Concept

Exhaustive search — where it appears

An enumeration that covers every candidate inside a stated bound, so that finding nothing establishes there is nothing — inside the bound. It settles a count only when some argument says nothing outside the bound could be new, and reporting the bound beside the result is what keeps the two claims apart.

Named by 6 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

classification · Decidability
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.

classification · Decidability
Two kinds of atom, and more ambiguity rather than less. Exhaustive searches on rings of four sizes. The third column counts homometric groups when every atom is identical; the fourth counts them when each atom may be one of two kinds. The fourth is larger at every size, and at nine and ten sites the third is nothing at all — there is no pair of arrangements of four identical atoms that a diffraction experiment cannot separate, and there are six and four once the atoms may differ. Distinguishing the atoms adds information to the structure and adds ambiguity to the measurement.

When the atoms are not all the same

Every homometric pair found so far is a pair of point sets, where an atom is a point and counts once. Give the atoms different scattering powers and the ambiguity does not go away — it grows. On a ring of nine there is no pair of four identical atoms that diffraction cannot separate, and there are six once two kinds of atom are allowed.

diffraction · Homometry
Where the sphere and the projective plane have no net. The number of different closed nets with three bonds at every atom and faces that are pentagons and hexagons only. On the sphere, with twelve pentagons and k hexagons for k up to 12, every count has at least one net except k = 1. On the projective plane, with six pentagons and h hexagons, each count sits under the sphere count it lifts to, since every hexagon of a projective net becomes two on the sphere. The projective counts for h = 0 to 6 are 1, 0, 0, 1, 1, 3, 3, so the projective plane has no net at h = 1 or 2: two gaps where the sphere has one. Every sphere count was found by enumeration and agrees with the published one.

A gap the sphere does not have

A net of pentagons and hexagons on the projective plane must have six pentagons, and the count permits any number of hexagons. Not every number happens. Lifting each net to the sphere turns the question into one about which cages have a centre — and the answer leaves two gaps where the sphere has one.

restriction · Curvature
Two symmetric structures a diffraction pattern cannot separate. Two arrangements of 6 atoms on a 6 × 6 torus, each invariant under the plane group p6m, drawn beside the Patterson they share. No translation and no inversion carries one onto the other, so they are different structures; every one of the thirty-six interatomic vector counts is the same, so every diffracted intensity is the same and no measurement at any resolution separates them. Of the 4 structures with this symmetry and this many atoms, there are only 3 Pattersons — so imposing the most symmetric of the seventeen plane groups has not removed the ambiguity.

Symmetry does not rescue a Patterson

Every homometric pair found so far sits on a bare ring with no operations imposed, and a real crystal sits in a space group. Impose one and the ambiguity does not go away: 12 of the 13 groups searched still have pairs, and at six atoms the hexagonal groups are indistinguishable two to three times as often as the general position.

diffraction · Homometry
Every vector realised, and not at the same hexagon count. Each row is a set of faces other than hexagons whose charge — the sum of 6 − k over them — comes to twelve, which is what a closed trivalent net on the sphere must pay. Each column is a number of hexagons added to that set, and the entry is how many different solids exist with exactly those faces, found by winding up every arrangement of them into a spiral. A dash means the search found none; a question mark means the planar reader declined the row and it is not evidence either way. Every row has an entry somewhere, which is Eberhard's theorem, and the first one is at 0, 2, 3, 4 hexagons depending on the row — so the charge decides everything except the number of hexagons, and the number of hexagons is not a function of the charge.

Everything except the hexagons

Three counts of what a closed net must carry end on the same admission: an arithmetic saying what a net must charge does not say that a net exists. Eberhard's theorem says how close the charge comes to being enough, and the answer has a shape nobody would guess — it fixes every face count except the hexagons, and the hexagons are exactly the entry it cannot see.

restriction · Curvature

Named alongside it

The objects these essays reach for when they reach for this one.

EnumerationCensusCombinatorial curvatureThe Euler characteristicHomometryOrbitThe Patterson functionPlane groupSemi decisionStructure factorAperiodic tile setAperiodicity

All concepts