Series

Decidability — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 8 tiles over 5 colours. Wang tiles: unit squares with a colour on each edge, which may be laid side by side only where the touching edges agree, and which may never be turned or reflected. That last restriction is what makes them a computational object rather than a jigsaw — an edge colour is a symbol passed from one tile to its neighbour, and turning a tile would let a symbol change direction. The set here was generated from a stated seed.

    Nothing decides whether a set of tiles tiles the plane

    This collection rests on decidability — generate a pattern, forget the group, rediscover it, compare. One question in the same subject has no procedure at all: given a finite set of tiles, whether they cover the plane cannot be decided by any algorithm whatever. What can be done is two half-searches, and measuring what they leave behind.

    part 1 · classification
  2. heesch-two: surrounded 2 times. A shape that tiles nothing, with the rings of copies it does accept: the seed in the first colour and 2 coronas of 7 and 16 copies round it. The search that built this finished, so the shape's Heesch number inside this box is exactly 2, and it cost 3,097 placements. Every cell touching a tile of one ring, corners included, is covered by the next.

    Surrounded twice over, and covering nothing

    A shape that tiles the plane can be surrounded by copies of itself for ever. A shape that tiles nothing cannot be surrounded for ever — but it can be surrounded once, and sometimes twice, and the number of times is a measurement of how much local success a global impossibility permits.

    part 2 · classification
  3. 656 sets, every one decided. Every set of one, two, three and four tiles over two colours — sixteen tiles exist in all, so these are complete lists rather than samples — reduced by relabelling the two colour alphabets, and each set decided by the two half-searches. The last column is the one that matters: it is empty. At these sizes there is no room for a set that tiles the plane and admits no periodic tiling, which is the residue undecidability lives in. The smallest aperiodic set is known to have eleven tiles and four colours.

    How much room a hard question needs

    No algorithm decides whether a set of tiles covers the plane. Every set of four or fewer tiles over two colours is nevertheless decided here, exhaustively, in under a second — because the sets that defeat the two half-searches have nowhere small to live.

    part 3 · classification
  4. The parity argument loses 36 pairs it had won alone. The argument that refutes ten of the twenty-one species walks round a polygon of odd size: the ring of polygons about it is a closed walk of odd length in a graph the species decides, and a bipartite graph has no such walk. With two species at a vertex the flanking pairs come from the union of two graphs, and a union of bipartite graphs need not be bipartite — so the walk stops being constrained. The fourth row is the cost: pairs whose members the argument kills on their own and which it cannot kill together.

    The argument that closes eleven

    Twenty-one vertex species satisfy the angle equation; a parity argument kills ten before anything is drawn, and the eleven survivors are all built. Asking the same question of tilings with two kinds of vertex, the parity argument evaporates — it constrains a walk in a graph one species decides, and two species decide the union of two graphs, which need not be bipartite. What is left is a search, and a search cannot close a count.

    part 4 · classification

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