取一个函数的负数<双[]，双倍>

Question

>我有一个数学函数-exp{-(x - 1)²} - exp{-0.5*(y - 2)²}使用该函数及其导数将其传递到BFGS算法中

Func<double[], double> f = (x) => 
    Math.Exp(-Math.Pow(x[0] - 1, 2)) + Math.Exp(-0.5 * Math.Pow(x[1] - 2, 2));
Func<double[], double[]> g = (x) => new double[] 
{
    // df/dx = -2 * e^(-(x - 1)²)(x - 1).
    -2 * Math.Exp(-Math.Pow(x[0] - 1, 2)) * (x[0] - 1),
    // df/dy = -e^(-1/2(y - 2)²) * (y - 2).
    -Math.Exp(-0.5 * Math.Pow(x[1] - 2, 2)) * (x[1] - 2)
};

现在，该算法已被编码以最小化传递给它的函数，这很好用。但是，我想添加最大化功能的功能。为此，我将仅通过否定函数和导数来包装最小化代码，并将结果传递给最小化代码。

算法的 ctor 是

public BroydenFletcherGoldfarbShanno(int numberOfVariables, 
    Func<double[], double> function, 
    Func<double[], double[]> gradient, 
    Random random = null)
        : this(numberOfVariables, random)
{
    if (function == null)
        throw new ArgumentNullException("function");
    if (gradient == null)
        throw new ArgumentNullException("gradient");
    this.Function = function;
    this.Gradient = gradient;
}

初始化Function/Gradient。这个类有一个Minimize方法，现在我添加了

public double Maximize()
{
     // Negate the function.
     Func<double[], double> f;
     f = (x) => -this.Function(x);
     this.Function = f;
     // Negate the derivatives.
     ...
}

我的问题是我怎么能仅仅否定this.Function/this.Gradient对象？我上面所做的抛出了StackOverflowException.

谢谢你的时间。

---

使用函数声明进行编辑

[TestMethod]
public void LBFGSMaximisationTest()
{
    // Suppose we would like to find the maximum of the function:
    // f(x, y) = exp{-(x - 1)²} + exp{-(y - 2)²/2}.
    // First we need write down the function either as a named
    // method, an anonymous method or as a lambda function.
    Func<double[], double> f = (x) =>
    Math.Exp(-Math.Pow(x[0] - 1, 2)) + Math.Exp(-0.5 * Math.Pow(x[1] - 2, 2));
    // Now, we need to write its gradient, which is just the
    // vector of first partial derivatives del_f / del_x.
    // g(x, y) = { del f / del x, del f / del y }.
    Func<double[], double[]> g = (x) => new double[] 
    {
        // df/dx = -2 * e^(-(x - 1)²)(x - 1).
        -2 * Math.Exp(-Math.Pow(x[0] - 1, 2)) * (x[0] - 1),
        // df/dy = -e^(-1/2(y - 2)²) * (y - 2).
        -Math.Exp(-0.5 * Math.Pow(x[1] - 2, 2)) * (x[1] - 2)
    };
    // Finally, we can create the L-BFGS solver, passing the functions as arguments.
    Random r = new SystemRandomSource(0, true);
    BroydenFletcherGoldfarbShanno lbfgs = new BroydenFletcherGoldfarbShanno(
        numberOfVariables: 2, function: f, gradient: g, random: r);
    // And then minimize the function.
    double maxValue = lbfgs.Maximize();
    ...

Answer 1

是的，它会：

f = (x) => -this.Function(x);
this.Function = f;

假设Function是类级别的Funct<double[],double>，你基本上是在写：

this.Function = (x) => -this.Function(x);

所以是的，那会爆炸。这是因为this.Function是通过捕获的范围延迟的。我怀疑你的意思是：

Func<double[], double> oldFunc = this.Function;
this.Function = (x) => -oldFunc(x);

现在，我们将旧函数捕获到委托中，并使用该捕获的委托而不是递归调用。