समस्या
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.
सीमाएँ
1 ≤ length ≤ 1,000
-50 ≤ each value ≤ 50
उदाहरण
इनपुट:
name = "sigmoid"
values = [0.0, 2.0]
अपेक्षित आउटपुट:
expected = [0.25, 0.104994]इनपुट:
std::string name = "sigmoid";
std::vector<double> values = {0.0, 2.0};
अपेक्षित आउटपुट:
std::vector<double> expected = {0.25, 0.104994};इनपुट:
string name = "sigmoid";
double[] values = new double[] {0.0, 2.0};
अपेक्षित आउटपुट:
double[] expected = new double[] {0.25, 0.104994};व्याख्या
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.
इनपुट:
name = "relu"
values = [-3.0, 0.0, 4.5]
अपेक्षित आउटपुट:
expected = [0.0, 0.0, 1.0]इनपुट:
std::string name = "relu";
std::vector<double> values = {-3.0, 0.0, 4.5};
अपेक्षित आउटपुट:
std::vector<double> expected = {0.0, 0.0, 1.0};इनपुट:
string name = "relu";
double[] values = new double[] {-3.0, 0.0, 4.5};
अपेक्षित आउटपुट:
double[] expected = new double[] {0.0, 0.0, 1.0};व्याख्या
ReLU is flat for x ≤ 0 (slope 0) and has slope 1 for x > 0, so the result is [0, 0, 1].
कैसे हल करें
पहले खुद कोशिश करें, फिर एक-एक करके संकेत खोलें।
संकेत 1
The derivative depends on the activation type, so handle each function separately for every input value.
संकेत 2
For sigmoid and tanh, reuse the activation output or its standard function value instead of recomputing from scratch in a complicated way.
संकेत 3
Watch the sign thresholds: ReLU uses a strict positive test, while leaky ReLU keeps a small slope for nonpositive values.
अपेक्षित जटिलता
O(n) time, O(n) extra space
संदर्भ
यह चुनौती CodeUtility के लिए अनुकूलित की गई है। ऊपर दिए गए स्टेटमेंट और सीमाओं का पालन करें।