# Antichains in the grid

In the previous post on this topic, we discussed Dilworth’s theorem on chains and antichains in a general partially ordered set. In particular, whatever the size of the largest antichain in a poset, it is possible to partition the poset into exactly that many chains. So for various specific posets, or the directed acyclic graphs associated to them, we are interested in the size of this largest antichain.

The following example turned out to be more interesting than I’d expected. At a conventional modern maths olympiad, there are typically three questions on each paper, and for reasons lost in the mists of time, each student receives an integer score between 0 and 7 per question. A natural question to ask is “how many students need to sit a paper before it’s guaranteed that one will scores at least as highly as another on every question?” (I’m posing this as a straight combinatorial problem – the correlation between scores on different questions will be non-zero and presumably positive, but that is not relevant here.)

The set of outcomes is clearly $\{0,1,\ldots,7\}^3$, with the usual weak domination partial order inherited from $\mathbb{R}^3$. Then an antichain corresponds to a set of triples of scores such that no triple dominates another triple. So the answer to the question posed is: “the size of the largest antichain in this poset, plus one.”

In general, we might ask about $\{1,2,\ldots,n\}^d$, again with the weak domination ordering. This directed graph, which generalises the hypercube as well as our example, is called the grid.

Heuristics for the largest antichain

Retaining the language of test scores on multiple questions is helpful. In the previous post, we constructed a partition of the poset into antichains, indexed by the elements of some maximal chain, by starting with the sources, then looking at everything descended only from sources, and so on. (Recall that the statement that this is possible was referred to as the dual of Dilworth’s theorem.) In the grid, there’s a lot of symmetry (in particular under the mapping $x\mapsto n+1-x$ in every coordinate), and so you end up with the same family of antichains whether you work upwards from the sources or downwards from the sinks. (Or vice versa depending on how you’ve oriented your diagram…) The layers of antichains also have a natural interpretation – each layer corresponds to a given total score. It’s clear a priori why each of these is an antichain. If A scores the same as B overall, but strictly more on the first question, this must be counterbalanced by a strictly lower score on another question.

So a natural guess for the largest antichain is the largest antichain corresponding to some fixed total score. Which total score should this be? It ought to be the middle layer, that is total score $\frac{(n+1)d}{2}$, or the two values directly on either side if this isn’t an integer. My intuition was probabilistic. The uniform distribution on the grid is achieved by IID uniform distributions in each coordinate, which you can think of as a random walk, especially if you subtract off the mean first. It feels that any symmetric random walk should have mode zero or next-to-zero. Certainly this works asymptotically in a rescaled sense by CLT, and in a slightly stronger sense by local CLT, but we don’t really want asymptotics here.

When I started writing the previous paragraph, I assumed there would be a simple justification for the claim that the middle layer(s) was largest, whether by straight enumeration, or some combinatorial argument, or even generating functions. Perhaps there is, and I didn’t spot it. Induction on d definitely works though, with a slightly stronger hypothesis that the layer sizes are symmetric around the median, and monotone on either side of the median. The details are simple and not especially interesting, so I won’t go into them.

From now on, the hypothesis is that this middle layer of the grid is the largest antichain. Why shouldn’t it, for example, be some mixture of middle-ish layers? (*) Well, heuristically, any score sequence in one layer removes several possibilities from a directly adjacent layer, and it seems unlikely that this effect is going to cancel out if you take some intermediate number of score sequences in the first layer. Also, the layers get smaller as you go away from the middle, so because of the large amount of symmetry (coordinates are exchangeable etc), it feels reasonable that there should be surjections between layers in the outward direction from the middle. The union of all these surjections gives a decomposition into chains.

This result is in fact true, and its proof by Bollobas and Leader, using shadows and compression can be found in the very readable Sections 0 and 1 of [1].

Most of the key ideas to a compression argument are present in the case n=2, for which some notes by Leader can be found here, starting with Proof 1 of Theorem 3, the approach of which is developed over subsequent sections. We treat the case n=2, but focusing on a particularly slick approach that does not generalise as successfully. We also return to the original case d=3 without using anything especially exotic.

Largest antichain in the hypercube – Sperner’s Theorem

The hypercube $\{0,1\}^d$ is the classical example. There is a natural correspondence between the vertices of the hypercube, and subsets of $[d]$. The ordering on the hypercube corresponds to the ordering given by containment on $\mathcal{P}([d])$. Almost by definition, the k-th layer corresponds to subsets of size k, and thus includes $\binom{d}{k}$ subsets. The claim is that the size of the largest antichain is $\binom{d}{\lfloor d/2 \rfloor}$, corresponding to the middle layer if d is even, and one of the two middle layers if d is odd. This result is true, and is called Sperner’s theorem.

I know a few proofs of this from the Combinatorics course I attended in my final year at Cambridge. As explained, I’m mostly going to ignore the arguments using compression and shadows, even though these generalise better.

As in the previous post, one approach is to exhibit a covering family of exactly this number of disjoint chains. Indeed, this can be done layer by layer, working outwards from the middle layer(s). The tool here is Hall’s Marriage Theorem, and we verify the relevant condition by double-counting. Probably the hardest case is demonstrating the existence of a matching between the middle pair of layers when d is odd.

Take d odd, and let $d':= \lfloor d/2\rfloor$. Now consider any subset S of the d’-th layer $\binom{[d]}{d'}$. We now let the upper shadow of S be

$\partial^+(S):= \{A\in \binom{[d]}{d'+1}\,:\, \exists B\in S, B\subset A\},$

the sets in the (d’+1)-th layer which lie above some set in S. To apply Hall’s Marriage theorem, we have to show that $|\partial^+(S)|\ge |S|$ for all choice of S.

We double-count the number of edges in the hypercube from $S$ to $\partial^+(S)$. Firstly, for every element $B\in S$, there are exactly d’ relevant edges. Secondly, for every element $A\in\partial^+(S)$, there are exactly d’ edges to some element of $\binom{[d]}{d'}$, and so in particular there are at most d’ edges to elements of S. Thus

$d' |S|=|\text{edges }S\leftrightarrow\partial^+(S)| \le d' |\partial^+(S)|,$

which is exactly what we require for Hall’s MT. The argument for the matching between other layers is the same, with a bit more notation, but also more flexibility, since it isn’t a perfect matching.

The second proof looks at maximal chains. Recall, in this context, a maximal chain is a sequence $\mathcal{C}=B_0\subset B_1\subset\ldots\subset B_d$ where each $B_k:= \binom{[d]}{k}$. We now consider some largest-possible antichain $\mathcal{A}$, and count how many maximal chains include an element $A\in\mathcal{A}$. If $|A|=k$, it’s easy to convince yourself that there are $\binom{d}{r}$ such maximal chains. However, given $A\ne A'\in\mathcal{A}$, the set of maximal chains containing A and the set of maximal chains containing A’ are disjoint, since $\mathcal{A}$ is an antichain. From this, we obtain

$\sum_{A\in\mathcal{A}} \binom{d}{|A|} \le d!.$ (**)

Normally after a change of notation, so that we are counting the size of the intersection of the antichain with each layer, this is called the LYM inequality after Lubell, Yamamoto and Meshalkin. The heuristic is that the sum of the proportions of layers taken up by the antichain is at most one. This is essentially the same as earlier at (*). This argument can also be phrased probabilistically, by choosing a *random* maximal chain, and considering the probability that it intersects the proposed largest antichain, which is, naturally, at most one. Of course, the content is the same as this deterministic combinatorial argument.

Either way, from (**), the statement of Sperner’s theorem follows rapidly, since we know that $\binom{d}{|A|}\le \binom{d}{\lfloor d/2\rfloor}$ for all A.

Largest antichain in the general grid

Instead of attempting a proof or even a digest of the argument in the general case, I’ll give a brief outline of why the previous arguments don’t transfer immediately. It’s pretty much the same reason for both approaches. In the hypercube, there is a lot of symmetry within each layer. Indeed, almost by definition, any vertex in the k-th layer can be obtained from any other vertex in the k-th layer just by permuting the labels (or permuting the coordinates if thinking as a vector).

The hypercube ‘looks the same’ from every vertex, but that is not true of the grid. Consider for clarity the n=8, d=3 case we discussed right at the beginning, and compare the scores (7,0,0) and (2,2,3). The number of maximal chains through (7,0,0) is $\binom{14}{7}$, while the number of maximal chains through (2,2,3) is $\binom{7}{2, 2,3}\binom{14}{4,5,5}$, and the latter is a lot larger, which means any attempt to use the second argument is going to be tricky, or at least require an extra layer of detail. Indeed, exactly the same problem arises when we try and use Hall’s condition to construct the optimal chain covering directly. In the double-counting section, it’s a lot more complicated than just multiplying by d’, as was the case in the middle of the hypercube.

Largest antichain in the d=3 grid

We can, however, do the d=3 case. As we will see, the main reason we can do the d=3 case is that the d=2 case is very tractable, and we have lots of choices for the chain coverings, and can choose one which is well-suited to the move to d=3. Indeed, when I set this problem to some students, an explicit listing of a maximal chain covering was the approach some of them went for, and the construction wasn’t too horrible to state.

[Another factor is that it computationally feasible to calculate the size of the middle layer, which is much more annoying in d>3.]

[I’m redefining the grid here as $\{0,1,\ldots,n-1\}^d$ rather than $\{1,2,\ldots,n\}^d$.]

The case distinction between n even and n odd is going to make both the calculation and the argument annoying, so I’m only going to treat the even case, since n=8 was the original problem posed. I should be honest and confess that I haven’t checked the n odd case, but I assume it’s similar.

So when n is even, there are two middle layers namely $\frac{3n}{2}-2, \frac{3n}{2}-1$ (corresponding to total score 10 and total score eleven in the original problem). I calculated the number of element in the $\frac{3n}{2}-1$ layer by splitting based on the value of the first coordinate. I found it helpful to decompose the resulting sum as

$\sum_{k=0}^{n-1} = \sum_{k=0}^{\frac{n}{2}-1} + \sum_{k=\frac{n}{2}}^{n-1},$

based on whether there is an upper bound, or a lower bound on the value taken by the second coordinate. This is not very interesting, and I obtained the answer $\frac{3n^2}{4}$, and of course this is an integer, since n is even.

Now to show that any antichain has size at most $\frac{3n^2}{4}$. Here we use our good control on the chain coverings in the case d=2. We note that there is a chain covering of the (n,d=2) grid where the chains have 2n-1, 2n-3,…, 3, 1 elements (%). We get this by starting with a maximal chain, then taking a maximal chain on what remains etc. It’s pretty much the first thing you’re likely to try.

Consider an antichain with size A in the (n,d=3) grid, and project into the second and third coordinates. The image sets are distinct, because otherwise a non-trivial pre-image would be a chain. So we have A sets in the (n,d=2) grid. How many can be in each chain in the decomposition (%). Well, if there are more than n in any chain in (%), then two must have been mapped from elements of the (n,d=3) grid with the same first coordinate, and so satisfy a containment relation. So in fact there are at most n image points in any of the chains of (%). So we now have a bound of $n^2$. But of course, some of the chains in (%) have length less than n, so we are throwing away information. Indeed, the number of images points in a given chain is at most

$\max(n,\text{length of chain}),$

and so the number of image points in total is bounded by

$n+\ldots+n+ (n-1)+(n-3)+\ldots+1,$

where there are n/2 copies of n in the first half of the sum. Evaluating this sum gives $\frac{3n^2}{4}$, exactly as we wanted.

References

[1] – Bollobas, Leader (1991) – Compressions and Isoperimetric Inequalities. Available open-access here.

# Chains and antichains

I’ve recently been at the UK-Hungary winter olympiad camp in Tata, for what is now my sixth time. As well as doing some of my own work, have enjoyed the rare diversion of some deterministic combinatorics. It seems to be a local variant of the pigeonhole principle that given six days at a mathematical event in Hungary, at least one element from {Ramsay theory, Erdos-Szekeres, antichains in the hypercube} will be discussed, with probability one. On this occasion, all were discussed, so I thought I’d write something about at least one of them.

Posets and directed acyclic graphs

This came up on the problem set constructed by the Hungarian leaders. The original formulation asked students to show that among any 17 positive integers, there are either five such that no one divides any other, or five such that among any pair, one divides the other.

It is fairly clear why number theory plays little role. We assign the given integers to the vertices of a graph, and whenever a divides b, we add a directed edge from the vertex corresponding to a to the vertex corresponding to b. Having translated the given situation into a purely combinatorial statement, fortunately we can translate the goal into the same language. If we can find a chain of four directed edges (hence five vertices – beware confusing use of the word ‘length’ here) then we have found the second possible option. Similarly, if we can find an antichain, a set of five vertices with no directed edges between them, then we have found the first possible option.

It’s worth noting that the directed graph we are working with with is transitive. That is, whenever there is an edge a->b and b->c, there will also be an edge a->c. This follows immediately from the divisibility condition. There are also no directed cycles in the graph, since otherwise there would be a cycle of integers where each divided its successor. But of course, when a divides b and these are distinct positive integers, this means that b is strictly larger than a, and so this relation cannot cycle.

In fact, among a set of positive integers, divisibility defines a partial order, which we might choose to define as any ordering whether the associated directed graph is transitive and acyclic, although obviously we could use language more naturally associated with orderings. Either way, from now on we consider posets and the associated DAGs (directed acyclic graphs) interchangeably.

Dilworth’s theorem

In the original problem, we are looking for either a large chain, or a large antichain. We are trying to prove that it’s not possible to have largest chain size at most four, and largest antichain size at most four when there are 17 vertices, so we suspect there may some underlying structure: in some sense perhaps the vertex set is the ‘product’ of a chain and an antichain, or at least a method of producing antichains from a single vertex.

Anyway, one statement of Dilworth’s theorem is as follows:

Statement 1: in a poset with nm+1 elements, there is either a chain of size n+1, or an antichain of size m+1.

Taking n=m=4 immediately finishes the original problem about families of divisors. While this is the most useful statement here, it’s probably not the original, which says the following:

Statement 2: in a poset, there exists $\mathcal{C}$ a decomposition into chains, and an antichain $A$ such that $|\mathcal{C}|=|A|$.

Remark 1: Note that for any decomposition into chains and any antichain, we have $|\mathcal{C}|\ge |A|$, since you can’t have more than one representative from any chain in the antichain. So Statement 2 is saying that equality does actually hold.

Remark 2: Statement 1 follows immediately from Statement 2. If all antichains had size at most m, then there’s a decomposition into at most m chains. But each chain has size n, so the total size of the graph is at most mn. Contradiction.

Unsuccessful proof strategies for Dilworth

Since various smart young people who didn’t know the statement or proof of Dilworth’s theorem attempted to find it (in the form of Statement 1, and in a special case) in finite time conditions, it’s easy to talk about what doesn’t work, and try to gain intellectual value by qualifying why.

• Forgetting directions: in general one might well attack a problem by asking whether we have more information than we need. But ignoring the directions of the edges is throwing away too much information. After doing this, antichains are fine, but maybe you need to exhibit some undirected ‘chains’. Unless these undirected chains are much longer than you are aiming for, you will struggle to reconstruct directed chains out of them.
• Where can the final vertex go?: in a classic trope, one might exhibit a directed graph on nm vertices with neither a chain of size n+1 nor an antichain of size m+1. We attempt to argue that this construction is essentially unique, and that it goes wrong when we add an extra vertex. As a general point, it seems unlikely to be easier to prove that exactly one class of configurations has a given property in the nm case, than to prove no configurations has the same property in the nm+1 case. A standalone proof of uniqueness is likely to be hard, or a disguised rehash of an actual proof of the original statement.
• Removing a chain: If you remove a chain of maximal length, then, for contradiction, what you have left is m(n-1)+1 vertices. If you have a long chain left, then you’re done, although maximality has gone wrong somewhere. So you have an antichain size n in what remains. But it’s totally unclear why it should be possible to extend the antichain with one of the vertices you’ve just removed.

An actual proof of Dilworth (Statement 1), and two consequences

This isn’t really a proof, instead a way of classifying the vertices in the directed graph so that this version of Dilworth. As we said earlier, we imagine there may be some product structure. In particular, we expect to be able to find a maximal chain, and a nice antichain associated to each element of the maximal chain.

We start by letting $V_0$ consist of all the vertices which are sources, that is, have zero indegree. These are minima in the partial ordering setting. Now let $V_1$ consist of all vertices whose in-neighbourhood is entirely contained in $V_0$, that is they are descendents only of $V_0$. Then let $V_2$ consist of all remaining vertices whose in-neighourhood is entirely contained in $V_0\cup V_1$ (but not entirely in $V_0$, otherwise it would have already been treated), and so on. We end up with what one might call an onion decomposition of the vertices based on how far they are from the sources. We end up with $V_0,V_1,\ldots,V_k$, and then we can find a chain of size k+1 by starting with any vertex in $V_k$ and constructing backwards towards the source. However, this is also the largest possible size of a chain, because every time we move up a level in the chain, we must move from $V_i$ to $V_j$ where j>i.

It’s easy to check that each $V_i$ is an antichain, and thus we can read off Statement 1. A little more care, and probably an inductive argument is required to settle Statement 2.

We have however proved what is often called the dual of Dilworth’s theorem, namely that in a poset there exists a chain C, and a decomposition into a collection $\mathcal{A}$ of antichains, for which $|C|=|\mathcal{A}|$.

Finally, as promised returning to Erdos-Szekeres, if not to positive integers. We apply Dilworth Statement 1 to a sequence of $m^2+1$ real numbers $a_0,a_1,\ldots,a_{m^2}$, with the ordering $a_i\rightarrow a_j$ if $i\le j$ and $a_i\le a_j$. Chains correspond to increasing subsequences, and antichains to decreasing subsequences, so we have shown that there is either a monotone subsequence of length m+1.