[Geee]

Is there an error in the visualization? It shows that every vector is rotated the same amount. My understanding was that they are randomized with different values, which results in a predictable distribution, which is easier to quantize.

[mesuvash]

That's actually correct and intentional. TurboQuant applies the same rotation matrix to every vector. The key insight is that any unit vector, when multiplied by a random orthogonal matrix, produces coordinates with a known distribution (Beta/arcsine in 2D, near-Gaussian in high-d). The randomness is in the matrix itself (generated once from a seed), not per-vector. Since the distribution is the same regardless of the input vector, a single precomputed quantization grid works for everything. I've updated the description to make this clearer.

[Geee]

Thanks. However, from this visualization it's not clear how the random rotation is beneficial. I guess it makes more sense on higher dimensional vectors.

[mesuvash]

Yes, this is important in high dimension. But sadly, very hard to visualize. In 2d it looks like unnecessary.

[fc417fc802]

I believe they are all rotated by the same random matrix, the purpose being (IIUC) to distribute the signal evenly across all dimensions. So effectively it drowns any structure that might be present in noise. That's essential for data efficiency in addition to avoiding bias related issues during the initial quantization step. However there are still some other issues due to bias that are addressed by a second quantization step involving the residual.

That said, I don't believe the visualization is correct. The grid for one doesn't seem to match what's described in the paper.

Also it's entirely possible I've misunderstood or neglected to notice key details.