[CPU shaders] Assertion failure in path_tiling stage when rendering Coat of Arms of Poland.svg about vello

[raphlinus]

Minimal repro case, based on extracting the offending line segment from the assert failure:

<svg width="1200" height="1200" viewBox="0 0 1200 1200" fill="none" xmlns="http://www.w3.org/2000/svg">
  <path d="M-0.36201912 1.7544264e-13L1354.9857 1.7544264e-13 100 100" fill="#ff0"/>
</svg>

This renders with artifact on GPU, which is an additional indication that the assert is detecting a real problem.

[raphlinus]

After a little more digging, I have a clearer picture what's going on. It's a horizontal line which is not exactly grid aligned (but obviously very close). If it were aligned, then the line would be discarded by the numerical robustness logic. On the CPU, a is being computed as 1, which is correct. However, b should be 1-epsilon, and it's being rounded up to 1. When computing floor(a*i+b) it should round down to i, but is instead i + 1. That in turn selects a grid square which is not part of the rasterization of the line, which triggers the assertion failure and other mischief.

The fix is not yet clear, though I've got the general outline. Obviously 0 <= b < 1 should be enforced, and if that's satisfied, then floor(a*i+b) should be 0 for the i=0 case. But I'm still worried about i=n-1 in particular.

Another thing that can go wrong is that a should be equal to 1 in the horizontal line case, but it can be <1 because it's calculated as dx * (dx + dy).recip(), which can have roundoff errors. As noted in the review for #374, if that's calculated on CPU using division, you get the right answer, but the guarantees of WGSL are weaker than IEEE floating point.

Obviously a patch to fix this specific error is to special case horizontal lines, but that's not satisfying, as it seems clear it wouldn't cover all failures. I think a good next step would be to carefully define what values for a, b are correct, then put in logic to enforce that, possibly some combination of clamping and verifying that floor(a * (n - 1) + b) is the right value. I'll think on it some more.

[raphlinus]

Ok, I've slept on it, and I believe the following will work. Compute b by taking the min with 0.99999994, which will guarantee it's strictly less than 1. Then compute floor(a * (n - 1) + b) and check whether it's equal to max(1, ceil(max(x0, x1)) - floor(min(x0, x1))) - 1. Note: that's the same value as for computing the count, and it's equal to the width in grid cells of the rasterized line minus one. If not equal, bump a by some epsilon (I believe 1e-9 is fine) with the sign needed to make that equal.

There's more work that can be done to validate this, but I believe that would result in fully robust rendering, and also reduces the pressure on the division operation to compute a. Note that this would also apply to multisampled rendering in fine (see the reviews for the msaa PR, linked above).

[raphlinus]

1e-9 is not fine, we're in f32 world, that's less than 1 ulp. I think 1e-7 is adequate, but the argument is subtle.

[raphlinus]

I believe this can be considered fixed as per the linked PR.