Problema
A dense layer is a group of neurons that all receive the same inputs. Neuron j has its own weights weights[j] and bias biases[j]:
z_j = inputs · weights[j] + biases[j]
Then the layer's activation is applied to each z_j:
"linear": output z_j unchanged"relu": max(0, z_j)"sigmoid": 1 / (1 + e^(-z_j))
Implement dense_forward(inputs, weights, biases, activation).
Input
inputs: The n input values.
weights: One row per neuron; each row has n weights.
biases: One bias per neuron.
activation: "linear", "relu" or "sigmoid".
Output
Return the list of outputs, one per neuron, in order.
Restrições
1 ≤ n ≤ 100, 1 ≤ number of neurons ≤ 100
-10 ≤ each value ≤ 10
Exemplos
Entrada:
biases = [0.0, -1.0]
inputs = [1.0, 2.0]
weights = [
[0.5, -1.0],
[1.5, 0.5]
]
activation = "relu"
Saída esperada:
expected = [0.0, 1.5]Entrada:
std::vector<double> biases = {0.0, -1.0};
std::vector<double> inputs = {1.0, 2.0};
std::vector<std::vector<double>> weights = {
{0.5, -1.0},
{1.5, 0.5}
};
std::string activation = "relu";
Saída esperada:
std::vector<double> expected = {0.0, 1.5};Entrada:
double[] biases = new double[] {0.0, -1.0};
double[] inputs = new double[] {1.0, 2.0};
double[][] weights = new double[][] {
new double[] {0.5, -1.0},
new double[] {1.5, 0.5}
};
string activation = "relu";
Saída esperada:
double[] expected = new double[] {0.0, 1.5};Explicação
Neuron 1: 1×0.5 + 2×(-1) + 0 = -1.5, and ReLU gives 0. Neuron 2: 1×1.5 + 2×0.5 - 1 = 1.5, which ReLU keeps. Result: [0, 1.5].
Entrada:
biases = [1.0, 0.0, -1.5]
inputs = [3.0]
weights = [
[2.0],
[-1.0],
[0.5]
]
activation = "linear"
Saída esperada:
expected = [7.0, -3.0, 0.0]Entrada:
std::vector<double> biases = {1.0, 0.0, -1.5};
std::vector<double> inputs = {3.0};
std::vector<std::vector<double>> weights = {
{2.0},
{-1.0},
{0.5}
};
std::string activation = "linear";
Saída esperada:
std::vector<double> expected = {7.0, -3.0, 0.0};Entrada:
double[] biases = new double[] {1.0, 0.0, -1.5};
double[] inputs = new double[] {3.0};
double[][] weights = new double[][] {
new double[] {2.0},
new double[] {-1.0},
new double[] {0.5}
};
string activation = "linear";
Saída esperada:
double[] expected = new double[] {7.0, -3.0, 0.0};Explicação
With a linear activation the outputs are just 3×2 + 1 = 7, 3×(-1) + 0 = -3 and 3×0.5 - 1.5 = 0.
Como abordar
Tente sozinho primeiro e depois abra uma dica de cada vez.
Dica 1
Treat each neuron separately: each one has its own weight row and bias.
Dica 2
Compute every neuron's dot product with the input vector, add its bias, then apply the chosen activation.
Dica 3
Make sure the activation choice is applied after the linear step, and preserve neuron order in the output.
Complexidade esperada
O(mn) time, O(m) extra space
Referências
Este desafio foi adaptado para o CodeUtility. Siga o enunciado e as restrições acima.