📘

v-Target from x0 and Noise

EasyDiffusion Models 📚 Diffusion Models 1

Problem Statement

The v-prediction target (Salimans & Ho, 2022) is a signal/noise-balanced combination. Compute it from x0 and eps.

Background

The velocity target is

$v = \sqrt{\bar{\alpha}}\, \varepsilon - \sqrt{1 - \bar{\alpha}}\, x_0$

It equals eps at high noise and -x0 at low noise, which is what makes v-prediction stable across the whole schedule and the default for progressive distillation.

Your Task

Implement:

def v_target(x0, eps, alpha_bar):

Return v as a list rounded to 4 decimals.

Input Format

x0, eps: lists of equal length D.
alpha_bar (float) in (0, 1).

Output Format

A list of D floats rounded to 4 decimals.

Sample

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

Output:

[-0.866, 0.5]

Example:

Input:

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

Output:

[-0.866, 0.5]

Reasoning:

Compute the scaling coefficients for the noise and signal components using $\bar{\alpha} = 0.25$ : the noise weight is $\sqrt{0.25} = 0.5$ and the signal weight is $\sqrt{1 - 0.25} = \sqrt{0.75} \approx 0.8660$ .
Calculate the v-prediction target for the first element ( $x_0 = 1.0$ , $\epsilon = 0.0$ ) by combining the weighted terms: $v_0 = 0.5 \cdot 0.0 - 0.8660 \cdot 1.0 = -0.8660$ .
Calculate the v-prediction target for the second element ( $x_0 = 0.0$ , $\epsilon = 1.0$ ) using the same weights: $v_1 = 0.5 \cdot 1.0 - 0.8660 \cdot 0.0 = 0.5$ .
Round each result to 4 decimal places to satisfy the output format, yielding $-0.866$ and $0.5$ .
The final output is [-0.866, 0.5]

Constraints:

len(x0) == len(eps), 0 < alpha_bar < 1.
v = sqrt(alpha_bar)*eps - sqrt(1 - alpha_bar)*x0.
Round 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 $T$ timesteps. At any timestep $t$ , the noisy sample $x_t$ can be expressed in closed form as:

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

where $\bar{\alpha}_t = \prod_{s=1}^{t} \alpha_s$ is the cumulative product of per-step signal retention factors, and $\varepsilon \sim \mathcal{N}(0, I)$ is standard Gaussian noise. The parameter $\bar{\alpha}_t$ controls the signal-to-noise ratio: when $\bar{\alpha}_t \approx 1$ (early timesteps), the sample is mostly signal; when $\bar{\alpha}_t \approx 0$ (late timesteps), it is mostly noise.

The v-prediction parameterization (Salimans & Ho, 2022) defines a target velocity:

$v = \sqrt{\bar{\alpha}}\, \varepsilon - \sqrt{1 - \bar{\alpha}}\, x_0$

This is a linear combination of the noise and the clean signal, weighted by the same coefficients that appear in the forward process. The key property is that $v$ smoothly transitions between $\varepsilon$ (when $\bar{\alpha} \to 1$ , high noise) and $-x_0$ (when $\bar{\alpha} \to 0$ , low noise). This balanced behavior makes v-prediction numerically stable across the entire diffusion schedule and is the default choice in progressive distillation pipelines.

Understanding this target requires recognizing that it is simply a reparameterization of the same underlying Gaussian distribution. Given $x_0$ and $\varepsilon$ , computing $v$ is a deterministic, element-wise operation—no optimization or iterative procedure is involved.

2. Algorithm Approach

This is a direct formula evaluation problem. The algorithm pattern is:

Parse the input vectors and scalar.
Compute the two scalar weights: $w_\varepsilon = \sqrt{\bar{\alpha}}$ and $w_{x_0} = \sqrt{1 - \bar{\alpha}}$ .
For each element index $i$ , compute $v_i = w_\varepsilon \cdot \varepsilon_i - w_{x_0} \cdot x_{0,i}$ .
Round each result to 4 decimal places.

There is no loop over timesteps, no gradient computation, and no iterative refinement. The entire computation is a single pass over the $D$ -dimensional vectors.

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.