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Layer Normalization

Implement layer normalization for a single vector.

Layer normalization normalizes across the feature dimension: $\text{LayerNorm}(x) = \frac{x - \mu}{\sqrt{\sigma^2 + \epsilon}} \cdot \gamma + \beta$

where μ and σ² are the mean and variance of x, γ (gain) and β (bias) are learnable parameters.

For this problem, use γ = 1 and β = 0 (no affine transform), and ε = 1e-5.

Input: Space-separated floats (the vector x) Output: Normalized vector, values rounded to 4 decimal places.

Input:

1.0 2.0 3.0

Output:

-1.2247 0.0000 1.2247

Reasoning:

The input vector $x$ is $[1.0, 2.0, 3.0]$ .
We calculate the mean $\mu$ and variance $\sigma^2$ of $x$ : $\mu = \frac{1.0 + 2.0 + 3.0}{3} = 2.0$ and $\sigma^2 = \frac{(1.0-2.0)^2 + (2.0-2.0)^2 + (3.0-2.0)^2}{3} = \frac{2}{3}$ .
Then, we apply the layer normalization formula: $\text{LayerNorm}(x) = \frac{x - \mu}{\sqrt{\sigma^2 + \epsilon}} \cdot \gamma + \beta = \frac{x - 2.0}{\sqrt{\frac{2}{3} + 1e-5}} \cdot 1 + 0$ .
The final output is calculated by applying the formula to each element of $x$ : $\left[\frac{1.0-2.0}{\sqrt{\frac{2}{3} + 1e-5}}, \frac{2.0-2.0}{\sqrt{\frac{2}{3} + 1e-5}}, \frac{3.0-2.0}{\sqrt{\frac{2}{3} + 1e-5}}\right]$ = $[-1.2247, 0.0000, 1.2247]$ .

Editor

Python 3.13.1

0/0

Run code to see test results.

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