問題
Backpropagation multiplies gradients by the derivative of each activation function. Implement activation_derivative(name, values) and return the derivative at every value:
| name | derivative |
|---|---|
"sigmoid" | s × (1 - s), where s = 1 / (1 + e^(-x)) |
"relu" | 1 if x > 0, otherwise 0 |
"leaky_relu" | 1 if x > 0, otherwise 0.01 |
"tanh" | 1 - tanh(x)² |
Input
name: One of "sigmoid", "relu", "leaky_relu" or "tanh".
values: The inputs x of the activation.
Output
Return the derivative at each value, in the same order.
制約
1 ≤ length ≤ 1,000
-50 ≤ each value ≤ 50
例
入力：
name = "sigmoid"
values = [0.0, 2.0]
期待される出力：
expected = [0.25, 0.104994]入力：
std::string name = "sigmoid";
std::vector<double> values = {0.0, 2.0};
期待される出力：
std::vector<double> expected = {0.25, 0.104994};入力：
string name = "sigmoid";
double[] values = new double[] {0.0, 2.0};
期待される出力：
double[] expected = new double[] {0.25, 0.104994};解説
At 0, s = 0.5 and the slope is 0.5 × 0.5 = 0.25, the steepest point of the sigmoid. At 2, s ≈ 0.880797 and the slope is about 0.104994.
入力：
name = "relu"
values = [-3.0, 0.0, 4.5]
期待される出力：
expected = [0.0, 0.0, 1.0]入力：
std::string name = "relu";
std::vector<double> values = {-3.0, 0.0, 4.5};
期待される出力：
std::vector<double> expected = {0.0, 0.0, 1.0};入力：
string name = "relu";
double[] values = new double[] {-3.0, 0.0, 4.5};
期待される出力：
double[] expected = new double[] {0.0, 0.0, 1.0};解説
ReLU is flat for x ≤ 0 (slope 0) and has slope 1 for x > 0, so the result is [0, 0, 1].
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
The derivative depends on the activation type, so handle each function separately for every input value.
ヒント 2
For sigmoid and tanh, reuse the activation output or its standard function value instead of recomputing from scratch in a complicated way.
ヒント 3
Watch the sign thresholds: ReLU uses a strict positive test, while leaky ReLU keeps a small slope for nonpositive values.
目安の計算量
O(n) time, O(n) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。