Exhaustive search
Named by 134 essays across 11 fields — each of them below, with the objects they name alongside it.
Three trisectors and a triangle nobody expected
Cut every angle of a triangle into three. The trisectors nearest each side meet in three points, and those three points are always the corners of an equilateral triangle — for every triangle there is, with no exceptions and no reason anybody finds obvious.
A game that decides what can be said
Two players take turns pointing at elements of two structures; if the second can survive k rounds, then no sentence with k quantifiers tells the structures apart — a statement about infinitely many formulas, settled by a finite search.
Every word once, around a cycle
A cyclic string of eight bits holds all eight three-bit words, each exactly once — and the reason such a thing exists is that the constraint linking overlapping windows is itself the construction.
The order everybody arrives in
Three people jointly earn nine, and the question is what each is owed. Ask instead what each adds on walking into a room the others are already in, average that over every order they could have arrived in, and four modest conditions leave no other answer.
A split nobody can walk away from
Every way of dividing what a group earns is a point of a triangle, and every coalition's threat to leave cuts a straight line across it. What survives all the cuts is the set of stable divisions — and for one three-player game there is nothing left.
The court that contradicts itself
Three judges each answer three questions, and each answers them consistently. Take the majority on each question separately and the answers no longer hang together — the body as a whole endorses a combination no member of it holds, and no rearrangement of the procedure removes the problem.
The planes a recurrence cannot leave
One multiplication and one addition, taken modulo a fixed number, produce a sequence that passes for random one value at a time. Taken two or three at a time it does not, and the reason is a whole-number relation that pins every point onto one of a small family of parallel lines.
The axiom is the shape of the graph
Add one operator meaning necessarily and the choice of which axioms to accept stops being a matter of taste. Each candidate axiom is true of exactly those worlds-and-arrows diagrams whose arrows have a stated property, and a logic is a class of graphs.
The edge that forces a triangle
A graph on six points can carry nine edges with no three of them closing a triangle. It cannot carry ten. The bound is n²/4, the graphs that achieve it are all the same shape, and both facts fall out of examining every graph there is.
Three in a row on the number line
Colour the numbers one to eight in two colours and it can be arranged that no three equally spaced numbers agree. Add the ninth and it cannot. The structure being forced is arithmetic rather than graphical, and the proof is a different proof.
The plane hiding in the squares
A complete family of orthogonal squares is not a collection of squares that happen to agree nowhere. It is a geometry — a plane with n² points in which every two points lie on exactly one line — and reading it that way is how the impossible orders were found.
Nine thousand four hundred and eight
There are four Latin squares of order four once the first row and column are fixed, fifty-six of order five, and nine thousand four hundred and eight of order six. The exact answer is known for eleven orders and for no more — and yet a half-finished square can always be finished.
Sixteen of five hundred and seventy-six
A Latin square is a multiplication table in which every equation has exactly one solution. Ask it to be associative as well and almost every square drops out — sixteen of the five hundred and seventy-six of order four survive, and they are the two groups.
The triangle nobody can settle
Does every triangular billiard table have a path that closes on itself? Acute triangles do, right triangles do, triangles with rational angles do — and for the rest the question has been open since it was asked.
Two diagrams the language cannot tell apart
A modal formula sees a diagram of worlds and arrows through a very narrow window. Exactly how narrow is settled by a game: where one player can answer every move, no formula whatever separates the two starting worlds, however different the diagrams look.
The axiom with no property of the arrows
Each axiom of modal logic can be matched by hand to a condition on the arrows between worlds. There is a recipe that does it for a whole class of axioms, and there is an axiom the recipe cannot reach — not because nobody has looked, but because no condition on the arrows defines it at all.
Nearly always, or nearly never
Toss a coin for every pair of points and ask whether the graph that results has some property. For a property a first-order sentence can state, the answer in the limit is never a genuine probability — it is zero or it is one, and the game is what proves it.
The distance a sentence can see
A first-order sentence with three quantifiers cannot notice anything about a graph beyond a fixed distance from the points it names. That single limitation is why it cannot say connected, and why the failure survives every attempt to add more quantifiers.
The only bit that survives
A shuffle can be called even or odd, and the label behaves under composition. Ask whether some cleverer label — a number out of three, or out of four — could behave the same way, and the answer is that nothing else can — one bit is exactly what a permutation gives up.
The puzzle that is exactly half solvable
A sliding puzzle sold with two tiles swapped is not a hard puzzle; it is an impossible one, and the proof is a quantity that no slide can change. The same argument, run three times at once, says that one arrangement of a scrambled cube in twelve is reachable.
Three points, however many there are
A point inside the hull of a thousand points is inside the hull of three of them. Any four points split into two groups whose hulls meet. And a family of convex sets, every three of which have a common point, has one common to all — three, in each case, being one more than the dimension.
Four ways out, and what each costs
An impossibility theorem lists conditions and says no rule has them all. That leaves exactly as many escapes as there are conditions, each of them a real institution — a dictator, a two-stage procedure, a supermajority, a restricted agenda — and each escape's price can be counted rather than argued about.
The best a code can be
A code is a set of words chosen far apart, and every construction answers "here is one" rather than "here is the best". The best can be computed at small lengths, and putting four classical bounds beside the exact answer shows which of them is doing the work and where none of them is.
![How many codewords lie within each radius, for a [7,3] code over 11 symbols. A bar for each decoding radius, its height the largest number of codewords found inside a ball of that radius around a randomly drawn received word, with the unique-decoding radius and the Johnson radius marked.](/f/how-many-codewords-lie-within-each-radius-for-a-7-3-code-e1fb4777.svg)
Past half the distance
A code of minimum distance five corrects two errors, and every account stops there. Two is the largest number for which the answer is unique — and a decoder that returns a short list instead of one answer reaches considerably further, which can be measured by counting the codewords in a ball.
Sixteen polygons with one dot inside
Fix one of Pick's two counts at one and ask what is left. The answer is a finite list, the list has exactly sixteen entries, each one is its own kind of object with a dual that is another entry, and the whole classification is a search a page can carry out.
The dots a circle catches
Pick's theorem gives a lattice polygon's area exactly, with no error term anywhere. Ask a circle the same question and the exactness is gone: the count is the area plus something, the something has been measured for two centuries, and nobody knows how big it is.
How short a cycle could be
The drift argument cannot see cycles at all, which is why it is not a proof. What can see them is arithmetic — a cycle's shape has to be a fraction that approximates the logarithm of three to base two extraordinarily well, and there are very few such fractions.
Every pattern happens exactly once
Choose any sequence of odds and evens and there is exactly one residue class whose orbit follows it, and exactly one fraction that cycles through it forever. The Collatz conjecture is then the statement that only one of those infinitely many cycles is made of whole numbers.
The game the algorithm was playing
Change what Duplicator has to offer — a whole bijection instead of one element — and the game stops measuring first-order logic and starts measuring colour refinement, the algorithm every practical graph-isomorphism test begins with. Two subjects that grew apart are one game with the moves relabelled.
A signal both can see
Two choosers who randomise privately can reach a set of outcomes that is smaller, and worse, than the set they reach when a device draws one cell and whispers each of them their half of it. Nothing is enforced and nobody is bound, and the arrangement is stable anyway.
The landscape nobody is looking at
Letting participants move one at a time to whatever is currently better can cycle forever, and on a network of congestible roads it cannot. The reason is a single number attached to each state that falls by exactly what the mover saves.
Finitely many, and nobody says how many
The theorem promises a dissection exists and the proof produces one. Running the proof on a hexagon produces thirty-nine pieces, ingenuity produces five, and there is no method for proving that five cannot be four.
The table inside every quota
Give seats to districts and parties at once, and every cell of the table has a fair share it ought to round from. A table rounding every cell to its floor or its ceiling, with every total exact, always exists. The biproportional method does not always choose one: here it gives a party 2 seats where its fair share is 3.088.
Where the rounding runs out
In two dimensions a table of seats inside every fair share always exists. Add a third family of totals — every district and party split between groups — and it need not. Sixteen halves in a three-by-three-by-three table meet every total, and no whole table does it without a seat where the fair share is nothing, because the halves close a loop of seven.
One tree for every cut
A network of six places has fifteen pairs, and each pair has its own cheapest cut. All fifteen can be read off a tree with five numbers on it: the cheapest cut between any two places is the smallest number on the tree's path between them. Gomory and Hu proved in 1961 that such a tree always exists, and building it takes five cuts, not fifteen.
A failed search is a proof
Search for an assignment by branching on variables and backing up whenever a clause turns false. If every branch fails, the tree the search leaves behind is itself a resolution refutation: write at each branch point the resolvent of the clauses below it, and the top of the tree is the empty clause. So every limit on short refutations is a limit on every such search — and the pigeonhole clauses, whose refutations are long, defeat them all.
As many cuts as colours
Two thieves steal a necklace and want half of every colour of bead each. However the beads are strung, they never need more cuts than there are colours — three cuts for three colours, four for four — and sometimes they need every one. The guarantee is the Borsuk–Ulam theorem again, with a point on a sphere read as a way of cutting the necklace, and every necklace of several small kinds has been checked against it.
The exponent that is smaller than Euler's
Euler's theorem raises every unit to the count of the units and gets one. The smallest exponent that works for all of them at once is often much smaller — and a composite is invisible to Fermat's test exactly when that smaller number divides n − 1.
Five weighings and the question is closed
Searching the triangle of splits can only ever fail to find a stable one, which is not the same as there being none. Weighing five families of coalitions against the whole settles the question outright — and the family that fails is the proof that nothing survives.
An objection one player makes to another
The core lets a coalition object to everybody at once. Narrow it to one player objecting to one other, require every such objection to be met by an equally loud one coming back, and exactly one split survives — with no dictionary order anywhere in the argument.
The bound is the answer to a search
Chebyshev's inequality is not a clever estimate that happens to be sharp. It is the exact answer to a maximisation over all distributions with a stated mean and variance, and the polynomial that proves nothing beats it is the certificate a search of that kind always produces.
No single input can move it far
Independence was never the hypothesis doing the work. A quantity built from many separately drawn inputs concentrates whenever changing one of them moves it only a little — and that covers quantities which are not sums of anything and have no formula at all.
Refutable in something small
A formula that is not a theorem of the constructive system fails in some finite algebra, and the algebra can be found by search. That single property is what makes the propositional logic decidable — and the predicate version loses the property and the decidability with it.
The boundary at three variables
Restrict a sentence to two variable names and it can still be arbitrarily long, because the names are reused. What it cannot be is deep: every satisfiable two-variable sentence has a small model, so asking whether one is satisfiable is a bounded search. Allow a third name and the question becomes undecidable.
A plane no field built
Every finite field builds a projective plane, and for a long time every known plane was built that way. The plane over Dickson's nearfield of order nine has ninety-one points, ninety-one lines and every incidence right — and Desargues' theorem fails in it on most configurations tried, except for one line, from which it never fails at all.
The curve that no three points in line define
In a finite plane, take as many points as possible with no three on a line. In odd order the largest such sets have one more point than the order — and every one of them, searched exhaustively in the small planes and proved by Segre for all odd orders, is a conic. In even order every tangent meets at one point, which can be added, and the curves stop being forced.
Five spokes squeezed into K5
The Petersen graph has no point with four neighbours, so no stretched copy of K5 can sit inside it. Contract its five spokes and K5 appears anyway. Kuratowski's theorem forbids stretched copies and Wagner's forbids squeezed ones, the two notions disagree on this graph — and they still name exactly the same planar graphs.
The people every stable answer leaves out
Let the lists be short and let one side take several partners. Stable matchings still exist and there can be many of them — but every one leaves out exactly the same people, and a member who is left with an empty place holds exactly the same partners in every one. A three-line count proves it.
A ring that no pairing can break
Put everybody in one pool and a stable pairing may not exist. Allow rings as well as pairs and something stable always exists — and the pairs-only answer fails exactly when that stable arrangement contains a ring of odd length. Two sides make every ring even, which is the whole reason the two-sided theorem holds.
The densest graph without a square
Forbid four points joined in a cycle and a graph can keep only about ½n^(3/2) of its edges — far fewer than the quarter of all pairs a triangle-free graph keeps. Counting pairs of neighbours proves the ceiling in two lines. What reaches it is not a random graph but a finite geometry: the points of a projective plane, joined when they are orthogonal.
No local rule can count the votes
A ring of cells, each holding 0 or 1, has to agree on whichever value is in the majority — every cell seeing only its neighbours. The best-known rule gets it right most of the time and wrong near a tie; no rule of any radius gets it right always. Yet two rules run one after the other do, on every ring, and the first of them is the traffic rule.
Three ordinary lines from a count
Kelly's proof finds one line through exactly two of the points by minimising a distance. Melchior, seven years earlier, had found three — by turning every point into a line and counting the corners, edges and regions of the picture that results. Euler's formula for the projective plane does the rest, and it says exactly which configurations have no more than three.
Infinitely many guessers, finitely many wrong
An infinite line of people each wears a black or white hat, sees every hat in front and none of their own, and must guess their own colour. With a finite line, each guesser is right half the time whatever they agree in advance. With an infinite line and the axiom of choice, they can agree a strategy under which all but finitely many are right — and nobody can carry it out.
One cell short of a transversal
A transversal of a Latin square picks one cell in every row and every column with every symbol different. The cyclic squares of even order have none, and that was settled by a parity argument centuries old. Whether every square of odd order has one is a conjecture from 1967 that nobody has proved; whether every square comes within one cell of having one was settled only in 2023, and only for squares large enough.
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
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 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.
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.
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.
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.
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.
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 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.
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.
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.
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 walk that may not step where it has been
Forbid a walk on the square grid from ever revisiting a site and the number of possible n-step walks grows like 2.638ⁿ instead of 4ⁿ — a number nobody can write down exactly. On the honeycomb it is exactly √(2 + √2), proved in 2010. And the walks spread out like n to the three-quarters, faster than any ordinary walk, which physicists have used since 1949 and mathematicians still cannot prove.
Cars that park, and trees that grow
Three cars arrive at a one-way street with three spaces; each has a favourite space, drives to it, and takes the first free one from there on. Of the 27 lists of favourites, exactly 16 let every car park — the same 16 as the labelled trees on four points. The reason is a circular street with one extra space, on which every list parks and exactly one rotation of it leaves the extra space empty.
A random tree is one part in e leaves
Choose a labelled tree on n points uniformly at random. A point is a leaf exactly when its label never appears in the tree's Prüfer code, so the share of leaves is (1 − 1/n)^(n − 2) — half the points for a tree on four, 36.8% for a large one, the reciprocal of e. The whole degree distribution follows the same way: one plus a Poisson count with mean one.
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.
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.
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.
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.
Three places cut apart
Separating two places as cheaply as possible is solved exactly by a flow. Separating three from one another is a different problem: no flow measures it, the pairwise answers do not add up to it, and the best shortcut known in 1994 — cut each place off on its own and throw the dearest cut away — is guaranteed only to within a third of the truth.
Divisors that add to three times the number
The divisors of 6 add up to 12, twice 6: a perfect number. The divisors of 120 add up to 360, three times 120, and those of 30,240 to four times it. Numbers like these were a sport for Fermat and Descartes, and they are held together by one fact — the ratio σ(n)/n is a product over the primes, and each prime can add only a little.
A ratio nobody else has
Divide the sum of a number's divisors by the number and you get its abundancy: 2 for every perfect number, 12/5 for both 30 and 140. Numbers that share an abundancy are called friends. Some numbers provably have no friend at all, most have friends only far away — and for 10, whose abundancy is 9/5, nobody knows whether a friend exists.
Three patterns in a circle
Race three coin patterns at once and the gamblers' accounting still gives each one's chance of arriving first — one fairness equation per pattern. What it does not give is any way to read the three-way result off the two-way ones. HHHT, TTHH and HTTH beat one another in a circle, and HHH loses both its head-to-head races and still finishes ahead of one of the patterns that beat it.
A proof that says which half
Classical logic proves p ∨ ¬p without any idea which half is true. The constructive system never does that: whenever it proves a disjunction, it proves one of the two halves. The reason is a picture — two countermodels placed side by side above a new first stage that forces neither — and the logics between the two lose the property exactly when their pictures are not allowed to be glued.
Several colours on every vertex
Give every pair from six points three colours, so that pairs with nothing in common share no colour. Counting says nine colours might do; ten are needed. Stahl conjectured in 1976 exactly how many colours every such problem needs — a formula that meets Lovász's topological answer at one colour a vertex and the obvious answer at k — and a search over stars and triangles confirms it in every case small enough to run.
At least as many lines as points
Sylvester's theorem says some line through two of the points misses all the rest. Remove one end of that line and the line itself disappears, taking at least one line away with one point. Run that backwards and it proves that n points not all in a line determine at least n lines — and the only sets that manage exactly n are a line of n − 1 points with one point off it.
A contradiction that is only a sum
Put a variable on every edge of a graph and ask each vertex for an odd or an even number of true edges, with the demands adding up to odd. Add all the demands and every edge is counted twice, so the left side is zero and the right side is one: the contradiction is a single sum. Resolution cannot add. It has to reach the same conclusion clause by clause, and on a graph where every group of vertices has many edges leaving it, that takes exponentially long.
The wait for the next sum of two squares
Sums of two squares thin out to a share of nought, and yet the gaps between them stay short. Take the largest square below any number; what is left over is small, and a small number is always close to a square. Two squarings in a row say the next sum of two squares is never more than about 2√2 times the fourth root of n away — a bound proved in 1947 that nobody has improved, sitting far above every gap anyone has found.
Six points in space and a pair that must link
Put six points anywhere in space and join every pair with a straight segment. Split the six into two triangles — there are ten ways — and at least one of the ten pairs of triangles is linked like two rings of a chain. No placement avoids it. The reason is a parity: moving an edge through another changes exactly two of the ten linking numbers, so their sum stays odd whatever is done.
Envy that any single item would cure
Envy-free up to one item lets a person's envy be excused if removing the envied bundle's best item would cure it. The stronger standard asks that removing any item would — even the one that person values least. Every allocation of three people's items can be searched, and an allocation meeting the stronger standard was there every time; for two people cut and choose finds one, for three it took until 2020 to prove, and for four nobody knows.
Thirty-one moves from solved
Parity settles which half of a sliding puzzle's arrangements can be reached and is silent about how far away any of them is. Searching every reachable arrangement of the three-by-three tray answers the second question exactly — two arrangements sit thirty-one moves out — and parity turns up again, this time as a law about distance.
Which graphs let the tokens go anywhere
A sliding puzzle is a graph with a token on every vertex but one. Richard Wilson found in 1974 what every such puzzle can reach, and the answer has a surprise in it — the half the tray is stuck with is not a fact about permutations at all, but about the board being two-coloured — and one exception, a graph of seven vertices that reaches exactly 120 of 720.
The order decides the colours
The simplest way to colour a graph is to take the vertices one at a time and give each the first colour its neighbours are not already using. It never needs more than one colour beyond the largest degree — and on a graph that needs only two colours it can be made to use as many as there are vertices on a side, depending on nothing but the order it is handed.
How many colours the plane needs
Colour every point of the plane so that no two points exactly one unit apart share a colour. Seven colours are enough, by a tiling with hexagons, and four are necessary, by a graph of seven points. For sixty-eight years nothing better was known on either side; then in 2018 a graph of 1,581 points showed four is not enough — and the answer is still somewhere from five to seven.
No algorithm can read what a program does
Whether a program halts cannot be decided by any algorithm. Henry Rice showed in 1953 that the halting problem is not special: no algorithm can decide any property of what a program does — whether it ever prints a 7, whether it computes the successor function, whether it is a virus — except the two properties that hold of every program or of none. Every one of the 20,736 smallest Turing machines can be checked by hand; the theorem says why that stops.
Every crowd holds a bowl or a dome
Among enough points in the plane, some k of them always bend upward like a bowl or some l bend downward like a dome. The number that forces it is a binomial coefficient, it is exactly right, and it proves that every large enough crowd contains a convex polygon — with a bound nobody could lower to the true answer for eighty years.
Two graphs the eigenvalues cannot tell apart
A graph's matrix has eigenvalues, and they count a surprising amount of the drawing: its edges, its triangles, every closed walk of every length. They do not count everything. A star with four arms and a square beside a lone point have the same eigenvalues exactly, although one of them is in two pieces — and on six points ten of the 156 graphs have a twin of this kind.
The corners are the orders of arrival
Line the players up, let each join in turn, and pay each what it adds on arrival: every order gives a split. When a newcomer always adds at least as much to a bigger group, those splits are exactly the corners of the core — so the core is never empty, it is the outline of the orders, and the average over all of them lies inside it. For a group worth the square of its size the outline is a hexagon whose corners are the six orderings of 1, 3 and 5.
Each user pays for its own last link
Several users must be connected to a source, and the cheapest network that does it is a tree. Dividing its cost so that no group of users would rather build its own looks like a hard search, and it has a one-line answer: each user pays for the link that joins it to the tree on its way to the source. No group is ever overcharged — while the average over orders of arrival, the rule that settles so much else, can charge a pair more than its own connection costs.
Every power of x that draws a hyperoval
In a plane of order 2^h, the graph of x^k plus two points at infinity is sometimes a hyperoval — as many points as a plane allows with no three in line. Searching every exponent in every plane from order 4 to 4096 finds hundreds that work, and once six symmetries of the problem are applied they fall into exactly the families already known: the conic, the translation curves, Segre's x⁶ and Glynn's two. Whether that list is complete in every order is open.
Windows counted like necklaces
The logistic map has one window of period three, two of period four, three of period five, five of period six, nine of period seven — and the sine map, which shares no algebra with it, has exactly the same numbers, in exactly the same order along the parameter. The counts are 1, 1, 1, 2, 3, 5, 9, 16, 28, 51, and they are the number of ways to thread a necklace of beads in two colours.
How few voters any majority needs
Any pattern of head-to-head majorities whatever — cycles within cycles, a candidate who beats the winner of every other contest and loses to its loser — can be produced by voters who each rank the candidates sensibly. McGarvey's recipe needs n(n − 1) of them for n candidates. The truth is far fewer: every pattern on five candidates takes three voters at most, a counting argument shows the number must eventually grow, and it grows only like n divided by its logarithm.
The fewest swaps to a winner
When no candidate beats every other head to head, Charles Dodgson proposed in 1876 to elect the one that is closest to doing so — the candidate that the fewest swaps of neighbouring names on the ballots would turn into a winner of every contest. The rule is easy to state and hard to compute: the count needs a search, and deciding the winner is provably among the hardest problems of its kind. A much simpler count, the votes still to be won, usually agrees, more often the larger the electorate.
The fewest triangles an edge density allows
Half of all possible edges can be drawn without a single triangle. One more, and triangles appear — not one but several at once. Push the density further and the question becomes a curve: for every share of edges, the fewest triangles a large graph can hold. The answer is a string of scallops, touching a simple parabola at the densities of the balanced multipartite graphs and bulging above it between them. It was guessed in the 1980s and proved in 2008 by a method that turns counting into positive-definite matrices.
As many points as two steps allow
In a graph where every point has d neighbours and every point is within two steps of every other, there can be at most d² + 1 points — one, its d neighbours, and d(d − 1) more reached through them. Graphs that meet the bound exactly are rare to the point of absurdity. The pentagon does it for d = 2, the Petersen graph for d = 3, a fifty-point graph found in 1960 for d = 7, and an eigenvalue argument proves there is nothing else — except possibly one graph with 3,250 points and 57 neighbours each, which nobody has found or ruled out.
Six numbers that need five pentagons
In 1638 Fermat wrote that every whole number is a sum of three triangular numbers, four squares, five pentagonal numbers, six hexagonal numbers, and so on for every polygon — and that he had a proof he would not write down. The claim is true; Cauchy proved it in 1813. What the claim hides is how unequal the cases are. Triangles and squares need their full count infinitely often. For pentagons, only six numbers ever need all five — 9, 21, 31, 43, 55 and 89 — and from hexagons on, two apiece.
Fifteen numbers decide every number
Lagrange proved that x² + y² + z² + w² takes every whole value. Ramanujan asked which other sums of four weighted squares do, and in 1917 listed fifty-five. One of them is wrong: x² + 2y² + 5z² + 5w² misses 15. The mistake points at a theorem found eighty years later — to know whether such a form reaches every number, it is enough to check that it reaches 1, 2, 3, 5, 6, 7, 10, 14 and 15. Nine checks decide infinitely many cases, and the nine numbers come out of an escalation any one can run.
There is a relation such that
A graph can be coloured with three colours exactly when there are three sets of its points such that every point lies in one and no edge joins two points of the same set. The sets are a guess; once guessed, the rest is a list of simple checks. Ronald Fagin proved in 1974 that this shape of sentence — 'there exist relations such that' followed by first-order conditions — describes exactly the problems whose solutions can be checked quickly, the class called NP. The central question of computer science is therefore also a question about what sentences can say.
The fewest moves to draw it
Every construction with ruler and compass is a sequence of moves — draw this line, draw that circle — and it is natural to ask for the shortest. Nobody can answer that by cleverness alone, but a machine can answer it by trying everything: the midpoint of a segment takes four moves and no fewer, the square on it five, a third of it five. Take the ruler away and the compass pays for it: six circles for the midpoint, seven for the square.
The longest code that survives every erasure
A Reed–Solomon code of k symbols can lose any n − k of its n and still be read. Over an alphabet of q symbols it can be at most q + 1 long — and it is conjectured that no code with the same perfect tolerance can ever be longer, apart from one family of exceptions in even characteristic. A search through every possible code for small alphabets confirms it cell by cell, a proof exists when q is prime, and for every other q the question is open.
A labelling every tree seems to have
Number the points of a tree 0 to n − 1 and write on each edge the difference of the numbers at its ends. The labelling is graceful if the edges then carry 1 to n − 1, each exactly once. Every tree anyone has ever checked — every one of the 551 trees on twelve points, and every tree up to about thirty-five — has such a labelling, and no one knows why. A graceful tree also tiles a complete graph by rotation, which is why the question was asked.
Every edge on exactly two cycles
Draw a graph on paper without crossings and its faces are cycles, each edge on exactly two of them. George Szekeres and Paul Seymour asked in the 1970s whether every graph without a bridge has such a set of cycles, drawn on paper or not. The Petersen graph needs five, the flower snark on twenty points five as well, every random graph tried has one, and nobody has proved it — though a minimal counterexample would have to be a snark.
How many gates a truth table needs
Every truth function can be built from gates, and the natural measure of a function is the fewest gates that build it. For three letters the whole answer can be computed: no function needs more than four. For n letters, a count shows that almost every function needs about 2ⁿ/n gates, and a construction shows that none needs more. And for any function anyone can actually write down, the best proof in fifty years says it needs 3.1n.
Perfect but for one factor
In 1638 Descartes wrote to Mersenne with an odd number whose divisors add up to exactly twice the number — provided one of its factors, 22021, is counted as a prime. It is not; it is 19² × 61. Nearly four centuries later that number is still the only odd one of its kind known, and a search of every odd number below ten million finds no other. It is the closest anyone has come to an odd perfect number, and what it shows is how an odd perfect number would have to be built.
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 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.
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 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.
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.
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.
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 rotation that hides the lattice
A congruential generator modulo a power of two has a lowest bit that alternates and pairs of outputs that lie on a few lines. Keep the generator exactly as it is, and show only eight bits of each state, rotated by an amount the state's own top bits choose: every output bit now runs the full cycle, the pairs fill the square as a random sequence would, and triples pass a test the state's own bits fail by a factor of six. Nothing about the state has changed, and four outputs still give it away.
The signs no completion can change
A quadratic in several variables can be completed square by square in many orders, and each order hands back different coefficients. What no order changes is how many come out positive and how many negative — and that count, read off a completion of A − tI, says how many eigenvalues lie below t without finding a single one.
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.
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.
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.
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.
Every function is a tree with two marks
There are n to the n functions from n points to themselves, and n to the n − 2 trees on those points. André Joyal noticed in 1981 that the missing factor of n² is a choice of two points — a head and a tail — and that a function, read the right way, simply is a tree with a head and a tail. The reading turns Cayley's formula into one line and hands over a fact about random trees from the birthday problem.
Almost every tree can be turned over
Cayley's n^(n−2) counts trees with labels on their points. Take the labels off and the count has no formula, because a symmetric shape absorbs labellings: the star on seven points can be labelled only seven ways, the one asymmetric shape 5,040. The bookkeeping that reconciles the two counts says something unexpected — almost every tree, labelled or not, can be turned over onto itself, where almost every graph cannot.
A total hung in a temple
Put a polygon's corners on a circle, cut it into triangles, and add up the radii of the circles inscribed in the triangles. Cut it a different way and the total is the same — for all fourteen ways of cutting a hexagon, to every digit. The fact was painted on a wooden tablet in a Japanese temple around 1800, and the reason for it is a theorem about one triangle and the distances from its circumcentre to its sides.
![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.](/f/six-formulas-on-the-interval-of-truth-values-01818ee4.svg)
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.
A contradiction that stays where it is
In classical logic one contradiction proves everything: from p and not-p, any q at all. Add a third truth value — both true and false — and count it as holding, and the explosion stops. The classical laws all survive as laws; what goes is exactly the reasoning that carries a contradiction somewhere it was not. Run the other way, the same three tables make a logic of gaps instead of gluts.
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.
Divisors that make every amount
Twelve's divisors — 1, 2, 3, 4, 6, 12 — can be chosen to add to every whole number from one to twenty-eight. Numbers like that are called practical, a test on their primes decides them, and they turn out to be counted like the primes: about 1.336 x / log x of them below x. For practical numbers Goldbach's conjecture and the twin conjecture are theorems.
Half the neighbours forces a tour
Deciding whether a graph has a closed tour through every point is one of the hardest problems computers are asked to solve. Gabriel Dirac found a condition that settles it in one glance: if every point is joined to at least half the others, a tour exists. The proof is a counting argument on a single path, it runs as an algorithm, and two small graphs show that the half cannot be lowered — while random graphs show how much stronger than necessary it is.
The boards a knight can tour
A knight can visit every square of a chessboard once and land a move from where it began. On a 5 × 5 board it cannot, on a 4 × 100 board it cannot, and on a 3 × 8 board it cannot, though on 3 × 10 it can in sixteen ways. Allen Schwenk found in 1991 the complete list of rectangles without a closed tour, and the reasons on it are of two kinds — a colouring that counts squares, and a second colouring that a tour would have to make agree with the first.
The centre a compass finds in six circles
A circle is drawn and its centre was never marked. A straightedge alone can never find it again. A compass alone can, from two points on the circle, in six circles — and a search through every construction of five circles or fewer shows that six is the least. The reason six works is an inversion that turns the circle into a straight line; the price of discarding the straightedge is exactly one move.
The arcs that make the centre cheap
Finding a circle's centre with the compass alone, from two points on it, costs six circles when the points are placed with no special relation — and as few as two when the arc between them is 60°. Searching every arc in steps of five degrees maps the price exactly, and it is not a smooth function of the arc: it is a list of coincidences, each one a small number of compass steps that happens to land on a chord equal to the radius.
Euclid's proof run as a machine
Euclid proved there is no last prime by multiplying the primes on any list, adding one, and noting that the result has a prime factor not on the list. Run the proof as a machine — start from 2, and each time take the smallest prime factor of one more than the product so far — and it produces 2, 3, 7, 43, 13, 53, 5, 6221671, … a sequence that never repeats, that reaches small primes late and large ones early, and that nobody can prove reaches every prime.
The best of thirty-four thousand scramblings
Verhoeff's check digit multiplies digits in the symmetry group of a pentagon after scrambling each one a different number of times, and 34,040 scramblings make it catch every swap of neighbours. Graded on the rarer errors, none of them catches everything, the best catch 86 of 90 doubled-digit slips and 848 of 900 swaps across a digit — and the scrambling Verhoeff published in 1969 is one of the forty that are best at both.
Named alongside it
The objects these essays reach for when they reach for this one.
CounterexampleParityInvariantCounting argumentModular arithmeticPermutationExistence proofLatin squareDualityProjective planeCoalitionConjecture