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Signal-to-Noise Ratio of a Noise Schedule

EasyDiffusion Models 📚 Diffusion Models 1

Problem Statement

Turn a cumulative-alpha schedule into the two quantities that actually describe "how hard is step $t$ ": the signal-to-noise ratio and its logarithm.

Background

Under the closed form $x_t = \sqrt{\bar{\alpha}_t}\,x_0 + \sqrt{1-\bar{\alpha}_t}\,\varepsilon$ , the signal power is $\bar{\alpha}_t$ and the noise power is $1-\bar{\alpha}_t$ . Hence

$\mathrm{SNR}(t) = \frac{\bar{\alpha}_t}{1 - \bar{\alpha}_t}, \qquad \lambda_t = \log \mathrm{SNR}(t)$

Two facts make this the right coordinate system for diffusion:

$\bar{\alpha}_t = 0.5$ is exactly the half-signal point: $\mathrm{SNR} = 1$ and $\lambda_t = 0$ .
$\lambda_t$ decreases roughly linearly over the useful part of a schedule, which is why continuous-time formulations parameterize by log-SNR rather than by $t$ , and why two schedules with the same $\lambda$ curve are the same model.

Your Task

Implement:

def snr_and_logsnr(alpha_bars):

Return (snr, log_snr), two 1-D NumPy arrays the same length as alpha_bars.

Input Format

alpha_bars: 1-D NumPy array with every entry strictly inside (0, 1).

Output Format

A tuple (snr, log_snr) of NumPy arrays.

Sample

ab = np.array([0.9, 0.5, 0.1])
snr, ls = snr_and_logsnr(ab)
print(np.round(snr, 4).tolist())
print(np.round(ls, 4).tolist())

At $\bar{\alpha} = 0.5$ signal and noise power are equal, so SNR = 1.0 and log SNR = 0.0; at 0.9 the SNR is 9, and at 0.1 it is one ninth.

Example:

Input:

ab = np.array([0.9, 0.5, 0.1])
snr, ls = snr_and_logsnr(ab)
print(np.round(snr, 4).tolist())
print(np.round(ls, 4).tolist())

Output:

[9.0, 1.0, 0.1111]
[2.1972, 0.0, -2.1972]

Reasoning:

0.9/(1-0.9) = 9, 0.5/0.5 = 1, 0.1/0.9 = 0.1111. Taking natural logs gives 2.1972, 0.0, -2.1972. The symmetry around alpha_bar = 0.5 is exact: log-SNR is an odd function of the signal fraction about that midpoint.

Constraints:

Every entry of alpha_bars is in the open interval (0, 1), so no division by zero and no log(0).
log_snr is the natural log.
Do not round inside the function.
Return arrays, not lists.

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1. Background Knowledge

In Diffusion Models, the forward process gradually adds Gaussian noise to data $x_0$ over $T$ steps. The state at step $t$ , denoted $x_t$ , can be expressed in closed form using the cumulative noise schedule $\bar{\alpha}_t$ :

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

Here, $\bar{\alpha}_t$ represents the signal power (variance of the original data component), and $1-\bar{\alpha}_t$ represents the noise power (variance of the added Gaussian noise $\varepsilon$ ). The Signal-to-Noise Ratio (SNR) is defined as the ratio of these two powers:

\mathrm{SNR}(t) = \frac{\bar{\alpha}_t}{1 - \bar{\alpha}_t}

This metric quantifies how much of the original signal remains relative to the accumulated noise. A high SNR indicates the data is still recognizable, while a low SNR indicates it is dominated by noise. The log-SNR ( $\lambda_t = \log \mathrm{SNR}(t)$ ) is often preferred in theoretical analyses and continuous-time formulations because it transforms the multiplicative relationship into an additive one, often resulting in a more linear progression over time steps.

2. Algorithm Approach

The problem requires a direct vectorized computation using NumPy. Since the input alpha_bars is a 1-D NumPy array, you should leverage NumPy's element-wise operations to compute the SNR and log-SNR for all elements simultaneously. This avoids explicit Python loops, ensuring efficiency and concise code. The approach involves two main mathematical transformations applied to the input array: division for SNR and logarithm for log-SNR.

3. Step-by-Step Strategy

Validate Input (Implicit): Assume the input alpha_bars is a valid NumPy array with values strictly between 0 and 1, as per the problem statement.
Compute Noise Power: Calculate the noise power component, which is $1 - \bar{\alpha}_t$ . In NumPy, this is simply 1 - alpha_bars.
Compute SNR: Divide the signal power (alpha_bars) by the noise power computed in the previous step. This yields the SNR array.

snr = alpha_bars / (1 - alpha_bars)

Compute Log-SNR: Apply the natural logarithm to the SNR array. Use np.log for this operation.

log_snr = np.log(snr)

Return Results: Return the tuple (snr, log_snr). Ensure both are NumPy arrays of the same shape as the input.

4. Common Pitfalls

Division by Zero: Although the problem states inputs are strictly inside $(0, 1)$ , be aware that if $\bar{\alpha}_t$ approaches 1, the denominator $1 - \bar{\alpha}_t$ approaches 0, leading to very large SNR values. If $\bar{\alpha}_t$ were exactly 1, this would cause a division-by-zero error. Similarly, if $\bar{\alpha}_t$ were 0, the SNR would be 0, and $\log(0)$ is undefined ( $-\infty$ ). The constraints prevent this, but understanding the limits is crucial.
Logarithm Base: The problem specifies $\lambda_t = \log \mathrm{SNR}(t)$ . In mathematics and most scientific computing contexts (including NumPy's np.log), this refers to the natural logarithm (base $e$ ). Do not use np.log10 unless explicitly requested.
Data Types: Ensure the output arrays are floating-point types. If alpha_bars is an integer array (unlikely given the range, but possible), the division might perform integer division in older Python versions or specific contexts. Using float conversion or ensuring alpha_bars is float64 is safe practice.
Vectorization: Avoid using for loops to iterate over alpha_bars. NumPy operations are optimized in C and will be significantly faster for large arrays.

5. Time & Space Complexity

Time Complexity: $O(N)$ , where $N$ is the length of the alpha_bars array. Each element undergoes a constant number of arithmetic operations (subtraction, division, logarithm). NumPy vectorization ensures these operations are performed efficiently in parallel.
Space Complexity: $O(N)$ , as we need to store two new arrays (snr and log_snr) of size $N$ . The input array is not modified in place.

solution.py

Python 3.13

Test Results

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Run code to see test results.