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Quadratic Beta Schedule

EasyDiffusion Models 📚 Diffusion Models 1

Problem Statement

Besides the linear schedule, early DDPM code offered a quadratic beta schedule: the square roots of the betas are linearly spaced. Build it.

Background

A quadratic schedule places sqrt(beta_t) on a linear grid from sqrt(beta_start) to sqrt(beta_end) across T steps, then squares:

$\beta_t = \left(\sqrt{\beta_{\text{start}}} + \frac{t}{T-1}\big(\sqrt{\beta_{\text{end}}} - \sqrt{\beta_{\text{start}}}\big)\right)^2, \quad t = 0, \dots, T-1$

This keeps the earliest betas smaller than a linear schedule, adding noise more gently at first.

Your Task

Implement:

def quadratic_beta_schedule(T, beta_start, beta_end):

Return a list of T betas rounded to 6 decimals.

Input Format

T (int): number of steps, T >= 2.
beta_start, beta_end (float): endpoints, 0 < beta_start < beta_end < 1.

Output Format

A list of T floats rounded to 6 decimals.

Sample

print(quadratic_beta_schedule(3, 0.0001, 0.04))

Output:

[0.0001, 0.011025, 0.04]

Example:

Input:

print(quadratic_beta_schedule(3, 0.0001, 0.04))

Output:

[0.0001, 0.011025, 0.04]

Reasoning:

Compute the square roots of the endpoints to establish the linear range for the schedule: $\sqrt{0.0001} = 0.01$ and $\sqrt{0.04} = 0.2$ .
Determine the step size for the linear spacing across $T=3$ steps: $\frac{0.2 - 0.01}{3 - 1} = \frac{0.19}{2} = 0.095$ .
Generate the linearly spaced values for the square roots of the betas:
- $t=0$ : $0.01$
- $t=1$ : $0.01 + 0.095 = 0.105$
- $t=2$ : $0.01 + 2(0.095) = 0.2$
Square each value to obtain the actual beta values:
- $\beta_0 = 0.01^2 = 0.0001$
- $\beta_1 = 0.105^2 = 0.011025$
- $\beta_2 = 0.2^2 = 0.04$
The final output is [0.0001, 0.011025, 0.04]

Constraints:

T >= 2, 0 < beta_start < beta_end < 1.
Linearly space the square roots, then square.
Round every beta to 6 decimals.

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1. Background Knowledge

In denoising diffusion probabilistic models (DDPM), the forward process gradually adds Gaussian noise to data over $T$ discrete timesteps. The amount of noise added at step $t$ is controlled by a scalar $\beta_t$ , which must satisfy $0 < \beta_t < 1$ . The choice of how $\beta_t$ varies with $t$ is called the noise schedule, and it significantly affects training stability and sample quality.

The most common schedule is linear, where $\beta_t$ increases uniformly from $\beta_{\text{start}}$ to $\beta_{\text{end}}$ . However, early DDPM implementations also offered a quadratic schedule: instead of spacing the $\beta_t$ values themselves linearly, the square roots of the betas are placed on a linear grid. Concretely, $\sqrt{\beta_t}$ is linearly interpolated between $\sqrt{\beta_{\text{start}}}$ and $\sqrt{\beta_{\text{end}}}$ , and then squared to recover $\beta_t$ . This makes the early betas smaller than in the linear case, so noise is introduced more gently at the beginning of the diffusion process.

Mathematically, for $t = 0, 1, \dots, T-1$ :

\beta_t = \left(\sqrt{\beta_{\text{start}}} + \frac{t}{T-1}\left(\sqrt{\beta_{\text{end}}} - \sqrt{\beta_{\text{start}}}\right)\right)^2

This is a simple linear interpolation in the square-root domain followed by a squaring operation.

2. Algorithm Approach

The problem reduces to a straightforward interpolation task:

Compute the square roots of the two endpoints.
For each index $t$ from $0$ to $T-1$ , compute the linearly interpolated value of $\sqrt{\beta_t}$ using the standard two-point interpolation formula.
Square the result to obtain $\beta_t$ .
Round to 6 decimal places and collect into a list.

No iterative or recursive logic is needed; a single loop over $T$ steps suffices.

3. Step-by-Step Strategy

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solution.py

Python 3.13

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