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Counting the same thing twice — page 10

One collection, counted by two different methods, and an identity that falls out because both answers have to agree. The proof is the pair of counts.
A planimeter's wheel measures an area by going round it. Polar planimeter with arms 2.3 and 2 traced round a closed curve; wheel roll 1.6478, times 2, equals the area 3.2955. Analysis

An area measured by walking round it

A surveyor's instrument from 1854 measures the area of any shape on a map by having its pointer steered once round the boundary; a small wheel rolls and slides, and its reading, times the length of one arm, is the area. Nothing touches the inside. The reason is that area can be written as an integral over the boundary — Green's theorem — and the instrument is that integral built in brass. Walk a curve that crosses itself and the same integral counts some regions twice.

Every majority pattern on 5 candidates, and the fewest voters that make it. 12 tournaments on 5 candidates: wins 22222 needs 3; wins 32221 needs 3; wins 32221 needs 3; wins 32221 needs 3; wins 33211 needs 3; wins 33211 needs 3; wins 42211 needs 3; wins 43111 needs 3; wins 33220 needs 3; wins 42220 needs 3; wins 33310 needs 3; wins 43210 needs 1. Applied

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 triangles a graph can have at each edge density. Razborov's minimum triangle density: 0.55: 0.0730, 0.6: 0.1415, 0.7: 0.2871, 0.75: 0.3750, 0.8: 0.4800, 0.9: 0.7200; it equals Goodman's bound at 1 − 1/t. Discrete

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.

The Hoffman–Singleton graph: five pentagons, five pentagrams. Fifty vertices of degree seven and girth five, built from five pentagons and five pentagrams joined by the rule j of pentagon h to h·i + j of pentagram i. Discrete

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.

The fewest pentagonal numbers adding to each number up to 120. Fewest pentagonal numbers summing to 1..120; the most ever needed up to 20000 is 5, by 9, 21, 31, 43, 55, 89. Geometry

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.

Tails of a sample drawn with and without replacement, against two bounds. t 0.000: without 1.00e+0, with 1.00e+0, Hoeffding 1.00e+0, Serfling 1.00e+0; t 0.075: without 2.65e-1, with 3.88e-1, Hoeffding 1.00e+0, Serfling 9.56e-1; t 0.150: without 1.35e-2, with 5.57e-2, Hoeffding 3.31e-1, Serfling 1.05e-1; t 0.225: without 1.10e-4, with 3.02e-3, Hoeffding 3.48e-2, Serfling 2.62e-3; t 0.300: without 1.19e-7, with 8.09e-5, Hoeffding 1.49e-3, Serfling 1.50e-5. Probability

Drawn without putting back

Every concentration bound on this shelf assumes the draws are independent. A real sample is not: a pollster does not ring the same person twice, and every ball taken from an urn changes what is left in it. The dependence runs the helpful way. Wassily Hoeffding proved in 1963 that a sample drawn without replacement is at least as concentrated as one drawn with it, for every convex measure of spread at once — and the variance falls by an exact factor that reaches zero when the whole urn is taken.

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