# Towards a General Direct Product Testing Theorem

The Direct Product encoding of a string a∈{0,1}^n on an underlying domain V⊆nk, is a function DP_V(a) which gets as input a set S∈ V and outputs a restricted to S. In the Direct Product Testing Problem, we are given a function F:V→{0,1}^k, and our goal is to test whether F is close to a direct product encoding, i.e., whether there exists some a∈{0,1}^n such that on most sets S, we have F(S)=DP_V(a)(S). A natural test is as follows: select a pair (S,S')∈ V according to some underlying distribution over V× V, query F on this pair, and check for consistency on their intersection. Note that the above distribution may be viewed as a weighted graph over the vertex set V and is referred to as a test graph. The testability of direct products was studied over various specific domains and test graphs (for example see Dinur-Steurer [CCC'14]; Dinur-Kaufman [FOCS'17]). In this paper, we study the testability of direct products in a general setting, addressing the question: what properties of the domain and the test graph allow one to prove a direct product testing theorem? Towards this goal we introduce the notion of coordinate expansion of a test graph. Roughly speaking a test graph is a coordinate expander if it has global and local expansion, and has certain nice intersection properties on sampling. We show that whenever the test graph has coordinate expansion then it admits a direct product testing theorem. Additionally, for every k and n we provide a direct product domain V⊆nk of size n, called the Sliding Window domain for which we prove direct product testability.

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## 1 Introduction

The direct product encoding of a function is a way to aggregate multiple values of the input function using a single query. Justifying the vague intuition that it is much harder to compute multiple values of a function rather then a single value of it, the direct product encoding has been successfully used in several contexts of hardness amplification. The hardness can either measure the fraction of inputs on which every reasonable-time algorithm fails to compute the input function, or the fraction of unsatisfied assignments of a given CNF-formula or the communication complexity of the function.

In most of the PCP constructions an assignment to the given input is broken into many tiny pieces. Each small piece is encoded individually and then one should be able to test whether these tiny pieces could be stitched together into a global assignment. This testability task is referred to as an agreement test, and instantiations of it include low degree tests such as the plane vs. plane [RS97], the line vs. line test [AS97] and the cube vs. cube test [BDN17], and the direct product test used in [DR06].

More concretely, we associate the direct product encoding of strings of size , with some underlying domain222For the ease of presentation, we only consider domains which are a subset of in this section. However, in the rest of the paper we consider which is a collection of subsets of , and all our results are proved for this more general case. which is a collection of subsets of of cardinality . Given a string its direct product encoding on the domain , denoted by , is defined as follows: For every set we define (where is the restriction of to the coordinates in ). In this paper we study the testability of this encoding, namely: Given we want to decide whether agrees with some on most sets while querying only on a few locations, specifically two. In other words, we focus on two-query tests in the paper where we pick a pair of subsets (both in the domain) according to some fixed distribution and then check if the two subsets agree on their intersection. We say that a domain admits a direct product testing theorem if there exists a two-query test satisfying the following: For every and if accepts

with probability

, then we have for some on -fraction of the sets in , where the constant behind the notation is independent of and .

This question was studied under various domains. Dinur and Steurer [DS14] analyzed a two-query test under the domain . Recently, Dinur and Kaufman [DK17] studied this question in a much shrunken domain, which is obtained by considering the set of the faces of a high dimensional expander. However, both of these proofs are tailored to the structure of their own domain and cannot be (trivially) generalized to other domains. It is natural to ask whether a more generalized argument can be applied covering both of these domains, and on which domains it may be applied. The main question we are investigating is as follows:

Which domains admit a two-query direct product testing theorem?

Let us elaborate more about the previous proofs. The proofs given by  [DS14] and [DK17] first analyze the testability in the high error regime, i.e. when the acceptance probability is slightly bounded away from . They show that any function that passes the test with non-negligible probability must agree with some legal codeword on fraction of sets. Then they analyze the test in the low error regime, i.e. when the acceptance probability of the test is close to . Finally they stitch local tiny agreements into a single codeword and show that the agreement is almost everywhere.

We would like to establish a direct product testing theorem using a more straightforward approach: we decode a string from the input function using the majority operator and then show that if the test passes with high probability then is close to the direct product encoding of the decoded string. More precisely, given the input function , we define a string as follows: for every coordinate we set to be the majority value of , where the majority is taken over the sets that contain . Next we show that if passes the test with probability then must be -close to . We remark that Dinur and Reingold [DR06] indeed followed this proof strategy, however, their proof admits only a relaxed notion of closeness between the input function and the direct product encoding of the decoded string (namely, that on most sets , and agree only on most of the coordinates in ).

Observe that any two-query test on a domain gives rise to a weighted graph whose vertex set is and the weight we assign for each pair is the probability of this pair being picked by the test333In this paper we analyze test graphs which are undirected.. We refer to this graph as the test graph. We say that a test graph yields a tester for the domain , if for every and every function the following holds: if the test accepts with probability , then must be -close to some . Here the test corresponds to picking an edge at random (according to the distribution of weights on the edges) and accepting if and only if .

Another proof insight that we desire is the explicit use of the properties of the underlying test graph. For example, one property that the test graph must satisfy to be a tester is that for most edges the intersection between and is linear in . Assume not, then we consider the following construction of : We start from for some and then for each we reset the value of for some random . Then for most sets with small intersection the test accepts but is far from any direct product codeword. Another property that the test graph must have is some notion of expansion. Summing up, our more refined question is as follows:

What properties of the test graph yields a tester for its underlying domain?

### 1.1 Our Results

Our conceptual contributions in this paper are two-fold. First, we introduce a notion called coordinate expansion which captures the properties of direct product testable domains. Second, we introduce the sliding window domain which is of size exactly equal to the universe and is direct product testable. Our main technical contribution is showing that domains having coordinate expansion with certain parameters admit a direct product theorem.

#### 1.1.1 A General Direct Product Theorem

We introduce below the notion of coordinate expansion. Informally, a coordinate expander has both global and local expansion properties, and has good intersection properties.

###### Definition 1.1 ((λ,ρ)-Coordinate Expander).

Let be a test graph, where . For let and be the subgraph of induced by the vertices in . The graph is called -coordinate expander if:

1. (where and is the normalized adjacency matrix of ).

2. For every we have that and for each the probability that a uniformly random neighbor of is in is at least .

3. For every subset and , satisfying , the probability that for a uniformly random neighbor of we have is upper bounded bounded by .

Notice that condition 1 implies that the test graph must be a good expander (in the traditional sense). Moreover, condition 2, implies that on certain local subsets (i.e., subsets containing a common coordinate) of vertices, the induced subgraph must be expanding as well. Finally, condition 3 implies that the neighbors of every subset samples well every subset of .

Observe that condition 2 is necessary for the test graph to be a direct product tester. To see this, consider a test graph that does not satisfy this property, namely, there exists a coordinate for which: there exits a set such that . Then, we show that the test graph does not yield a tester. Indeed, consider the following construction of : we first choose for some . Then for every we change the value of to . Clearly, the distance of from a direct product encoding equals . However, the rejection probability equals:

 2⋅PrS′∼S[S∈Bi and S′∈Vi∖Bi]≤2⋅Pr[S∈Bi]⋅Pr[S′∈Vi]⋅PrS′∈Vi[S′∉Bi|S∈Bi]=o(1)⋅δ.

Then, we show our main technical result that coordinate expansion implies direct product testing (for a certain range of parameters).

###### Theorem 1.2.

Let and . Let be a test graph, , let , and . Let be a -coordinate expander. If passes the test implied by the test graph with probability then is -close to for some .

The overview of the above proof is given in Section 1.2. Also, as an application of the above theorem, we show444The claim as written here is slightly inaccurate. Please refer to Appendix B for a precise statement. a direct product theorem for the test graph isomorphic to the Johnson graph when is close to , where is a graph whose vertex set is the set of all subsets of of cardinality , and two subsets have an edge if their intersection is equal to . This should be compared to [DS14], where they show the direct product for the Johnson graph for all the layers up to (i.e., for all where ).

The main open problem stemming from our work is to improve the parameters in Theorem 1.2. In particular, does the following hold?

###### Open Problem 1.3.

Does -coordinate expansion imply a direct product theorem?

A positive resolution of the above open question would imply direct product testability on the test graph isomorphic to the Johnson graph for every layer of the Boolean hypercube (completely recovering the results in [DS14]). It even implies a direct product testability on a new domain: where the subsets are stemming from -dimensional subspaces of and two subsets are connected by an edge if they intersect on a -dimensional subspace (this is referred to as the Grassmann graph). Finally, we would like to recall that Theorem 1.2 states that -coordinate expansion implies a direct product theorem, i.e., in order to positively resolve Open Problem 1.3, we might need to improve the analysis in the proof of Theorem 1.2 to accommodate test graphs with weaker expansion properties.

In fact, if we can resolve Open Problem 1.3 in a slightly stronger way i.e., if we show that for some small enough constant , we have -coordinate expansion implies a direct product theorem then we recover the testability result of [DK17] on Ramanujan complexes. Summarizing, we view the study of coordinate expansion as providing a unified framework to prove direct product theorems. Also, it might be useful in the future to establish direct product testability for new domains (in a black-box manner).

#### 1.1.2 Sliding Window Domain

In this subsubsection, we define a new direct product testable domain which we call the sliding window domain, and also discuss about the necessary and sufficient structure that a domain (and test graph) should have, in order to admit direct product testing.

For every , the sliding window domain is the collection of all contiguous -sized subsets (windows) of , i.e., , where the addition is done modulo . Two vertices (i.e., subsets in ) have an edge in the test graph, if their intersection is non-empty. Notice that and yet we show that it admits a direct product theorem (see Theorem 5.1 for a simple proof).

Let us put the above result in context with the recent breakthrough of Dinur and Kaufman [DK17]. In [DK17], the authors obtain a direct product testable domain (subset of ) of size . The domain arises from the highly non-trivial object called Ramanujan complex. Such a domain is studied because apart from admitting a direct product theorem over a domain of size linear in the universe (i.e., ), it also has other desirable properties such as distance amplification which are needed for applications in gap and hardness amplification. Thus, our direct product testing result (Theorem 5.1) provides a conceptual clarification that if one is only interested in direct product testing as a property testing question, then there is a very simple domain of size , namely the sliding window domain, which is testable.

Roughly speaking, a domain (subset of ) has distance amplification if for every two strings of relative distance , the relative distance between their direct product encoding is . This seems to be a crucial property for PCP applications of direct product testing. Thus, the construction of the sliding window domain provides a conceptual clarification as to why we need high dimensional expanders: we can obtain direct product testing from simple constructions like the sliding window domain and we can obtain distance amplification from known constructions of vertex expanders (see Appendix D); but to obtain both simultaneously, [DK17] needed high dimensional expanders. We leave it as an open question whether there exists a simple construction admitting both direct product testability and distance amplification.

###### Open Problem 1.4.

Is there a (relatively) simple domain of linear size in the universe (i.e., ) for which we have both direct product testing and distance amplification?

Lack of Global Expansion. We would like to now briefly discuss about the minimal structure of the domain (and the test graph) sufficient to prove a direct product theorem. This is highlighted by the sliding window domain, an in particular by the proof of its testability (Lemma 5.2 to be precise). Notice that has very bad edge-expansion/vertex-expansion but is a very good local expander, i.e., the induced subgraph containing any particular coordinate has good expansion (in fact is a clique). Lemma 5.2 guarantees that in such situations555Lemma 5.2 can be generalized to accommodate test graphs which are locally subgraphs that strongly satisfy the expander mixing lemma. the domain admits direct product testing if for every vertex in the test graph, and every element in that vertex, the probability of retaining that element when moving to a uniformly random neighbor is bounded from below by a positive constant. The probability of retaining a coordinate when moving to a random neighbor is in , and thus admits a direct product theorem. Therefore, demonstrates that direct product testing does not require the test graph to be an expander (like the Johnson/Ramanujan graph) but only needs to have certain local expansion properties. Finally, recall that we had earlier argued that local expansion is necessary (to justify the need for condition 2 in Definition 1.1) for direct product testing.

Finally, it seems that conditions 1 and 3 in coordinate expansion are not (necessarily) needed for direct product testing, but are merely artifacts of our proof (Theorem 1.2). However, these conditions might imply distance amplification666This would be an interesting question to resolve in either direction. and are typically guaranteed in structured domains of interest (namely, Johnson, Grassmannian, and Ramanujan).

### 1.2 Technical Contribution: Proof Overview of Theorem 1.2

For the sake of convenience, through out this subsection, we fix and the test would pick pairs that intersect on elements and checks for agreement. As suggested above there is a natural way to decode any function using the majority operator: define a string by setting to be the majority value of for all . We define , i.e., is the subset of the domain that disagrees with the direct product encoding of the decoded string. Also for we call conflicting if . Our goal is to show that the test rejects with probability as is the relative distance between and .

Indeed fix , then it must contain at least one conflicting coordinate, say . Observe that with probability we also have that . Now if were a random set containing , then since at least half of the elements that contain agree with the majority value, the test rejects with probability . And the overall rejection probability of the test would be at least and we are done.

However, is not a random set that contains , it intersects with on further coordinates. Therefore, it may well be that among the neighbors of that contain we do not see the majority value so often. A natural way to overcome this is by aggregating all s’ that contain and disagree with the majority value on . We could try to show that if we start from some member of this set then with constant probability we reach that contains and resides outside of this aggregated set (by using the local expansion property). But this leads into another problem: using this argument sets that contain many conflicting coordinates are counted many times, whereas sets that contain few conflicting coordinates are counted much less.

Our analysis proceeds by studying the variance of the number of conflicting coordinates in the following manner. We first sort the set

based on the number of their conflicting coordinates. Let (resp. ) be the first (resp. last) third of the elements in according the sorting. We first show that if the number of conflicting coordinates of each member in is much smaller than it is in , then the test rejects with probability . To show this, we prove that whenever the test picks then with constant probability is in (by using the global expansion property). Moreover, there is a large subset of conflicting coordinates in which are also in (follows from condition 3 in Definition 1.1). However, has few conflicting coordinates in total (by our choice of ), and thus, there must be a coordinate in that agrees with the majority value on but disagrees on it on and hence the test rejects the edge .

On the other hand, if the number of conflicting coordinates does not vary a lot among these sets, then we analyze the test by selecting (at random) a single conflicting coordinate in and focusing on the rejection probability based only on the value of the selected coordinate.

### 1.3 Related Work

The question of testing the direct product was studied extensively when the underlying domain  [GS00, DR06, DG08, DS14, IKW12]. In this setting, Goldreich and Safra [GS00] proposed a constant query test. Dinur and Reingold [DR06] suggested the two-query test mentioned above and analyzed it in the high acceptance regime but with a relaxed distance measure.

The state of the art in this context is the result of Dinur and Steurer777The result in [DS14] is stated in the language of tuples, i.e., the domain is a subset of , but their result also holds when the domain is a collection of -sized subsets of . See [DDG17] for more details. [DS14] dealing with the domain where varies between and . They analyze the aforementioned two-query test with -intersection size. They analyze it in the high acceptance regime and show that indeed admits a direct product testing theorem. The proof is quite involved and in particular analyzes first the low acceptance regime. Recently, in a breakthrough paper, Dinur and Kaufman [DK17] analyzed the two-query test when the underlying domain is obtained from the set of faces of a Ramanujan complex. Their approach crucially relies on the result of [DS14]. More recently Dinur et al. [DHK18] introduced the notion of double samplers and remarked that it might admit a direct product theorem.

We remark that the direct product testability question was further analyzed in the low acceptance regime under the domain , see [DG08, IKW12, DN17] and also under the domain where the universe is , and the domain is the set of all subspaces of [IKW12].

### 1.4 Organization of the Paper

Section 2 lists the notations and technical tools that we use in the paper. In Section 3 we formalize the notion of direct products and their testing. In Section 4 we prove our main technical result, namely, that whenever the underlying test graph is a -coordinate expander it admits a direct product testing theorem. Finally, in Section 5 we introduce the sliding window domain for which we show a direct product theorem.

## 2 Preliminaries

In this section, we list the notations and technical tools used in this paper.

Notations. We use the following notations throughout the paper. We denote the set by . For any , with , we denote by , all subsets of of cardinality . For any set , we denote by the power set of , i.e., the set of all subsets of . For any graph and any two subsets , we denote by the set of all edges between and . For any , we denote by the relative Hamming distance between and given by the fraction of coordinates in which and differ.

Johnson Graph Family. For every such that , is a graph which is a member of the Johnson graph family, whose vertex set is , and whose edge set is .

Expander Mixing Lemma. The following is a standard claim concerning the expansion of two sets in expander graphs. For completeness we include a proof in Section A:

###### Claim 2.1.

Let be a -regular graph and be its adjacency matrix. Let

be its second largest eigenvalue in absolute value. Let

satisfying: then:

where the probability is given by first picking uniformly at random from , and then picking according to . Furthermore, let be a distribution on satisfying that for every two elements : , then:

 Pr(u,v)[v∈T∣u∼μ]≤|T||V|+λd⋅√c|T||S|

## 3 Direct Product Testing: The Setting

In this section, we formalize the notion of direct products and their testing. Specifically, we formalize the notion of direct product testing through test graphs, which is slightly non-standard but it helps in introducing the notion of coordinate expansion in a later section succinctly.

For every subset of , let be the class of all functions whose domain is and range is . Let be the domain of the direct product. Let be the class of all functions whose domain is and maps every subset in to a function in . The direct product encoding is a function defined as follows: for every string , and every subset , let be defined as the projection function which maps to , the string restricted to only the coordinates in .

###### Definition 3.1.

For two functions we define their relative distance as:

 Δ(F,G)=|{S∈V|F(S)≠G(S)}||V|.

For a function and a set of functions we define the distance between and as the minimal distance between and some function . If , we say that is -close to , otherwise, it is -far from .

For every function , we define as follows: Given construct in the following way,

 aFi=majS∈VS∋i(F(S)i).

Then, we define .

Let be a graph whose vertex set is . Then we interpret as a test graph on functions defined on in the following sense:

Test : Input: A function . Procedure: Pick an edge in uniformly at random. Output: Accept if and only if .

It is important to note that we allow self loops and multiple edges between a pair of vertices. Also, we can generalize the above direct product testing setting to the case when is a multiset of , and the results in this paper still hold. However, we choose not to handle this more general setting for the sake of clarity of presentation. The above remark also applies to the case of studying test graphs which are not regular in degree, that are not considered in this paper. Finally, throughout the paper, we drop the subscript in , if is clear from the context.

## 4 Direct Product Testing: Coordinate Expansion

In this section we prove our main technical result, namely, that whenever the underlying test graph is a -Coordinate Expander (defined next) it admits a direct product testing theorem.

###### Definition 4.1 ((λ,ρ)-Coordinate Expander).

Let and let be a test graph, where . For let and be the subgraph of induced by the vertices in . Let , where is the normalized adjacency matrix of . The graph is called -coordinate expander if:

1. and for every we have .

2. For every and for each we have .

3. For every subset and , satisfying , we have .

Informally, a domain is a coordinate expander if the test graph is an expander and every induced subgraph of the test graph containing a fixed coordinate is also an expander888Actually, the property of an expander that we need is that for any two sets of vertices in the graph, the number of edges between and is roughly equal to , where is the density of the edge set of the graph., and it has good correlation/intersection properties – i.e., for any subset and coordinate , an uniformly random neighbor of contains with constant probability (say ), and for every in the domain, and any subset of , the number of elements of that we see in a random neighbor of is close to the expected number, which is . Below, we see that coordinate expansion of the test graph implies a direct product theorem for the underlying domain.

###### Theorem 4.2.

Let , and let and be some constants. Let be a graph, , let , and . Let be a -coordinate expander. If passes with probability then is -close to .

###### Proof.

Let . We define as follows:

 B={S∣F(S)≠F∗(S)} and C=V∖B.

Let . Given a subset we say that a coordinate is conflicting if the value of at does not equal . For a set denote by the set of conflicting coordinates in . We show that rejects with probability at least .

Let us sort in ascending order the elements of based on the number of coordinates on which they disagree with . For a parameter we define the set as the set of last elements of (and similarly the set is the set of the first elements of ). We denote by the number of conflicting coordinates of the -th element of .

Let . We consider two cases based on and .

##### Case 1: m1−c>2ρm1/2 or m1/2>2ρmc:

For both the possibilities we have similar arguments, which is why they are clubbed under one case, but will be handled separately for ease of presentation.

##### Case 1A: m1−c>2ρm1/2:

The probability that an uniformly random is in is . Now by Claim 2.1, we get that

 Pr[S′∈B>1/2|S∈B≥1−c]<β/2+λ√12c,

so with probability at least if then .

Now, by the third property of -coordinate expander, the probability that is at most . Notice that the probability that is at least the probability that (because ). Hence we have that the probability that is at most .

Overall, using union bound, conditioned on , the probability that and is at least . But in such a case since we get , so there exists at least one coordinate on which but , so the test rejects. In total rejects with probability at least (where we used a trivial bound that ). Notice that holds for whenever .

##### Case 1B: m1/2≥2ρmc:

In this case we would like to mimic the proof strategy of the previous case. That is we would like to show that with non-zero constant probability a random neighbor in is in . By an application of Claim 2.1, we get:

 Pr[S′∈B>c|S∈B≥1/2]<(1−c)β+λ√2−2c,

so with probability at least if then .

Now, by the third property of -coordinate expander, the is at most . Notice that and thus . Therefore we have .

Overall, using union bound, conditioned on , the probability that and is at least . But in such a case since we get , so there exists at least one coordinate on which but , so the test rejects. In total rejects with probability at least (where we used a trivial bound that ). Notice that holds for whenever .

##### Case 2: m1−c≤4ρ2mc:

Define as the set . Observe that in the number of conflicting coordinates is between and . Now we would like to consider a different test that selects according to . If then accepts. Otherwise, it picks uniformly at random and checks for consistency only on , namely: It rejects iff and . Clearly the rejection probability of is at most the rejection probability of . We conclude the proof by showing that rejects with probability .

With probability the test selects and we would like to analyze the rejection probability conditioned on that. For this sake we bound the probability of the following events:

• is the event where .

• is the event where and where .

If the event occurs but does not, then it must be the case that . Hence rejects. As a consequence . Thus it suffices to show that is a positive constant bounded away from 0.

To bound the probability for the event we use Claim 2.1: The probability of conditioned on is at most .

Since the graph is a -coordinate expander then for each , we have that , in particular this is true for , hence: .

Now we divide the event into disjoint events depending on the value of and bound the rejection probability of conditioned on specific value of . Fix and assume that selects and sets (so ). We denote by the fraction . Observe that , since otherwise the majority value would become the value of , but we have .

Note, that under the assumption that selects and , sets with few conflicting coordinates are more likely to be chosen than those who have many of them. However, since by our assumption the number of conflicting coordinates is between and , then sets with conflicting coordinates are only -times more probable than those having -conflicting coordinates. Denote by the distribution of picking assuming that selects . By an application of Claim 2.1 we get:

 PrS∼μ,S′[S′∈~Bi]≤∣∣~Bi∣∣|Vi|+λ√4ρ2≤12+2λ/ρ

So we get that,

 Pr[E2|S∈B(c,1−c)]=(1−PrS∼μ,S′[S′∈~Bi∣i0∈S′])⋅Pr[i0∈S′]≥ρ2−2λ.

Summing up, we get:

 Pr[T′ rejects|S∈B(c,1−c)] ≥ Pr[E2|S∈B(c,1−c)]−Pr[E1|S∈B(c,1−c)] ≥ ρ2−2λ−(2c+λ√2c1−2c) ≥ 14−2c−λ(2+√2c1−2c),

a constant bounded away from 0 for whenever . ∎

###### Remark 4.3.

Notice that the above theorem holds for any satisfying the condition that the below three expressions are a constant bounded away from 0:

 1/2−λ(√12c+1), c−λ(√2−2c+1), ρ2−2λ−(2c+λ√2c1−2c).

In Appendix B, we consider the test graph and using the above remark show a direct product theorem when is close to and is close to .

## 5 Sliding Window Domain

In this section, we introduce the sliding window domain for which we show a direct product theorem.

Construction. Let such that . Let be a collection of subsets of of Hamming weight .

 A={{i,…,i+k−1}∣i∈[n]},

where the addition is done999Strictly speaking, the addition is done modulo and then the resulting number is incremented by one. modulo .

Testability. The domain of our direct product test is . The corresponding test is as follows:

Test : Input: A function . Procedure: Pick uniformly at random . Then pick uniformly at random such that . Output: Accept if and only if .

The test graph of the above is given by the vertex set and the edge set . The correctness of the above test is shown below. We would like to emphasize that and yet admits a direct product theorem.

###### Theorem 5.1.

Let and . If passes with probability then is -close to .

###### Proof.

We will in fact prove a more general direct product testing result.

###### Lemma 5.2.

Let and be a -regular graph where , let , and . For every , let the induced subgraph of in be a clique (with self loops). Additionally, let be a constant such that for every and every , the probability that a uniformly random neighbor of in contains is at least . If passes with probability then is -close to .

Now we show that the above lemma gives the proof of the theorem. Let . Note that for every , the induced subgraph of in is a clique (with self loops) because any two subsets in have in their intersection and thus have non-empty intersection. Also for every and every , the probability that a uniformly random neighbor of in contains is at least . Thus, from Lemma 5.2 the theorem follows. ∎

We complete the proof of the above theorem by showing Lemma 5.2 below.

###### Proof of Lemma 5.2.

Let . Let be defined as follows:

 B={S∣F(S)≠F∗(S)}.

Let be defined as follows:

 Ci={S∈Vi∣F(S)i=aFi}.

By definition of , it is clear that .

Since passes with probability this implies that the number of edges that fail is at most .

Fix . Fix (arbitrarily) such that . Now observe that whenever , the test rejects the edge in because . This implies that there are at least many edges incident on that fail the test . Therefore, there are in total at least edges that fail the test. Recall that the total number of rejected edges is at most . Thus we have that . The proof is concluded by noting that the distance between and is exactly . ∎

Note that Lemma 5.2 holds even when the induced subgraph of in is a clique without self loops. In Appendix C, we provide a couple of direct product theorems on domains that are known in literature as an immediate consequence of this lemma.

Lack of Global Expansion. Notice that has very bad edge-expansion/vertex-expansion but is a very good local expander, i.e., the induced subgraph containing any particular coordinate has good expansion (in fact is a clique). Lemma 5.2 guarantees that and thus admits a direct product theorem. Therefore, demonstrates that direct product testing does not require the test graph to be an expander (like the Johnson/Ramanujan graph) but only to have certain local expansion properties.

Sub-linear Size Domains. We remark here that we could consider subsets of of size smaller than which still admit a direct product theorem. For example consider as follows:

 ~A={{\nicefracik2,…,\nicefracik2+k−1}∣i∈[2n/k]},

and the test graph is given by the vertex set and the edge set . It is easy to see that admits a direct product theorem by applying Lemma 5.2. Again, we emphasize that and yet admits a direct product theorem.

Comparison with Dinur and Kaufman. One might wonder that if direct product testing results can be established on linear sized direct product domains using simple constructions such as the sliding window domain then, why did [DK17] work so hard and use extremely heavy objects such as high dimensional expanders to obtain linear sized direct product domains. This is because for applications to gap and hardness amplification, it is desirable that a direct product domain also has distance amplification (defined below) and high dimensional expanders have distance amplification whereas the sliding window domain does not.

###### Definition 5.3 (Distance Amplification, [Dk17]).

A direct product domain is said to have distance amplification if for every such that , we have that .

Thus, the construction of the sliding window domain provides a conceptual clarification as to why we need high dimensional expanders: we can obtain direct product testing from simple constructions like the sliding window domain and we can obtain distance amplification from known constructions of vertex expanders (see Appendix D); but to obtain both simultaneously, [DK17] needed high dimensional expanders.

### Acknowledgments

We are truly grateful to Irit Dinur for her constant support throughout this project and for her many illuminating and helpful discussions and comments. We also thank the anonymous reviewers for their detailed and useful feedback.

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## Appendix A Missing Proofs

###### Proof of Claim 2.1.

We prove only the furthermore part. The first part follows by plugging . For a set we denote by

the characteristic vector of

. Let be the distribution vector that describes . First observe that:

 Pr(u,v)[v∈T∣u∼μ]=1d⋅(pμ)t⋅A⋅1T,

where the probability is taken over that is drawn according to and is a uniformly random neighbor of . Note that

is an eigenvector of

corresponding to the largest eigenvalue (in absolute value) of . We decompose the vectors: as follows:

 pμ=1|V|1V+→p and 1T