Đề bài
XOR (exclusive or) is the classic problem a single neuron cannot solve, but a network with one hidden layer can. Train a network with 2 inputs, 2 hidden neurons and 1 output, all sigmoid (σ(z) = 1 / (1 + e^(-z))), using the loss L = (y - target)².
The training data, always used in this order in every epoch:
| input | target |
|---|---|
| [0, 0] | 0 |
| [0, 1] | 1 |
| [1, 0] | 1 |
| [1, 1] | 0 |
For each example:
- Forward: a_j = σ(x₁ × hidden_weights[j][0] + x₂ × hidden_weights[j][1] + hidden_biases[j]), y = σ(a₁ × output_weights[0] + a₂ × output_weights[1] + output_bias)
- delta_out = 2 × (y - target) × y × (1 - y), and delta_j = delta_out × output_weights[j] × a_j × (1 - a_j), using the weights before this example's update
- Update every parameter with learning rate lr: output_weights[j] -= lr × delta_out × a_j, output_bias -= lr × delta_out, hidden_weights[j][i] -= lr × delta_j × x_i, hidden_biases[j] -= lr × delta_j
Implement train_xor(hidden_weights, hidden_biases, output_weights, output_bias, learning_rate, epochs).
Output
After training, return the network's predictions y for the inputs [0, 0], [0, 1], [1, 0] and [1, 1], in that order.
Ràng buộc
0 ≤ epochs ≤ 5,000
0 < learning_rate ≤ 5
-2 ≤ each starting weight and bias ≤ 2
Ví dụ
Đầu vào:
epochs = 0
output_bias = 0.0
hidden_biases = [0.0, 0.3]
learning_rate = 1.5
hidden_weights = [
[0.2, 0.8],
[0.9, -0.7]
]
output_weights = [-0.6, 0.4]
Kết quả mong đợi:
expected = [0.482451, 0.436972, 0.494378, 0.452729]Đầu vào:
long long epochs = 0;
double output_bias = 0.0;
std::vector<double> hidden_biases = {0.0, 0.3};
double learning_rate = 1.5;
std::vector<std::vector<double>> hidden_weights = {
{0.2, 0.8},
{0.9, -0.7}
};
std::vector<double> output_weights = {-0.6, 0.4};
Kết quả mong đợi:
std::vector<double> expected = {0.482451, 0.436972, 0.494378, 0.452729};Đầu vào:
long epochs = 0;
double output_bias = 0.0;
double[] hidden_biases = new double[] {0.0, 0.3};
double learning_rate = 1.5;
double[][] hidden_weights = new double[][] {
new double[] {0.2, 0.8},
new double[] {0.9, -0.7}
};
double[] output_weights = new double[] {-0.6, 0.4};
Kết quả mong đợi:
double[] expected = new double[] {0.482451, 0.436972, 0.494378, 0.452729};Giải thích
With 0 epochs nothing is trained, so the result is just the forward pass with the starting weights: about [0.482, 0.437, 0.494, 0.453]. The untrained network gives every input nearly the same answer.
Đầu vào:
epochs = 2000
output_bias = 0.0
hidden_biases = [0.0, 0.3]
learning_rate = 1.5
hidden_weights = [
[0.2, 0.8],
[0.9, -0.7]
]
output_weights = [-0.6, 0.4]
Kết quả mong đợi:
expected = [0.015727, 0.982119, 0.985515, 0.013682]Đầu vào:
long long epochs = 2000;
double output_bias = 0.0;
std::vector<double> hidden_biases = {0.0, 0.3};
double learning_rate = 1.5;
std::vector<std::vector<double>> hidden_weights = {
{0.2, 0.8},
{0.9, -0.7}
};
std::vector<double> output_weights = {-0.6, 0.4};
Kết quả mong đợi:
std::vector<double> expected = {0.015727, 0.982119, 0.985515, 0.013682};Đầu vào:
long epochs = 2000;
double output_bias = 0.0;
double[] hidden_biases = new double[] {0.0, 0.3};
double learning_rate = 1.5;
double[][] hidden_weights = new double[][] {
new double[] {0.2, 0.8},
new double[] {0.9, -0.7}
};
double[] output_weights = new double[] {-0.6, 0.4};
Kết quả mong đợi:
double[] expected = new double[] {0.015727, 0.982119, 0.985515, 0.013682};Giải thích
After 2,000 epochs the predictions are about [0.016, 0.982, 0.986, 0.014]: low for [0, 0] and [1, 1], high for [0, 1] and [1, 0]. The network has learned XOR.
Hướng tiếp cận
Hãy tự thử trước, sau đó mở từng gợi ý một.
Gợi ý 1
Process the four XOR samples in the same fixed order for every epoch, and update parameters immediately after each one.
Gợi ý 2
Use forward propagation, then backpropagate the output error into both hidden neurons with the current weights before applying updates.
Gợi ý 3
Watch for the sigmoid derivative and the fact that hidden-layer updates depend on the input values, so zero inputs leave some weights unchanged.
Độ phức tạp mong đợi
O(epochs) time, O(1) extra space
Tham khảo
Thử thách này đã được điều chỉnh cho CodeUtility. Hãy làm theo đề bài và các ràng buộc ở trên.