Full DDPM Ancestral Step

mediumPRODDPM and DDIM SamplingDiffusion ModelsDiffusion Models 1

Problem Statement

Implement one complete DDPM ancestral sampling step: the eps-parameterized mean plus the injected posterior noise sigma_t * z.

Background

The reverse mean from a noise prediction is

$\mu_\theta = \frac{1}{\sqrt{\alpha_t}}\left(x_t - \frac{\beta_t}{\sqrt{1 - \bar{\alpha}_t}}\, \hat{\varepsilon}\right), \qquad \alpha_t = 1 - \beta_t$

Then a sample is drawn as x_{t-1} = mu + sigma_t * z where z is supplied and sigma_t = sqrt(beta_tilde_t) with the posterior variance beta_tilde_t = (1 - alpha_bar_prev)/(1 - alpha_bar_t) * beta_t. At the final step (alpha_bar_prev == 1) no noise is added.

Your Task

Implement:

def ddpm_step(x_t, eps, z, beta_t, alpha_bar_t, alpha_bar_prev):

Return x_{t-1} as a list rounded to 4 decimals. If alpha_bar_prev == 1.0, return the mean only.

Input Format

x_t, eps, z: lists of length D.
beta_t, alpha_bar_t, alpha_bar_prev (float).

Output Format

A list of D floats rounded to 4 decimals.

Sample

print(ddpm_step([1.0, 1.0], [0.0, 0.0], [1.0, -1.0], 0.2, 0.5, 0.64))

Output:

[1.4975, 0.7386]

Example

Input

print(ddpm_step([1.0, 1.0], [0.0, 0.0], [1.0, -1.0], 0.2, 0.5, 0.64))

Output

[1.4975, 0.7386]

Reasoning

eps=0 so mean = x_t/sqrt(0.8) = [1.1180,1.1180]. sigma = sqrt(0.36/0.50.2) = sqrt(0.144) = 0.3795. x = mean + 0.3795[1,-1] = [1.4975, 0.7386].

Constraints

len(x_t) == len(eps) == len(z); schedule values in (0, 1).
sigma_t = sqrt((1-alpha_bar_prev)/(1-alpha_bar_t)*beta_t).
No noise when alpha_bar_prev == 1.0; round to 4 decimals; avoid -0.0.

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Concept primerPREVIEW

1. Background Knowledge

DDPM (Denoising Diffusion Probabilistic Models) reverse a forward noising process by learning a noise-prediction network $\epsilon_\theta(x_t, t)$ . Given a noisy sample $x_t$ and the predicted noise $\hat{\varepsilon}$ , the model estimates the posterior mean $\mu_\theta(x_t, t)$ , which points toward the cleaner sample $x_{t-1}$ . The mean formula is:

\mu_\theta = \frac{1}{\sqrt{\alpha_t}}\left(x_t - \frac{\beta_t}{\sqrt{1 - \bar{\alpha}_t}}\, \hat{\varepsilon}\right)

where $\alpha_t = 1 - \beta_t$ and $\bar{\alpha}_t = \prod_{s=1}^{t} \alpha_s$ is the cumulative product of $\alpha$ values.

The full ancestral step does not just return the mean. It also injects posterior noise to preserve the stochastic nature of the reverse process. The posterior variance is:

\tilde{\beta}_t = \frac{1 - \bar{\alpha}_{t-1}}{1 - \bar{\alpha}_t}\, \beta_t

and the standard deviation is $\sigma_t = \sqrt{\tilde{\beta}_t}$ . The sampled output is $x_{t-1} = \mu_\theta + \sigma_t \cdot z$ , where $z \sim \mathcal{N}(0, I)$ is supplied externally. At the final step ( $t=1$ ), $\bar{\alpha}_0 = 1$ , so $\tilde{\beta}_1 = 0$ and no noise is added.

2. Algorithm Approach

This is a direct formula evaluation problem. There is no iteration or search. The approach is:

Compute $\alpha_t = 1 - \beta_t$ .
Compute the scalar coefficient for the noise term: $\frac{\beta_t}{\sqrt{1 - \bar{\alpha}_t}}$ .
Compute the mean vector $\mu_\theta$ element-wise.
Compute $\sigma_t$ from the posterior variance formula.
If $\bar{\alpha}_{t-1} = 1.0$ , return $\mu_\theta$ ; otherwise add $\sigma_t \cdot z$ element-wise.
Round each element to 4 decimal places.

3. Step-by-Step Strategy

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solution.py

Python 3.13

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