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Forward Propagation in a Two-Layer Network

Run inputs through a hidden ReLU layer and a sigmoid output layer.

Ce que vous allez apprendre

  • Chain a hidden layer and an output layer
  • Feed one layer's output into the next
  • Run a full network prediction

Problème

Afficher la traduction

A small neural network has a hidden layer and an output layer. Data flows forward through them:

  1. Hidden layer: h_j = ReLU(inputs · hidden_weights[j] + hidden_biases[j]), where ReLU(z) = max(0, z)
  2. Output layer: y_k = sigmoid(h · output_weights[k] + output_biases[k]), where sigmoid(z) = 1 / (1 + e^(-z))

Implement forward_pass(inputs, hidden_weights, hidden_biases, output_weights, output_biases).

Input

inputs: The n input values.

hidden_weights, hidden_biases: The hidden layer, one row and one bias per hidden neuron.

output_weights, output_biases: The output layer, one row and one bias per output; each row has one weight per hidden neuron.

Output

Return the list of network outputs y.

Contraintes

1 ≤ inputs, hidden neurons, outputs ≤ 50

-10 ≤ each value ≤ 10

Exemples

Exemple 1
Entrée :
inputs = [1.0, 0.5]
hidden_biases = [0.0, 0.5]
output_biases = [0.0]
hidden_weights = [
    [1.0, -2.0],
    [0.5, 1.0]
]
output_weights = [
    [1.0, -1.0]
]

Sortie attendue :
expected = [0.182426]
Entrée :
std::vector<double> inputs = {1.0, 0.5};
std::vector<double> hidden_biases = {0.0, 0.5};
std::vector<double> output_biases = {0.0};
std::vector<std::vector<double>> hidden_weights = {
    {1.0, -2.0},
    {0.5, 1.0}
};
std::vector<std::vector<double>> output_weights = {
    {1.0, -1.0}
};

Sortie attendue :
std::vector<double> expected = {0.182426};
Entrée :
double[] inputs = new double[] {1.0, 0.5};
double[] hidden_biases = new double[] {0.0, 0.5};
double[] output_biases = new double[] {0.0};
double[][] hidden_weights = new double[][] {
    new double[] {1.0, -2.0},
    new double[] {0.5, 1.0}
};
double[][] output_weights = new double[][] {
    new double[] {1.0, -1.0}
};

Sortie attendue :
double[] expected = new double[] {0.182426};

Explication

Hidden: ReLU(1×1 + 0.5×(-2) + 0) = ReLU(0) = 0 and ReLU(1×0.5 + 0.5×1 + 0.5) = 1.5. Output: sigmoid(0×1 + 1.5×(-1) + 0) = sigmoid(-1.5) ≈ 0.182426.

Exemple 2
Entrée :
inputs = [2.0]
hidden_biases = [0.0, 0.0]
output_biases = [-1.0, 0.0]
hidden_weights = [
    [1.0],
    [-1.0]
]
output_weights = [
    [0.5, 0.5],
    [-0.5, 2.0]
]

Sortie attendue :
expected = [0.5, 0.268941]
Entrée :
std::vector<double> inputs = {2.0};
std::vector<double> hidden_biases = {0.0, 0.0};
std::vector<double> output_biases = {-1.0, 0.0};
std::vector<std::vector<double>> hidden_weights = {
    {1.0},
    {-1.0}
};
std::vector<std::vector<double>> output_weights = {
    {0.5, 0.5},
    {-0.5, 2.0}
};

Sortie attendue :
std::vector<double> expected = {0.5, 0.268941};
Entrée :
double[] inputs = new double[] {2.0};
double[] hidden_biases = new double[] {0.0, 0.0};
double[] output_biases = new double[] {-1.0, 0.0};
double[][] hidden_weights = new double[][] {
    new double[] {1.0},
    new double[] {-1.0}
};
double[][] output_weights = new double[][] {
    new double[] {0.5, 0.5},
    new double[] {-0.5, 2.0}
};

Sortie attendue :
double[] expected = new double[] {0.5, 0.268941};

Explication

Hidden: ReLU(2) = 2 and ReLU(-2) = 0. Output 1: sigmoid(2×0.5 + 0 - 1) = sigmoid(0) = 0.5. Output 2: sigmoid(2×(-0.5) + 0) = sigmoid(-1) ≈ 0.268941.

Comment l'aborder

Essayez d'abord seul, puis ouvrez les indices un par un.

Indice 1

First compute all hidden activations from the inputs, then use those hidden outputs for the final layer.

Indice 2

Each layer is a batch of dot products plus biases, with ReLU in the hidden layer and sigmoid at the output.

Indice 3

Do not mix original inputs into the second layer; it only sees the hidden layer outputs.

Complexité attendue

O(hn + oh) time, O(h + o) extra space

Références

Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.

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