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Decided by exhaustion — page 5

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 signal both can see, and neither wants to disobey. A two-by-two game with a distribution over its four cells, drawn as the weight on each. Obeying the recommendation is a best reply for both choosers, and the pair collects 21/2 between them. Applied

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.

 A landscape nobody is looking at, and every move goes downhill on it. The 8 states of a congestion game with 3 participants and two resources, ordered by Rosenthal's potential, with every improving unilateral move drawn as an arrow. Every arrow points downward. Applied

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.

 What the chain costs on a 6-gon: 39 pieces. A regular 6-gon fanned into 4 triangles, each with the three cuts that turn it into a rectangle, beside the running count of the pieces the whole chain produces — 39 of them. Computation

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.

 One sign decides which curve it is. 3 conics drawn from the general quadratic, each labelled with its discriminant B² − 4AC and the curve that sign names, checked against how many times the curve meets a large circle. Geometry

One sign decides which curve

The general quadratic in two variables has six coefficients and draws a conic. Which of the four it draws is settled by a single combination of three of them, and the other three cannot change the answer however they are chosen.

 58.6% at 46 candidates, against 37% without the values. The chance of ending with the best candidate when the values are shown, against the number of candidates, for 10 sizes. It falls towards 0.5802 rather than towards 1/e. Probability

When the numbers are shown

The secretary rule wins a third of the time and cannot do better, because it is told only who is ahead. Show the actual values and say where they came from, and the same problem is won three times in five — by a standard that falls as the end approaches.

 An online rule taking nine tenths of what an oracle takes. The share of the oracle's expected maximum secured by the best single threshold, and by the threshold at the median of the maximum, for 8 field sizes of independent uniform values. Probability

Half of what an oracle takes

Compare an online rule not against the best it could have done but against a rule that has seen every value in advance. One fixed threshold secures half of what the oracle collects, whatever the distributions are — and there is an example on which half is all there is.

 6 vertices folded to 4, and a graph that decides. The graph built from 3 generator words, folded until no vertex has two edges of one label leaving it. Reading a word from the base vertex decides membership, and 6 words are tested. Topology

Folding a graph until it decides

A subgroup of a free group usually arrives as a list of words, and almost nothing about it is readable from the list. Draw the words as loops, merge every pair of edges with the same label leaving one point, and what is left is a machine that decides membership by reading.

 Volume and one more number decide what a solid can be cut into. A table of a cube, a prism, a sixth of a cube, a regular tetrahedron, a regular octahedron and a collection of two tetrahedra with one octahedron, giving each one's volume, its Dehn invariant computed from its measured dihedral angles, and whether it can be cut into a box of equal volume. Computation

The obstruction that was the only one

Dehn showed in 1901 that a cube cannot be cut into a regular tetrahedron of the same volume, because a number built from edges and angles disagrees. For sixty-four years nobody knew whether that number was the whole story. Sydler proved in 1965 that it is: volume and Dehn's number together decide every case.

 Up to 4 choices among 60, and the thresholds that nest. For 60 candidates in random order and 1, 2, 3, 4 acceptances, the chance of ending with the best among those accepted — 37.3%, 60.0%, 74.3%, 83.5% — and where along the sequence the rule starts accepting for each number of choices in hand. Probability

The thresholds that nest

Allow a second acceptance in the secretary problem and the chance of holding the best rises from about 37 per cent to about 59. The best rule is still a threshold — but one threshold for each number of choices still in hand, the earlier ones starting sooner, and each additional choice buying less than the one before.

 Add the odds from the end: start at event 5 of 12. The odds of 12 independent events with chances 1/k — the secretary problem, the sum of the odds from the end reaching one at event 5, and the chance of stopping on the last success from each starting point, highest at 39.6%. Probability

Add the odds from the end

Watch a sequence of independent events and try to stop exactly on the last one that happens. Add up the odds of the events from the end backwards until the total reaches one, and stop at the first success from there. That rule is the best possible for any probabilities whatever, and the secretary problem is the special case in which the chances are one over the position.

 The first 60 powers of (3 + 4i)/5. The powers of (3 + 4i)/5 marked on the unit circle, each a further turn by the same angle, with the power that comes closest to returning to 1 marked. Algebra

On the circle and never home

Every root of unity lies on the unit circle, and so does the point (3 + 4i)/5 — yet no power of it ever returns to 1. A root of a whole-number polynomial of degree ten does the same. What forces a point home is a condition on the numbers its polynomial ties it to, and how far one of them may stray is a question open since 1933.

 Biproportional seats against their fair shares. A table of votes for 3 districts and 4 parties beside the seats the biproportional method gives, each with the fair share from the continuous fit, and the cell whose seats fall outside its quota marked. Applied

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.

 Sixteen halves in a three-by-three-by-three table of seats. A three-way table of fair shares drawn as three slices, one per group, with sixteen cells holding a half and every line total, along districts, parties and groups, equal to zero or one. Applied

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.

 The zeros of x² + y² + z² over GF(5), and of x² + y² over GF(7). Grids of every point over a small prime field with the solutions of a quadratic equation filled in: the three-variable equation drawn as one slice per value of z, beside a two-variable equation with far fewer solutions. Computation

Solutions that come in multiples of p

Count the solutions of x² + y² + z² = 0 in the field with five elements and there are 25; with seven, there are 49. Whenever a system of equations has more unknowns than its total degree, its number of solutions is a multiple of the characteristic — which forces a solution besides zero, and the reason is a sum over the field that vanishes because its non-zero elements form one cycle.

 The cubic curves over GF(43) with the most and the fewest points. The solutions of two equations y squared equals x cubed plus ax plus b over the field with 43 elements, drawn as dots on a square grid: the curve with the most points and the curve with the fewest. Computation

Give or take twice the square root

A cubic curve over the integers mod 43 should have about 44 points — one for each value of x, on average, and one at infinity. No curve misses by more than 13, the largest whole number below 2√43, and every count from 31 to 57 belongs to some curve. The first fact is Hasse's theorem, the second Deuring's, and the way the counts spread between the limits is a semicircle.

 A network of six places and the tree that holds all fifteen of its cheapest cuts. A network with capacities on its roads beside a tree on the same places, whose edge numbers give the cheapest cut between any two places as the smallest number on the path joining them. Discrete

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.

 5 two-literal clauses drawn as 10 arrows. The literals of four variables arranged in a circle with an arrow for each implication a two-literal clause contains, the arrows forming a path from one literal to its negation and back highlighted, beside the list of clauses and their arrows. Logic

Two literals make an arrow

A clause of two literals, p ∨ q, says that if p is false then q is true, and if q is false then p is true — two arrows. A set of such clauses is a directed graph on the literals, and it is unsatisfiable exactly when some variable and its negation reach each other. Each of those two paths is a chain of resolution steps, and finding them takes time proportional to the size of the input.

 A failed search on four clauses, read as a resolution refutation. A binary search tree branching on variables, each branch ending at a clause the partial assignment makes false, with every branch point labelled by the resolvent of the clauses below it, the top label being the empty clause. Logic

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.

 Bhāskara's cyclic method on x² − 61y² = 1, step by step. A table of the cyclic method's rows: the helper m chosen at each step and the near miss a² − Db² = k it produces, ending at k = 1 with the fundamental solution. Number

A method that is allowed to miss

Bhāskara's cyclic method solves x² − Dy² = 1 by aiming at the wrong target. It keeps a pair a, b with a² − Db² = k for some small k, combines it with a helper chosen so that k can be divided out, and repeats until k is 1. For D = 61 it reaches the ten-digit fundamental solution in 13 steps, where walking the convergents of √61 takes 22 — and for every D up to 100 it is faster.

 The states the continued fraction of √61 can be in. A grid of whole-number pairs with the band of reduced states shaded, the pairs that qualify marked, and the cycle of states visited by the expansion of the square root numbered in order. Number

Why the expansion has to repeat

The continued fraction of √61 runs 7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 and then starts again. It must: each step's state is a pair of whole numbers trapped in a small band, and only 14 pairs fit. The expansion of √61 visits 11 of them in a cycle, the other 3 form a cycle of their own, and the period reads the same backwards before its last term, which is twice the first.

 Two opposite points on a globe with the same temperature and the same pressure. A world map in longitude and latitude with one curve where each point's temperature matches its opposite point's and another where the pressures match, crossing at a pair of opposite points that are marked. Topology

Two opposite points that agree twice

At any moment there are two points on opposite sides of the Earth with the same temperature and the same pressure. On a seeded globe they sit at 11.9°N 44.6°E and 11.9°S 135.4°W. The reason is the circle argument that halved two shapes, run one dimension up: the differences between opposite readings, walked round the equator, wind round zero an odd number of times — and an odd number cannot be zero.

 A necklace of 4 orange, 4 blue, 2 green beads, shared fairly with 3 cuts. A row of coloured beads cut at marked places into pieces, each piece labelled with the thief who receives it, so that both thieves get half of every colour. Topology

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 powers of 2 modulo 13, as a ring of 12. The non-zero residues modulo 13 placed on a circle, with the successive powers of 2 joined by straight lines into a closed walk of 12 steps. Number

One residue whose powers are all of them

Fermat's theorem says every order divides p − 1. It does not say that anything has order exactly p − 1, which is a separate and stronger claim — and what forces it is a count of how many numbers share each divisor with p − 1.

 The orders of the 8 units modulo 15. A strip of the units modulo 15 with each one's multiplicative order beneath it, the largest order marked at 4 against φ(15) = 8. Number

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.

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