โจทย์
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.
เงื่อนไข
1 ≤ n ≤ 100, 1 ≤ number of neurons ≤ 100
-10 ≤ each value ≤ 10
ตัวอย่าง
อินพุต:
biases = [0.0, -1.0]
inputs = [1.0, 2.0]
weights = [
[0.5, -1.0],
[1.5, 0.5]
]
activation = "relu"
ผลลัพธ์ที่คาดหวัง:
expected = [0.0, 1.5]อินพุต:
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";
ผลลัพธ์ที่คาดหวัง:
std::vector<double> expected = {0.0, 1.5};อินพุต:
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";
ผลลัพธ์ที่คาดหวัง:
double[] expected = new double[] {0.0, 1.5};คำอธิบาย
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].
อินพุต:
biases = [1.0, 0.0, -1.5]
inputs = [3.0]
weights = [
[2.0],
[-1.0],
[0.5]
]
activation = "linear"
ผลลัพธ์ที่คาดหวัง:
expected = [7.0, -3.0, 0.0]อินพุต:
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";
ผลลัพธ์ที่คาดหวัง:
std::vector<double> expected = {7.0, -3.0, 0.0};อินพุต:
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";
ผลลัพธ์ที่คาดหวัง:
double[] expected = new double[] {7.0, -3.0, 0.0};คำอธิบาย
With a linear activation the outputs are just 3×2 + 1 = 7, 3×(-1) + 0 = -3 and 3×0.5 - 1.5 = 0.
แนวทางการแก้
ลองทำเองก่อน แล้วค่อยเปิดคำใบ้ทีละข้อ
คำใบ้ 1
Treat each neuron separately: each one has its own weight row and bias.
คำใบ้ 2
Compute every neuron's dot product with the input vector, add its bias, then apply the chosen activation.
คำใบ้ 3
Make sure the activation choice is applied after the linear step, and preserve neuron order in the output.
ความซับซ้อนที่คาดหวัง
O(mn) time, O(m) extra space
แหล่งอ้างอิง
โจทย์นี้ปรับให้เหมาะกับ CodeUtility กรุณาทำตามคำชี้แจงและข้อจำกัดด้านบน