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Decided by exhaustion — page 10

Questions with finitely many cases, settled by going through all of them — and what changes when a claim about every argument becomes a count.
A random walk across the regions to the one that satisfies every clause. Four overlapping ellipses with a dot in each of their sixteen regions, one marked as the only assignment satisfying the clauses, and a path of arrows from a random starting region to it, each arrow crossing one ellipse. Logic

A walk that beats trying everything

To decide whether clauses of three letters can all be satisfied, the obvious method tries all 2ⁿ assignments. Uwe Schöning's method, from 1999, starts at a random assignment and wanders: pick a clause that is false, flip one of its letters at random, and repeat three times as many times as there are letters. A single try usually fails, but it succeeds with chance at least about (3/4)ⁿ, so about (4/3)ⁿ tries are enough — and the reason is a walk on a line that goes the wrong way two times in three.

Class numbers of x² + ny², with Euler's idoneal numbers marked. A scatter of class number against n up to 2000, on a logarithmic scale, with the 65 idoneal numbers marked on the power-of-two levels. Number

Euler's sixty-five convenient numbers

For some n, whether a prime can be written as x² + ny² is settled by its remainder on division by 4n alone, the way a prime's remainder on division by four settles whether it is a sum of two squares. Euler found sixty-five such n, from 1 to 1,848, called them convenient, and used the largest to prove that 18,518,809 is prime. Every one of them is a number whose class group has no element of order more than two — and whether the list is complete is still not known.

A single table for 9 guests over 4 nights. Small circles of 9 guests, one per night, each showing the night's seating as a closed zigzag path; every pair of guests is adjacent in exactly one of them. Probability

Every pair side by side, once

Seat an odd number of guests at round tables for as many nights as it takes, the same table sizes every night, so that every two guests sit side by side on exactly one night. For a single table a zigzag turned a notch each night does it for any number of guests. For other table plans the answer is almost always yes — and for six guests at two tables of three, nine at tables of four and five, and eleven at three, three and five, an exhaustive search proves it is no.

What survives on each agenda, 3 judges. A table of four agendas with, for each, the size of its largest inconsistent set, the number of independent unanimous rules for 3 judges, how many are consistent on every profile, and how many of those are dictatorships, oligarchies and other rules: two unconnected questions 324 of 324; a chain of two thresholds 129 of 324; two premises and their conjunction 7 of 5,832; a ranking of three options 3 of 5,832. Applied

What the agenda leaves standing

Ask for a rule that settles each question from the votes on that question, follows a unanimous court and never contradicts itself, and search every such rule for three judges. On a ranking of three options, three survive: one dictator per judge. On two premises and their conjunction, seven survive: every rule in which a fixed set of judges must all agree. On a chain of thresholds, a hundred and twenty-nine, majority among them. The difference is not in the rules. It is in which answers force which, and whether that forcing ever runs back.

Three splits that between them beat every other. The triangle of splits of one pound among three players, with the three half-and-half splits marked and every other lattice point coloured by which of the three dominates it: 134, 134, 134 beaten by exactly one, 91 by more. Applied

Three splits that answer every other

When any two of three people can take a pound, no division of it is safe from a pair walking off. Von Neumann and Morgenstern's answer was not a division but a set of them: three half-and-half splits that never beat one another and between them beat everything else. It is a solution — and so is the line on which one player is held at any fixed share below a half, so the same game has infinitely many, each describing a different settled way of treating the third player.

Six permutations as six arrows between three symbols. A directed graph with 3 vertices and 6 arrows, one per permutation of 3 symbols in shorthand, every vertex with two arrows in and two out, numbered in the order of an Euler circuit spelling 213231. Computation

Every ordering once, around a cycle

No cycle can show every ordering of three symbols as a window of three: a window holding each symbol once forces the next symbol to repeat the one just dropped, so the sequence has period three and shows three orderings of six. Two repairs work. Write each ordering by its first two entries and the transitions form a balanced graph, so Euler's theorem hands over the cycle at once. Or add a fourth symbol and ask only that each window keep a different relative order — which works too, but no graph explains why.

Which Paley graphs answer every small request. size 2: first passes at 13; size 3: first passes at 29; size 4: first passes at 89; primes checked up to 113. Logic

The squares that answer every request

Rado's graph has, for any finite sets U and V, a vertex joined to all of U and none of V. A finite graph can only answer the small requests, and the Paley graphs — residues modulo a prime, joined when their difference is a square — are the classic way to build one. Checking every request exactly finds the thresholds: 13 residues answer every request of two, 29 every request of three, 89 every request of four. The theorem that guarantees it asks for 64, 576 and 4,096. Coin-toss graphs of the same sizes almost never manage it.

One try: letters set in a random order, forced or guessed. Ten rows, one per letter in the order processed, marking 5 guessed and 5 forced letters, each forced one with the clause that forced it. Logic

A letter the clauses already decide

Set the letters of a formula one at a time, in a random order, and guess each one with a coin — unless some clause has already had its other two letters set to false, in which case the clause decides. A try succeeds when every guess is right, so what matters is how many letters get guessed. On a formula with one solution, every letter has a clause that forces it in at least one order out of three, so at most two thirds of the letters are guessed on average, and about 1.587ⁿ tries suffice. It is slower than the random walk. Refined, it is the fastest method known.

The seven graphs of the Petersen family. Seven small graph drawings, K₆, G7, K₃,₃,₁, K₄,₄ minus an edge, G8, G9, Petersen graph, arranged by vertex count with 6 arrows for the triangle-to-star exchanges between them. Topology

Seven graphs that must link

Six points joined in every way cannot be placed in space without two disjoint triangles hooking together. Trade any triangle for a three-pointed star, or a star back for a triangle, and the property survives; doing it in every possible way from K₆ reaches exactly seven graphs, the Petersen graph among them, and stops. Every placement of every one of them links, by the same parity count. Remove any edge from any of them, and placements that link nothing appear at once. The seven are the whole answer: a graph must link exactly when it contains one of them.

Petersen's graph and a tour that cannot close. The words of length 5 with 2 ones, joined when disjoint (10 vertices, 15 edges), with an open tour through every vertex drawn heavy. Discrete

The symmetric graphs no tour can close

Take the words of five places with two ones and join two when they share no one. The ten words look exactly alike to the graph, and no closed tour runs through them — because every way of leaving out one edge at each word leaves two pentagons. Only four connected graphs this symmetric are known to fail like that, and nobody knows whether a fifth exists.

Parity and majority on the cube, and the edges between true and false. Two cubes of three-letter assignments, parity with all twelve edges joining a true and a false corner and majority with 6; Khrapchenko's bound at its best 9.00 against an exact size of 10 and 4.00 against an exact size of 5. Logic

The formula that cannot share

A circuit computes the parity of n letters with a chain of n − 1 exclusive-or gates. A formula in and, or and not, which may use each result only once, needs n² letters at least — a bound Khrapchenko proved by counting the cube's edges between true and false — and exhaustive search over every formula of four letters finds parity sitting exactly on it.

Seven points split into three groups whose hulls share a point. Seven random points in a square split into three groups of sizes 2, 2, 3 whose convex hulls share a common point; 4 of the 301 splits work. Topology

Seven points split three ways

Any seven points in the plane can be divided into three groups whose convex hulls share a point, and six points in general position never can. The theorem is Helge Tverberg's; its topological version, in which straight lines may bend, is true when the number of groups is a power of a prime and false otherwise — and the proof that decides which is the same antipodal argument that halves a sandwich.

Every three-state machine and what becomes of it. halt: 1379; no halting entry: 12492; exact repeat: 361; shifted repeat: 2254; dies backwards: 36; unexplained: 27. Logic

The machines that need a reason never to stop

Run every three-state machine on a blank tape and the ones that halt announce themselves: the longest stops after 21 steps. The work is in the others. Each needs a reason it will never stop, and four simple kinds of reason settle all but 27 of them — machines that sweep back and forth over a growing tape and defeat every check that looks for a repeat.

Three polynomials of signs around the circle. The size of the polynomial around the unit circle divided by √64 for all-plus, random and Rudin–Shapiro sign polynomials of length 64; maxima 8.00, 2.50, 1.41. Algebra

The flattest polynomials of signs

A polynomial whose coefficients are all +1 or −1 has average size √n on the unit circle. Keeping it near √n everywhere is the problem Littlewood posed: the Rudin–Shapiro polynomials never exceed √2 times it, searching every sign pattern up to length 22 finds the best are the Barker sequences, and whether the maximum can come arbitrarily close to √n is still open.

The most nearly equal dissections of a square into 3, 5 and 7 triangles found. 3 triangles, areas 0.25000, 0.50000, 0.25000, spread 2.5000e-1; 5 triangles, areas 0.19098, 0.21353, 0.19098, 0.19098, 0.21353, spread 2.2543e-2; 7 triangles, areas 0.14151, 0.14458, 0.14151, 0.14458, 0.14151, 0.14174, 0.14458, spread 3.0711e-3. Computation

How nearly equal an odd number of triangles can be

A square cannot be cut into an odd number of triangles of equal area. It can be cut into five triangles whose areas differ by about two hundredths, seven that differ by three thousandths, nine by a ten-thousandth and a half — the spread falling by a factor of seven or more with every two triangles added, closing on equality and never reaching it. A search for the closest finds that the best five it can make have the golden ratio in their areas.

A random Latin square of order ten with an orthogonal mate. A 10 by 10 grid with a uniformly random Latin square and an orthogonal mate written into each cell; the square has 848 transversals. Computation

A mate is rare until order ten

Euler asked whether a Latin square can have an orthogonal mate, and for every order but two and six the answer is yes for some square. For a square chosen at random the answer is different. Exactly 6 of the 56 reduced squares of order five have a mate and none of order six does; of squares drawn at random, about one in a hundred of order seven has one, one in three hundred of order eight, one in seventy of order nine — and three in five of order ten.

Langford pairings of order 3, 4 and 7. Rows of boxes with arcs joining equal numbers: order 3: 3 1 2 1 3 2; order 4: 4 1 3 1 2 4 3 2; order 7: 7 3 6 2 5 3 2 4 7 6 5 1 4 1. Discrete

A sum that forbids half the pairings

Put two 1s, two 2s, …, two ns in a row so that between the two ks there are exactly k other numbers. For n = 3 there is one way, 312132; for n = 4 one way; for n = 5 and 6 none at all, and the reason is a single sum — adding up the positions of every entry in two ways forces n(3n − 1)/2 to be even. A search confirms the count forbids nothing that exists and finds 26 arrangements at n = 7 and 108,144 at n = 12, but the sum is all anyone knows about why.

A Lights Out board and the presses that clear it. A 5 by 5 board with 11 lights on, cleared by 7 presses; it has four solutions in all. Computation

The boards on which every light goes out

In Lights Out, pressing a light toggles it and its neighbours, and the goal is to turn every light off. Over the field with two elements the puzzle is a system of linear equations, and whether every pattern can be cleared depends only on whether one matrix is invertible. On a 3 × 3 board it is; on 5 × 5 a quarter of patterns can be cleared; on 39 × 39 one in four billion. Which sizes fail is decided by a common factor of two polynomials.

Luxembourg's votes and Luxembourg's power, 1958 to 1995. Paired bars for Luxembourg at five enlargements of the Council: vote share falling and power share rising from nothing in 1958 to 0.95% in 1973 and 3.02% in 1981. Applied

A vote worth nothing until it was outnumbered

From 1958 to 1973 Luxembourg held one vote of seventeen in the Council of the European Communities and could never once change an outcome. When the Council grew, Luxembourg's share of the votes fell and its share of the power rose from nothing. Power is not a quantity a member holds; it is a property of the whole assembly, and changing the assembly moves it in directions nobody would guess.

Three kinds of member, and a triple that breaks a grouping. Two copies of one market with three members of each of three kinds and cyclic preferences: on the left a grouping blocked by the triple B, 1, p; on the right a stable grouping. Applied

A third kind of member

Two sides always have a stable matching, and the proof is a procedure. Add a third kind of member — capitals who rank numbers, numbers who rank letters, letters who rank capitals — and every market small enough to check still has a stable grouping. Ten million markets of three have at least ten each. Nobody can prove it continues, because the structure the two-sided proof stands on is gone.

No pair of weights modulo ten catches everything. Two 10 × 10 grids over pairs of alternating weights modulo ten, shaded by the share of single errors and of swaps each catches. No pair is full in both; weights 3 and 1, ringed, catch every single error and 88.9% of swaps. Computation

Ten digits need a symmetry that does not commute

A check digit should catch one wrong digit and two neighbours swapped. Over eleven symbols a weighted sum does both; over the ten decimal digits no weighted sum can, no scheme of any shape built on adding modulo ten can, and the reason is the same as the reason Euler's thirty-six officers cannot be paraded. What works is the ten symmetries of a pentagon, which do not commute.

Four seeds of the middle-square method, each until it repeats. Values against step for the four-digit seeds 6239, 1234, 5735, 4100 under the middle-square rule, each ending where a value repeats: after 111, 57, 31, 4 steps. Computation

Square it and keep the middle

John von Neumann's first generator of random numbers squared a number and kept its middle digits. Followed from every four-digit seed, it never lasts more than 111 steps before a value repeats — a third of what the birthday problem allows a truly random rule — and one seed in five ends at zero for ever. A counter added before each squaring cures the collapse.

Six formulas on the interval of truth values. Graphs on [0, 1] of six one-variable formulas of Łukasiewicz logic, each a zigzag of whole-number slopes with values 0 or 1 at the ends. Logic

The zigzags a formula can draw

Let truth be any number from 0 to 1, and Łukasiewicz's connectives turn every formula in one variable into a graph. Every graph that appears is a zigzag of straight pieces with whole-number slopes, ending at 0 or 1 — and McNaughton proved in 1951 that every such zigzag appears. A logic of degrees of truth turns out to be a theory of piecewise-linear functions with integer coefficients.

Seventy, and the totals its divisors can make. A strip of the totals 0 to 74 marking those reachable as sums of distinct proper divisors of 70; 70 itself and the excess 4 are not. Number

Abundant, and still not a sum of its parts

Seventy's proper divisors add to seventy-four, more than seventy itself — and yet no selection of them adds to exactly seventy. The reason is that the excess, four, cannot be made from the small divisors one, two and five. Numbers like seventy are called weird; there are thirty-six below twenty thousand, every one of them even, and nobody knows whether an odd one exists.

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