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Linear Beta Schedule and Alpha-Bar

EasyDiffusion Models 📚 Diffusion Models 1

Problem Statement

Build the linear noise schedule used by the original DDPM paper and derive the two quantities every later formula depends on: the per-step signal retention alpha_t and the cumulative signal retention alpha_bar_t.

Background

The forward diffusion process corrupts data one small step at a time:

$q(x_t \mid x_{t-1}) = \mathcal{N}\!\left(x_t;\ \sqrt{1-\beta_t}\,x_{t-1},\ \beta_t I\right)$

The variances $\beta_1, \dots, \beta_T$ are the noise schedule. DDPM uses a linear schedule: $T$ values spaced evenly (endpoints included) from beta_start to beta_end, with the defaults $\beta_1 = 10^{-4}$ and $\beta_T = 0.02$ for $T = 1000$ .

From the betas we define

$\alpha_t = 1 - \beta_t, \qquad \bar{\alpha}_t = \prod_{s=1}^{t} \alpha_s$

$\bar{\alpha}_t$ is the running product, not a running sum. It is what makes the forward process collapse into a single closed-form Gaussian, so getting it right is the foundation of everything else in diffusion.

Your Task

Implement:

def linear_schedule(T, beta_start=1e-4, beta_end=0.02):

Return a tuple (betas, alphas, alpha_bars) of three 1-D NumPy arrays, each of length T, using 0-based indexing (element i corresponds to timestep $t = i+1$ in the maths above).

Input Format

T (int): number of diffusion steps, T >= 1.
beta_start, beta_end (float): first and last beta. Both endpoints must appear in the array.

Output Format

A tuple of three NumPy arrays (betas, alphas, alpha_bars).

Sample

b, a, ab = linear_schedule(5, 0.1, 0.5)
print(np.round(b, 4).tolist())
print(np.round(a, 4).tolist())
print(np.round(ab, 4).tolist())

[0.1, 0.2, 0.3, 0.4, 0.5]
[0.9, 0.8, 0.7, 0.6, 0.5]
[0.9, 0.72, 0.504, 0.3024, 0.1512]

Betas are evenly spaced from 0.1 to 0.5, alphas are 1 - beta, and alpha_bars is their cumulative product, so it decays quickly toward zero.

Example:

Input:

b, a, ab = linear_schedule(5, 0.1, 0.5)
print(np.round(b, 4).tolist())
print(np.round(a, 4).tolist())
print(np.round(ab, 4).tolist())

Output:

[0.1, 0.2, 0.3, 0.4, 0.5]
[0.9, 0.8, 0.7, 0.6, 0.5]
[0.9, 0.72, 0.504, 0.3024, 0.1512]

Reasoning:

np.linspace(0.1, 0.5, 5) gives betas [0.1, 0.2, 0.3, 0.4, 0.5], so alphas are [0.9, 0.8, 0.7, 0.6, 0.5]. The cumulative product runs 0.9, 0.9*0.8 = 0.72, 0.72*0.7 = 0.504, 0.504*0.6 = 0.3024, 0.3024*0.5 = 0.1512 -- the signal that survives to each step. Using a cumulative sum instead would give values above 1, which is impossible for a variance-preserving process.

Constraints:

T >= 1; when T == 1 the array holds beta_start only.
betas must include both endpoints (np.linspace, not np.arange).
alpha_bars is a cumulative product, not a cumulative sum.
Use float (float64) arrays; do not round inside the function.

More from Diffusion Models 1

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1. Background Knowledge

Diffusion models are generative models that learn to reverse a gradual noising process. The forward process adds Gaussian noise to data $x_0$ over $T$ steps until it becomes pure noise $x_T$ . This process is defined by a noise schedule, typically a sequence of variances $\beta_1, \dots, \beta_T$ . In the original DDPM paper, a linear schedule is used, where $\beta_t$ increases linearly from a small value $\beta_{\text{start}}$ to a larger value $\beta_{\text{end}}$ .

The quantity $\alpha_t = 1 - \beta_t$ represents the signal retention at step $t$ . It indicates how much of the previous signal $x_{t-1}$ is preserved in $x_t$ . The cumulative signal retention $\bar{\alpha}_t = \prod_{s=1}^{t} \alpha_s$ is crucial because it allows sampling $x_t$ directly from $x_0$ using the reparameterization trick: $x_t = \sqrt{\bar{\alpha}_t} x_0 + \sqrt{1 - \bar{\alpha}_t} \epsilon$ . This closed-form expression relies entirely on the correct computation of $\bar{\alpha}_t$ .

In implementation, we use 0-based indexing for arrays, so index $i$ corresponds to timestep $t = i+1$ . The array betas contains $\beta_1, \dots, \beta_T$ , alphas contains $\alpha_1, \dots, \alpha_T$ , and alpha_bars contains $\bar{\alpha}_1, \dots, \bar{\alpha}_T$ . Understanding the difference between element-wise operations and cumulative operations is key.

2. Algorithm Approach

The problem requires generating three related arrays based on a linear interpolation and cumulative product. The approach involves three main stages:

Generate Betas: Create an array of $T$ values linearly spaced between beta_start and beta_end. This can be done using NumPy's linspace function.
Compute Alphas: Apply the element-wise transformation $\alpha_t = 1 - \beta_t$ to the betas array.
Compute Alpha Bars: Calculate the cumulative product of the alphas array. This means each element at index $i$ is the product of all elements from index $0$ to $i$ .

This approach leverages vectorized operations in NumPy for efficiency and clarity.

3. Step-by-Step Strategy

Import NumPy: Ensure numpy is imported as np.
Create betas: Use np.linspace(beta_start, beta_end, T) to generate the linear schedule. This function returns an array of $T$ evenly spaced values, including both endpoints.
Create alphas: Subtract the betas array from 1 using vectorized subtraction: alphas = 1 - betas.
Create alpha_bars: Use np.cumprod(alphas) to compute the cumulative product. This function returns an array where the $i$ -th element is the product of all elements up to and including index $i$ .
Return Tuple: Return the tuple (betas, alphas, alpha_bars).

Example code structure:

import numpy as np

def linear_schedule(T, beta_start=1e-4, beta_end=0.02):
 betas = np.linspace(beta_start, beta_end, T)
 alphas = 1 - betas
 alpha_bars = np.cumprod(alphas)
 return betas, alphas, alpha_bars

4. Common Pitfalls

Indexing Confusion: The problem uses 0-based indexing for arrays but 1-based indexing for mathematical timesteps. Ensure that betas corresponds to $\beta_1$ , betas to $\beta_2$ , etc. The linspace function naturally aligns with this if you request $T$ points.
Cumulative Sum vs. Product: Do not use np.cumsum for alpha_bars. The definition is a product $\prod_{s=1}^{t} \alpha_s$ , so np.cumprod is required. Using sum will yield incorrect results.
Endpoint Inclusion: Ensure that both beta_start and beta_end are included in the betas array. np.linspace includes endpoints by default, but be cautious if using other methods like np.arange or manual loops.
Data Types: Ensure the arrays are of floating-point type. While np.linspace returns floats by default, explicit casting might be needed in some contexts, though usually not for this problem.
Off-by-One Errors: When implementing manually without cumprod, ensure the loop starts correctly and accumulates the product properly. Using cumprod avoids this issue.

5. Time & Space Complexity

Time Complexity: $O(T)$ . Generating betas with linspace takes $O(T)$ . Computing alphas is an element-wise operation taking $O(T)$ . Computing alpha_bars with cumprod is also $O(T)$ . Thus, the total time complexity is linear in $T$ .
Space Complexity: $O(T)$ . We store three arrays of length $T$ : betas, alphas, and alpha_bars. The space required is proportional to $T$ .

solution.py

Python 3.13

Test Results

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