Discrete — page 3

The few points that cut a flat graph

Any graph that can be drawn without crossings, however large, falls into pieces of at most two thirds once a few points are removed — about the square root of its size, never more than 2.83 times it. A grid shows the square root cannot be beaten, a ring of breadth-first neighbours comes close, and a cycle through a shallow tree finishes the job.

Moves that only ever add edges

Turán's theorem says the densest graph avoiding a complete graph on r + 1 points is the balanced r-part graph. Zykov's proof finds it by a sequence of moves — turn a point into a copy of a better-connected one it is not joined to — each of which adds edges and none of which can create the forbidden clique. A second proof spreads a unit of weight over the points and gets the same bound from a maximum.

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.

A walk that splices in its own detours

Euler proved that a walk crossing every bridge once needs every landmass to have an even number of bridges, and then stated, without proof, that this was enough. The missing half took 137 years, and it is not an argument but a procedure: walk until stuck, notice that stuck can only mean home, and splice in a detour from anywhere with edges left. The procedure never fails, and the reason fits in one sentence about arriving and leaving.

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.

Every place changes back

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

The walk through the middle levels

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

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.

Multiplying every number on the dial at once

Join every residue on a dial to twice itself and the chords draw a heart-shaped curve with one cusp; join each to three times itself and the curve has two. The picture is the whole multiplication map at once, and it holds three facts: the map splits the dial into cycles whose lengths are orders, those cycles on a dial of 2ⁿ − 1 are the binary necklaces of length n, and the curve is the caustic light draws inside a cup.

A root lifted one digit at a time

On a dial of seven, 3 × 3 is 2. On a dial of forty-nine the square root of 2 must reduce to 3, so there are only seven candidates, and exactly one of them works: 10. On a dial of 343 exactly one lift of 10 works: 108. Each step adds one digit on the left, found by solving a linear equation, and the digits go on for ever — a number …21216213 whose square is 2, in a world where closeness means divisibility by seven.

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.

The cheapest way to send

Put a price on every road as well as a capacity and ask for the cheapest way to send four units. Twenty-eight ways exist and one is cheapest, and two certificates prove it without comparing it with the other twenty-seven: no cycle of roads it leaves unused costs less than nothing to push round, and there are prices at the places that every usable road fails to beat.

Enough partners in every finite group

Hall's theorem says a matching exists unless some group has too few partners between them. On an infinite graph that is false: one vertex that can be paired with anybody, and every other with exactly one, passes the test on every finite group and has no matching at all. The failure needs a vertex with infinitely many choices. Forbid that, and the theorem comes back, by an argument about trees rather than about matchings.

Matching as they arrive

Applicants arrive one at a time, and each must be given a post or turned away on the spot. Any sensible rule matches at least half as many as the best assignment chosen with hindsight, and an adversary arranging the arrivals can hold every fixed rule to exactly half. Shuffle the posts once, at random, and give each arrival its best-ranked free post: the guarantee rises to 1 − 1/e, about 63%, and no rule of any kind can do better.

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.

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.

The longest climb of a shuffle

Erdős and Szekeres guarantee that any n distinct numbers hold a climb or a fall of √n, and there are orders that allow nothing more. Shuffle a deck at random instead and the longest climb is almost exactly 2√n — with fluctuations of size n to the power one-sixth, distributed exactly as the largest eigenvalue of a large random matrix.

The crossings a graph cannot avoid

Five points all joined need one crossing, six need three, seven need nine. Euler's formula gives a lower bound that grows like the square of the number of points and is badly wrong; a sampling trick turns the same formula into a bound that grows like the fourth power and is right to within a constant. And for drawings with straight edges the count turns out to be something else entirely: the number of quadrilaterals the points make.

Every point at the average of its neighbours

A planar graph can be drawn without crossings, but finding such a drawing looks like a search. Tutte found in 1963 that it is not: pin one face to a convex polygon, put every other point at the average of its neighbours, and the drawing that results has straight edges, convex faces and no crossings at all — provided no two points can cut the graph apart. It is the position a network of equal springs settles into, and it is also where a random walk expects to leave.

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.

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.