समस्या
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.
सीमाएँ
1 ≤ n ≤ 100, 1 ≤ number of neurons ≤ 100
-10 ≤ each value ≤ 10
उदाहरण
इनपुट:
biases = [0.0, -1.0]
inputs = [1.0, 2.0]
weights = [
[0.5, -1.0],
[1.5, 0.5]
]
activation = "relu"
अपेक्षित आउटपुट:
expected = [0.0, 1.5]इनपुट:
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";
अपेक्षित आउटपुट:
std::vector<double> expected = {0.0, 1.5};इनपुट:
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";
अपेक्षित आउटपुट:
double[] expected = new double[] {0.0, 1.5};व्याख्या
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].
इनपुट:
biases = [1.0, 0.0, -1.5]
inputs = [3.0]
weights = [
[2.0],
[-1.0],
[0.5]
]
activation = "linear"
अपेक्षित आउटपुट:
expected = [7.0, -3.0, 0.0]इनपुट:
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";
अपेक्षित आउटपुट:
std::vector<double> expected = {7.0, -3.0, 0.0};इनपुट:
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";
अपेक्षित आउटपुट:
double[] expected = new double[] {7.0, -3.0, 0.0};व्याख्या
With a linear activation the outputs are just 3×2 + 1 = 7, 3×(-1) + 0 = -3 and 3×0.5 - 1.5 = 0.
कैसे हल करें
पहले खुद कोशिश करें, फिर एक-एक करके संकेत खोलें।
संकेत 1
Treat each neuron separately: each one has its own weight row and bias.
संकेत 2
Compute every neuron's dot product with the input vector, add its bias, then apply the chosen activation.
संकेत 3
Make sure the activation choice is applied after the linear step, and preserve neuron order in the output.
अपेक्षित जटिलता
O(mn) time, O(m) extra space
संदर्भ
यह चुनौती CodeUtility के लिए अनुकूलित की गई है। ऊपर दिए गए स्टेटमेंट और सीमाओं का पालन करें।