https://stackoverflow.com/questions/3706219/algorithm-for-iterating-over-an-outward-spiral-on-a-discrete-2d-grid-from-the-or

spiral.js

// Utils
const range = n => [...Array(n).keys()];
const add = ([x0, y0]) => ([x1, y1]) => [x0 + x1, y0 + y1];
const rotate = θ => ([x, y]) => [
  Math.round(x * Math.cos(θ) - y * Math.sin(θ)),
  Math.round(x * Math.sin(θ) + y * Math.cos(θ))
];
const map = f => g =>
  function*() {
    for (const v of g()) {
      yield f(v);
    }
  };

// Implementation
const spiralOut = i => {
  const n = Math.floor(i / 2) + 1;
  const leg = range(n).map(x => [x, 0]);
  const transform = p => add([n, 0])(rotate(-Math.PI / 2)(p));

  return function*() {
    yield* leg;
    yield* map(transform)(spiralOut(i + 1))();
  };
};

const spiralIn = ([w, h]) => {
  const leg = range(w).map(x => [x, 0]);
  const transform = p => add([w - 1, 1])(rotate(Math.PI / 2)(p));

  return w * h === 0
    ? function*() {}
    : function*() {
        yield* leg;
        yield* map(transform)(spiralIn([h - 1, w]))();
      };
};

let i = 0;
for (let p of spiralOut(0)()) {
  if (i++ > 30) break;
  console.log(p);
}