At least |S|k-1 · (|S| - 1) frontiers: a graph theory problem

At least | _S | ^{k − 1} · (| _S | − ₁ ) frontiers:

a graph theory problem

Darij Grinberg

October 18, 2015 (modified version of a note from 2009)

1. Problem

LetSbe a finite set. Letkbe a positive integer. Let Abe a subset ofS^k satisfying

|A|=|S|^k−1. Let B=S^k\ A.

For everyv∈ S^kand everyi∈ {1, 2, . . . ,k}, we denote byv_ithei-th component of thek-tuplev(remember thatvis an element ofS^k, that is, ak-tuple of elements ofS). Then, everyv ∈S^k satisfies v= (v₁,v₂, . . . ,v_k).

Let F be the set of all pairs (a,b) ∈ A×B for which there exists an i ∈ {1, 2, . . . ,k} satisfying aj =bj for all j 6=i ₁

. (Speaking less formally, let F be the set of all pairs(a,b) ∈ A×Bfor which the k-tuples aand b differ in at most one position.)

Prove that |F| ≥ |S|^k−1·(|S| −1).

2. Remark

In the case |S| = 2, this is an old problem (which appeared, for example, in a Moscow MO 1962 preparation booklet, and which is a particular case of Cheeger’s inequality for the hypercube).

I use to call the elements of F“frontiers” between the sets A and B.

3. Solution

Since A ⊆ S^k and B = S^k\ A, we have A∩ B = _∅ and A∪B = S^k. Hence, S^k\B= A.

Define a map φ: A×B →F as follows:

1Of course, “for allj6=i” means “for allj∈ {1, 2, . . . ,k}satisfyingj6=i” here.

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At least|S|^k−1·(|S| −1) frontiers October 18, 2015

Let (u,v) ∈ A×B be a pair. Then,u ∈ A and v ∈ B, so thatu ∈/ Band v ∈/ A (since A∩B =_∅).

We define a subset T of{0, 1, 2, . . . ,k} by

T={i ∈ {0, 1, 2, . . . ,k} | (v₁,v₂, . . . ,v_i,u_i+1,u_i+2, . . . ,u_k) ∈ B}

2. Then, 0 /∈ T (since (u1,u2, . . . ,u_k) =u ∈/ B) and k∈ T (since(v1,v2, . . . ,v_k) = v ∈ B). In particular, k ∈ T yields T 6= ∅. Thus, the set T has a minimal element (since T is a finite set). Letα be this minimal element. Then, α ∈ T and α−1 /∈ T. We have α 6=0 (since α ∈ T but 0 /∈ T). Thus, α−1 ∈ {0, 1, 2, . . . ,k} (sinceα ∈ T ⊆ {0, 1, 2, . . . ,k}).

Now, α ∈ T yields (v₁,v2, . . . ,vα,u_α+1,u_α+2, . . . ,u_k) ∈ B, while α −1 /∈ T yields(v1,v2, . . . ,vα−1,uα,uα+1, . . . ,u_k) ∈/ B, so that(v1,v2, . . . ,vα−1,uα,uα+1, . . . ,u_k) ∈ S^k\B= A. Set a= (v₁,v₂, . . . ,v_α−1,u_α,u_α+1, . . . ,u_k)and

b = (v₁,v2, . . . ,vα,u_α+1,u_α+2, . . . ,u_k). Then,a= (v₁,v2, . . . ,v_α−1,uα,u_α+1, . . . ,u_k)∈ Aand

b = (v₁,v₂, . . . ,v_α,u_α+1,u_α+2, . . . ,u_k) ∈ B, so that (a,b) ∈ A×B. Besides, there exists an i ∈ {1, 2, . . . ,k} satisfying a_j =b_j for all j 6=i

(namely, i = α ³).

Hence,(a,b)∈ F(by the definition of F).

Now set φ(u,v) = (a,b). Thus we have defined a map φ: A×B→ F.

Next, we will prove that

φ⁻¹({(a,b)}) ≤ |S|^k−1 for every (a,b) ∈ F. In fact, let(a,b) ∈ F. Since (a,b) ∈ F, we have (a,b) ∈ A×B, so that a ∈ A and b ∈ B, so thata6=b (since A∩B =∅). But since(a,b) ∈ F,

there exists ani ∈ {1, 2, . . . ,k} satisfying a_j =b_j for all j 6=i

. (1)

Consider thisi.

We must have ai 6=bi 4.

Now, consider some (u,v) ∈ _φ⁻¹({(a,b)}). Then, φ(u,v) = (a,b). Thus, by the definition ofφ, there exists an α ∈ {0, 1, 2, . . . ,k} such that

a= (_{v1,v2, . . . ,vα−1,uα,uα+1, . . . ,uk}) and b = (_{v1,v2, . . . ,vα,uα+1,uα+2, . . . ,uk}). (2)

Consider thisα.

2Fori=0, the notation(v1,v2, . . . ,vi,ui+1,ui+2, . . . ,uk)means(u1,u2, . . . ,uk). Fori=k, the notation(v1,v2, . . . ,vi,ui+1,ui+2, . . . ,uk)means(v1,v2, . . . ,vk).

3In fact, a_j = b_j for all j 6= α (in fact, for any j, we have a_j =

vj, ifj<α;

u_j, ifj≥α and b_j =

v_j, ifj≤α;

u_j, ifj>α ; thus, if j 6= α, this simplifies to a_j =

v_j, ifj<α;

u_j, ifj>α and b_j = v_j, ifj<α;

uj, ifj>α , so thataj=bj for allj6=α).

4In fact, otherwise, we would have a_i = b_i, what, combined witha_j = b_j for all j 6= i, would yielda_j =b_jfor allj∈ {1, 2, . . . ,k}, so thata=b, contradictinga6=b.

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At least|S|^k−1·(|S| −1) frontiers October 18, 2015

We must have uα 6= vα 5. Since aα = uα and bα = vα, this yields aα 6= bα. Hence, α = i (since otherwise, we would have α 6= i, so that aα = bα by (1), contradictingaα 6=_bα). Thus, (2) becomes

a= (v₁,v2, . . . ,v_i−1,u_i,u_i+1, . . . ,u_k) and b = (v₁,v2, . . . ,v_i,u_i+1,u_i+2, . . . ,u_k). Now, a= (_{v1,v2, . . . ,vi−1,ui,ui+1, . . . ,uk})yields aj =_{uj for all j}≥_{i. Hence,}

u= (u₁,u2, . . . ,u_i−1,u_i,u_i+1, . . . ,u_k) = (u₁,u2, . . . ,u_i−1,a_i,a_i+1, . . . ,a_k)

∈ _Sⁱ⁻¹× {_ai} × {ai+1} × · · · × {ak}. (3)

6

Also, b= (v1,v2, . . . ,vi,ui+₁,ui+₂, . . . ,u_k)yieldsbj =vj for all j ≤i. Hence, v = (v₁,v₂, . . . ,v_i,v_i+1,v_i+2, . . . ,v_k) = (b₁,b₂, . . . ,b_i,v_i+1,v_i+2, . . . ,v_k)

∈ {b₁} × {b2} × · · · × {b_i} ×S^k−i. (4)

7

By (3) and (4), we have

(u,v) ∈Sⁱ⁻¹× {a_i} × {a_i+1} × · · · × {a_k}×{b₁} × {b₂} × · · · × {b_i} ×S^k−i

for every(u,v)∈ _φ⁻¹({(a,b)}). Thus,

φ⁻¹({(a,b)})

≤

Sⁱ⁻¹× {_ai} × {ai+1} × · · · × {ak}×{b1} × {b2} × · · · × {bi} ×S^k−i

=

Sⁱ⁻¹× {ai} × {ai+1} × · · · × {a_k}·{b1} × {b2} × · · · × {bi} ×S^k−i

=









 Sⁱ⁻¹

| {z }

=|S|ⁱ⁻¹

· |{a_i}|

| {z }

=1

· |{a_i+1}|

| {z }

=1

· · · |{a_k}|

| {z }

=1











·











|{b₁}|

| {z }

=1

· |{b₂}|

| {z }

=1

· · · |{b_i}|

| {z }

=1

·S^k−i

| {z }

=|S|^k−i











=|S|ⁱ⁻¹· |S|^k−i =|S|^k−1 (5)

5because otherwise, we would haveu_α =v_αand thus

a=



v₁,v2, . . . ,vα−1, u_α

|{z}=vα

,u_α+1, . . . ,u_k



= (v₁,v2, . . . ,v_α,u_α+1,uα+2, . . . ,u_k) =b,

contradictinga6=b.

6By abuse of notation, we are writing Sⁱ⁻¹ × {a_i} × {a_i+1} × · · · × {a_k} for S×S× · · · ×S

| {z }

i−1 factors

× {a_i} × {a_i+1} × · · · × {a_k}here.

7By abuse of notation, we are writing{b1} × {b2} × · · · × {bi} ×S^k−i for {b1} × {b2} × · · · × {bi} ×S×S× · · · ×S

| {z }

k−ifactors

here.

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At least|S|^k−1·(|S| −1) frontiers October 18, 2015

for every(a,b) ∈ F.

Thus,

|A×B|=

∑

(a,b)∈F

|{(u,v)∈ A×B | _φ(u,v) = (a,b)}|

=

∑

(a,b)∈F

φ⁻¹({(a,b)})

| {z }

≤|S|^k−1 (by (5))

≤

∑

(a,b)∈F

|S|^k−1

=|F| · |S|^k−1. But

|A×_B| =|A| · |B| =|A| ·S^k\ _A

=|A| ·S^k

− |_A|

=|S|^k−1·S^k

− |S|^k−1=|S|^k−1·|S|^k− |S|^k−1, so this becomes

|S|^k−1·|S|^k− |S|^k−1≤ |F| · |S|^k−1, thus|S|^k− |S|^k−1≤ |F|, so that

|F| ≥ |S|^k− |S|^k−1=|S|^k−1·(|S| −1), qed.

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