Math.NET能解决任何矩阵吗?

时间:2014-08-04 14:16:31

标签: .net math matrix linear-algebra mathnet

我正在尝试使用Math.NET来解决以下系统:

系数矩阵A:

var matrixA = DenseMatrix.OfArray(new[,] {
    { 20000, 0, 0, -20000, 0, 0, 0, 0, 0 },
    { 0, 666.66666666666663, 2000, 0, -666.66666666666663, 2000, 0, 0, 0 }, 
    { 0, 2000,  8000,  0,  -2000, 4000, 0, 0, 0 },
    { -20000, 0, 0, 20666.66666666666663, 0, 2000, -666.66666666666663, 0, 2000 },
    { 0, -666.66666666666663, -2000, 0, 20666.66666666666663, -2000, 0, -20000, 0 },
    { 0, 2000, 4000, 2000, -2000, 16000, -2000, 0, 4000 },
    { 0, 0, 0, -666.66666666666663, 0, -2000, 666.66666666666663, 0, -2000 },
    { 0, 0, 0, 0, -20000, 0, 0, 20000, 0 },
    { 0, 0, 0, 2000, 0, 4000, -2000, 0, 7999.9999999999991 }});

结果矢量b:

double[] loadVector = { 0, 0, 0, 5, 0, 0, 0, 0, 0 };
var vectorB = MathNet.Numerics.LinearAlgebra.Vector<double>.Build.Dense(loadVector);

我从有限元分析示例问题中提取了这些数字,因此基于该示例我期望的答案是:

[0.01316, 0, 0.0009199, 0.01316, -0.00009355, -0.00188, 0, 0, 0]

然而,我发现Math.NET和online Matrix Calculator主要给我零(来自迭代求解器),NaN或大的不正确数字(来自直接求解器)作为解决方案。

在Math.NET中,我尝试将我的矩阵插入到提供的示例中,包括:

迭代示例:

namespace Examples.LinearAlgebra.IterativeSolversExamples
{
/// <summary>
/// Composite matrix solver
/// </summary>
public class CompositeSolverExample : IExample
{
    public void Run()
    {
        // Format matrix output to console
        var formatProvider = (CultureInfo)CultureInfo.InvariantCulture.Clone();
        formatProvider.TextInfo.ListSeparator = " ";

        // Solve next system of linear equations (Ax=b):
        // 5*x + 2*y - 4*z = -7
        // 3*x - 7*y + 6*z = 38
        // 4*x + 1*y + 5*z = 43

        // Create matrix "A" with coefficients
        var matrixA = DenseMatrix.OfArray(new[,] { { 20000, 0, 0, -20000, 0, 0, 0, 0, 0 }, { 0, 666.66666666666663, 2000, 0, -666.66666666666663, 2000, 0, 0, 0 }, 
                                                { 0, 2000,  8000,  0,  -2000, 4000, 0, 0, 0 }, { -20000, 0, 0, 20666.66666666666663, 0, 2000, -666.66666666666663, 0, 2000 },
                                                {0, -666.66666666666663, -2000, 0, 20666.66666666666663, -2000, 0, -20000, 0 }, { 0, 2000, 4000, 2000, -2000, 16000, -2000, 0, 4000 },
                                                { 0, 0, 0, -666.66666666666663, 0, -2000, 666.66666666666663, 0, -2000 }, { 0, 0, 0, 0, -20000, 0, 0, 20000, 0 },
                                                 {0, 0, 0, 2000, 0, 4000, -2000, 0, 7999.9999999999991 }});


        Console.WriteLine(@"Matrix 'A' with coefficients");
        Console.WriteLine(matrixA.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // Create vector "b" with the constant terms.
        double[] loadVector = {0,0,0,5,0,0,0,0,0};
        var vectorB = MathNet.Numerics.LinearAlgebra.Vector<double>.Build.Dense(loadVector);
        Console.WriteLine(@"Vector 'b' with the constant terms");
        Console.WriteLine(vectorB.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // Create stop criteria to monitor an iterative calculation. There are next available stop criteria:
        // - DivergenceStopCriterion: monitors an iterative calculation for signs of divergence;
        // - FailureStopCriterion: monitors residuals for NaN's;
        // - IterationCountStopCriterion: monitors the numbers of iteration steps;
        // - ResidualStopCriterion: monitors residuals if calculation is considered converged;

        // Stop calculation if 1000 iterations reached during calculation
        var iterationCountStopCriterion = new IterationCountStopCriterion<double>(500000);

        // Stop calculation if residuals are below 1E-10 --> the calculation is considered converged
        var residualStopCriterion = new ResidualStopCriterion<double>(1e-10);

        // Create monitor with defined stop criteria
        var monitor = new Iterator<double>(iterationCountStopCriterion, residualStopCriterion);

        // Load all suitable solvers from current assembly. Below in this example, there is user-defined solver
        // "class UserBiCgStab : IIterativeSolverSetup<double>" which uses regular BiCgStab solver. But user may create any other solver
        // and solver setup classes which implement IIterativeSolverSetup<T> and pass assembly to next function:
        var solver = new CompositeSolver(SolverSetup<double>.LoadFromAssembly(Assembly.GetExecutingAssembly()));

        // 1. Solve the matrix equation
        var resultX = matrixA.SolveIterative(vectorB, solver, monitor);
        Console.WriteLine(@"1. Solve the matrix equation");
        Console.WriteLine();

        // 2. Check solver status of the iterations.
        // Solver has property IterationResult which contains the status of the iteration once the calculation is finished.
        // Possible values are:
        // - CalculationCancelled: calculation was cancelled by the user;
        // - CalculationConverged: calculation has converged to the desired convergence levels;
        // - CalculationDiverged: calculation diverged;
        // - CalculationFailure: calculation has failed for some reason;
        // - CalculationIndetermined: calculation is indetermined, not started or stopped;
        // - CalculationRunning: calculation is running and no results are yet known;
        // - CalculationStoppedWithoutConvergence: calculation has been stopped due to reaching the stopping limits, but that convergence was not achieved;
        Console.WriteLine(@"2. Solver status of the iterations");
        Console.WriteLine(monitor.Status);
        Console.WriteLine();

        // 3. Solution result vector of the matrix equation
        Console.WriteLine(@"3. Solution result vector of the matrix equation");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 4. Verify result. Multiply coefficient matrix "A" by result vector "x"
        var reconstructVecorB = matrixA*resultX;
        Console.WriteLine(@"4. Multiply coefficient matrix 'A' by result vector 'x'");
        Console.WriteLine(reconstructVecorB.ToString("#0.00\t", formatProvider));
        Console.WriteLine();
        Console.Read();
    }
}
}

直接示例:

namespace Examples.LinearAlgebraExamples
{
/// <summary>
/// Direct solvers (using matrix decompositions)
/// </summary>
/// <seealso cref="http://en.wikipedia.org/wiki/Numerical_analysis#Direct_and_iterative_methods"/>
public class DirectSolvers : IExample
{
    /// <summary>
    /// Gets the name of this example
    /// </summary>
    public string Name
    {
        get
        {
            return "Direct solvers";
        }
    }

    /// <summary>
    /// Gets the description of this example
    /// </summary>
    public string Description
    {
        get
        {
            return "Solve linear equations using matrix decompositions";
        }
    }

    /// <summary>
    /// Run example
    /// </summary>
    public void Run()
    {
        // Format matrix output to console
        var formatProvider = (CultureInfo) CultureInfo.InvariantCulture.Clone();
        formatProvider.TextInfo.ListSeparator = " ";

        // Solve next system of linear equations (Ax=b):
        // 5*x + 2*y - 4*z = -7
        // 3*x - 7*y + 6*z = 38
        // 4*x + 1*y + 5*z = 43

         matrixA = DenseMatrix.OfArray(new[,] { { 20000, 0, 0, -20000, 0, 0, 0, 0, 0 }, { 0, 666.66666666666663, 2000, 0, -666.66666666666663, 2000, 0, 0, 0 }, 
                                                { 0, 2000,  8000,  0,  -2000, 4000, 0, 0, 0 }, { -20000, 0, 0, 20666.66666666666663, 0, 2000, -666.66666666666663, 0, 2000 },
                                                {0, -666.66666666666663, -2000, 0, 20666.66666666666663, -2000, 0, -20000, 0 }, { 0, 2000, 4000, 2000, -2000, 16000, -2000, 0, 4000 },
                                                { 0, 0, 0, -666.66666666666663, 0, -2000, 666.66666666666663, 0, -2000 }, { 0, 0, 0, 0, -20000, 0, 0, 20000, 0 },
                                                 {0, 0, 0, 2000, 0, 4000, -2000, 0, 7999.9999999999991 }});

        Console.WriteLine(@"Matrix 'A' with coefficients");
        Console.WriteLine(matrixA.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // Create vector "b" with the constant terms.
        double[] loadVector = { 0, 0, 0, 5000, 0, 0, 0, 0, 0 };
        var vectorB = MathNet.Numerics.LinearAlgebra.Vector<double>.Build.Dense(loadVector);
        Console.WriteLine(@"Vector 'b' with the constant terms");
        Console.WriteLine(vectorB.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 1. Solve linear equations using LU decomposition
        var resultX = matrixA.LU().Solve(vectorB);
        Console.WriteLine(@"1. Solution using LU decomposition");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 2. Solve linear equations using QR decomposition
        resultX = matrixA.QR().Solve(vectorB);
        Console.WriteLine(@"2. Solution using QR decomposition");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 3. Solve linear equations using SVD decomposition
        matrixA.Svd().Solve(vectorB, resultX);
        Console.WriteLine(@"3. Solution using SVD decomposition");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 4. Solve linear equations using Gram-Shmidt decomposition
        matrixA.GramSchmidt().Solve(vectorB, resultX);
        Console.WriteLine(@"4. Solution using Gram-Shmidt decomposition");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 5. Verify result. Multiply coefficient matrix "A" by result vector "x"
        var reconstructVecorB = matrixA*resultX;
        Console.WriteLine(@"5. Multiply coefficient matrix 'A' by result vector 'x'");
        Console.WriteLine(reconstructVecorB.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // To use Cholesky or Eigenvalue decomposition coefficient matrix must be 
        // symmetric (for Evd and Cholesky) and positive definite (for Cholesky)
        // Multipy matrix "A" by its transpose - the result will be symmetric and positive definite matrix
        var newMatrixA = matrixA.TransposeAndMultiply(matrixA);
        Console.WriteLine(@"Symmetric positive definite matrix");
        Console.WriteLine(newMatrixA.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 6. Solve linear equations using Cholesky decomposition
        newMatrixA.Cholesky().Solve(vectorB, resultX);
        Console.WriteLine(@"6. Solution using Cholesky decomposition");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 7. Solve linear equations using eigen value decomposition
        newMatrixA.Evd().Solve(vectorB, resultX);
        Console.WriteLine(@"7. Solution using eigen value decomposition");
        Console.WriteLine(resultX.ToString("#0.00\t", formatProvider));
        Console.WriteLine();

        // 8. Verify result. Multiply new coefficient matrix "A" by result vector "x"
        reconstructVecorB = newMatrixA*resultX;
        Console.WriteLine(@"8. Multiply new coefficient matrix 'A' by result vector 'x'");
        Console.WriteLine(reconstructVecorB.ToString("#0.00\t", formatProvider));
        Console.WriteLine();
        Console.Read();
    }
}
}

示例问题中的数字可能是错误的,但我需要确保我在继续之前正确使用Math.NET。我是否按照它们的使用方式使用这些求解器示例?还有什么我可以尝试这些例子不包括在内吗?

The Finite Element Analysis Example Problem (p.8, Example 1):

他们似乎搞砸了某个地方的单位,所以为了让我的矩阵与之匹配,我们不得不使用以下输入:

Member  A (mm^2)    E (N/mm^2)  I (mm^4)    L (mm)
AB     600000000    0.0002      60000000      6
BC     600000000    0.0002      60000000      6

另请注意,他们已经删除了一些在计算过程中自然会消失的行和列。这些行和列仍然存在于我正在使用的矩阵中

3 个答案:

答案 0 :(得分:5)

  

Math.NET能解决任何矩阵吗?

不,它不能。具体来说,它不能解决没有解决方案的方程组,也不能解决任何其他求解器。

在这种情况下,你的矩阵A是单数的,即它没有反转。这意味着你的方程系统要么没有解,也就是说它是不一致的,或者它有无限的解(见例Introduction to Numerical Methods中的6.5节)。奇异矩阵具有零行列式。你可以使用Determinant方法在mathnet中看到这个:

Console.WriteLine("Determinant {0}", matrixA.Determinant());

这给出了

Determinant 0

A是单数的条件是当其行(或列)的线性组合为零时。例如,这里第2行,第5行和第8行的总和为零。这些不是唯一加在一起得到零的行。 (你以后会看到另一个例子。实际上有三种不同的方法,从技术上讲,这个9x9矩阵是&#34;等级6&#34;而不是&#34;等级9&#34; )。

请记住,当您尝试解决Ax=b时,您所做的就是解决一组联立方程式。在二维中,您可能有一个系统,如

A = [1 1   b = [1 
     2 2],      2]

并解决此问题相当于找到x0x1以便

  x0 +   x1 = 1
2*x0 + 2*x1 = 2

这里有无限的解决方案满足x1 = 1 - x0,即沿着x0 + x1 = 1行。

的替代方案
A = [1 1   b = [1 
     1 1],      2]

相当于

  x0 +  x1 = 1
  x0 +  x1 = 2

显然没有解决方案,因为我们可以从第二个方程中减去第一个方程式来获得0 = 1

在你的情况下,第1,第4和第7个方程是

 20000*x0 -20000               *x3                                          = 0
-20000*x0 +20666.66666666666663*x3 +2000*x5 -666.66666666666663*x6 +2000*x8 = 5
            -666.66666666666663*x3 -2000*x5 +666.66666666666663*x6 -2000*x8 = 0

将这些加在一起会产生0=5,因此您的系统没有解决方案。

在Matlab或R等交互式环境中探索矩阵最容易。因为Python在Visual Studio中可用,它通过numpy提供了类似Matlab的环境,我已经用上面的代码演示了上面的代码。蟒蛇。我建议Python tools for visual studio,我在Visual Studio 2012和2013中都成功使用过。

# numpy is a Matlab-like environment for linear algebra in Python
import numpy as np

# matrix A
A = np.matrix ([
    [ 20000, 0, 0, -20000, 0, 0, 0, 0, 0 ],
    [ 0, 666.66666666666663, 2000, 0, -666.66666666666663, 2000, 0, 0, 0 ], 
    [ 0, 2000,  8000,  0,  -2000, 4000, 0, 0, 0 ],
    [ -20000, 0, 0, 20666.66666666666663, 0, 2000, -666.66666666666663, 0, 2000 ],
    [ 0, -666.66666666666663, -2000, 0, 20666.66666666666663, -2000, 0, -20000, 0 ],
    [ 0, 2000, 4000, 2000, -2000, 16000, -2000, 0, 4000 ],
    [ 0, 0, 0, -666.66666666666663, 0, -2000, 666.66666666666663, 0, -2000 ],
    [ 0, 0, 0, 0, -20000, 0, 0, 20000, 0 ],
    [ 0, 0, 0, 2000, 0, 4000, -2000, 0, 7999.9999999999991 ]])

# vector b
b = np.array([0, 0, 0, 5, 0, 0, 0, 0, 0])
b.shape = (9,1)

# attempt to solve Ax=b
np.linalg.solve(A,b)

此操作失败并显示错误信息:LinAlgError: Singular matrix。您可以看到A是单数的,例如,显示第2行,第5行和第8行的总和为零

A[1,]+A[4,]+A[7,]

注意到行是零索引的。

通过将柱状向量0=5附加到b,然后添加相应的(0-)来证明第1,第4和第7个方程导致A形成增广矩阵索引)一起排行

Aaug = np.append(A,b,1)

Aaug[0,] + Aaug[3,] + Aaug[6,]

最后,即使你的矩阵不是单数,你仍然可以有一个数值不稳定的问题:在这种情况下,问题被称为病态。检查矩阵的条件编号,了解如何执行此操作(wikipedianp.linalg.cond(A)matrixA.ConditionNumber())。

答案 1 :(得分:4)

您问题中的最后两句话是问题的根源:

  

另请注意,他们已经删除了一些在计算过程中自然会消失的行和列。这些行和列仍然存在于我正在使用的矩阵中。

在您的示例问题中,您的关节被固定以防止在某些方向上移动(称为边界条件)。有时在进行有限元分析时,如果没有根据这些边界条件从刚度矩阵和加载矩阵中删除适当的行和列,最终会得到一个无法求解的系统,这就是这里的情况。 / p>

再次尝试使用DirectSolver:

var matrixA = DenseMatrix.OfArray(new[,] { {20000,  0,  -20000, 0,  0}, {0, 8000    ,0, -2000   ,4000},
                                                   {-20000, 0,  20666.667   ,0, 2000}, {0,  -2000   ,0, 20666.67,   -2000},
                                                    {0, 4000    ,2000   ,-2000, 16000}});

double[] loadVector = { 0, 0, 5, 0, 0 };
var vectorB = MathNet.Numerics.LinearAlgebra.Vector<double>.Build.Dense(loadVector);

要回答你的问题,是的,你正确使用这些方法,但是你正在解决错误的系统。修复你的输入,你应该得到你正在寻找的输出。

我还应该指出,我建议使用Direct Solver示例的原因是因为您似乎正在寻找一个精确的解决方案。迭代求解器仅通过近似解决方案来节省计算时间。

答案 2 :(得分:0)

不,它无法解决奇异矩阵。但是这里没有任何其他代码,因为这里没有解决方案。

对于您的特定情况,发布的矩阵A是单数。大小为9×9,但等级为6. MATLAB报告的条件为1.9e17。因此,请在检查合理答案之前检查您的刚度矩阵组成。也许你需要对矩阵进行标准化,即提取E I系数,使数字从1e5下降到1附近,这在数值上更合适。

<强> FYI

如果您不喜欢Math.NET,或者您想验证代码,请使用纯c#。通过MSDN Magazine article阅读此James McCaffrey并使用列出的代码。

var A = new [,] { ... };
var b = new [] { ... };
var x = LU.SustemSolve(A,b);