Условие
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.
Ограничения
1 ≤ n ≤ 100, 1 ≤ number of neurons ≤ 100
-10 ≤ each value ≤ 10
Примеры
Ввод:
biases = [0.0, -1.0]
inputs = [1.0, 2.0]
weights = [
[0.5, -1.0],
[1.5, 0.5]
]
activation = "relu"
Ожидаемый вывод:
expected = [0.0, 1.5]Ввод:
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";
Ожидаемый вывод:
std::vector<double> expected = {0.0, 1.5};Ввод:
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";
Ожидаемый вывод:
double[] expected = new double[] {0.0, 1.5};Пояснение
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].
Ввод:
biases = [1.0, 0.0, -1.5]
inputs = [3.0]
weights = [
[2.0],
[-1.0],
[0.5]
]
activation = "linear"
Ожидаемый вывод:
expected = [7.0, -3.0, 0.0]Ввод:
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";
Ожидаемый вывод:
std::vector<double> expected = {7.0, -3.0, 0.0};Ввод:
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";
Ожидаемый вывод:
double[] expected = new double[] {7.0, -3.0, 0.0};Пояснение
With a linear activation the outputs are just 3×2 + 1 = 7, 3×(-1) + 0 = -3 and 3×0.5 - 1.5 = 0.
Как подступиться
Сначала попробуйте сами, затем открывайте подсказки по одной.
Подсказка 1
Treat each neuron separately: each one has its own weight row and bias.
Подсказка 2
Compute every neuron's dot product with the input vector, add its bias, then apply the chosen activation.
Подсказка 3
Make sure the activation choice is applied after the linear step, and preserve neuron order in the output.
Ожидаемая сложность
O(mn) time, O(m) extra space
Ссылки
Это задание адаптировано для CodeUtility. Следуйте условию и ограничениям выше.