/// <summary>
/// Converts conic representation of an ellipse (given the coefficients vector) to the parametric representation.
/// </summary>
/// <param name="conic">The implicit equation coefficients</param>
/// <returns>Parametric representation</returns>
private static EllipseParams Conic2Parametric(double[] conic)
{
var A = new SymmetricMatrix(2);
A[0, 0] = conic[0]; // a
A[1, 1] = conic[2]; // c
A[0, 1] = A[1, 0] = 0.5 * conic[1]; // half b
var B = new ColumnVector(conic[3], conic[4]);
var C = conic[5];
var eig = A.Eigensystem();
Debug.Assert(eig.Eigenvalue(0) * eig.Eigenvalue(1) > 0);
var D = new double[] { eig.Eigenvalue(0), eig.Eigenvalue(1) };
var Q = eig.Eigentransformation().Transpose();
var t = -0.5 * A.Inverse() * B;
var c_h = t.Transpose() * A * t + B.Transpose() * t + C;
return(new EllipseParams
{
Center = new Point(t[0], t[1]),
XRadius = Math.Sqrt(-c_h / D[0]),
YRadius = Math.Sqrt(-c_h / D[1]),
Degrees = 180 * Math.Atan2(Q[0, 1], Q[0, 0]) / Math.PI,
});
}
public void SymmetricRandomMatrixInverse()
{
for (int d = 1; d <= 100; d = d + 11)
{
SymmetricMatrix M = TestUtilities.CreateSymmetricRandomMatrix(d, 1);
SymmetricMatrix MI = M.Inverse();
Assert.IsTrue(TestUtilities.IsNearlyEqual(MI * M, UnitMatrix.OfDimension(d)));
}
}
public void SymmetricHilbertMatrixInverse()
{
for (int d = 1; d < 4; d++)
{
SymmetricMatrix H = TestUtilities.CreateSymmetricHilbertMatrix(d);
SymmetricMatrix HI = H.Inverse();
Assert.IsTrue(TestUtilities.IsNearlyEqual(HI * H, UnitMatrix.OfDimension(d)));
}
// fails for d >= 4; look into this
}
public void SymmetricRandomMatrixInverse()
{
for (int d = 1; d <= 100; d = d + 11)
{
Console.WriteLine("d={0}", d);
SquareMatrix I = TestUtilities.CreateSquareUnitMatrix(d);
SymmetricMatrix M = TestUtilities.CreateSymmetricRandomMatrix(d, 1);
SymmetricMatrix MI = M.Inverse();
Assert.IsTrue(TestUtilities.IsNearlyEqual(M * MI, I));
}
}
public void inverse()
{
block(OriginA, DiagRowCount);
//ParallelInverseOfSymmetric2 pppp = new ParallelInverseOfSymmetric2(SymmetricMatrix);
//var aa = pppp.invA.Multiply(SymmetricMatrix);
//invA = uniteBlockOfSymmetric(DiagonalMatrix, pppp.invA);
DiagonalMatrix = DiagonalMatrix.Inverse();
invA = uniteBlockOfSymmetric(DiagonalMatrix, SymmetricMatrix.Inverse());
//var sss = OriginA.Multiply(invA);
}
public void SymmetricHilbertMatrixInverse()
{
for (int d = 1; d <= 4; d++)
{
Console.WriteLine("d={0}", d);
SquareMatrix I = TestUtilities.CreateSquareUnitMatrix(d);
SymmetricMatrix H = TestUtilities.CreateSymmetricHilbertMatrix(d);
SymmetricMatrix HI = H.Inverse();
Assert.IsTrue(TestUtilities.IsNearlyEqual(H * HI, I));
}
// fails for d > 4! look into this
}
public void inverse()
{
DateTime start = DateTime.Now;
var span = DateTime.Now - start;
//1级分裂
cA = new cyMatrix(orignA, 2, 2);
SymmetricMatrix q22 = new SymmetricMatrix(cA.blockMatrix1[2].Clone().Array);
IMatrix q21 = cA.blockMatrix1[1].Clone();
SymmetricMatrix Q = q22.Inverse();
SymmetricMatrix q11 = new SymmetricMatrix(cA.blockMatrix1[0].Minus(q21.Transposition.Multiply(Q).Multiply(q21)).Array);
Q11 = q11.Inverse(); //.GetInverse();
var task0 = new Task(() => GetQ21(q21, Q));
var task1 = new Task(() => GetQ22(q21, Q));
task0.Start(); task1.Start();
Task.WaitAll(task0, task1);
cA.blockMatrix1[0] = Q11;
cA.blockMatrix1[1] = Q21;
cA.blockMatrix1[2] = Q22;
invA = cA.uniteBlockOfSymmetric(cA, 1);
var sssss0 = invA.Multiply(orignA);
#region hide
////start = DateTime.Now;
////2级分裂
//cA1 = new cyMatrix(cA.blockMatrix1[0], 2, 2);
//cA2 = new cyMatrix(cA.blockMatrix1[2], 2, 2);
//DiagM = new SymmetricMatrix[coreNum];
//DiagM[0] = cA1.blockMatrix1[0].Clone();
//DiagM[1] = cA1.blockMatrix1[2].Clone();
//DiagM[2] = cA2.blockMatrix1[0].Clone();
//DiagM[3] = cA2.blockMatrix1[2].Clone();
////var s11 = DiagM[0].Clone();
//Parallel.For(0, coreNum, (int i) =>
//{
// DiagM[i] = DiagM[i].GetInverse();//.Inverse;
//});
////var sssss = s11.Multiply(DiagM[0]);
////并行求对角线上的块矩阵
//DiagQ = new SymmetricMatrix[coreNum];
////通过使用任务来对代码进行并行化
////创建任务
//var t0 = new Task(() => GenerateDiagQ0());
//var t1 = new Task(() => GenerateDiagQ1());
//var t2 = new Task(() => GenerateDiagQ2());
//var t3 = new Task(() => GenerateDiagQ3());
//t0.Start(); t1.Start(); t2.Start(); t3.Start();
//Task.WaitAll(t0, t1, t2, t3);//等待所有任务的完成
////var q0 = DiagQ[0].Clone();
////var q1 = DiagQ[1].Clone();
////var q2 = DiagQ[2].Clone();
////var q3 = DiagQ[3].Clone();
////求逆
//Parallel.For(0, coreNum, (int i) =>
//{
// DiagQ[i] = DiagQ[i].GetInverse();//.Inverse;
//});
////var r0 = q0.Multiply(DiagQ[0]);
////var r1 = q1.Multiply(DiagQ[1]);
////var r2 = q2.Multiply(DiagQ[2]);
////var r3 = q3.Multiply(DiagQ[3]);
//#region 比下面这个多5ms，故省去
////start = DateTime.Now; var b0 = new Task(() => GenenateBlock0());
////var b1 = new Task(() => GenenateBlock1());
////var b2 = new Task(() => GenenateBlock2());
////var b3 = new Task(() => GenenateBlock3());
////b0.Start(); b1.Start(); b2.Start(); b3.Start();
////Task.WaitAll(b0, b1, b2, b3);//等待所有任务的完成
////span = DateTime.Now - start;
////Console.WriteLine(span.TotalMilliseconds + "ms融合3----");
//#endregion
//var xxx = DiagM[1].Plus(DiagM[1].Multiply(cA1.blockMatrix1[1]).Multiply(DiagQ[0]).Multiply(cA1.blockMatrix1[1].Transposition).Multiply(DiagM[1]));
//var xxx2 = DiagM[3].Plus(DiagM[3].Multiply(cA2.blockMatrix1[1]).Multiply(DiagQ[2]).Multiply(cA2.blockMatrix1[1].Transposition).Multiply(DiagM[3]));
//var c0 = new Task(() => GenenateBlock0());
//var c2 = new Task(() => GenenateBlock2());
//c0.Start(); c2.Start(); Task.WaitAll(c0, c2);
//cA1.blockMatrix1[0] = DiagQ[0];
////cA1.blockMatrix1[2] = DiagQ[1];
//cA2.blockMatrix1[0] = DiagQ[2];
////cA2.blockMatrix1[2] = DiagQ[3];
////var ooo = cA1.blockMatrix1[2].Minus(xxx);
////DiagM[0] = cA1.uniteBlockOfSymmetric(cA1,1);
////DiagM[1] = cA2.uniteBlockOfSymmetric(cA2,1);
////var ss = DiagM[0].Multiply(cA.blockMatrix1[0]);
////var ss1 = DiagM[1].Multiply(cA.blockMatrix1[2]);
//cA1.blockMatrix1[2] = xxx;
//cA2.blockMatrix1[2] = xxx2;
//DiagM[0] = cA1.uniteBlockOfSymmetric(cA1, 1);
//DiagM[1] = cA2.uniteBlockOfSymmetric(cA2, 1);
//var sss = DiagM[0].Multiply(cA.blockMatrix1[0]);
//var nt0 = new Task(() => GeneratenewDiagQ0());
//var nt1 = new Task(() => GeneratenewDiagQ1());
//nt0.Start();
//nt1.Start();
//Task.WaitAll(nt0, nt1);//等待所有任务的完成
//Parallel.For(0, 2, (int i) =>
//{
// DiagQ[i] = DiagQ[i].GetInverse();//.Inverse;
//});
//#region 比下面这个直接按顺序计算多15ms，故省去
////start = DateTime.Now;
////var nb0 = new Task(() => GeneratenewBlock0());
////var nb1 = new Task(() => GeneratenewBlock1());
////nb0.Start(); nb1.Start();
////Task.WaitAll(nb0, nb1);//等待所有任务的完成
////span = DateTime.Now - start;
////Console.WriteLine(span.TotalMilliseconds + "ms融合30000");
//#endregion
//var xxxx = DiagM[1].Plus(DiagM[1].Multiply(cA.blockMatrix1[1]).Multiply(DiagQ[0]).Multiply(cA.blockMatrix1[1].Transposition).Multiply(DiagM[1]));
//GeneratenewBlock0();
//cA.blockMatrix1[0] = DiagQ[0];
////cA.blockMatrix1[2] = DiagQ[1];
////invA = cA.uniteBlockOfSymmetric(cA, 1);
////var ssss = invA.Multiply(orignA);
////var oooo = cA1.blockMatrix1[2].Minus(xxx);
//cA.blockMatrix1[2] = xxxx;
//invA = cA.uniteBlockOfSymmetric(cA, 1);
////var sssss0 = invA.Multiply(orignA);
#endregion
}
/// <summary>
/// Finds the Weibull distribution that best fits the given sample.
/// </summary>
/// <param name="sample">The sample to fit.</param>
/// <returns>The best fit parameters.</returns>
/// <exception cref="ArgumentNullException"><paramref name="sample"/> is null.</exception>
/// <exception cref="InvalidOperationException"><paramref name="sample"/> contains non-positive values.</exception>
/// <exception cref="InsufficientDataException"><paramref name="sample"/> contains fewer than three values.</exception>
public static WeibullFitResult FitToWeibull(this IReadOnlyList <double> sample)
{
if (sample == null)
{
throw new ArgumentNullException(nameof(sample));
}
if (sample.Count < 3)
{
throw new InsufficientDataException();
}
foreach (double value in sample)
{
if (value <= 0.0)
{
throw new InvalidOperationException();
}
}
// The log likelihood function is
// \log L = N \log k + (k-1) \sum_i \log x_i - N K \log \lambda - \sum_i \left(\frac{x_i}{\lambda}\right)^k
// Taking derivatives, we get
// \frac{\partial \log L}{\partial \lambda} = - \frac{N k}{\lambda} + \sum_i \frac{k}{\lambda} \left(\frac{x_i}{\lambda}\right)^k
// \frac{\partial \log L}{\partial k} =\frac{N}{k} + \sum_i \left[ 1 - \left(\frac{x_i}{\lambda}\right)^k \right] \log \left(\frac{x_i}{\lambda}\right)
// Setting the first expression to zero and solving for \lambda gives
// \lambda = \left( N^{-1} \sum_i x_i^k \right)^{1/k} = ( < x^k > )^{1/k}
// which allows us to reduce the problem from 2D to 1D.
// By the way, using the expression for the moment < x^k > of the Weibull distribution, you can show there is
// no bias to this result even for finite samples.
// Setting the second expression to zero gives
// \frac{1}{k} = \frac{1}{N} \sum_i \left[ \left( \frac{x_i}{\lambda} \right)^k - 1 \right] \log \left(\frac{x_i}{\lambda}\right)
// which, given the equation for \lambda as a function of k derived from the first expression, is an implicit equation for k.
// It cannot be solved in closed form, but we have now reduced our problem to finding a root in one-dimension.
// We need a starting guess for k.
// The method of moments equations are not solvable for the parameters in closed form
// but the scale parameter drops out of the ratio of the 1/3 and 2/3 quantile points
// and the result is easily solved for the shape parameter
// k = \frac{\log 2}{\log\left(\frac{x_{2/3}}{x_{1/3}}\right)}
double x1 = sample.InverseLeftProbability(1.0 / 3.0);
double x2 = sample.InverseLeftProbability(2.0 / 3.0);
double k0 = Global.LogTwo / Math.Log(x2 / x1);
// Given the shape paramter, we could invert the expression for the mean to get
// the scale parameter, but since we have an expression for \lambda from k, we
// don't need it.
//double s0 = sample.Mean / AdvancedMath.Gamma(1.0 + 1.0 / k0);
// Simply handing our 1D function to a root-finder works fine until we start to encounter large k. For large k,
// even just computing \lambda goes wrong because we are taking x_i^k which overflows. Horst Rinne, "The Weibull
// Distribution: A Handbook" describes a way out. Basically, we first move to variables z_i = \log(x_i) and
// then w_i = z_i - \bar{z}. Then lots of factors of e^{k \bar{z}} cancel out and, even though we still do
// have some e^{k w_i}, the w_i are small and centered around 0 instead of large and centered around \lambda.
//Sample transformedSample = sample.Copy();
//transformedSample.Transform(x => Math.Log(x));
double[] transformedSample = new double[sample.Count];
for (int j = 0; j < sample.Count; j++)
{
transformedSample[j] = Math.Log(sample[j]);
}
double zbar = transformedSample.Mean();
for (int j = 0; j < transformedSample.Length; j++)
{
transformedSample[j] -= zbar;
}
// After this change of variable the 1D function to zero becomes
// g(k) = \sum_i ( 1 - k w_i ) e^{k w_i}
// It's easy to show that g(0) = n and g(\infinity) = -\infinity, so it must cross zero. It's also easy to take
// a derivative
// g'(k) = - k \sum_i w_i^2 e^{k w_i}
// so we can apply Newton's method.
int i = 0;
double k1 = k0;
while (true)
{
i++;
double g = 0.0;
double gp = 0.0;
foreach (double w in transformedSample)
{
double e = Math.Exp(k1 * w);
g += (1.0 - k1 * w) * e;
gp -= k1 * w * w * e;
}
double dk = -g / gp;
k1 += dk;
if (Math.Abs(dk) <= Global.Accuracy * Math.Abs(k1))
{
break;
}
if (i >= Global.SeriesMax)
{
throw new NonconvergenceException();
}
}
// The corresponding lambda can also be expressed in terms of zbar and w's.
double t = 0.0;
foreach (double w in transformedSample)
{
t += Math.Exp(k1 * w);
}
t /= transformedSample.Length;
double lambda1 = Math.Exp(zbar) * Math.Pow(t, 1.0 / k1);
// We need the curvature matrix at the minimum of our log likelihood function
// to determine the covariance matrix. Taking more derivatives...
// \frac{\partial^2 \log L} = \frac{N k}{\lambda^2} - \sum_i \frac{k(k+1) x_i^k}{\lambda^{k+2}}
// = - \frac{N k^2}{\lambda^2}
// The second expression follows by inserting the first-derivative-equal-zero relation into the first.
// For k=1, this agrees with the variance formula for the mean of the best-fit exponential.
// Derivatives involving k are less simple.
// We end up needing the means < (x/lambda)^k log(x/lambda) > and < (x/lambda)^k log^2(x/lambda) >
double mpl = 0.0; double mpl2 = 0.0;
foreach (double x in sample)
{
double r = x / lambda1;
double p = Math.Pow(r, k1);
double l = Math.Log(r);
double pl = p * l;
double pl2 = pl * l;
mpl += pl;
mpl2 += pl2;
}
mpl = mpl / sample.Count;
mpl2 = mpl2 / sample.Count;
// See if we can't do any better here. Transforming to zbar and w's looked ugly, but perhaps it
// can be simplified? One interesting observation: if we take expectation values (which gives
// the Fisher information matrix) the entries become simple:
// B_{\lambda \lambda} = \frac{N k^2}{\lambda^2}
// B_{\lambda k} = -\Gamma'(2) \frac{N}{\lambda}
// B_{k k } = [1 + \Gamma''(2)] \frac{N}{k^2}
// Would it be bad to just use these directly?
// Construct the curvature matrix and invert it.
SymmetricMatrix C = new SymmetricMatrix(2);
C[0, 0] = sample.Count * MoreMath.Sqr(k1 / lambda1);
C[0, 1] = -sample.Count * k1 / lambda1 * mpl;
C[1, 1] = sample.Count * (1.0 / MoreMath.Sqr(k1) + mpl2);
CholeskyDecomposition CD = C.CholeskyDecomposition();
if (CD == null)
{
throw new DivideByZeroException();
}
C = C.Inverse();
// Do a KS test to compare sample to best-fit distribution
WeibullDistribution distribution = new WeibullDistribution(lambda1, k1);
TestResult test = sample.KolmogorovSmirnovTest(distribution);
// return the result
return(new WeibullFitResult(lambda1, k1, C, distribution, test));
}