Problem
An artificial neuron multiplies each input by a weight, adds a bias, and passes the result through an activation function:
z = x₁×w₁ + x₂×w₂ + … + xₙ×wₙ + bias
output = sigmoid(z) = 1 / (1 + e^(-z))
Implement neuron_output(inputs, weights, bias).
Input
inputs: The input values.
weights: One weight per input (same length as inputs).
bias: The neuron's bias.
Output
Return the neuron's output, a decimal number between 0 and 1.
Constraints
1 ≤ length of inputs = length of weights ≤ 100
-10 ≤ each input, weight and bias ≤ 10
Examples
Input:
bias = 0.5
inputs = [1.0, 2.0, 3.0]
weights = [0.2, 0.4, 0.6]
Expected output:
expected = 0.964429
Input:
double bias = 0.5;
std::vector<double> inputs = {1.0, 2.0, 3.0};
std::vector<double> weights = {0.2, 0.4, 0.6};
Expected output:
double expected = 0.964429;
Input:
double bias = 0.5;
double[] inputs = new double[] {1.0, 2.0, 3.0};
double[] weights = new double[] {0.2, 0.4, 0.6};
Expected output:
double expected = 0.964429;
Explanation
z = 1×0.2 + 2×0.4 + 3×0.6 + 0.5 = 3.3, and sigmoid(3.3) ≈ 0.964429.
Input:
bias = 0.0
inputs = [0.5, -1.0]
weights = [2.0, 1.0]
Expected output:
expected = 0.5
Input:
double bias = 0.0;
std::vector<double> inputs = {0.5, -1.0};
std::vector<double> weights = {2.0, 1.0};
Expected output:
double expected = 0.5;
Input:
double bias = 0.0;
double[] inputs = new double[] {0.5, -1.0};
double[] weights = new double[] {2.0, 1.0};
Expected output:
double expected = 0.5;
Explanation
z = 0.5×2 + (-1)×1 + 0 = 0, and sigmoid(0) = 0.5: the neuron is exactly undecided.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
This is just a weighted sum of the inputs plus a bias, followed by one sigmoid.
Hint 2
Pair each input with its corresponding weight and accumulate the products before adding the bias.
Hint 3
Watch that the final output is passed through sigmoid, not returned as the raw sum.
Expected complexity
O(n) time, O(1) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.