Why every level behaves the same
Inside a supertile the ribbons only notice whether each side has an odd or an even number of edges. Going up a level makes every side longer, but never turns odd into even, so to the ribbons every level looks like the smallest one, which the computer checked. Here is why, in pictures.
This is the idea behind step 7 of the proof; the proof itself is on Prove, step 7.
The stepWhat happens to a supertile’s sides from one level to the next.
What changes when you go up a level?
A supertile is built from its children, laid side by side. Here are the eight children of a real Δ supertile, each side labelled with its number of edges. Go up a level and every number grows, but no side changes from odd to even, or back.
How can one check cover every level?
Level 1 was computed outright: all 18 ways of assembling it (nine kinds, with the crossover either way round) give the same look from outside, the signature that every supertile shares (Prove, step 4). Each next level is the same assembly with sides longer by even amounts, so it has the signature too. Each domino knocks over the next.
The reason: black and white cornersWhy odd stays odd, seen in the corners of the tiles.
Why does a side stay odd or even?
Every tile has 14 corners, an even number, so they can be coloured black and white alternately all the way round: a corner is black when it is an even-numbered corner of its tile (v0, v2, …, v12) and white when it is odd-numbered. Where two tiles meet along an edge, an even-numbered edge always meets an odd-numbered one (the same fact that puts their ports on the same point), so the colours agree. The corners of the whole tiling are coloured like a chessboard: every edge joins a black corner to a white one. (A computer check found no clash in a supertile of every kind four levels up, about 28,000 corners each.)
Walk along edges and the colour flips at every step. So a path has an even number of edges exactly when it starts and ends on the same colour. A supertile’s side is such a path: it is even when its two end corners match and odd when they differ.
So a side’s odd-or-even is settled by the colours of its two end corners and nothing else. Its parts, its children’s sides, always add up right: an odd part is one change of colour and an even part is none, so an even side has its odd parts in pairs and an odd side has an odd number of them. What the colours do not say on their own is which corners match. That is the next three questions.
How do the colours steer the zigzag?
Where two supertiles share a run of edges, the ribbons zigzag along it (Prove, step 5). Colour the corners of the run: they alternate, …, because every edge joins black to white. The blue ribbon is above the run at every black corner and below it at every white corner ; the pink one the other way round. So which side a ribbon comes out on is just the colour of the last corner.
Do the colours match from level to level?
The colouring comes from the tiles themselves, so colours can be compared between levels. Here are a tile, a supertile and a supertile of supertiles, with their key corners: the six main corners, numbered 0 to 5 (the corners of the hexagon a supertile looks like from far away), and the two crossover corners, A and B.
They match. Every kind has exactly the same colours at its six corners at every level (checked directly up to level 4), and the crossover corners are always A black and B white, like the tile’s own crossover corners v0 and v1. For A and B the reason is simple: a supertile’s A is its Γ child’s A, and so on down to the first tile of a Mystic, where it is that tile’s v0, which is black; B is the corner next to A, so it is white. The next question shows why the six corners keep their colours too.
What is the pattern’s DNA?
A parent’s corner is the same point as a corner of one of its children, and which child, and which of its corners, is the same at every level: it is part of how every level is assembled. So a corner can be followed down. Δ’s corner 1 is corner 4 of its Π child; that is corner 3 of the Π’s Ξ child; that is corner 4 of the Ξ’s Π child; and so on, down to a single tile, where it is one of the tile’s own corners. A point never changes colour, so the corner has that tile corner’s colour at every level.
Every corner of every kind ends up in one of four loops: corners that are their own child’s corner. This is the pattern’s DNA.
Pick any corner to follow it down:
So the colours of a supertile’s corners are copies of the colours of five corners of the tile itself, v0, v5, v6, v7 and v10, and the odd-or-even of every side, which is just whether its two corners match, is fixed once and for all. The only thing the tile has to get right is the one loop of two, Π’s corner 4 and Ξ’s corner 3: they swap at every level, and they sit on tile corners v10 and v6, both even, so both black. Had one of them been odd, those corners would change colour at every level, and the sides next to them would switch between odd and even.
What do the loops look like on the real tiles?
Each loop is a set of nested supertiles pinned at one point. A Ψ supertile’s corner 2 is the corner 2 of its Ψ child, which is the corner 2 of that one’s Ψ child, and so on: Ψs inside Ψs, all touching at the same point, down to a single tile. Zoom in on the point and every level looks the same. For Π and Ξ the supertiles alternate: a Π, inside it a Ξ, inside that a Π, …
The numbers behind itAn aside: the side lengths themselves are Fibonacci numbers, and that is how the paper shows odd stays odd (Prove, step 6).
Where do the Fibonacci numbers come in?
Every side length is a Fibonacci number in disguise. In the Fibonacci numbers each is the sum of the two before, so their odd-or-even goes odd, odd, even (odd + odd), and then the same again, for ever: every third one is even.
Why does it have to be Fibonacci?
Think of a side as a row of bricks of two sizes: a long brick, L, of 8 edges (an ε side), and a short brick, S, of 5 edges (half an η side, which has 10). Let the row grow by one rule: every long brick becomes a long and a short, L → L S, and every short brick grows into a long one, S → L.
That rule is the whole reason for Fibonacci. After a round, the long bricks are all the bricks of the round before (each brick leaves one long brick behind), and the short bricks are the long ones of the round before. So each new row is the row above it followed by the row above that, and its length is the sum of theirs: 8, 13, 21, 34, 55, … It cannot be anything but Fibonacci.
Three rounds make one level of supertiles. A long brick becomes L S L L S: 3 long and 2 short, the same as 3 ε sides and one η side, which is exactly how a parent’s ε side is made from its children’s sides. A short brick becomes L S L, so an η side (two shorts) becomes 4 ε sides and an η side: again exactly the real recipe. Then the whole row counts as one long brick of the next level, as eight children count as one parent.
What is a side made of?
Here is a real Δ supertile three levels up: 559 tiles, with its eight children outlined. Pick one of its sides. It is made of its children’s sides, and each of those is made of its own children’s sides, by the same recipe every time: an ε side is three ε sides and one η side, and an η side is four ε sides and one η side. Those are the long and short bricks above, with ε as one long brick and η as two short ones.
Do the recipes always add up?
A parent’s side is made of its children’s sides, laid end to end. So its length is a sum, and odd + odd is even, odd + even is odd: the sum’s odd-or-even is fixed by how many odd parts it has. Pick one of the Δ’s six sides:
Every side has a type, a small Greek letter, and a side of a given type is always built from the same types of child side, in every kind of supertile and at every level. So eight recipes cover everything. Check each one: the parts always add up to the same odd-or-even as the side itself. That copies level 1’s odd and even to level 2, level 2’s to level 3, and so on for ever (the paper’s Corollary 6.3 (Parities are constant; argued Lean)).
The signature is what a supertile looks like from outside (Prove, step 4): a hook at every corner but two, one crossover, and no ring inside. Since no supertile at any level has a ring, no strand ever closes into a loop, in any spectre tiling whose supertiles fill the plane: the Daedalus Theorem. The proofs have not been peer reviewed: what has been checked.