2D Gaussian Splat Evaluation

mediumPRONeural Rendering3DGSCV: Image-Based Rendering

Compute the 2D Gaussian splat value at a pixel.

3D Gaussian Splatting represents scenes as millions of 3D Gaussians. When projected to 2D, each becomes an elliptical Gaussian:

$G(x, y) = \exp\left(-\frac{(x-\mu_x)^2}{2\sigma_x^2} - \frac{(y-\mu_y)^2}{2\sigma_y^2}\right)$

where:

$(x, y)$ is the pixel position
$(\mu_x, \mu_y)$ is the Gaussian center
$(\sigma_x, \sigma_y)$ are the standard deviations (scale)

This axis-aligned version is simplified - real 3DGS uses full 2D covariance for rotated ellipses.

Example

Input

gaussian_splat((0, 0), (0, 0), (1, 1))

Output

1.0

Reasoning

Inputs: pixel is the query point $(x, y) = (0, 0)$ , center is the splat's mean $(\mu_x, \mu_y) = (0, 0)$ , and sigma gives the spread along each axis, $(\sigma_x, \sigma_y) = (1, 1)$ , in pixel units.
Offsets: the pixel sits exactly on the center, so $x - \mu_x = 0$ and $y - \mu_y = 0$ .
Exponent terms: each axis contributes its squared offset scaled by twice its variance: $\frac{0^2}{2 \cdot 1^2} = 0$ for $x$ and $\frac{0^2}{2 \cdot 1^2} = 0$ for $y$ . Their sum is $0$ .
Value: $G = e^{-0} = 1$ , which rounded to 4 decimal places is the output 1.0.
Notice that $1$ is the largest value a splat can take: any offset makes the exponent negative and the value falls toward $0$ . One standard deviation away along $x$ , e.g. pixel $(1, 0)$ , gives $e^{-1/2} \approx 0.6065$ , and a larger $\sigma_x$ would make that falloff slower.

Constraints

pixel: (x, y) query position
center: (μx, μy) Gaussian center
sigma: (σx, σy) standard deviations
Return Gaussian value, rounded to 4 decimal places

Learn this in

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Concept primerPREVIEW

To compute the 2D Gaussian splat value at a pixel, you are just evaluating the Gaussian formula at a given $(x,y)$ using the provided mean and standard deviations. The core of the task is plugging numbers into the exponent and applying exp in a numerically safe way.

1. Background Knowledge

A 2D Gaussian (or 2D normal) is a smooth bell-shaped function over the plane. In this simplified, axis-aligned form, the value at pixel $(x,y)$ is:

G(x, y) = \exp\left( -\frac{(x-\mu_x)^2}{2\sigma_x^2} - \frac{(y-\mu_y)^2}{2\sigma_y^2} \right)

Here:

$($ \mu $_x,$ \mu $_y)$ is the center of the splat in image coordinates.
$\sigma_x,$ \sigma $_y$ control the spread (scale) horizontally and vertically.
The value is in $(0, 1]$ : it is 1 at the center and decays smoothly as you move away.

In neural rendering / Gaussian splatting, millions of such Gaussians are projected into the image plane; each one contributes a soft footprint (splat) to nearby pixels. Rendering is then done by accumulating these contributions (often with blending or alpha compositing), but this problem isolates just the evaluation of one splat at one pixel.

2. Algorithm / Approach

This is a direct formula evaluation problem:

Compute horizontal and vertical offsets from the Gaussian center.
Normalize by the standard deviations $\sigma_x,$ \sigma $_y$ .
Build the exponent term (sum of squared normalized distances, with $-\frac{1}{2}$ factor).
Apply the exponential function.

This matches a common pattern: evaluate a parametric function at a given input. No loops or advanced data structures are needed.

3. Step-by-Step Strategy

Assume you are given:

x, y (pixel coordinates)
mu_x, mu_y (Gaussian center)
sigma_x, sigma_y (standard deviations; strictly positive)

Steps:

Compute offsets:

dx = x - mu_x
dy = y - mu_y

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solution.py

Python 3.13

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