本文整理汇总了C#中DivergenceStopCriterium.DetermineStatus方法的典型用法代码示例。如果您正苦于以下问题：C# DivergenceStopCriterium.DetermineStatus方法的具体用法？C# DivergenceStopCriterium.DetermineStatus怎么用？C# DivergenceStopCriterium.DetermineStatus使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类DivergenceStopCriterium的用法示例。
在下文中一共展示了DivergenceStopCriterium.DetermineStatus方法的15个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: DetermineStatusWithDivergence
public void DetermineStatusWithDivergence()
{
const float Increase = 0.5f;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we'll have one to few iterations
float previous = 1;
for (var i = 0; i < Iterations - 1; i++)
{
previous *= (1 + Increase + 0.01f);
criterium.DetermineStatus(i,
new DenseVector(new[] { 1.0f }),
new DenseVector(new[] { 1.0f }),
new DenseVector(new[] { previous }));
Assert.IsInstanceOfType(typeof(CalculationRunning), criterium.Status, "Status check fail.");
}
// Add the final residual. Now we should have divergence
previous *= (1 + Increase + 0.01f);
criterium.DetermineStatus(Iterations - 1,
new DenseVector(new[] { 1.0f }),
new DenseVector(new[] { 1.0f }),
new DenseVector(new[] { previous }));
Assert.IsInstanceOfType(typeof(CalculationDiverged), criterium.Status, "Status check fail.");
}示例2: ResetCalculationState
public void ResetCalculationState()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. Blow it up instantly
var status = criterium.DetermineStatus(
1,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(double.NaN, 0)}));
Assert.AreEqual(IterationStatus.Diverged, status, "Status check fail.");
// Reset the state
criterium.Reset();
Assert.AreEqual(Increase, criterium.MaximumRelativeIncrease, "Incorrect maximum");
Assert.AreEqual(Iterations, criterium.MinimumNumberOfIterations, "Incorrect iteration count");
Assert.AreEqual(IterationStatus.Continue, criterium.Status, "Status check fail.");
}示例3: DetermineStatusWithTooFewIterations
public void DetermineStatusWithTooFewIterations()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we'll have to few iterations
for (var i = 0; i < Iterations - 1; i++)
{
var status = criterium.DetermineStatus(
i,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex((i + 1)*(Increase + 0.1), 0)}));
Assert.AreEqual(IterationStatus.Continue, status, "Status check fail.");
}
}示例4: DetermineStatusWithIllegalIterationNumberThrowsArgumentOutOfRangeException
public void DetermineStatusWithIllegalIterationNumberThrowsArgumentOutOfRangeException()
{
var criterium = new DivergenceStopCriterium(0.5, 15);
Assert.Throws<ArgumentOutOfRangeException>(() => criterium.DetermineStatus(
-1,
DenseVector.Create(3, i => 4),
DenseVector.Create(3, i => 5),
DenseVector.Create(3, i => 6)));
}示例5: DetermineStatusWithDivergenceThroughNaN
public void DetermineStatusWithDivergenceThroughNaN()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we'll have to few iterations
for (var i = 0; i < Iterations - 5; i++)
{
var status = criterium.DetermineStatus(
i,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex((i + 1)*(Increase - 0.01), 0)}));
Assert.AreEqual(IterationStatus.Continue, status, "Status check fail.");
}
// Now make it fail by throwing in a NaN
var status2 = criterium.DetermineStatus(
Iterations,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(double.NaN, 0)}));
Assert.AreEqual(IterationStatus.Diverged, status2, "Status check fail.");
}示例6: DetermineStatusWithDivergence
public void DetermineStatusWithDivergence()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we'll have one to few iterations
double previous = 1;
for (var i = 0; i < Iterations - 1; i++)
{
previous *= 1 + Increase + 0.01;
var status = criterium.DetermineStatus(
i,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(previous, 0)}));
Assert.AreEqual(IterationStatus.Continue, status, "Status check fail.");
}
// Add the final residual. Now we should have divergence
previous *= 1 + Increase + 0.01;
var status2 = criterium.DetermineStatus(
Iterations - 1,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(previous, 0)}));
Assert.AreEqual(IterationStatus.Diverged, status2, "Status check fail.");
}示例7: DetermineStatusWithNoDivergence
public void DetermineStatusWithNoDivergence()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium<Complex>(Increase, Iterations);
// Add residuals. We should not diverge because we won't have enough increase
for (var i = 0; i < Iterations*2; i++)
{
var status = criterium.DetermineStatus(
i,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex((i + 1)*(Increase - 0.01), 0)}));
Assert.AreEqual(IterationStatus.Continue, status, "Status check fail.");
}
}示例8: DetermineStatusWithIllegalIterationNumberThrowsArgumentOutOfRangeException
public void DetermineStatusWithIllegalIterationNumberThrowsArgumentOutOfRangeException()
{
var criterium = new DivergenceStopCriterium<float>(0.5, 15);
Assert.Throws<ArgumentOutOfRangeException>(() => criterium.DetermineStatus(
-1,
Vector<float>.Build.Dense(3, 4),
Vector<float>.Build.Dense(3, 5),
Vector<float>.Build.Dense(3, 6)));
}示例9: DetermineStatusWithDivergenceThroughNaN
public void DetermineStatusWithDivergenceThroughNaN()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we'll have to few iterations
for (var i = 0; i < Iterations - 5; i++)
{
criterium.DetermineStatus(
i,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex((i + 1)*(Increase - 0.01), 0)}));
Assert.IsInstanceOf(typeof (CalculationRunning), criterium.Status, "Status check fail.");
}
// Now make it fail by throwing in a NaN
criterium.DetermineStatus(
Iterations,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(double.NaN, 0)}));
Assert.IsInstanceOf(typeof (CalculationDiverged), criterium.Status, "Status check fail.");
}示例10: DetermineStatusWithNoDivergence
public void DetermineStatusWithNoDivergence()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we won't have enough increase
for (var i = 0; i < Iterations*2; i++)
{
criterium.DetermineStatus(
i,
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex(1.0, 0)}),
new DenseVector(new[] {new Complex((i + 1)*(Increase - 0.01), 0)}));
Assert.IsInstanceOf(typeof (CalculationRunning), criterium.Status, "Status check fail.");
}
}示例11: DetermineStatusWithNullResidualVectorThrowsArgumentNullException
public void DetermineStatusWithNullResidualVectorThrowsArgumentNullException()
{
var criterium = new DivergenceStopCriterium(0.5, 15);
Assert.Throws<ArgumentNullException>(() => criterium.DetermineStatus(
1,
DenseVector.Create(3, i => 4),
DenseVector.Create(3, i => 5),
null));
}示例12: ResetCalculationState
public void ResetCalculationState()
{
const double Increase = 0.5;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. Blow it up instantly
criterium.DetermineStatus(1,
new DenseVector(new [] { 1.0f }),
new DenseVector(new [] { 1.0f }),
new DenseVector(new [] { float.NaN }));
Assert.IsInstanceOfType(typeof(CalculationDiverged), criterium.Status, "Status check fail.");
// Reset the state
criterium.ResetToPrecalculationState();
Assert.AreEqual(Increase, criterium.MaximumRelativeIncrease, "Incorrect maximum");
Assert.AreEqual(Iterations, criterium.MinimumNumberOfIterations, "Incorrect iteration count");
Assert.IsInstanceOfType(typeof(CalculationIndetermined), criterium.Status, "Status check fail.");
}示例13: DetermineStatusWithTooFewIterations
public void DetermineStatusWithTooFewIterations()
{
const float Increase = 0.5f;
const int Iterations = 10;
var criterium = new DivergenceStopCriterium(Increase, Iterations);
// Add residuals. We should not diverge because we'll have to few iterations
for (var i = 0; i < Iterations - 1; i++)
{
criterium.DetermineStatus(i,
new DenseVector(new [] { 1.0f }),
new DenseVector(new [] { 1.0f }),
new DenseVector(new [] { (i + 1) * (Increase + 0.1f) }));
Assert.IsInstanceOfType(typeof(CalculationRunning), criterium.Status, "Status check fail.");
}
}示例14: DetermineStatusWithNullResidualVector
public void DetermineStatusWithNullResidualVector()
{
var criterium = new DivergenceStopCriterium(0.5, 15);
criterium.DetermineStatus(1,
new DenseVector(3, 4),
new DenseVector(3, 5),
null);
}示例15: DetermineStatusWithIllegalIterationNumber
public void DetermineStatusWithIllegalIterationNumber()
{
var criterium = new DivergenceStopCriterium(0.5, 15);
criterium.DetermineStatus(-1,
new DenseVector(3, 4),
new DenseVector(3, 5),
new DenseVector(3, 6));
}