Task 4 of 5

Two Halves Make a Sweep

The red half alone is not a sweep — half the grid has not been touched. The black half is the same kernel with its parity test flipped, and the ordering that matters is in the chaining: black(red(u)). The black cells read the grid the red half produced, so they see this sweep's reds, not last sweep's. That single fact is all that separates Gauss-Seidel from Jacobi.

Write black(u) instead and both halves read the same old grid. The code still runs, the answer still looks plausible, and you have written Jacobi with an extra kernel launch — which is the most expensive way to be wrong in this module, because nothing about the output shouts.

Goal: write the black half-sweep and chain the two halves into one full red-black Gauss-Seidel sweep.

Requirements

The black kernel is the red kernel with its parity test flipped: cells with (x + y) % 2 === 1 update, everything else passes through
Create the red kernel first and the black kernel second
One full sweep is await black(await red(u)) — the black half reads what the red half wrote
The boundary is untouched by both halves

Hint 1 — the black kernel

Copy the red kernel and change one character:

const parity = (x + y) % 2;
if (parity !== 1) return u[y][x];

Still a number, never a boolean — the WebGL backend rejects const isBlack = … exactly as it rejects isRed.

Hint 2 — the chain is the lesson

const afterRed = await red(guess);
const afterBoth = await black(afterRed);   // NOT black(guess)

Or, in one line: await black(await red(guess)) — the inner await is not optional, because an un-awaited kernel call hands the next kernel a Promise instead of a grid.

Same idea elsewhere

Two dispatches with a dependency between them is the ordinary shape of GPU work: WebGPU inserts a barrier between compute passes, CUDA orders them on a stream, Vulkan wants an explicit pipeline barrier. What you cannot do — on any of them — is order threads inside one dispatch, which is exactly why the sequential Gauss-Seidel had to be split into two of them in the first place.

All tasks in Iterative Linear Solvers

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