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Decided by exhaustion — page 7

Questions with finitely many cases, settled by going through all of them — and what changes when a claim about every argument becomes a count.
The postman's route: 24 blocks of street, walked in 28. A street network with its odd-degree vertices marked and the streets a shortest closed route must walk twice drawn doubled, dashed in a second colour, pairing up the odd vertices. Discrete

The streets a postman walks twice

A postman must walk every street of a district and come back. If every corner has an even number of streets, no street needs walking twice. If not, some must — and the ones repeated always join the odd corners in pairs. Pricing every way of pairing them finds the shortest round; pairing the nearest corners first does not.

 Six sentences from two quantifiers, and which imply which. A diagram of the six sentences that can be built from two quantifiers and a relation, arranged from strongest to weakest with arrows for implication, each labelled with how many of the 512 relations on three points satisfy it. Logic

Six sentences from two quantifiers

One relation, two variables, 'for every' and 'there is': there are eight ways to arrange them and six different sentences come out. Which of them imply which is a small, complete diagram, found by checking all 512 relations on three points — and the diagram crosses over in the middle, which is where every confusion about the order of quantifiers lives.

 The pairs from five, joined when disjoint, need three colours. The Petersen graph drawn with its ten vertices labelled by pairs from one to five, edges joining disjoint pairs, and a proper colouring with three colours. Topology

The colours a circle forces

Take every pair from five things and join two pairs when they share nothing. Three colours are enough to colour the result so joined pairs differ, and two are not — but no triangle, no dense cluster and no counting argument explains why. The reason is five points on a circle and a direction that cannot be told apart from its opposite, and the same reason, one sphere at a time, settles Kneser's question for every size.

 36 3-tuples with product e, and the 3 that no turn moves. Every ordered choice of 3 elements of the 6 symmetries of a 3-gon whose product is the identity, in cards grouped by cyclic turning. 3 cards hold a single tuple repeating one element; the other 11 hold 3 each. Algebra

Necklaces made of symmetries

Lagrange's theorem says a subgroup's size divides the group's, and the converse is false. One piece of the converse is true: every prime that divides the size is the order of some element. The proof threads the group's own elements onto a necklace whose product is nothing, turns it, and counts — the argument that proved Fermat's little theorem with beads, with the beads replaced by motions.

 The nearest consistent verdict: a 3-way tie at distance 4. A table of the 4 consistent judgement sets on the agenda p, q, and p and q, each with its number of disagreements with each of 3 judges and the total; the smallest total is marked. Applied

The nearest consistent verdict

When a court's majorities contradict each other, one repair is to announce the consistent verdict that disagrees with the judges least. It treats the premises and the conclusion alike, which neither of the two standard procedures does. On the classic case it returns a three-way tie; on five judges, with every question weighted equally, it never returns a single answer on a troubled profile at all — and what breaks the tie is a decision about which question matters more.

 Which agendas majority can vote on safely. A table of 7 agendas with the size of their largest minimally inconsistent set and the count of inconsistent majority outcomes over all profiles of three and five judges. Applied

Agendas that cannot contradict themselves

A court voting on two unconnected questions never contradicts itself, and neither does one voting on a chain of thresholds. A court voting on two premises and their conjunction sometimes does. What separates them is the size of the smallest sets of judgements that cannot all be true: pairs are harmless, because two majorities always share a judge, and triples are not. The same count says exactly how large a supermajority has to be to stay consistent on any agenda.

 The 1,344 tours of the 4-cube, by how often each place changes. A bar for each pattern of change counts among all closed walks through the 4-cube, with the number of tours having it; the reflected code's pattern and the perfectly even one are marked. Discrete

Every place changes back

A closed walk through every corner of a cube changes one place at each step, and each place, having changed, must change back before the walk returns home. So every place changes an even number of times — which is why no walk on three places can share the work evenly, why perfect sharing is possible only when the number of places is a power of two, and what sorts the 1,344 walks on the 4-cube into exactly four kinds.

 A cycle through the middle two levels of the 5-cube: 20 words. The words of length 5 with 2 or 3 ones placed round a ring in the order of a Hamiltonian cycle, alternating between the two levels, each step changing a single place. Discrete

The walk through the middle levels

On seven places, the words with three ones and the words with four number thirty-five each. Is there a closed walk through all seventy, changing one place at a time and never leaving those two levels? On five places the answer is 24 walks, on seven and nine a search finds one in moments — and whether one exists for every odd length was open for thirty years, until Torsten Mütze proved in 2016 that it always does.

 Twenty-seven choices of trisector, and the eighteen equilateral triangles. Twenty-seven small panels, one for each choice of trisecting line at each corner of a triangle, each drawing the triangle and the triangle the chosen lines cut out; the eighteen equilateral ones are marked. Geometry

Eighteen equilateral triangles

Every angle of a triangle has three trisectors, not one, once the angle and its outside are both counted. Choosing one at each corner gives twenty-seven ways to cut out a triangle, and eighteen of them give an equilateral one. The nine that fail are exactly the choices whose labels add to 2, 5 or 8 — and all eighteen equilateral triangles have their sides in the same three directions, fixed by a third of the difference between two angles.

 Weighted votes: two functions with a cut, and parity without one. For three functions of three letters, the eight assignments placed on a line by a weighted count, with true assignments filled and the threshold marked where one exists. Logic

A plane through the cube

Some truth functions are weighted votes: give each letter a weight, add the weights of the true letters, and say yes when the total passes a threshold. On the cube of assignments, such a function is a plane cutting the true corners from the false. Majority is one. Exclusive-or is not, and never can be — and of the 65,536 functions of four letters, only 1,882 are. The ones that are are exactly what a single artificial neuron can compute.

 9 corners of the 4-cube: some corner always has 2 chosen neighbours. The 4-dimensional cube with 9 of its corners chosen so that no chosen corner has more than 2 chosen neighbours, the fewest possible, with the edges between chosen corners drawn heavy. Logic

Half the cube and √n neighbours

Choose more than half the corners of an n-dimensional cube, any way at all, and some chosen corner has at least √n chosen neighbours. That statement about a cube settled a thirty-year question about how sensitive a truth function must be to its inputs, and its proof is a matrix of plus and minus ones whose square is n times the identity. A search over every choice for the 4-cube finds the bound exactly: nine corners, and some corner always has two chosen neighbours.

 The solutions of x² − 2y² = N, class by class. Rows for several right-hand sides N, each marking the solutions of x² − 2y² = N at the logarithm of x + y√2, coloured by class, over alternately shaded windows one unit-step wide. Number

Two families of solutions, and a box that holds both

Replace the 1 in Pell's equation by 7 and x² − 2y² = 7 still has infinitely many solutions — but they fall into exactly two families, each one an orbit of the same multiplication, and every family has a member inside a box whose size is fixed in advance. How many families there are is then a count of factors, and 3 has none.

 Euclid's game on 34 and 21. A 34 by 21 rectangle tiled by the squares of Euclid's algorithm, each run of equal squares shaded by the player who faces it, with the deciding run outlined. Geometry

The player who meets the first long run

Turn Euclid's algorithm into a game: two players take turns cutting squares off the rectangle, any number from the current run, and whoever cuts the last one wins. The whole game is decided before it starts — by whether the ratio of the sides is more or less than the golden ratio, which is the same thing as how many runs of length one come first.

 A smallest model of ∀x (Ax → ∃y (By ∧ ¬Cy)) ∧ ∃x (Ax ∧ Cx) ∧ ∀x (Bx → ¬Ax). Three overlapping circles with some regions shaded as empty and a single dot in each occupied region, forming a model of a sentence of monadic first-order logic. Logic

One thing in each region is enough

Give first-order logic its full apparatus of nested quantifiers but only one-place predicates, and every question about truth is still settled by the regions of a diagram. A predicate cannot tell apart two things in the same region, so no model ever needs more than one thing per region — and with three predicates there are only 255 models to try.

 Carroll's babies and crocodiles, on four ellipses. Four overlapping ellipses labelled with the four classes of a sorites, the regions emptied by its premises shaded, and the regions its conclusion requires to be empty outlined. Logic

The conclusion is what survives the erasing

Lewis Carroll's puzzles give three premises about four classes — babies, logical people, the despised, crocodile-managers — and ask what follows. Draw all four, shade what the premises rule out, then erase the classes the conclusion is not about: a region survives as empty only if everything above it was. What is left is the conclusion, and erasing a class turns out to be exactly one step of resolution.

 Six thirtieths of a turn that add to nothing. On the left a unit circle with the roots of unity in a vanishing sum marked and coloured by origin; on the right the same unit vectors placed head to tail, returning to their start. Algebra

The sums of roots of unity that add to nothing

All n of the n-th roots of unity add to zero, and so does any regular polygon among them, turned. Those are not the only vanishing sums: six thirtieths of a turn close into a loop with no polygon in them. Which counts of roots can close at all is decided by the prime factors of n — no seven fifteenths ever add to zero — and the same question counts where the diagonals of a regular polygon cross.

 A line in every direction in the plane over GF(7), in 31 points. A square grid of the points of a small finite plane with the points of a Kakeya set filled, beside a list of the lines it contains, one for each direction. Computation

No set with a line in every direction is small

In the plane over the integers modulo 7 there are 49 points and lines in 8 directions. A set holding a whole line in every direction needs 31 of the points — more than half — and in any dimension such a set fills a fixed share of the space. In the real plane the same sets can have area zero. Over a finite field one polynomial of low degree shows they cannot be small.

 A + B modulo 13: 4 and 3 residues make 9. A clock face of residues with two sets marked on an inner ring and their sumset marked on an outer ring. Computation

A sum of two sets modulo a prime cannot be small

Add every element of one set of residues to every element of another. Over the whole numbers the sums always number at least |A| + |B| − 1. Modulo a prime the sums can wrap round and collide, and still they never number fewer — the theorem Cauchy proved in 1813 and Davenport again in 1935. Modulo 12 they can. A polynomial of low degree explains the difference in a paragraph.

 Shares on two triangles and a go-between. A network of players with each node labelled by its share of what the whole network earns, and its number of links beneath it. Applied

Cutting a link costs both of its ends the same

Three players, any two of whom can earn 1 together — but only if they are linked. Link all three and each is due a third. Remove one link and the player holding both of the others is due two thirds. Averaging over orders on the game the network allows is the one rule under which breaking any link costs the two players it joined exactly the same, and it pays go-betweens more than their links.

 How often three candidates' majorities go in a circle. The exact probability that pairwise majority among three candidates is cyclic, for every odd number of voters from 1 to 41, under two models of how ballots are drawn, with their limits of about 8.77% and 6.25%. Applied

How often the majority goes in a circle

Three voters and three candidates give 216 profiles, and 12 of them are cycles. Count every electorate up to 41 voters exactly and the share climbs towards 8.77%, a number Guilbaud found in 1952 as the solid angle where three half-spaces at the tetrahedral angle overlap. Add candidates and a winner goes missing half the time; let voters share one axis and cycles vanish. The number is always a property of the model of how ballots are drawn.

 Latin squares of order 4, and the ones that are also Sudoku grids. Two 4×4 Latin squares with their 2×2 boxes outlined: one in which every box also holds 1 to 4, and the cyclic square, whose top-left box repeats a symbol. Of all 576 Latin squares of order 4, 288 pass the box test. Computation

A Latin square with boxes

A finished Sudoku is a Latin square of order nine with one extra rule: each 3×3 box holds every digit once. At order four the extra rule keeps exactly half of the 576 Latin squares, the 288 survivors are two grids in disguise, and no puzzle can be pinned down by fewer than four clues. At order nine every one of those questions needed a computer, and the answers are 6.67 × 10²¹ grids, 5.47 billion essentially different ones, and seventeen clues.

 Whole-number points in a strip, and the shadow they cast. The lattice points satisfying 2x ≤ 5y ≤ 2x + 1 for x from 0 to 30, and their projection onto the x-axis, which repeats every 5. Logic

Arithmetic with addition alone

Over the real numbers, a quantifier's shadow is described by inequalities. Over the whole numbers with addition and multiplication, a shadow can be any set a computer can list. In between lies arithmetic with addition and no multiplication, and there the shadows are always the same kind of thing: a finite exception, then a pattern that repeats. The whole numbers made from coins worth 6, 9 and 20 are every number from 44 on; the squares, which need multiplication, never repeat at all.

 15 stable matchings, by what each side pays. A scatter plot of every stable matching of one instance by the total rank each side receives, running from side one's best matching to side two's, with the median, the least-total and the most even matchings marked. Applied

The matching in the middle

List every stable matching of a market, give each member their stable partners sorted from best to worst, and hand each the one in the middle. Nothing says the result should even be a matching — two people might pick the same partner — and yet it always is one, it is always stable, and the other side gets its median partners too.

 Eighty-one cards and twenty with no SET among them. A three-by-three arrangement of three-by-three grids covering the eighty-one points of four-dimensional space modulo three, with twenty cells marked that contain no three on a line. Computation

Twenty cards with no set among them

The card game SET is a four-dimensional space over the integers mod 3, and a set is a line in it. Twenty cards can avoid every line and twenty-one cannot — a fact that took a proof in 1970 — while laying cards down at random and stopping when nothing more fits reaches twenty about once in two thousand tries.

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