Lower Bound

For the lower bound, we make use of the identification of each vertex of the n-cube with a binary string of lengthn. A vertex in the hypercube is even if its corresponding binary string has an even number of 1’s. In the same way a vertex is odd if its corresponding binary string has an odd number of 1’s. Figure 4.1 illustratesn-cubes with 1 ≤n ≤ 3. Denote the set of even vertices of a given n-cube by E and the set of odd vertices by ¯E. The sets E and ¯E are are illustrated in Figures 4.2 and 4.3, respectively. 0 1 00 10 01 11 000 100 010 110 001 101 011 111

Figure 4.1 The n-cubes for 1≤n ≤3

Lemma 4.4. Suppose thatS ⊆V(Qn), be a subset of even vertices i.e. each element of S contains an even number of 1’s. Then

d(S)≥ |S|+|S| 2

0 00 11 000 110 101 011

Figure 4.2 The even vertices of the n-cube where 1≤n ≤3 1 10 01 100 010 001 111

Figure 4.3 The odd vertices of then-cube where 1≤n≤3

Proof. Suppose thatS ⊆E, is a subset of the even vertices of Qn. Let ¯S be obtained fromS by switching the value of first entry of vertex inS. We have thatd(S) =d( ¯S) by symmetry. Suppose that T = (V(T), E(T)) and ¯T = (V( ¯T), E( ¯T)) are Steiner trees ofSand ¯S, respectively. Naively, we have thatd(S∪S¯)≥2|S|−1. Furthermore, connecting T and ¯T yields a subgraph of Qn spanning S ∪S¯. Hence, d(S ∪S¯) ≤

|E(T)∪E( ¯T)|+ 1. Then,

2|S| −1≤ |E(T)∪E( ¯T)|+ 1

=|E(T)|+|E( ¯T)| − |E(T)∩E( ¯T)|+ 1 = 2d(S)− |E(T)∩E( ¯T)|+ 1,

which implies that

2|S| −2≤2d(S)− |E(T ∩T¯)|. (4.1) Let Γ =hα, βi,jibe the subgroup of automorphisms ofQn generated by the following automorphisms

α :v0v1· · ·vn−1 7→v1· · ·vn−1v0

βi,j :v0v1· · ·vi· · ·vj· · ·vn−1 7→v0v1· · ·v¯i· · ·v¯j· · ·vn−1.

Here, α cycles through a binary stringv while each βi,j “flips” i’th and j’th bits of a binary string v.

We now show that for any two edges e1, e2 ∈E(Qn), there exists a unique γ ∈Γ such that γ(e2) = e1. Suppose that e1 = ab and e2 = uv where a and u are even vertices while b and v are odd vertices. Without loss of generality, we may assume that a=0, the vertex of all zeros. This implies that the string b contains a single 1. We shall first prove existence.

First, using a composition of automorphisms of the form bi,j we may map uv to

0vˆ. Using some power of the automorphism α, we may then map the edge 0vˆto the edge 0b=e1. Hence, there exists γ ∈Γ such that γ(e2) =e1.

To show that this map is unique, we provide a combinatorial argument. Enu- merate the images of a binary string v =v0v1· · ·vn−1 under automorphisms in Γ. If

γ ∈Γ, then

γ(v) = uiui+1· · ·unu1· · ·ui−1,

where ui ∈ {vi,v¯i} for each 1 ≤ i≤ n. Now there are n choices for i. Furthermore,

γ(v) contains an even number of “flipped” bits. So there are 2n−1 _{choices for the} number of 1’s. So there aren2n−1 _{possible images ofv under aγ ∈Γ. Since there are}

n2n−1 _{edges in the n-cube, this implies that |γ| ≥ n2}n−1_{. Hence, any automorphism}

We now consider the experiment of selecting elements λ1, λ2 ∈ Γ with uniform probability and independently applying them toT and ¯T, respectively. Since for any two edgesf, g∈E(Qn) there there is a unique automorphism mapping anf tog, we have that for any two automorphisms λ1, λ2 ∈Γ, it holds that

P[f ∈λ1(T)] =P[f ∈λ2(T)] =

|E(T)|

n2n−1 .

We now consider the random variable X =|E(λ1(T)∩λ2( ¯T))|. For the expected value ofX, _E(X), we have that

max λ1,λ2 {|E(λ1(T)∩λ2( ¯T))|} ≥E(X). We observe that E(X) = X f∈E(Qn) P[f ∈λ1(T)∧f ∈λ2( ¯T)] = X f∈E(Qn) |E(T)| n2n−1 · |E( ¯T)| n2n−1 = |E(T)| 2 n2n−1 = d(S) 2 n2n−1, which implies max

λ1,λ2

{e(λ1(T)∩λ2( ¯T))} ≥ d(S)

2

n2n−1. Applying this to equation (4.1), we

see that

2d(S)− d(S)

2

n2n−1 ≥2|S| −2. Now, d(S) =|S|+x for some x≥0. So,

2(|S|+x)−(|S|+x) 2 n2n−1 ≥2|S| −2 2|S|+ 2x− |S| 2_{+ 2|S|x+x}2 n2n−1 ≥2|S| −2 2x− 2|S|x n2n−1 + 2|S| − |S|2_+x2 n2n−1 ≥2|S| −2 2x(1− |S| n2n−1)≥ |S|2_+x2 n2n−1 −2 x≥ |S| 2 n2n −1. Hence, d(S)≥ |S|+ _n2|S|n2 −1.

With these preliminary results in hand, we can determine the asymptotic growth of sdiamk(Qn) for largek.

Theorem 4.5. If k =k(n), then 1. If k = Ω(2n_{), then sdiam} k(Qn) =k+O(2n/n), and 2. If lim n→∞ k 2n_/n =∞, then sdiamk(Qn)∼k. Proof. If k ≤ 2n−1_{, let S ⊂ V(Q}

n) be a subset of the even vertices of size k. If

k > 2n−1_{, let S contain all even vertices and choose the remaining odd vertices} randomly. Applying the bounds determined in Lemmas 4.3 and 4.4, we see that

k+ k 2

n2n −1≤d(S)≤sdiamk(Qn)≤k+ 2n

n (1 +o(1)).

This immediately gives that sdiamk(Qn)≥k−1 +O(2n/n) and sdiamk(Qn) = Ω(k). If lim

n→∞

k

2n_/n =∞, then we have that 2

n_{/n=o(k), so sdiam(Q}

n) =k(1 +o(1)), giving sdiamk(Qn)∼k.