Task 1 of 5

One Sweep of Jacobi

A square metal plate, its edges clamped at fixed temperatures. What does the inside settle to? Not "what happens next" — the answer once nothing happens any more. That steady state solves ∇²u = 0, and the 5-point stencil turns it into one tiny equation per interior cell: every cell equals the average of its four neighbours. Nine hundred equations, nine hundred unknowns, all tangled together.

Nobody inverts that matrix. You guess, and improve the guess. Jacobi's method is that made literal: set every interior cell to the average of its neighbours as they were before this sweep, and repeat. Because every cell reads the previous iterate and nothing else, all 1,024 threads are independent — it is the pure gather Thinking in Parallel calls the shape that always parallelises, and one whole sweep is one kernel call.

The edges never move: they are the known values that pin the answer down. (Reaction–Diffusion steps this same stencil forward in time; here we are solving for the state where time has stopped.)

everyone reads yesterday — which is exactly why everyone can go at once

Goal: finish the sweep kernel — interior cells return the average of their four neighbours in u, boundary cells return their own value unchanged.

Requirements

Boundary cells — x or y equal to 0 or size − 1 — return u[y][x] untouched
Interior cells return (left + right + up + down) / 4: four neighbours, no centre
Read only u — no cell may see a value written during this sweep

Hint 1 — the edges come first

Guard the boundary before you do any arithmetic, so the neighbour reads below can never leave the grid:

if (x === 0 || y === 0 || x === this.constants.size - 1
    || y === this.constants.size - 1) {
  return u[y][x];
}

Hint 2 — the four neighbours

Only ever vary one coordinate at a time: u[y][x - 1] and u[y][x + 1] along the row, u[y - 1][x] and u[y + 1][x] down the column. The centre u[y][x] is not part of a Jacobi average.

Hint 3 — the whole return

return (u[y][x - 1] + u[y][x + 1]
      + u[y - 1][x] + u[y + 1][x]) / 4;

Same idea elsewhere

Jacobi is the starting point of every multigrid solver on every platform — a CUDA or WGSL version of this kernel is line-for-line the same gather, with a buffer swap where your JavaScript assignment is. It is also why "matrix-free" is a phrase: nobody stores the 900×900 matrix this stencil stands for, because the kernel is the matrix.

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