📘

x0 and eps from a v-Prediction

MediumDiffusion Models 📚 Diffusion Models 1

Problem Statement

A v-prediction network outputs v; the sampler needs x0 and eps back. Given x_t, the predicted v, and alpha_bar, recover both.

Background

With a = sqrt(alpha_bar) and b = sqrt(1 - alpha_bar), the inverse relations are

$\hat{x}_0 = a\, x_t - b\, v, \qquad \hat{\varepsilon} = b\, x_t + a\, v$

These follow from the orthogonal rotation that defines v; note it is an exact linear map, so no clipping is required here.

Your Task

Implement:

def x0_eps_from_v(x_t, v, alpha_bar):

Return a dict with "x0" and "eps", each a list rounded to 4 decimals.

Input Format

x_t, v: lists of equal length D.
alpha_bar (float) in (0, 1).

Output Format

A dict with two lists.

Sample

print(x0_eps_from_v([1.0, 0.0], [0.0, 1.0], 0.25))

Output:

{'x0': [0.5, -0.866], 'eps': [0.866, 0.5]}

Example:

Input:

print(x0_eps_from_v([1.0, 0.0], [0.0, 1.0], 0.25))

Output:

{'x0': [0.5, -0.866], 'eps': [0.866, 0.5]}

Reasoning:

Compute the scaling factors from $\alpha_{\bar{}} = 0.25$ : $a = \sqrt{0.25} = 0.5$ and $b = \sqrt{1 - 0.25} = \sqrt{0.75} \approx 0.8660$ .
Calculate the first element of $\hat{x}_0$ using $\hat{x}_0 = a x_t - b v$ : $0.5(1.0) - 0.8660(0.0) = 0.5$ .
Calculate the second element of $\hat{x}_0$ : $0.5(0.0) - 0.8660(1.0) = -0.8660$ , which rounds to $-0.866$ .
Calculate the first element of $\hat{\varepsilon}$ using $\hat{\varepsilon} = b x_t + a v$ : $0.8660(1.0) + 0.5(0.0) = 0.8660$ , which rounds to $0.866$ .
Calculate the second element of $\hat{\varepsilon}$ : $0.8660(0.0) + 0.5(1.0) = 0.5$ .
The final output is {'x0': [0.5, -0.866], 'eps': [0.866, 0.5]}

Constraints:

len(x_t) == len(v), 0 < alpha_bar < 1.
x0 = a*x_t - b*v, eps = b*x_t + a*v, with a=sqrt(alpha_bar), b=sqrt(1-alpha_bar).
Round both to 4 decimals; avoid -0.0.

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

1. Background Knowledge

In diffusion models, the forward process gradually adds Gaussian noise to a data point $x_0$ over $T$ steps. At any discrete step $t$ , the noisy sample $x_t$ can be expressed as a linear combination of the original data and the noise:

x_t = \sqrt{\bar{\alpha}_t}\, x_0 + \sqrt{1 - \bar{\alpha}_t}\, \varepsilon

where $\bar{\alpha}_t \in (0, 1)$ is the cumulative product of per-step signal retention factors, and $\varepsilon \sim \mathcal{N}(0, I)$ . The coefficients $a = \sqrt{\bar{\alpha}_t}$ and $b = \sqrt{1 - \bar{\alpha}_t)$ form an orthonormal pair ( $a^2 + b^2 = 1$ ), which means the mapping between $(x_0, \varepsilon)$ and $(x_t, v)$ is a rotation in the 2D subspace spanned by each coordinate.

The v-prediction parameterization, introduced by Kingma & Gao (2023), predicts the vector $v = b\, x_0 - a\, \varepsilon$ instead of directly predicting $x_0$ or $\varepsilon$ . This parameterization has been shown to improve training stability, especially at high noise levels. Because the transformation is a rotation, it is invertible: given $x_t$ and $v$ , you can recover both $x_0$ and $\varepsilon$ through a simple linear combination.

The inverse relations are:

\hat{x}_0 = a\, x_t - b\, v, \qquad \hat{\varepsilon} = b\, x_t + a\, v

These follow directly from substituting the forward equation into the definition of $v$ and solving the resulting $2 \times 2$ linear system. Since the rotation matrix is orthogonal, no clipping or normalization is needed—the recovered values are exact (up to floating-point precision).

2. Algorithm Approach

This is a direct algebraic inversion problem. The core idea is:

Compute the scalar coefficients $a = \sqrt{\bar{\alpha}_t}$ and $b = \sqrt{1 - \bar{\alpha}_t}$ .
Apply the two linear combinations element-wise across all $D$ dimensions.
Round results to the required precision.

There is no iterative procedure, no optimization, and no branching logic. The entire computation is a vectorized linear map applied independently to each coordinate.

3. Step-by-Step Strategy

👑

Continue the full explanation

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

🔒

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

0/0

Run code to see test results.