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Terminal SNR and the Zero-SNR Fix

MediumDiffusion Models 📚 Diffusion Models 1

Problem Statement

Common schedules never fully destroy the signal: at the last step alpha_bar_T is slightly above 0, leaking image information the sampler can never remove. Detect this and apply the "enforce zero terminal SNR" rescaling from Lin et al. (2024).

Background

The signal-to-noise ratio at step t is SNR_t = alpha_bar_t / (1 - alpha_bar_t). A schedule has zero terminal SNR iff alpha_bar_T = 0. The fix rescales the sqrt(alpha_bar) curve linearly so its first value is unchanged and its last value becomes 0:

$\sqrt{\bar{\alpha}_t}' = \frac{\sqrt{\bar{\alpha}_t} - \sqrt{\bar{\alpha}_{T}}}{\sqrt{\bar{\alpha}_0} - \sqrt{\bar{\alpha}_{T}}}\cdot \sqrt{\bar{\alpha}_0}$

then square back to get the corrected alpha_bar.

Your Task

Implement:

def fix_zero_terminal_snr(alpha_bar):

Return the corrected alpha_bar list rounded to 6 decimals. The last value must be exactly 0.0.

Input Format

alpha_bar: list of T cumulative products (decreasing).

Output Format

A list of T floats rounded to 6 decimals.

Sample

print(fix_zero_terminal_snr([0.81, 0.36, 0.04]))

Output:

[0.81, 0.26449, 0.0]

Example:

Input:

print(fix_zero_terminal_snr([0.81, 0.36, 0.04]))

Output:

[0.81, 0.26449, 0.0]

Reasoning:

Compute the square roots of the input values to work in the $\sqrt{\bar{\alpha}}$ domain: $\sqrt{0.81} = 0.9$ , $\sqrt{0.36} = 0.6$ , and $\sqrt{0.04} = 0.2$ .
Identify the first ( $s_0 = 0.9$ ) and last ( $s_T = 0.2$ ) values to determine the linear rescaling range, which is $s_0 - s_T = 0.9 - 0.2 = 0.7$ .
Apply the linear transformation $\frac{s - s_T}{s_0 - s_T} \cdot s_0$ $\frac{s - s _{T}}{s _{0} - s _{T}} \cdot s_{0}$ to each element to shift the curve so the last value becomes zero while keeping the first value unchanged:
- For the first element: $\frac{0.9 - 0.2}{0.7} \cdot 0.9 = 1 \cdot 0.9 = 0.9$
- For the second element: $\frac{0.6 - 0.2}{0.7} \cdot 0.9 = \frac{0.4}{0.7} \cdot 0.9 \approx 0.514286$
- For the last element: $\frac{0.2 - 0.2}{0.7} \cdot 0.9 = 0$
Square the rescaled values to return to the $\bar{\alpha}$ $\overset{α}{ˉ}$ domain and round to 6 decimal places:
- $0.9^2 = 0.81$
- $(0.514286)^2 \approx 0.26449$
- $0^2 = 0.0$
The final output is [0.81, 0.26449, 0.0]

Constraints:

2 <= T <= 100000.
Rescale the sqrt(alpha_bar) curve so the first value is preserved and the last hits 0.
Square back; the final value is exactly 0.0; round to 6 decimals.

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1. Background Knowledge

In diffusion models, the forward process gradually adds Gaussian noise to a data sample $x_0$ over $T$ discrete steps. The state at step $t$ is defined as $x_t = \sqrt{\bar{\alpha}_t} x_0 + \sqrt{1 - \bar{\alpha}_t} \epsilon$ , where $\epsilon \sim \mathcal{N}(0, I)$ . The term $\bar{\alpha}_t$ represents the cumulative product of noise schedule parameters $\alpha_i = 1 - \beta_i$ up to step $t$ . As $t$ increases, $\bar{\alpha}_t$ decreases, meaning less signal and more noise.

The Signal-to-Noise Ratio (SNR) at step $t$ is given by $\text{SNR}_t = \frac{\bar{\alpha}_t}{1 - \bar{\alpha}_t}$ . Ideally, at the final step $T$ , the SNR should be zero, implying $\bar{\alpha}_T = 0$ . This ensures that $x_T$ is pure noise, containing no information about the original image $x_0$ . However, many standard noise schedules (like cosine or linear) result in $\bar{\alpha}_T > 0$ , leaving a small amount of signal leakage. This leakage can degrade the quality of generated samples because the sampler cannot fully "forget" the original data.

The Zero-Terminal SNR fix, proposed by Lin et al. (2024), corrects this by rescaling the $\sqrt{\bar{\alpha}_t}$ curve. The goal is to preserve the shape of the noise schedule while forcing the final value to zero. This is done by linearly interpolating the square root of $\bar{\alpha}_t$ between its initial value and zero, then squaring the result back to obtain the corrected $\bar{\alpha}_t$ .

2. Algorithm Approach

The approach involves a simple linear transformation on the square root of the input array:

Extract endpoints: Identify $\sqrt{\bar{\alpha}_0}$ (first element) and $\sqrt{\bar{\alpha}_T}$ (last element).
Linear rescaling: For each element $\sqrt{\bar{\alpha}_t}$ , apply the formula:

\sqrt{\bar{\alpha}_t}' = \frac{\sqrt{\bar{\alpha}_t} - \sqrt{\bar{\alpha}_T}}{\sqrt{\bar{\alpha}_0} - \sqrt{\bar{\alpha}_T}} \cdot \sqrt{\bar{\alpha}_0}

This maps $\sqrt{\bar{\alpha}_0}$ to itself and $\sqrt{\bar{\alpha}_T}$ to $0$ . 3. Square back: Compute $\bar{\alpha}_t' = (\sqrt{\bar{\alpha}_t}')^2$ . 4. Round and enforce: Round each value to 6 decimal places and explicitly set the last element to 0.0.

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