Problem
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.
Einschränkungen
1 ≤ n ≤ 100, 1 ≤ number of neurons ≤ 100
-10 ≤ each value ≤ 10
Beispiele
Eingabe:
biases = [0.0, -1.0]
inputs = [1.0, 2.0]
weights = [
[0.5, -1.0],
[1.5, 0.5]
]
activation = "relu"
Erwartete Ausgabe:
expected = [0.0, 1.5]Eingabe:
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";
Erwartete Ausgabe:
std::vector<double> expected = {0.0, 1.5};Eingabe:
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";
Erwartete Ausgabe:
double[] expected = new double[] {0.0, 1.5};Erläuterung
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].
Eingabe:
biases = [1.0, 0.0, -1.5]
inputs = [3.0]
weights = [
[2.0],
[-1.0],
[0.5]
]
activation = "linear"
Erwartete Ausgabe:
expected = [7.0, -3.0, 0.0]Eingabe:
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";
Erwartete Ausgabe:
std::vector<double> expected = {7.0, -3.0, 0.0};Eingabe:
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";
Erwartete Ausgabe:
double[] expected = new double[] {7.0, -3.0, 0.0};Erläuterung
With a linear activation the outputs are just 3×2 + 1 = 7, 3×(-1) + 0 = -3 and 3×0.5 - 1.5 = 0.
So gehst du vor
Versuch es zuerst allein und öffne dann einen Hinweis nach dem anderen.
Hinweis 1
Treat each neuron separately: each one has its own weight row and bias.
Hinweis 2
Compute every neuron's dot product with the input vector, add its bias, then apply the chosen activation.
Hinweis 3
Make sure the activation choice is applied after the linear step, and preserve neuron order in the output.
Erwartete Komplexität
O(mn) time, O(m) extra space
Referenzen
Diese Aufgabe wurde für CodeUtility angepasst. Beachte die oben genannten Vorgaben und Einschränkungen.