% A room: a square enclosure with a doorway in one wall.
%
% Close the doorway (gap = 0) and it is a sealed cavity, where a pulse never
% leaves and the sound settles into the room's own modes. Open it and the room
% becomes a resonator that leaks: sound escapes through the aperture, and the
% ringing decays. This is the scene the microphone is for. Put the source and
% the microphone inside, listen to a click, and what comes back is a direct
% arrival followed by echoes closing up into reverberation.
%
% `side`, `gap` and `thick` are in metres — a real room, not a fraction of the
% domain — so `side` is the half-width and the room is `2*side` across.
%
% The walls are *slower* than the background rather than faster, and that is a
% deliberate trade. Reflection at an interface goes as |c2 - c1|/(c2 + c1)
% with the density constant, so a wall at cwall = 0.15 reflects about 75% of
% the pressure amplitude, as one at cwall = 7 would. But the timestep is set
% by the *fastest* speed anywhere on the grid, so a hard wall would make every
% step of the whole simulation several times smaller, while a slow one is
% free. What a slow wall costs instead is resolution inside itself: the
% wavelength there is shorter by the same factor, and the grid barely
% resolves it. That is what the absorption is for. It damps what gets into
% the wall so that the badly resolved part never comes back out, which is
% also what a real wall does.
%
% Turn `cwall` above 1 for a genuinely rigid room and watch the timestep in
% the readout drop to match.
function [c, sig] = medium(x, y, L, h, c0, cwall, side, gap, thick, absorb)
  s = 1.5*h;

  % The shell between two concentric squares: |x| and |y| both within `side`
  % is the room, and out to side + thick is the wall.
  r = max(abs(x), abs(y));
  outer = 0.5 * (1 - tanh((r - (side + thick)) / s));
  inner = 0.5 * (1 - tanh((r - side) / s));
  wall = outer - inner;

  % The doorway, cut out of the right-hand wall. Written so that gap = 0 shuts
  % it exactly: max(0, ...) is identically zero there, where a tanh of the same
  % thing would leave a half-open cell behind and quietly ruin the seal.
  through = min(1, max(0, gap/2 - abs(y)) / s);
  right = 0.5 * (1 + tanh((x - side) / s));
  wall = max(0, wall - through .* right);

  c = c0 * (1 + (cwall - 1) * wall);
  sig = sponge(x, y, L, 0.2*L, 1700) + absorb * wall;
end