Problem
Backpropagation applies the chain rule layer by layer, from the output back to the inputs. The network has n inputs, h hidden neurons and one output, all with sigmoid activation (σ(z) = 1 / (1 + e^(-z))):
- hidden: a_j = σ(inputs · hidden_weights[j] + hidden_biases[j])
- output: y = σ(a · output_weights + output_bias)
- loss: L = (y - target)²
Gradients:
- delta_out = 2 × (y - target) × y × (1 - y)
- Output layer: dL/doutput_weights[j] = delta_out × a_j and dL/doutput_bias = delta_out
- delta_j = delta_out × output_weights[j] × a_j × (1 - a_j)
- Hidden layer: dL/dhidden_weights[j][i] = delta_j × inputs[i] and dL/dhidden_biases[j] = delta_j
Implement backprop_gradients(inputs, hidden_weights, hidden_biases, output_weights, output_bias, target).
Output
Return a matrix with h + 1 rows:
- Row j (for each hidden neuron j, in order): the n gradients of
hidden_weights[j], then the gradient ofhidden_biases[j]. - Last row: the h gradients of
output_weights, then the gradient ofoutput_bias.
Constraints
1 ≤ n, h ≤ 20
-5 ≤ each input, weight and bias ≤ 5
0 ≤ target ≤ 1
Examples
Input:
inputs = [1.0]
target = 1.0
output_bias = 0.0
hidden_biases = [0.0]
hidden_weights = [
[0.0]
]
output_weights = [0.0]
Expected output:
expected = [
[0.0, 0.0],
[-0.125, -0.25]
]Input:
std::vector<double> inputs = {1.0};
double target = 1.0;
double output_bias = 0.0;
std::vector<double> hidden_biases = {0.0};
std::vector<std::vector<double>> hidden_weights = {
{0.0}
};
std::vector<double> output_weights = {0.0};
Expected output:
std::vector<std::vector<double>> expected = {
{0.0, 0.0},
{-0.125, -0.25}
};Input:
double[] inputs = new double[] {1.0};
double target = 1.0;
double output_bias = 0.0;
double[] hidden_biases = new double[] {0.0};
double[][] hidden_weights = new double[][] {
new double[] {0.0}
};
double[] output_weights = new double[] {0.0};
Expected output:
double[][] expected = new double[][] {
new double[] {0.0, 0.0},
new double[] {-0.125, -0.25}
};Explanation
a = σ(0) = 0.5 and y = σ(0) = 0.5. delta_out = 2 × (0.5 - 1) × 0.25 = -0.25. Output row: [-0.25 × 0.5, -0.25] = [-0.125, -0.25]. delta_1 = -0.25 × 0 × 0.25 = 0, so the hidden row is [0, 0]: a zero output weight blocks the gradient.
Input:
inputs = [1.0, 0.0]
target = 0.0
output_bias = -1.0
hidden_biases = [0.0, 0.0]
hidden_weights = [
[0.0, 1.0],
[0.0, -1.0]
]
output_weights = [1.0, 1.0]
Expected output:
expected = [
[0.0625, 0.0, 0.0625],
[0.0625, 0.0, 0.0625],
[0.125, 0.125, 0.25]
]Input:
std::vector<double> inputs = {1.0, 0.0};
double target = 0.0;
double output_bias = -1.0;
std::vector<double> hidden_biases = {0.0, 0.0};
std::vector<std::vector<double>> hidden_weights = {
{0.0, 1.0},
{0.0, -1.0}
};
std::vector<double> output_weights = {1.0, 1.0};
Expected output:
std::vector<std::vector<double>> expected = {
{0.0625, 0.0, 0.0625},
{0.0625, 0.0, 0.0625},
{0.125, 0.125, 0.25}
};Input:
double[] inputs = new double[] {1.0, 0.0};
double target = 0.0;
double output_bias = -1.0;
double[] hidden_biases = new double[] {0.0, 0.0};
double[][] hidden_weights = new double[][] {
new double[] {0.0, 1.0},
new double[] {0.0, -1.0}
};
double[] output_weights = new double[] {1.0, 1.0};
Expected output:
double[][] expected = new double[][] {
new double[] {0.0625, 0.0, 0.0625},
new double[] {0.0625, 0.0, 0.0625},
new double[] {0.125, 0.125, 0.25}
};Explanation
Both hidden values are σ(0) = 0.5, so y = σ(0.5 + 0.5 - 1) = 0.5 and delta_out = 2 × 0.5 × 0.25 = 0.25. Output row: [0.125, 0.125, 0.25]. Each delta_j = 0.25 × 1 × 0.25 = 0.0625, giving hidden rows [0.0625, 0, 0.0625] (the second input is 0).
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Work backward from the output neuron first, then reuse that signal to compute each hidden neuron's contribution.
Hint 2
You need one delta for the output layer and one delta per hidden neuron, both based on sigmoid derivatives.
Hint 3
Be careful to use each hidden neuron's own output weight when propagating error back to that neuron.
Expected complexity
O(hn) time, O(hn) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.