MathLabs

Worked solution: Smith–Myers–Kaplan–Goodman-Strauss's aperiodic monotiles: the hat and the Spectre (2023)

Step 2 of 8: The hat sits inside a whole continuum Tile(a,b)\mathrm{Tile}(a,b) of related shapes
In plain words

The hat's boundary is made of edges of exactly two lengths, coming in matching parallel pairs — the way a person's arms come in a matching pair even though the two arms point different directions. If you are allowed to stretch or shrink one of the two lengths independently of the other, you get a whole family of related shapes that all fit together the same combinatorial way; the family is written Tile(a,b)\mathrm{Tile}(a,b) for the two edge lengths aa and bb, and the hat is just one specific member of it.

Almost every member of this family turns out to be an aperiodic monotile just like the hat. Only three special members — where the two lengths make an unusually symmetric shape — happen to tile periodically instead, and those three exceptions turn out to be exactly the shapes needed to prove the rest are aperiodic.

Tile(a,b),hat=Tile(1,3),turtle=Tile(3,1)\mathrm{Tile}(a,b), \quad \text{hat} = \mathrm{Tile}(1, \sqrt{3}), \quad \text{turtle} = \mathrm{Tile}(\sqrt{3}, 1)
Detailed analysis

The hat is built from kites of the [3.4.6.4][3.4.6.4] tiling, whose edges come in two lengths that can be taken to be 11 and 3\sqrt{3} (an edge of length 22 counts as two consecutive edges of length 11), and these edges occur in parallel pairs. Allowing the two lengths a,b≥0a, b \ge 0 (not both zero) to vary independently gives a continuum of shapes Tile(a,b)\mathrm{Tile}(a,b), all combinatorially identical (Smith, Myers, Kaplan & Goodman-Strauss 2023, §2). The hat is Tile(1,3)\mathrm{Tile}(1, \sqrt{3}); a second shape Smith found independently, the "turtle" (a 1010-kite polykite), is Tile(3,1)\mathrm{Tile}(\sqrt{3}, 1) in the same family.

Three special members admit simple periodic tilings instead of aperiodic ones: the "chevron" Tile(0,1)\mathrm{Tile}(0,1) (a tetriamond made of 44 equilateral triangles), the "comet" Tile(1,0)\mathrm{Tile}(1,0) (an octiamond of 88 equilateral triangles), and the equilateral Tile(1,1)\mathrm{Tile}(1,1). Every other member of the continuum, including the hat itself, is an aperiodic monotile.

The chevron and comet are not a side note: because contracting all edges of one length to zero in any hat-tiling turns it into a chevron-tiling, and contracting the other length turns it into a comet-tiling, these two periodic relatives become the tools used in the next step's aperiodicity proof.

Knowledge used in this step