Why the right side is instant

Slime pulls itself into the shape with the least surface. In this cell the walls wrap around, so the glass shows repeating copies, and a blob can join up with itself through a wall. The usual way to find the settled shape is to simulate it: smear the slime a little, keep the most crowded spots, repeat until nothing moves, first on a coarse grid and then a fine one. A new paper proves which shape always wins for each amount of slime: a ball, then a tube that runs through the walls, then a flat layer, then the same shapes as air holes. So the right side just draws that shape once. The pillar scene is outside the paper.

What's measured here

• Solver ms: performance.now() around the solver's own work only (left: every blur + keep-the-top step; right: three warm-up builds, then the mean of a timed batch of builds). Drawing, the area check and shape naming are not timed.
• Surface: one estimator for both sides (blur the voxels by one voxel, add up the slope). Its error on the exact shape shows on the right as a small ± next to 0 %.
• Best possible (REFERENCE): the paper's formula I(fill). The left solver cannot know it is done without it; we only use it to stop the clock.

Where the prior technique still wins / limits

• Anything the paper doesn't cover: a pillar, a non-cubic cell, gravity. Only the simulation can answer there (scenes pillar, pillar-drop; the warm start is a measured guess, not a theorem).
• It gives you an animation of the slime flowing for free; the formula only gives the end state.
• With a tuned coarse stage, the simulation is already fast and rarely stuck at these grid sizes; the gain is a constant factor (about ×10–15 here) plus a guarantee.