public void ComputeTest()
{
// Example AND problem
double[][] inputs =
{
new double[] { 0, 0 }, // 0 and 0: 0 (label -1)
new double[] { 0, 1 }, // 0 and 1: 0 (label -1)
new double[] { 1, 0 }, // 1 and 0: 0 (label -1)
new double[] { 1, 1 } // 1 and 1: 1 (label +1)
};
// Dichotomy SVM outputs should be given as [-1;+1]
int[] labels =
{
// 0, 0, 0, 1
-1, -1, -1, 1
};
// Create a Support Vector Machine for the given inputs
KernelSupportVectorMachine machine = new KernelSupportVectorMachine(new Linear(0), inputs[0].Length);
// Instantiate a new learning algorithm for SVMs
SequentialMinimalOptimization smo = new SequentialMinimalOptimization(machine, inputs, labels);
// Set up the learning algorithm
smo.Complexity = 100.0;
// Run
double error = smo.Run();
Assert.AreEqual(0, error);
Assert.AreEqual(-1, Math.Sign(machine.Compute(inputs[0])));
Assert.AreEqual(-1, Math.Sign(machine.Compute(inputs[1])));
Assert.AreEqual(-1, Math.Sign(machine.Compute(inputs[2])));
Assert.AreEqual(+1, Math.Sign(machine.Compute(inputs[3])));
// At this point we have the weighted support vectors
// w sv b
// (+4) * (1,1) -3
// (-2) * (1,0)
// (-2) * (0,1)
//
// However, it can be seen that the last SV can be written
// as a linear combination of the two first vectors:
//
// (0,1) = (1,1) - (1,0)
//
// Since we have a linear space (we are using a linear kernel)
// this vector could be removed from the support vector set.
//
// f(x) = sum(alpha_i * x * x_i) + b
// = 4*(1,1)*x - 2*(1,0)*x - 2*(0,1)*x - 3
// = 4*(1,1)*x - 2*(1,0)*x - 2*((1,1) - (1,0))*x - 3
// = 4*(1,1)*x - 2*(1,0)*x - 2*(1,1)*x + 2*(1,0)*x - 3
// = 4*(1,1)*x - 2*(1,0)*x - 2*(1,1)*x + 2*(1,0)*x - 3
// = 2*(1,1)*x - 3
// = 2*x1 + 2*x2 - 3
//
SupportVectorReduction svr = new SupportVectorReduction(machine);
double error2 = svr.Run();
Assert.AreEqual(-1, Math.Sign(machine.Compute(inputs[0])));
Assert.AreEqual(-1, Math.Sign(machine.Compute(inputs[1])));
Assert.AreEqual(-1, Math.Sign(machine.Compute(inputs[2])));
Assert.AreEqual(+1, Math.Sign(machine.Compute(inputs[3])));
}
