Fifty Steps or Five

The number of sampling steps is yours to choose, which means a sampler has to be able to jump from step 60 straight to step 40 without visiting anything in between. Doing that correctly is the whole trick.

Write $\bar{\alpha}_t$ for the signal remaining at step $t$, with $\bar{\alpha}_0 = 1$ (nothing destroyed). One jump, from step $t$ down to an earlier step $p$, is three moves:

1. Predict the noise $\hat{\varepsilon}$ in the current latent $x_t$.
2. Strip all of it out, giving the denoiser's guess at the clean image:

$\hat{x}_0 = \frac{x_t - \sqrt{1 - \bar{\alpha}_t}\; \hat{\varepsilon}}{\sqrt{\bar{\alpha}_t}}$

1. Put back exactly as much noise as step $p$ is supposed to have:

$x_p = \sqrt{\bar{\alpha}_p}\; \hat{x}_0 \; + \; \sqrt{1 - \bar{\alpha}_p}\; \hat{\varepsilon}$

Move 3 is what makes a jump of any size legal: the latent is rebuilt at whatever noise level you asked for, instead of being nudged down by one.

The denoiser here is a toy one, so the arithmetic is reproducible: at step $t$ it predicts the noise to be c[t] times whatever latent it is shown, elementwise.

Task: write denoise(x, abar, c, steps), returning the finished result at step 0 as a plain list, rounded to 4 decimal places — and rounded only there, never between jumps.

• x is the starting latent, sitting at step steps[0].
• abar[t] is the signal remaining at step t, and abar[0] is 1.0.
• c[t] is the toy denoiser's multiplier at step t.
• steps is strictly decreasing and every entry is at least 1. Visit those steps in order; after the last one, the next stop is step 0.

The two visible tests are the same latent and the same schedule, differing only in which steps get visited — and they do not agree. Skipping is an approximation, and the gap between those two answers is the speed-versus-quality dial every diffusion tool puts in front of you.