Quadratic Beta Schedule

easyPROForward Process and Noise SchedulesDiffusion ModelsDiffusion 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.

More from Diffusion Models 1

Classifier-Free Guidance: Uncond/Cond Mixingeasy DDPM Posterior Mean and Variance from x0medium Adaptive CFG Scheduling over the Trajectoryhard Classifier Gradient Guidance Stepmedium Negative Prompt as the Unconditional Branchmedium

Concept primerPREVIEW

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

PRO

Continue the full explanation

You're reading the free preview. Unlock the complete walkthrough, the code editor, test runner and reference solution with Premium.

PRO

Editor locked

The code editor is locked for Pro problems. It is only available for free problems. Please upgrade to gain access to the code editor for all problems.

solution.py

Python 3.13

Test results