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