Maximum density of copies of a graph in the n-cube

John Goldwasser Ryan Hansen

West Virginia University

Qn is the n-cube.

V(Qn) all binary n-tuples.

Two vertices joined by an edge iff differ in precisely one coordinate.A sub-d-cube has d coordinates that vary, n fixed.

Example with n=10 and d=4: 1 0 * 1 * * 0 1 * 1

There are C(n,d)2(n-d) sub-d-cubes of Qn

Let d and n be integers (d much smaller than n)Let H be a set of vertices in a d-cube (which we calla configuration)MAIN QUESTION: How do you choose vertices in an n-cube so that the maximum possible fraction of sub-d-cubes have an “exact copy of H”? Example with d=3

H Not copy of H Copy of H

Formally, given a subset H of the vertices of a d-cube, if Sn is a subset of V(Qn), we let G(H,d,n,Sn) be the number of sub-d-cubes whose intersection with Sn is a copy of H. We let

g(H,d,n) = max G(H,d,n,Sn)/total # sub-d-cubes

(max over all subsets Sn of V(Qn)

An averaging argument shows that g(H,d,n) is a nonincreasing function of n, so we let π(H,d) denote its limit as n goes to infinity, and refer to it as the d-cube density of H.

So π(H,d) is the limit as n goes to infinity of the maximum possible fraction of sub-d-cubes which can have a copy of H.

For the given H above, our construction shows that F(H,3) ≥ ¾. We conjecture that equality holds.

Example: H We’ll show π(H,2) = ½ Lower bound construction:Partition the set {1,2,3,…,n} of coordinates into sets A and B. Let Sn be the set of all vertices in Qn such that the number of 1’s in A is even.

If a Q2 subgraph has one coordinate in A and one in B then it will have an exact copy of H. If A and B are each n/2, that’s half of the Q2’s.

So ½ ≤ π(H,2)

For the upper bound, each vertex in a copy of H is adjacent to one vertex in H which is in Sn and one which is not.

Each vertex v in the Qn is in C(n,2) sub-2-cubes. Only those in which one neighbor is in Sn and one is not have a chance of being a “good” Q2. This is at most roughly half of the sub-2-cubes containing v.

Since the maximum fraction of “good” sub-2-cubes at any vertex is ½, certainly the maximum fraction overall is at most ½.

We define πlocal(H,d) to be the limit as n goes to infinity of the maximum fraction of “good” sub-d-cubes at any vertex v for any subset Sn of V(Qn). Certainly it is at least as big as π(H,d). We have:

½ ≤ π(H,2) ≤ πlocal(H,2) ≤ ½

So π(H,2) = ½.

If K is the subset of the 2-cube shownat the right, then π(K,2) = 1. (Just chooseall vertices the sum of whose coordinatesis even. )

In general, π(H,d) is hard to determine. We have found its value in only a few special cases. For the vertex configuration R in Q₃ shown at the right, π(R,3) = 4/9.

Let Pd be a single vertex in Qd. we have been unable to determine f(R,d) for any d≥2.

A simple construction (choose all vertices with weight a multiple of 3) shows that f(P2,2) ≥ 2/3.

Using flag algebras, Rahil Baber has shown f(P2,2) ≤ .6858

f(P3,3) ≤ .6101

f(P4,4) ≤ .6026

(Is this a monotone sequence? Can you prove it?)

My Conjecture: For sufficiently large d, π(Pd,d) = [(d-1)/d](d-1)

(what you get if you choose each vertex with probability 1/2d)

Of course this is close to 1/e.

Easy Proposition: For any configuration H in Qd,

π(H,d) ≥ d!/dd

(obtained from a “blow-up” of H)

Example with n = 4, H = {1001, 1101, 0010}

Partition {1,2,3,…,n} into 4 sets of equal size and put the following vertices into S: A B C D odd even even odd odd odd even odd even even odd even

If a sub-4-cube has one vertex in each of A,B,C,D then it will have a copy of H.Probability is 4!/44.

If d=3, there are 14 isomorphically distinct vertex configurations H with at most 4 vertices. We have been able to determine π(H,3) for 4 of them. The smallest upper bound is .30476 for two adjacent vertices (3!/33 = .2222)

If d=4, there are 238 isomorphically distinct vertex configurations H with at most 8 vertices. We have been able to determine π(H,4) for about 6 of them. All but two of the flag algebra upper bounds are at least .100.

The other two have flag algebra upper bounds of .09421 and .09375They are both 8-cycles.

There are three isomorphically distinct configurations of 8 vertices in Q₄ which induce an 8-cycle. Of these, one is a “perfect” 8-cycle: 4 pairs of vertices distance 4 apart. The perfect 8-cycle C8

0000 1000 1100 1110 1111 0111 0011 0001

Main Theorem: π(C8, 4) = 4!/44 = 3/32 = .09375

Conjecture: π(C2d, d) = d!/dd for all d>3.

The conjecture can’t hold for d=3. Letting S be all vertices in Qn with weight not a multiple of 3 is a construction which shows that π(C6,3) is at least 3/9, and 3!/33 = 2/9.

To prove the theorem it suffices to prove πlocal(C8,4) ≤ 3/32,

Because then:

3/32 ≤ π(C8, 4) ≤ πlocal(C8, 4) ≤ 3/32

Let Pd+1 denote a “perfect path” in Qd:

d+1 vertices in Qd where the Hamming distance between the two end-vertices is d.

Theorem: π(P4, 3) = 3/8

(Proof is using sequence with “Property V”)

Conjecture: π(Pd+1, d) = d!/(d+1)(d-1) if d ≥ 3.

To do this we considered the set of all sequences of d distinct elements from {1,2,3,…,n}. We showed that the limit, as n goes to infinity, of the fraction of these that satisfy a certain condition we call Property U is equal to πlocal(C8, 4).

Then we showed that the maximum fraction of these sequences that have Property U is equal to the maximum fraction of sets of 4 vertices in a bipartite graph which span 2 disjoint edges.

Theorem: Let G be a bipartite graph with n vertices. Then the number of sets of 4 vertices in G which span 2 disjoint edges is at most n4/256. Equality holds iff G is 2 disjoint copies of Kn/2, n/2.

The fraction of “good” sets of 4 vertices is the limit as n goes to infinity of [n4/256]/C(n,4), which is 3/32.

The problem of finding the graph (not necessarily bipartite) on n vertices which has the greatest number of sets of 4 vertices which span 2 disjoint edges is a special case of a problem solved about 20 years ago by Bollobás, J. Brown and Siderenko.

They showed that among the host graphs on n vertices which span the most copies of a complete bipartite graph is one which itself is complete bipartite. Two disjoint edges is the complement of K2,2 and a simple calculation shows that Kn/2, n/2 has the most induced copies of K2,2 so two copies of Kn/2 has the most induced copies of 2 disjoint edges.

Comments and questions

For what other vertex configurations H in the d-cube can π(H,d) be calculated?

What about edge configurations, rather than vertex configurations? (We have some results.)

Which bipartite graph with n vertices has the most sets of six vertices which induce three disjoint edges? Is the obvious conjecture correct?

Thank you!