public void lbfgsTest2()
{
Accord.Math.Tools.SetupGenerator(0);
// Suppose we would like to find the minimum 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, as:
//
// 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
var lbfgs = new BoundedBroydenFletcherGoldfarbShanno(numberOfVariables: 2, function: f, gradient: g);
// And then minimize the function:
Assert.IsTrue(lbfgs.Minimize());
double minValue = lbfgs.Value;
double[] solution = lbfgs.Solution;
// The resultant minimum value should be -2, and the solution
// vector should be { 1.0, 2.0 }. The answer can be checked on
// Wolfram Alpha by clicking the following the link:
// http://www.wolframalpha.com/input/?i=maximize+%28exp%28-%28x-1%29%C2%B2%29+%2B+exp%28-%28y-2%29%C2%B2%2F2%29%29
double expected = -2;
Assert.AreEqual(expected, minValue, 1e-10);
Assert.AreEqual(1, solution[0], 1e-3);
Assert.AreEqual(2, solution[1], 1e-3);
}
