Field

Discrete — page 2

Counting, graphs, and things that come in whole pieces.

A sequence of 3² with no climb and no fall longer than 3. 10 terms plotted in order, each labelled with the longest climb and the longest fall ending at it; the first 9 keep both counters at 3 or below and the last one cannot.

The sequence that cannot avoid a staircase

Any ten numbers in a row contain four that climb or four that fall. The proof gives every term a pair of counters, notices that no two terms can share a pair, and is finished — with a bound that is exactly right.

$Everybody's share of the 24 chains. The subsets of a set of 4, each labelled with the fraction of maximal chains it lies on; the shares of any antichain add to at most one, and to exactly one only for a whole layer.$

Everybody's share of the chains

There are twenty-four ways to build a four-element set one element at a time. Every subset lies on some of them, and no two incomparable subsets share one — so an antichain is a set of disjoint shares of a single whole.

The cube cut into 6 symmetric chains. The subsets of a set of 4 partitioned into 6 chains by the bracket rule, each chain running from size k to size 4 − k and passing once through the middle layer.

The cube cut into chains

Write a subset as a string of brackets, match them the ordinary way, and the unmatched ones say which chain it is on. Six chains cover all sixteen subsets of a four-element set, and the bound and the example arrive together.

At most 3 of the 7 arcs can pairwise meet. The 7 elements arranged round a circle with the 7 arcs of 3 consecutive drawn, and the largest collection of arcs that pairwise intersect picked out.

The largest family that always meets

Change the question from "no two comparable" to "every two share an element" and the answer changes shape. The best antichain is a whole layer; the best intersecting family is a star, and the proof is a circle.

$16 ways to sort it. A small order with its 16 linear extensions counted, and for each incomparable pair the fraction of extensions putting one before the other.$

How many ways to sort it

An order says some things come before others and leaves the rest open. Counting the orderings consistent with it measures how much is still unknown — and the counting is as hard as any counting problem gets.

A bad path, and the path it reflects to. Two grids, 6 by 5. On the left a monotone path that dips below the diagonal, with its first offending step marked; on the right the same path with everything after that step reflected, which ends one square right and one square below the corner.

Counting the paths that go wrong

The number of good paths across a grid has no obvious formula. The number of bad ones does, because every bad path can be reflected into a path to a different corner, and that reflection is a perfect matching between two sets nobody chose to relate.

One word, four objects. The balanced word (()())() drawn as a lattice path, as nested brackets, as a triangulation of a 6-gon and as a binary tree. The four are the same object in four notations, and each is built here from the word itself.

One word, and four objects

A balanced string of brackets, a lattice path, a triangulated polygon and a binary tree are four different-looking things counted by the same numbers. They are not four things that happen to agree — each is a way of writing the others down, and the translation is mechanical.

The equation the objects satisfy. A diagram of the decomposition C = 1 + xC², with a table of the first several coefficients computed two ways: by the convolution the equation prescribes, and from the closed form.

The equation a sequence satisfies

Write the whole sequence as the coefficients of one series, and the recursion becomes an equation with a square in it. Solving the equation by the ordinary quadratic formula produces the closed form, the growth rate and the correction term, none of which the recursion offers.

The 14 triangulations, joined by single flips. The flip graph of a 6-gon: 14 triangulations drawn as small polygons and joined by 21 edges, one for each pair differing in a single diagonal.

The solid whose corners are triangulations

Take the triangulations of a hexagon as points and join two of them when a single diagonal can be swapped for another. The result is not merely a graph — it is the edge skeleton of a genuine convex polyhedron, with fourteen corners, three square faces and six pentagonal ones.

Four solids with the same counts and every volume. The Reeve tetrahedra at heights 1, 2, 3, 5, drawn in wireframe with a table of their lattice-point counts and volumes. All have four boundary points and none inside; their volumes run from 0.17 to 0.83.

The theorem that has no version in space

A lattice polygon's area is decided completely by two counts of dots. The obvious guess is that a lattice solid's volume is decided by the same two counts in three dimensions, and there is a family of tetrahedra with identical counts and every volume that says otherwise.

The sixteen lattice polygons with a single point inside. A grid of sixteen small lattice polygons, each drawn on its own patch of grid with the single interior point marked, labelled with its number of boundary points.

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 lattice-point count inside a circle, less its area, out to radius 160. A plot of the difference between the number of lattice points in a disc and the disc's area, against radius, with envelopes proportional to the square root and the two-thirds power drawn.

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.

A run down a diagonal, and the entry it adds to. 9 rows of Pascal's triangle with 5 entries shaded and the entry they add to marked. The claim is checked by adding the shaded entries: 1 + 3 + 6 + 10 + 15 = 35.

The run that lands one place along

Add up a run of entries down one of Pascal's diagonals and the total is another entry of the triangle — one row further down and one place along. The same triangle holds four more sums of that kind, and each is a different question answered by the same additive rule.

35 routes across a 4 by 3 grid. A grid with each cell holding the number of monotone routes reaching it. The far corner holds 35, which is the binomial coefficient of 7 choose 3.

Every entry counts the routes to it

Turn Pascal's triangle forty-five degrees and it becomes a grid of street corners, with each entry counting the ways of walking there. Identities between the entries then become statements about routes, and the statements are proved by cutting the routes in one place.

3 carries in base 2, and 2 divides it 3 times. The addition of 5 and 7 written in base 2, column by column, with the carries marked. There are 3, and 2 divides the binomial coefficient 792 exactly 3 times.

The carries decide the divisibility

How many times a prime divides a binomial coefficient is not a fact about the coefficient at all. It is a count of the carries that happen when two numbers are added in that prime's base, which is a question about column addition and has nothing to do with choosing anything.

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