x0 and eps from a v-Prediction

mediumPROParameterizations and the Reverse PosteriorDiffusion ModelsDiffusion 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.

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Concept primerPREVIEW

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

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solution.py

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