = 0;

                //2 D1 = [x .ˆ 2, x .* y, y .ˆ 2]; % quadratic part of the design matrix
                //3 D2 = [x, y, ones(size(x))]; % linear part of the design matrix
                for (int xx = 0; xx < numPoints; xx++)
                {
                    Point p = pointList[xx];
                    D1[xx, 0] = p.X * p.X;
                    D1[xx, 1] = p.X * p.Y;
                    D1[xx, 2] = p.Y * p.Y;

                    D2[xx, 0] = p.X;
                    D2[xx, 1] = p.Y;
                    D2[xx, 2] = 1;
                }

                //4 S1 = D1’ * D1; % quadratic part of the scatter matrix
                temp = D1.Transpose() * D1;
                for (int xx = 0; xx < 3; xx++)
                {
                    for (int yy = 0; yy < 3; yy++)
                    {
                        S1[xx, yy] = temp[xx, yy];
                    }
                }

                //5 S2 = D1’ * D2; % combined part of the scatter matrix
                temp = D1.Transpose() * D2;
                for (int xx = 0; xx < 3; xx++)
                {
                    for (int yy = 0; yy < 3; yy++)
                    {
                        S2[xx, yy] = temp[xx, yy];
                    }
                }

                //6 S3 = D2’ * D2; % linear part of the scatter matrix
                temp = D2.Transpose() * D2;
                for (int xx = 0; xx < 3; xx++)
                {
                    for (int yy = 0; yy < 3; yy++)
                    {
                        S3[xx, yy] = temp[xx, yy];
                    }
                }

                //7 T = - inv(S3) * S2’; % for getting a2 from a1
                if (Determinant(S3) > 0)
                {
                    T = -1 * S3.Inverse() * S2.Transpose();

                    //8 M = S1 + S2 * T; % reduced scatter matrix
                    M = S1 + S2 * T;

                    //9 M = [M(3, :) ./ 2; - M(2, :); M(1, :) ./ 2]; % premultiply by inv(C1)
                    M = C1 * M;

                    //10 [evec, eval] = eig(M); % solve eigensystem
                    ComplexEigensystem eigenSystem = M.Eigensystem();

                    //11 cond = 4 * evec(1, :) .* evec(3, :) - evec(2, :) .ˆ 2; % evaluate a’Ca
                    //12 a1 = evec(:, find(cond > 0)); % eigenvector for min. pos. eigenvalue
                    for (int xx = 0; xx < eigenSystem.Dimension; xx++)
                    {
                        Vector <Complex> vector    = eigenSystem.Eigenvector(xx);
                        Complex          condition = 4 * vector[0] * vector[2] - vector[1] * vector[1];
                        if (condition.Im == 0 && condition.Re > 0)
                        {
                            // Solution is found
                            for (int yy = 0; yy < vector.Count(); yy++)
                            {
                                a1[yy, 0] = vector[yy].Re;
                            }
                        }
                    }

                    //13 a2 = T * a1; % ellipse coefficients
                    a2 = T * a1;

                    //14 a = [a1; a2]; % ellipse coefficients
                    par[0] = a1[0, 0];
                    par[1] = a1[1, 0];
                    par[2] = a1[2, 0];
                    par[3] = a2[0, 0];
                    par[4] = a2[1, 0];
                    par[5] = a2[2, 0];
                }
            }

            for (int i = 0; i < par.Count(); i++)
            {
                if (!double.IsNaN(par[i]) && !double.IsInfinity(par[i]))
                {
                    Parameters[i] = par[i];
                }
                else
                {
                    Parameters[i] = 0;
                }
            }
        }
예제 #4
0
        private static void Main(string[] args)
        {
            // define squares
            BoardSpace[] spaces = new BoardSpace[] {
                new BoardSpace("Go", 0.0, Double.PositiveInfinity),
                new BoardSpace("Mediterranean Avenue", 2.0, 60.0),
                new BoardSpace("Community Chest", 0.0, Double.PositiveInfinity),
                new BoardSpace("Baltic Avenue", 4.0, 60.0),
                new BoardSpace("Income Tax", 0.0, Double.PositiveInfinity),
                new BoardSpace("Reading Railroad", 25.0, 200.0),
                new BoardSpace("Oriental Avenue", 6.0, 100.0),
                new BoardSpace("Chance", 0.0, Double.PositiveInfinity),
                new BoardSpace("Vermont Avenue", 6.0, 100.0),
                new BoardSpace("Connecticut Avenue", 8.0, 120.0),
                new BoardSpace("Jail", 0.0, Double.PositiveInfinity),
                new BoardSpace("St. Charles Place", 10.0, 140.0),
                new BoardSpace("Electric Company", 4.0 * 6.0, 150.0),
                new BoardSpace("States Avenue", 10.0, 140.0),
                new BoardSpace("Virginia Avenue", 12.0, 160.0),
                new BoardSpace("Pennsylvania Railroad", 25.0, 200.0),
                new BoardSpace("St. James Place", 14.0, 180.0),
                new BoardSpace("Community Chest", 0.0, Double.PositiveInfinity),
                new BoardSpace("Tennessee Avenue", 14.0, 180.0),
                new BoardSpace("New York Avenue", 16.0, 200.0),
                new BoardSpace("Free Parking", 0.0, Double.PositiveInfinity),
                new BoardSpace("Kentucky Avenue", 18.0, 220.0),
                new BoardSpace("Chance", 0.0, Double.PositiveInfinity),
                new BoardSpace("Indiana Avenue", 18.0, 220.0),
                new BoardSpace("Illinois Avenue", 20.0, 240.0),
                new BoardSpace("B & O Railroad", 25.0, 200.0),
                new BoardSpace("Atlantic Avenue", 22.0, 260.0),
                new BoardSpace("Ventnor Avenue", 22.0, 260.0),
                new BoardSpace("Water Works", 4.0 * 6.0, 150.0),
                new BoardSpace("Marvin Gardens", 24.0, 280.0),
                new BoardSpace("Go To Jail", 0.0, Double.PositiveInfinity),
                new BoardSpace("Pacific Avenue", 26.0, 300.0),
                new BoardSpace("North Carolina Avenue", 26.0, 300.0),
                new BoardSpace("Community Chest", 0.0, Double.PositiveInfinity),
                new BoardSpace("Pennsylvania Avenue", 28.0, 320.0),
                new BoardSpace("Short Line", 25.0, 200.0),
                new BoardSpace("Chance", 0.0, Double.PositiveInfinity),
                new BoardSpace("Park Place", 35.0, 350.0),
                new BoardSpace("Luxury Tax", 0.0, Double.PositiveInfinity),
                new BoardSpace("Boardwalk", 50.0, 400.0)
            };

            // number of squares
            int n = spaces.Length;

            // compute the transition matrix which takes the dice roll
            // into account: Moves us ahead by two spaces with
            // probability 1/36 (“snake-eyes”), three spaces with
            // probability 2/36, and so on up to twelve spaces with
            // probability 1/36 (“double-sixes”), etc.
            SquareMatrix R = new SquareMatrix(n);

            for (int c = 0; c < n; c++)
            {
                R[(c + 2) % n, c]  = 1.0 / 36.0;
                R[(c + 3) % n, c]  = 2.0 / 36.0;
                R[(c + 4) % n, c]  = 3.0 / 36.0;
                R[(c + 5) % n, c]  = 4.0 / 36.0;
                R[(c + 6) % n, c]  = 5.0 / 36.0;
                R[(c + 7) % n, c]  = 6.0 / 36.0;
                R[(c + 8) % n, c]  = 5.0 / 36.0;
                R[(c + 9) % n, c]  = 4.0 / 36.0;
                R[(c + 10) % n, c] = 3.0 / 36.0;
                R[(c + 11) % n, c] = 2.0 / 36.0;
                R[(c + 12) % n, c] = 1.0 / 36.0;
            }

            // compute the special matrix, which takes board into account:
            // The column for space 30, “Go To Jail”, transitions with
            // certainty to space 10, the jail.
            // A token that lands on a community chest space sometimes
            // transitions to go and sometimes transitions to jail, but
            // most often just stays put
            SquareMatrix S = new SquareMatrix(n);

            for (int c = 0; c < n; c++)
            {
                if (c == 30)
                {
                    // go to jail
                    S[10, 30] = 1.0;
                }
                else if ((c == 7) || (c == 22) || (c == 36))
                {
                    // chance
                    // advance to go
                    S[0, c] = 1.0 / 16.0;
                    // advance to illinois avenue
                    S[24, c] = 1.0 / 16.0;
                    // take a walk on the boardwalk
                    S[39, c] = 1.0 / 16.0;
                    // go to jail
                    S[10, c] = 1.0 / 16.0;
                    // take a ride on the reading
                    S[5, c] = 1.0 / 16.0;
                    // advance to St. Charles place
                    S[11, c] = 1.0 / 16.0;
                    // go back 3 spaces
                    S[(c - 3) % S.Dimension, c] = 1.0 / 16.0;
                    // advance token to the nearest utility
                    if ((c < 12) || (c > 28))
                    {
                        S[12, c] = 1.0 / 16.0;
                    }
                    else
                    {
                        S[28, c] = 1.0 / 16.0;
                    }
                    // advance token to the nearest railroad
                    if (c < 5)
                    {
                        S[5, c] += 1.0 / 16.0;
                    }
                    else if (c < 15)
                    {
                        S[15, c] = 1.0 / 16.0;
                    }
                    else if (c < 25)
                    {
                        S[25, c] = 1.0 / 16.0;
                    }
                    else if (c < 35)
                    {
                        S[35, c] = 1.0 / 16.0;
                    }
                    else
                    {
                        S[5, c] += 1.0 / 16.0;
                    }
                    // stay put
                    S[c, c] = 7.0 / 16.0;
                }
                else if ((c == 2) || (c == 17) || (c == 33))
                {
                    // community chest
                    // advance to go
                    S[0, c] = 1.0 / 16.0;
                    // go to jail
                    S[10, c] = 1.0 / 16.0;
                    // stay put
                    S[c, c] = 14.0 / 16.0;
                }
                else
                {
                    // all other spaces are no-ops
                    S[c, c] = 1.0;
                }
            }

            // compute the complete transition matrix
            SquareMatrix P = S * R;

            // verify Markov conditions
            for (int c = 0; c < P.Dimension; c++)
            {
                double sr = 0.0;
                for (int r = 0; r < P.Dimension; r++)
                {
                    sr += P[r, c];
                }
                Debug.Assert(Math.Abs(sr - 1.0) < 1.0e-15);
            }

            // An alternative now is to begin with some initial state vector
            // guess, and repeatedly apply P until the resultant vector
            // no longer changes. The resultant vector will be the dominant
            // eigenvector. That's how PageRank is evaluated by Google,
            // using MapReduce.

            // Another alternative is to compute the eigenvalues and
            // eigenvectors of the transition matrix
            ComplexEigensystem E = P.Eigensystem();

            // get the dominant eigenvector: The one with the largest
            // eigenvalue
            Complex         e = -1.0;
            IList <Complex> v = null;

            for (int i = 0; i < E.Dimension; i++)
            {
                Complex ei = E.Eigenvalue(i);
                if (ComplexMath.Abs(ei) > ComplexMath.Abs(e))
                {
                    e = ei;
                    v = E.Eigenvector(i);
                }
            }

            // verify that it has eigenvalue 1 and is real
            Debug.Assert(Math.Abs(e.Re - 1.0) < 1.0e-15);
            Debug.Assert(e.Im == 0.0);
            for (int i = 0; i < n; i++)
            {
                Debug.Assert(v[i].Im == 0.0);
            }

            // normalize the probabilities
            double sv = 0.0;

            for (int i = 0; i < E.Dimension; i++)
            {
                sv += v[i].Re;
            }
            double[] p = new double[E.Dimension];
            for (int i = 0; i < E.Dimension; i++)
            {
                p[i] = v[i].Re / sv;
            }

            // Having found the dominant eigenvector, we can examine its components
            // to find the long-run relative probabilities of a token landing on each
            // space.
            // Print the probabilities

            // First column: Which rent-generating properties are most landed upon

            // Second column: A rent-seeking player would be willing to accept a slightly
            // less frequently landed upon space, if it generates a much higher rent when
            // landed upon --> Multiply the landing probability per turn by the rent per landing

            // Third column: A clever investor cares not only about cash flow, but
            // also about the return on his investment (ROI), that is the rent generated
            // per dollar paid --> Divide the rent per turn by the property price

            for (int i = 0; i < n; i++)
            {
                Console.WriteLine("{0,-24} {1:f4} {2:f4} {3:f4}", spaces[i].Name, p[i], p[i] * spaces[i].Rent, p[i] * spaces[i].Rent / spaces[i].Price);
            }

            // First column: Which rent-generating properties are most landed upon:
            // Illinois Avenue (3.2%) and New York Avenue (3.1%), followed by Tennessee, the Reading, and the B & O (2.9% each).

            // Second column: A rent-seeking player would be willing to accept a slightly less frequently landed upon space,
            // if it generates a much higher rent when landed upon. To know which spaces maximize rent per turn,
            // we need to multiply the landing probability per turn by the rent per landing.
            // Boardwalk ($1.32/turn) is far and away the most rent-generating property to own

            // Third column: Finally, a clever investor cares not only about cash flow, but also about the return on his
            // investment (ROI), that is the rent generated per dollar paid. Dividing the rent per turn by
            // the property price changes the picture yet again. By this metric, the best values on the board
            // are the utilities, followed by the railways.

            Console.ReadLine();
        }
예제 #5
0
        static public AnyMatrix <Double> GetBestFitEllipse(List <MyShortPoint> Points)
        {
            int numPoints            = Points.Count;
            RectangularMatrix D1     = new RectangularMatrix(numPoints, 3);
            RectangularMatrix D2     = new RectangularMatrix(numPoints, 3);
            SquareMatrix      S1     = new SquareMatrix(3);
            SquareMatrix      S2     = new SquareMatrix(3);
            SquareMatrix      S3     = new SquareMatrix(3);
            SquareMatrix      T      = new SquareMatrix(3);
            SquareMatrix      M      = new SquareMatrix(3);
            SquareMatrix      C1     = new SquareMatrix(3);
            RectangularMatrix a1     = new RectangularMatrix(3, 1);
            RectangularMatrix a2     = new RectangularMatrix(3, 1);
            RectangularMatrix result = new RectangularMatrix(6, 1);
            RectangularMatrix temp;

            C1[0, 0] = 0;
            C1[0, 1] = 0;
            C1[0, 2] = 0.5;
            C1[1, 0] = 0;
            C1[1, 1] = -1;
            C1[1, 2] = 0;
            C1[2, 0] = 0.5;
            C1[2, 1] = 0;
            C1[2, 2] = 0;
            //2 D1 = [x .? 2, x .* y, y .? 2]; % quadratic part of the design matrix
            //3 D2 = [x, y, ones(size(x))]; % linear part of the design matrix
            for (int xx = 0; xx < Points.Count; xx++)
            {
                MyShortPoint p = Points[xx];
                D1[xx, 0] = p.x * p.x;
                D1[xx, 1] = p.x * p.y;
                D1[xx, 2] = p.y * p.y;
                D2[xx, 0] = p.x;
                D2[xx, 1] = p.y;
                D2[xx, 2] = 1;
            }
            //4 S1 = D1’ * D1; % quadratic part of the scatter matrix
            temp = D1.Transpose() * D1;
            for (int xx = 0; xx < 3; xx++)
            {
                for (int yy = 0; yy < 3; yy++)
                {
                    S1[xx, yy] = temp[xx, yy];
                }
            }
            //5 S2 = D1’ * D2; % combined part of the scatter matrix
            temp = D1.Transpose() * D2;
            for (int xx = 0; xx < 3; xx++)
            {
                for (int yy = 0; yy < 3; yy++)
                {
                    S2[xx, yy] = temp[xx, yy];
                }
            }
            //6 S3 = D2’ * D2; % linear part of the scatter matrix
            temp = D2.Transpose() * D2;
            for (int xx = 0; xx < 3; xx++)
            {
                for (int yy = 0; yy < 3; yy++)
                {
                    S3[xx, yy] = temp[xx, yy];
                }
            }
            //7 T = – inv(S3) * S2’; % for getting a2 from a1
            T = -1 * S3.Inverse() * S2.Transpose();
            //8 M = S1 + S2 * T; % reduced scatter matrix
            M = S1 + S2 * T;
            //9 M = [M(3,  ./ 2; - M(2,  ; M(1,  ./ 2]; % premultiply by inv(C1)
            M = C1 * M;
            //10 [evec, eval] = eig(M); % solve eigensystem
            ComplexEigensystem eigenSystem = M.Eigensystem();

            //11 cond = 4 * evec(1,  .* evec(3,  – evec(2,  .? 2; % evaluate a’Ca
            //12 a1 = evec(:, find(cond > 0)); % eigenvector for min. pos. eigenvalue
            for (int xx = 0; xx < eigenSystem.Dimension; xx++)
            {
                Complex[] vector = eigenSystem.Eigenvector(xx);

                Complex condition = 4 * vector[0] * vector[2] - vector[1] * vector[1];
                if (condition.Im == 0 && condition.Re > 0)
                {
                    // Solution is found
                    // Console.WriteLine(“\nSolution Found!”);

                    //System.Windows.Forms.MessageBox.Show("Ellipse gOT");

                    for (int yy = 0; yy < vector.Length; yy++)
                    {
                        a1[yy, 0] = vector[yy].Re;
                    }
                }
            }
            //13 a2 = T * a1; % ellipse coefficients
            a2 = T * a1;
            //14 a = [a1; a2]; % ellipse coefficients
            result[0, 0] = a1[0, 0];
            result[1, 0] = a1[1, 0];
            result[2, 0] = a1[2, 0];
            result[3, 0] = a2[0, 0];
            result[4, 0] = a2[1, 0];
            result[5, 0] = a2[2, 0];
            return(result);
        }
예제 #6
0
        static void Main(string[] args)
        {
            SquareMatrix T = new SquareMatrix(3);

            // Construct the transition matrix for the weather.
            T[0, 0] = 0.5; // sunny to sunny
            T[1, 0] = 0.4; // sunny to rainy
            T[2, 0] = 0.1; // sunny to cloudy

            T[0, 1] = 0.3; // rainy to sunny
            T[1, 1] = 0.4; // rainy to rainy
            T[2, 1] = 0.3; // rainy to cloudy

            T[0, 2] = 0.2; // cloudy to sunny
            T[1, 2] = 0.3; // cloudy to rainy
            T[2, 2] = 0.5; // cloudy to cloudy

            // verify Markov conditions
            for (int c = 0; c < T.Dimension; c++)
            {
                double sr = 0.0;
                for (int r = 0; r < T.Dimension; r++)
                {
                    sr += T[r, c];
                }
                Debug.Assert(Math.Abs(sr - 1.0) < 1.0e-15);
            }

            // Begin with some initial state vector, and repeatedly
            // apply P until the resultant vector no longer changes.
            // The resultant vector will be the dominant eigenvector.
            ComplexEigensystem E = T.Eigensystem();

            Complex         e     = -1.0;
            IList <Complex> v     = null;
            Complex         maxEI = E.Eigenvalue(0);

            v = E.Eigenvector(0);

            // Find the dominant eigenvector
            for (int i = 1; i < E.Dimension; i++)
            {
                Complex ei = E.Eigenvalue(i);
                if (ComplexMath.Abs(ei) > ComplexMath.Abs(maxEI))
                {
                    maxEI = ei;
                    v     = E.Eigenvector(i);
                }
            }

            // verify that it has eigenvalue 1 and is real
            Debug.Assert(Math.Abs(maxEI.Re - 1.0) < 1.0e-15);
            Debug.Assert(maxEI.Im == 0.0);
            for (int i = 0; i < 3; i++)
            {
                Debug.Assert(v[i].Im == 0.0);
            }

            // normalize the probabilities
            double sv = 0.0;

            for (int i = 0; i < E.Dimension; i++)
            {
                sv += v[i].Re;
            }
            double[] p = new double[E.Dimension];
            for (int i = 0; i < E.Dimension; i++)
            {
                p[i] = v[i].Re / sv;
            }

            // print the probabilities (steady state of the weather)
            Console.Write("Sunny = ");
            Console.WriteLine(p[0]);
            Console.Write("Rainy = ");
            Console.WriteLine(p[1]);
            Console.Write("Cloudy = ");
            Console.WriteLine(p[2]);
            Console.ReadLine();
        }