    # On Mappings on the Hypercube with Small Average Stretch

Let A ⊆{0,1}^n be a set of size 2^n-1, and let ϕ{0,1}^n-1→ A be a bijection. We define the average stretch of ϕ as avgStretch(ϕ) = E[ dist(ϕ(x),ϕ(x'))], where the expectation is taken over uniformly random x,x' ∈{0,1}^n-1 that differ in exactly one coordinate. In this paper we continue the line of research studying mappings on the discrete hypercube with small average stretch. We prove the following results. (1) For any set A ⊆{0,1}^n of density 1/2 there exists a bijection ϕ_A {0,1}^n-1→ A such that avgstretch(ϕ_A) = O(√(n)). (2) For n = 3^k let A_ rec-maj = {x ∈{0,1}^n : rec-maj(x) = 1}, where rec-maj : {0,1}^n →{0,1} is the function recursive majority of 3's. There exists a bijection ϕ_ rec-maj{0,1}^n-1→ A_ rec-maj such that avgstretch(ϕ_ rec-maj) = O(1). (3) Let A_ tribes = {x ∈{0,1}^n : tribes(x) = 1}. There exists a bijection ϕ_ tribes{0,1}^n-1→ A_ tribes such that avgstretch(ϕ_ tribes) = O((n)). These results answer the questions raised by Benjamini et al. (FOCS 2014).

## Authors

03/01/2022

### Two Classes of Power Mappings with Boomerang Uniformity 2

Let q be an odd prime power. Let F_1(x)=x^d_1 and F_2(x)=x^d_2 be power ...
03/17/2022

### A recursive function coding number theoretic functions

We show that there exists a fixed recursive function e such that for all...
10/29/2018

### Location and scale behaviour of the quantiles of a natural exponential family

Let P_0 be a probability on the real line generating a natural exponenti...
03/08/2018

### A note on two-colorability of nonuniform hypergraphs

For a hypergraph H, let q(H) denote the expected number of monochromatic...
03/11/2018

### Paths between colourings of sparse graphs

The reconfiguration graph R_k(G) of the k-colourings of a graph G has as...
05/10/2020

### Plurality in Spatial Voting Games with constant β

Consider a set of voters V, represented by a multiset in a metric space ...
07/12/2019

### Equiprobable mappings in weighted constraint grammars

We show that MaxEnt is so rich that it can distinguish between any two d...
##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

In this paper we continue the line of research from [BCS14, RS16, JS18] studying geometric similarities between different subsets of the hypercube . Given a set of size and a bijection from , we define the average stretch of as

 avgStretch(ϕ)=Ex∼x′∈Hn−1[dist(ϕ(x),ϕ(x′))],

where the expectation is taken over a uniformly random that differ in exactly one coordinate.111Note that any -Lipschitz function satisfies . That is, the notion of average stretch is a relaxation of the Lipschitz property.

The origin of this notion is motivated by the study of complexity of distributions [GGN10, Vio12, LV12]. In this line of research given a distribution on the goal is to find a mapping such that if

is the uniform distribution over

, then is (close to) the distribution , and each output bit of the function is computable efficiently (e.g., computable in , i.e., by polynomial size circuits of constant depth).

Motivated by the goal of proving lower bounds for sampling from the uniform distribution on some set , Lovett and Viola [LV12] suggested the restricted problem of proving that no bijection from to can be computed in . Toward this goal they noted that it suffices to prove that any such bijection requires large average stretch. Indeed, by the structural results of [Hås86, Bop97, LMN93] it is known that any such mapping that is computable by a polynomial size circuit of depth has , and hence proving that any bijection requires super-polylogarithmic average stretch implies that it cannot be computed in .

Studying this problem, [BCS14]

have shown that for odd

if is the hamming ball of density , i.e. , then there is a -bi-Lipschitz mapping form to , thus suggesting that proving a lower bound for a bijection from to requires new ideas beyond the sensitivity-based structural results of [Hås86, Bop97, LMN93] mentioned above. In [RS16] it has been shown that if a subset of density

is chosen uniformly at random then with high probability there is a bijection

with . This result has been recently improved by [JS18], who showed that for a random set of density with high probability there exists a -Lipschitz bijection from to .

The following problem was posed in [BCS14], and repeated in [RS16, JS18].

###### Problem 1.1.

Exhibit a subset of density such that any bijection has , or prove that no such subset exists.222Throughout the paper, the density of a set is defined as .

To rephrase creftype 1.1, we are interested to determine a tight upper bound on the that holds uniformly for all sets of density . Note that since the diameter of is , for any set of density any bijection has . It is natural to ask how tight this bound is, i.e., whether there exists of density such that any bijection requires linear average stretch.

It is consistent with our current knowledge (though hard to believe) that for any set of density there is a mapping with . The strongest lower bound we are aware of is for the set . Note that the distance between any two points in is at least 2, and hence for any mapping . Proving a lower bound strictly greater than 2 for any set is an open problem, and prior to this work we are not aware of any sublinear upper bounds that apply uniformly to all sets.

### 1.1 A uniform upper bound on the average stretch

We prove a non-trivial uniform upper bound on the average stretch of a mapping that applies to all sets of density . Specifically, we show that for any set there exists a bijection with .

###### Theorem 1.

For any set of density there exists a bijection such that .

Toward this goal we prove a stronger result bounding the average transportation distance between two arbitrary sets of density . Specifically, we prove the following theorem.

###### Theorem 2.

For any two sets of density there exists a bijection such that .

Note that creftype 1 follows immediately from creftype 2 by the following simple argument.

###### Proposition 1.2.

Let be a bijection. Then .

###### Proof.

Using the triangle inequality we have

 avgStretch(ϕ) = Ex∼Hn−1i∼[n−1][dist(ϕ(x),ϕ(x+ei))] ≤ E[dist(x,ϕ(x))+dist(x,x+ei)+dist(x+ei,ϕ(x+ei))] = E[dist(x,ϕ(x))]+1+E[dist(x+ei,ϕ(x+ei))] = 2E[dist(x,ϕ(x))]+1,

as required. ∎

### 1.2 Bounds on the average stretch for specific sets

Next, we study two specific subsets of defined by Boolean functions commonly studied in the field “Analysis of Boolean functions” [O’D14]. Specifically, we study two monotone noise-sensitive functions: the recursive majority of 3’s, and the tribes function.

It was suggested in [BCS14] that the set of ones of these functions may be such that any mapping requires large . We show that for the recursive majority function there is such a mapping with . For the tribes function we show a mapping with . Below we formally define the functions, and discuss our results.

#### 1.2.1 Recursive majority of 3’s

The recursive majority of 3’s function is defined as follows.

###### Definition 1.3.

Let be a positive integer. Define the function recursive majority of 3’s as follows.

• For the function is the majority function on the input bits.

• For the function is defined recursively as follows. For each write , where each for each . Then, .

Note that for all , and hence the density of the set is . We prove the following result regarding the set .

###### Theorem 3.

For a positive let , and let . There exists a mapping such that .

#### 1.2.2 The tribes function

The tribes function is defined as follows.

###### Definition 1.4.

Let be two positive integers, and let . The function is defined as a DNF consisting of disjoint clauses of width .

 tribes(x1,x2,…,xw;…;x(s−1)w+1…xsw)=s⋁i=1(x(i−1)w+1∧x(i−1)w+2∧⋯∧xiw).

That is, the function partitions inputs into disjoint “tribes” each of size , and returns 1 if and only if at least one of the tribes “votes” 1 unanimously.

It is clear that . The interesting settings of parameters and are such that the function is close to balanced, i.e., this probability is close to . Given let be the largest integer such that . For such choice of the parameters we have (see, e.g., [O’D14, Section 4.2]).

Consider the set . Since the density of is not necessarily equal to , we cannot talk about a bijection from to . In order to overcome this technical issue, let be an arbitrary superset of of density . We prove that there is a mapping from to with average stretch . In fact, we prove a stronger result, namely that the average transportation distance of is .

###### Theorem 4.

Let be a positive integer, and let be the largest integer such that . Let , and let be an arbitrary superset of of density . Then, there exists a bijection such that . In particular .

The rest of the paper is organized as follows. We prove creftype 2 in Section 2. In Section 3 we prove creftype 3, and in Section 4 we prove creftype 4.

## 2 Proof of creftype 2

We provide two different proofs of creftype 2. The first proof, in Section 2.1 shows a slightly weaker bound of on the average stretch using the Gale-Shapley result on the stable marriage problem. The idea of using the stable marriage problem has been suggested in [BCS14], and we implement this approach. Then, in Section 2.2, we show the bound of

by relating the average stretch of a mapping between two sets to known estimates on the Wasserstein distance on the hypercube.

### 2.1 Upper bound on the average transportation distance using stable marriage

Recall the Gale-Shapley theorem on the stable marriage problem. In the stable marriage problem we are given two sets of elements and each of size . For each element (reps. ) we have a ranking of the elements of (reps. ) given as an bijection () representing the preferences of each (resp. ). A matching (or a bijection) said to be unstable if there are some , and such that , , but , and ; that is, both and prefer to be mapped to each other before their matchings given by . We say that a matching is stable otherwise.

###### Theorem 2.1 (Gale-Shapley theorem).

For any two sets and any rankings for each and there exists a stable matching .

Consider the stable marriage problem on the sets and with preferences induced by the distances in the graph. That is, for each we have if and only if with ties broken arbitrarily. Similarly, for each we have if and only if with ties are broken arbitrarily.

Let be a bijection. We show below that if , then is not a stable matching. Let , and consider the set

 F:={x∈X∣dist(x,ϕ(x))≥k}.

Note that since the diameter of is , and , it follows that . Indeed, we have , and thus .

Next, we use Talagrand’s concentration inequality.

###### Theorem 2.2 ([Tal95, Proposition 2.1.1]).

Let be two positive integers, and let . Denote by the set of all whose distance from is at least , i.e., . Then .

By Theorem 2.2 we have , and hence, for it holds that

 μn(F≥k)≤e−ln(n)/μn(F)≤(1/n)/(k/n)=1/k.

In particular, since , there is some that does not belong to . That is, there is some and such that . On the other hand, for , by definition of we have and , and hence is not stable, as and prefer each other to their current matching. Therefore, in a stable matching , and by the Gale-Shapley theorem such a matching, indeed, exists. ∎

### 2.2 Proof of creftype 2 using transportation theory

Next we prove creftype 2, by relating our problem to a known estimate on the Wasserstein distance between two measures on the hypercube. Recall that the -Wasserstein distance between two measures and on is defined as

 W1(μ,ν)=infq∑x,ydist(x,y)q(x,y),

where the infimum is taken over all couplings of and , i.e., and for all . That is, we consider an optimal coupling of and minimizing , the expected distance between and , where is distributed according to and is distributed according to .

We prove the theorem using the following two claims.

###### Claim 2.3.

Let and be uniform measures over the sets and respectively. Then, there exists a bijection from to such that .

###### Claim 2.4.

Let and be uniform measures over the sets and respectively. Then

###### Proof of creftype 2.3.

Observe that any bijection from to naturally defines a coupling of and , where for all . Therefore, .

For the other direction note that in the definition of we are looking for the infimum of the linear function , where the infimum is taken over the Birkhoff polytope of all doubly stochastic matrices. By the The Birkhoff-von Neumann theorem [Bir46, vN53, Kőn36] this polytope is the convex hull whose extremal points are precisely the permutation matrices. Therefore, the optimum is obtained on such an extremal point, and hence there exists a bijection from to such that . ∎

###### Proof of creftype 2.4.

The proof of the claim follows rather directly from the techniques in transportation theory (see [RS13, Section 3.4]). Specifically, using Definition 3.4.2 and combining Proposition 3.4.1, Equation 3.4.42, and Proposition 3.4.3, where , and is the uniform distribution on we have the following theorem.

###### Theorem 2.5.

Let be an arbitrary distribution on the discrete hypercube , and let be the uniform distribution on . Then

 W1(ν,μn)≤√12n⋅D(ν∣∣μn).

In particular, by letting be the uniform distribution over the set of cardinality , we have , and hence . Analogously, we have . Therefore, by triangle inequality, we conclude that , as required. ∎

This completes the proof of creftype 2.

## 3 Average stretch for recursive majority of 3’s

In this section we prove creftype 3, showing a mapping from to with constant average stretch. The key step in the proof is the following lemma.

###### Lemma 3.1.

For a positive let . There exists satisfying the following properties.

1. for all .

2. For each there is a unique such that .

3. For every we have .

We postpone the proof of Lemma 3.1 for now, and show how it implies creftype 3.

###### Proof of creftype 3.

Let be the mapping from Lemma 3.1. Define as , where is the string obtained from by appending to it as the ’th coordinate.

The mappings naturally induce a bipartite graph , where and , possibly, containing parallel edges. Note that by the first two items of Lemma 3.1 the graph is 2-regular. Indeed, for each the neighbours of are , and for each there is a unique and a unique such that , and hence .

Since the bipartite graph is 2-regular, it has a perfect matching. Let be the bijection from to induced by a perfect matching in , and for each let be such that . We claim that . Let be uniformly random in that differ in exactly one coordinate, and let be uniformly random. Then

 E[dist(ϕ(x),ϕ(x′))] = E[dist(fk(x∘bx),fk(x′∘bx′))] ≤ E[dist(fk(x∘bx),fk(x∘r))]+E[dist(fk(x∘r),fk(x′∘r))] +E[dist(fk(x′∘r),fk(x′∘bx′))].

For the first term, since is equal to with probability by Lemma 3.1 Item 3 we get that . Analogously the third term is bounded by . In the second term we consider the expected distance between applied on inputs that differ in a random coordinate , which is at most , again, by Lemma 3.1 Item 3. Therefore . ∎

###### Proof of Lemma 3.1.

Define by induction on . For define as

 000 ↦110 100 ↦101 010 ↦011 001 ↦111 x ↦xfor all x∈{110,101,011,111}.

That is, acts as the identity map for all , and maps all inputs in to in a one-to-one way. Note that is a non-decreasing mapping, i.e., for all and .

For define recursively using as follows. For each let be the ’th third of the interval . For write , where is the ’th third of . Let be defined as , and let . Define

 f(r)k−1(x(r))={fk−1(x(r)),if wr≠yrx(r)otherwise.

Finally, the mapping is defined as

 fk(x)=f(1)k−1(x(1))∘f(2)k−1(x(2))∘f(3)k−1(x(3)).

That is, if then , and hence , and otherwise, for all where and .

Next we prove that satisfies the properties stated in Lemma 3.1.

1. It is clear from the definition that if then , and hence .

2. Next, we prove by induction on that the restriction of to induces a bijection. For the statement clearly holds. For suppose that the restriction of to induces a bijection. We show that for every the mapping has a preimage of in . Write , where is the ’th third of . Let be defined as . Since it follows that . Let such that .

For each such that and it must be the case that , and hence, by the induction hypothesis, there is some such that . For each such that define . Since , it follows that . It is immediate by the construction that, indeed, .

3. Fix . In order to prove consider the following events.

 E1={rec-majk(x)=1=rec-majk(x+ei)}, E2={rec-majk(x)=0,rec-majk(x+ei)=1}, E3={rec-majk(x)=1,rec-majk(x+ei)=0}, E4={rec-majk(x)=0=rec-majk(x+ei)}.

Then . The following three claims prove an upper bound on .

.

.

###### Claim 3.4.

.

By symmetry, it is clear that . Therefore, using the fact that , and noting that , the claims above imply that