Disjointness through the Lens of Vapnik-Chervonenkis Dimension: Sparsity and Beyond
Anup Bhattacharya, Sourav Chakraborty, Arijit Ghosh, Gopinath Mishra, Manaswi Paraashar
aa r X i v : . [ c s . CC ] J un Disjointness through the Lens of VapnikChervonenkisDimension: Sparsity and Beyond
Anup Bhattacharya ∗ Sourav Chakraborty ∗ Arijit Ghosh ∗ Gopinath Mishra ∗ Manaswi Paraashar ∗ Abstract
The disjointness problem - where Alice and Bob are given two subsets of { , . . . , n } andthey have to check if their sets intersect - is a central problem in the world of communicationcomplexity. While both deterministic and randomized communication complexities for thisproblem are known to be Θ( n ), it is also known that if the sets are assumed to be drawn fromsome restricted set systems then the communication complexity can be much lower. In this workwe explore how communication complexity measures change with respect to the complexity of theunderlying set system. The complexity measure for the set system that we use in this work is theVapnik-Chervonenkis (VC) dimension. More precisely, on any set system with VC dimensionbounded by d , we analyze how large can the deterministic and randomized communicationcomplexities be, as a function of d and n . The d -sparse set disjointness problem, where thesets have size at most d , is one such set system with VC dimension d . The deterministicand the randomized communication complexities of the d -sparse set disjointness problem havebeen well studied and is known to be Θ ( d log ( n/d )) and Θ( d ), respectively, in the multi-roundcommunication setting. In this paper, we address the question of whether the randomizedcommunication complexity is always upper bounded by a function of the VC dimension ofthe set system, and does there always exist a gap between the deterministic and randomizedcommunication complexity for set systems with small VC dimension.In this paper, we construct two natural set systems of VC dimension d , motivated from ge-ometry. Using these set systems we show that the deterministic and randomized communicationcomplexity can be e Θ ( d log ( n/d )) for set systems of VC dimension d and this matches the deter-ministic upper bound for all set systems of VC dimension d . We also study the deterministic andrandomized communication complexities of the set intersection problem when sets belong to aset system of bounded VC dimension. We show that there exists set systems of VC dimension d such that both deterministic and randomized (one-way and multi-round) complexity for the setintersection problem can be as high as Θ ( d log ( n/d )), and this is tight among all set systemsof VC dimension d . Keywords.
Communication complexity, VC dimension, Sparsity, and Geometric Set System ∗ Indian Statistical Institute, Kolkata, India
Introduction
Since its introduction by Yao [Yao79], communication complexity occupies a central positionin theoretical computer science. A striking feature of communication complexity is its interplaywith other diverse areas like analysis, combinatorics, and geometry [KN96, Rou16]. Vapnik andChervonenkis [VC71] introduced the measure
Vapnik-Chervonenkis dimension or VC dimension for set systems in the context of statistical learning theory. As was the case with communicationcomplexity, VC dimension has found numerous connections and applications in many differentareas like approximation algorithms, discrete and combinatorial geometry, computational geometry,discrepancy theory and many other areas [Mat09, Cha01, PA11, Mat13]. In this work we studyboth of them under the same lens: of restricted systems and, for the first time, prove that geometricsimplicity does not necessarily imply efficient communication complexity.Lets start with recollecting some definitions from VapnikChervonenkis theory. Let S be acollection of subsets of a universe U . For a subset y of U , we define S| y := { y ∩ x : x ∈ S} . We say a subset y of U is shattered by S if S| y = 2 y , where 2 y denotes the power set of y . Vap-nikChervonenkis (VC) dimension of S , denoted as VC-dim( S ), is the size of the largest subset y of U shattered by S . VC dimension has been one of the fundamental measures for quantifyingcomplexity of a collection of subsets.Now let us revisit the world of communication complexity. Let f : Ω × Ω → Ω. In commu-nication complexity , two players Alice and Bob get as inputs x ∈ Ω and y ∈ Ω respectively, andthe goal for the players is to device a protocol to compute f ( x, y ) by exchanging as few bits ofinformation between themselves as possible.The deterministic communication complexity D ( f ) of a function f is the minimum number ofbits Alice and Bob will exchange in the worst case to deterministically compute the function f . Inthe randomized setting, both Alice and Bob share an infinite random source and the goal is togive the correct answer with probability at least 2 /
3. The randomized communication complexity R ( f ) of f denotes the minimum number of bits exchanged by the players in the worst case input bythe best randomized protocol computing f . In both deterministic and randomized settings, Aliceand Bob are allowed to make multiple rounds of interaction. Communication complexity when thenumber of rounds of interaction is bounded is also often studied. An important special case is whenonly one round of communication is allowed, that is, only Alice is allowed to send messages to Boband Bob computes the output. We will denote by D → ( f ) and R → ( f ) the one way deterministiccommunication complexity and one way randomized communication complexity respectively, of f .One of the most well studied functions in communication complexity is the disjointness function.Given a universe U known to both Alice and Bob, the disjointness function , Disj U : 2 U × U →{ , } , where 2 U denotes the power set of U , is defined as follows: Disj U ( x, y ) = ( , if x ∩ y = ∅ , o/w (1) This is the communication complexity setting with shared random coins. If no random string is shared, it iscalled the private random coins setting. By [New91] we know that the communication complexity in both the settingdiffers by at most a logarithmic additive factor.
1e also define the intersection function . Given a universe U known to both Alice and Bob, the intersection function , Int U : 2 U × U → U , where 2 U denotes the power set of U , is defined as Int U ( x, y ) = x ∩ y . It is easy to see that Int U is harder function to compute than Disj U . The Disj U function and its different variants, like Int U , have been one of the most important problems incommunication complexity and have found numerous applications in areas like streaming algorithmsfor proving lower bounds [Rou16, RY20]. By abuse of the notation, when U = [ n ], where [ n ] denotesthe set { , . . . , n } , we will denote the functions Disj [ n ] and Int [ n ] by Disj n and Int n respectively.Using the standard rank argument [KN96, RY20] one can show that D ( Disj n ) = Θ( n ). Ina breakthrough paper, Kalyanasundaram and Schnitger [KS92] proved that R ( Disj n ) = Ω( n ).Razborov [Raz92] and Bar-Yossef et al. [BJKS04] gave alternate proofs for the above result. Fromthe above cited results we can also see the D ( Int n ) = R ( Int n ) = Θ( n ).Naturally, one would also like to ask what happens to the deterministic and randomized com-munication complexity (one way or multiple round) of Disj n , when both Alice and Bob know thattheir inputs have more structure. In particular what can we say if the inputs are guaranteed tobe from a subset of S ⊆ U , where S is known to both players. Let Disj U functions restricted to S × S be denoted by
Disj U | S×S . This problem has also been studied extensively, mostly for certainspecial classes of subsets
S ⊆ U . For example, the sparse set disjointness function, where the set S contains all the subsets of U of size at most d , is an important special case of these works.We will denote by d -SparseDisj n and d -SparseInt n , the functions Disj n | S ×S and Int n | S × S respectively, where S is the collections of all subsets of [ n ] of size at most d . Using the rankargument [KN96, RY20], one can again show that, for all d ≤ n , the deterministic communica-tion complexity of d -SparseDisj n is Ω (cid:0) d log nd (cid:1) . H˚astad and Wigderson [HW07], and Dasgupta etal. [DKS12] showed that the randomized communication complexity and one round randomized com-munication complexity of d -SparseDisj n is R ( d -SparseDisj n ) = Θ( d ) and R → ( d -SparseDisj n ) =Θ( d log d ) respectively. In a follow up work, Saglam and Tardos [ST13] proved that with O (log ∗ d )rounds of communication and O ( d ) bits of communication it is possible to compute d -SparseDisj n .More recently, Brody et al. [BCK +
14] proved that R → ( d -SparseInt n ) = Θ ( d log d ) and R ( d -SparseInt n ) =Θ( d ). These results show that in the d -sparse setting, there is a separation between randomizedand deterministic communication complexity of Disj n and Int n functions.One would like to ask what happens to the communication complexity for other restrictionsto the disjointness (and intersection) problem. The following are two natural problems, with ageometric flavor, for which one would like to study the communication complexity. Problem 1 ( Discrete Line Disj ) . Let G ⊂ Z be a set of n points in Z and L be the setof all lines in R . Also, let L = L d denote the collection of all d -size subsets of L . The Dis-crete Line Disj on G and L is a function, Disj G | L×L : L × L → { , } defined as Disj G | L×L ( { ℓ , . . . , ℓ d } , { ℓ ′ , . . . , ℓ ′ d } ) is 1 if and only if there exists a line in Alice’s set that intersects someline in Bob’s set at some point in G . Formally, Disj G | L×L (cid:0) { ℓ , . . . , ℓ d } , { ℓ ′ , . . . , ℓ ′ d } (cid:1) = ( , if ∃ i, j ∈ [ d ] s.t. ℓ i ∩ ℓ ′ j ∩ G = ∅ , o/w (2) Problem 2 ( Discrete Interval Disj ) . Let X ⊂ Z be a set of n points in Z and Int be theset of all possible intervals. Also, let I = Int d denote the collection of all d -size subsets of Int .The
Discrete Interval Disj on X and I is a function, Disj X | I×I : I × I → { , } defined as We assume that Alice has the set { ℓ , . . . , ℓ d } and Bob has the set { ℓ ′ , . . . , ℓ ′ d } . isj X | I×I ( { I , . . . , I d } , { I ′ , . . . , I ′ d } ) is 1 if and only if there exists an interval in Alice’s set thatintersects some interval in Bob’s set at some point in X . Disj X | I×I (cid:0) { I , . . . , I d } , { I ′ , . . . , I ′ d } (cid:1) = ( , if ∃ i, j ∈ [ d ] s.t. I i ∩ I ′ j ∩ X = ∅ , o/w (3)Note that both the Discrete Line Disj and
Discrete Interval Disj functions are general-izations of sparse set disjointness function. Although it may not be obvious at first look, but boththe
Discrete Line Disj function and the
Discrete Interval Disj functions are disjointnessfunctions restricted to a suitable subset. In fact, the connection between the
Sparse set disjointness function ( d -SparseDisj n ), the Discrete Line Disj function and the
Discrete Interval Disj function run deep - all the three subsets of the domain which help to define the functions as restric-tion of the disjointness function have VC dimension Θ( d ), see Appendix A. Naturally one wouldlike to know, if the fact that the collection of subsets S has VC dimension d has any implicationon the communication complexity of Disj U | S×S . For example, is the randomized communicationcomplexity of
Discrete Line Disj function and the
Discrete Interval Disj function upperbounded by a function of d (independent of n )? And, do the Discrete Line Disj functionand the
Discrete Interval Disj function also have a separation between their randomized anddeterministic communication complexities similar to that of the
Sparse set disjointness function( d -SparseDisj n )? We show that these are not necessarily the cases. Theorem 3.
For
Discrete Interval Disj : there exists a X ⊂ Z with n points such that D ( Disj X | I×I ) = D → ( Disj X | I×I ) = R → ( Disj X | I×I ) = Θ (cid:16) d log nd (cid:17) . Theorem 4.
For
Discrete Line Disj : there exists a G ⊂ Z with n points such that D ( Disj G | L×L ) = D → ( Disj G | L×L ) = Θ (cid:0) d log nd (cid:1) and, for the randomized setting, R ( Disj G | L×L ) = Ω (cid:18) d log( n/d )log log( n/d ) (cid:19) Discrete Line Int , that is, the intersection finding version of
Discrete Line Disj is definedas follows : the objective is to compute a function
Int G | L×L : L × L → G that is defined as Int G | L×L ( { ℓ , . . . , ℓ d } , { ℓ ′ , . . . , ℓ ′ d } ) = [ i,j ∈ [ d ] (cid:0) ℓ i ∩ ℓ ′ j ∩ G (cid:1) . As we have already mentioned, Brody et al. [BCK +
14] proved that R ( d -SparseInt n ) = Θ( d ),whereas D ( d -SparseInt n ) = Θ (cid:0) d log nd (cid:1) . We show that Discrete Line Int does not demonstratesuch a separation between the deterministic and randomized communication complexity. We assume that Alice has the set { I , . . . , I d } and Bob has the set { I ′ , . . . , I ′ d } . Take n integer points on the x -axis. For Discrete Line Disj setting, restrict only to lines orthogonal to x -axis.For Discrete Interval Disj setting, take n integer points on Z and only restrict to intervals containing one integerpoint. Both of these restriction will give the disjointness problem in the d -sparse setting. Agian, we will assume that Alice has the set { ℓ , . . . , ℓ d } and Bob has the set { ℓ ′ , . . . , ℓ ′ d } . heorem 5. For
Discrete Line Int : there exists a G ⊂ Z with n points such that D ( Int G | L×L ) = D → ( Int G | L×L ) = R → ( Int G | L×L ) = R ( Int G | L×L ) = Θ (cid:16) d log nd (cid:17) . The upper bound for all the above three theorems can be obtained from the fact that thecorresponding sets have VC dimension Θ( d ), see Appendix A. Sauer-Shelah Lemma [Sau72, She72,VC71] states that if
S ⊆ [ n ] and VC-dim( S ) ≤ d then | S |≤ (cid:0) end (cid:1) d . Thus if VC-dim( S ) ≤ d , thenthe Sauer-Shelah Lemma implies that D → ( Int n | S×S ) = O (cid:0) d log nd (cid:1) . So, O (cid:0) d log nd (cid:1) is a upperbound to the above questions, both for randomized and deterministic and also for the one-waycommunication. But can the randomized communication complexity of Disj U | S×S and
Int U | S×S be even lower when S has VC dimension d ? The following result, which is a direct consequence ofTheorems 3, 4 and 5, enables us to we answer the question in the negative: Theorem 6.
Let ≤ d ≤ n .1. There exists S ⊆ [ n ] with VC-dim ( S ) ≤ d and R → ( Disj n | S×S ) = Ω (cid:0) d log nd (cid:1) .2. There exists S ⊆ [ n ] with VC-dim ( S ) ≤ d and R ( Disj n | S×S ) = Ω (cid:16) d log( n/d )log log( n/d ) (cid:17) .3. There exists S ⊆ [ n ] with VC-dim ( S ) ≤ d and R ( Int n | S×S ) = Ω (cid:0) d log nd (cid:1) . The following table compares our result with the previous best known lower bound for
Disj U | S×S and
Int U | S×S among all sets S ⊂ U of VC dimension d .Problems R ( Disj n | S×S ) R → ( Disj n | S×S ) R ( Int n | S×S ) R → ( Int n | S×S )Previously Known Ω( d ) Ω( d log d ) Ω( d ) Ω( d log d )[HW07] [DKS12] [BCK +
14] [BCK + (cid:16) d log( n/d )log log( n/d ) (cid:17) Ω (cid:0) d log nd (cid:1) Ω (cid:0) d log nd (cid:1) Ω (cid:0) d log nd (cid:1) Table 1: The largest communication complexity, for the functions
Disj n | S×S and
Int n | S×S ,among all S ⊆ [ n ] of VC dimension d , that was previously known and what we prove in this paper.Tight bounds of Ω (cid:0) d log nd (cid:1) for the largest D ( Disj n | S×S ), D → ( Disj n | S×S ), D ( Int n | S×S ) and D → ( Int n | S×S ), among all S ⊂ [ n ] of VC dimension d , follows directly from the fact that if S is acollection of all subsets of [ n ] of size at most d then D ( Disj n | S×S ) = D ( Int n | S×S ) = Ω (cid:0) d log (cid:0) nd (cid:1)(cid:1) ,see [KN96, RY20]. Notations
We denote the set { , . . . , n } by [ n ]. For a binary number x , decimal ( x ) denotes the decimalvalue of x . For two vectors x and y in { , } n , x ∩ y = { i ∈ [ n ] : x i = y i = 1 } , and x ⊆ y when x i ≤ y i for each i ∈ [ n ]. For a finite set X , 2 X denotes the power set of X . For x, y ∈ R with x < y ,[ x, y ] denotes the closed interval is the set of all real numbers that lies between x and y One way communication complexity (Theorems 3 and 6 (1))
In this section we will prove the following result.
Theorem 7.
For all n ≥ d , there exists X ⊂ Z with | X | = n and R ⊆ X with VC-dim ( R ) = 2 d ,such that R ⊆ X ∩ [ ≤ j ≤ d I j | { I , . . . , I d } ∈ I and R → ( Disj X | R×R ) = Ω (cid:16) d log nd (cid:17) . Note that the set I is defined in Problem 2. Remark 1.
The above result takes care of the proofs of Theorem 3 and Theorem 6 (1).The hard instance, for the proof of the above theorem, is inspired by the interval set systemsin combinatorial geometry and is constructed in Section 2.1. In Section 2.2, we proof Theorem 7by using a reduction from
Augmented Indexing , which we denote by
AugIndex ℓ . Formally theproblem AugIndex ℓ is defined as follows: Alice gets a string x ∈ { , } ℓ and Bob gets an index j ∈ [ ℓ ] and all x j ′ for j ′ < j . Bob reports x j as the output. Proposition 8. (Miltersen et al. [MNSW98]) R → ( AugIndex ℓ ) = Ω( ℓ ). We construct a set X ⊂ Z with | X | = n and R ⊆ X with VC-dim( R ) = 2 d . Informally, X is the union of the set of points present in the union of d pairwise disjoint intervals, in Z , eachcontaining nd points. Each set in R is the union of the set of points in the subintervals anchoredeither at the left or the right end point of each of the above d intervals. Formally, the descriptionof X and R are given below along with some of its properties that are desired to show Theorem 7. The ground set X : Let m = nd −
2. Without loss of generality we can assume that m = 2 k ,where k ∈ N . Let J = { , . . . , m + 1 } be the set of m + 2 consecutive integers that starts from theorigin and ends at m + 1. Similarly, let J p be the set of m + 1 consecutive integers that starts at p ∈ Z and ends at p + m + 1. Let p , . . . , p d be d points in Z such that the sets J p , . . . , J p d arepairwise disjoint. Let the ground set X be d S i =1 J p i . Note that X ⊂ Z and | X | = ( m + 2) d = n . The subsets of X in R : R ⊆ X contains two types of sets R and R m +1 , where • Take any d intervals R , . . . , R d of integer lengths such that, for all i ∈ [ d ], length of R i is atmost m + 1, R i ⊆ [ p i , p i + m + 1], and R i starts at p i . Note that R i does not intersect withany X \ J p i . The set A = d S i =1 ( R i ∩ X ) is an element in R . We say that A is generated by R , . . . , R d . • Take any d intervals R ′ , . . . , R ′ d of integer lengths such that, for all i ∈ [ d ], length of R ′ i isat most m + 1, R ′ i ⊆ [ p i , p i + m + 1] and R ′ i ends at p i + m + 1. Note that R ′ i does notintersect with any X \ J p i . The set B = d S i =1 ( R ′ i ∩ X ) is an element in R m +1 . We say that B is generated by R ′ , . . . , R ′ d . 5he following claim bounds the VC dimension of R , constructed as above. Claim 9.
For X ⊂ Z with | X | = n and R ⊂ X as described above, VC-dim ( R ) = 2 d ,Proof. The proof follows from the fact that any subset of of X containing 2 d + 1 points will containat least three points from some J p i , where i ∈ [ d ]. These points in J p i can not be shattered by thesets in R . Also, observe that there exists 2 d points, with two from each J p j , that can be shatteredby the sets in R .Now, we give a claim about X and R constructed above that will be required for our proof ofTheorem 7. Claim 10.
Let A ∈ R and B ∈ R m +1 be such that A is generated by R , . . . , R d and B is generatedby R ′ , . . . , R ′ d . Then A and B intersects if and only if there exists an i ∈ [ d ] such that R i intersects R ′ i at a point in J p i . The proof of Claim 10 follows directly from our construction of X ⊂ Z and R ⊆ X , as J p , . . . , J p d are pairwise disjoint. AugIndex d log m to Disj X | R×R
Before presenting the reduction we recall the definitions of
AugIndex d log m and Disj X | R×R .In
AugIndex d log m , Alice gets x ∈ { , } d log m and Bob gets an index j and x j ′ for each j ′ < j .The objective of Bob is to report x j as the output. In Disj X | R×R , Alice gets A ∈ R and Bobgets B ∈ R m +1 . The objective of Bob is to determine whether A ∩ B = ∅ . Note that X, R , R and R m +1 are as discussed in the Section 2.1.Let P be an one-way protocol that solves Disj X | R×R with o (cid:0) d log nd (cid:1) = o ( d log m ) bits ofcommunication. Now, we consider the following protocol P ′ for AugIndex d log m that has thesame one way communication cost as that of Disj X | R×R . Then we will be done with the proof ofTheorem 7.
Protocol P ′ for AugIndex d log m problemStep-1 Let x ∈ { , } d log m be the input of Alice. Bob gets an index j ∈ [ d log m ] and bits x j ′ foreach j ′ < j . Step-2
Alice will form d strings a , . . . , a d ∈ { , } log m by partitioning the string x into d partssuch that a i = x ( i −
1) log m +1 . . . x i log m , where i ∈ [ d ]. Bob first forms a string y ∈ { , } d log m ,where y j ′ = x j ′ for each j ′ < j , y j = 1, and y j ′ = 0 for each j ′ > j . Then Bob finds b , . . . , b d ∈ { , } log m by partitioning the string y in to d parts such that b i = y ( i −
1) log m +1 . . . y i log m ,where i ∈ [ d ]. Step-3
For each i ∈ [ d ], let R i and R ′ i be the intervals that starts at p i and ends at p i + m + 1,respectively, where R i = [ p i , m + p i − decimal ( a i )] and R ′ i = [ p i + m +1 − decimal ( b i ) , p i + m +1].Alice finds the set A ∈ R generated by R , . . . , R d and Bob finds the set B ∈ R m +1 generatedby R ′ , . . . , R ′ d , i.e., A = S i ∈ [ d ] ( R i ∩ X ) and B = S i ∈ [ d ] ( R ′ i ∩ X ). Step-4
Alice and Bob solves
Disj X | R×R on inputs A and B , and report x j = 0 if and only if Disj X | R×R ( A, B ) = 0. Note that x j is the output of AugIndex d log m problem.6he following observation follows from the description of the protocol P ′ and from the construc-tion of X ⊂ Z and R ⊆ X . Observation 11.
Let i ∗ ∈ [ d ] such that j ∈ { ( i ∗ −
1) log m + 1 , i ∗ log m } . Then(i) R i ∩ R ′ i = ∅ for all i = i ∗ .(ii) R i ∗ ∩ R ′ i ∗ = ∅ if and only if decimal ( b i ∗ ) ≤ decimal ( a i ∗ ).(iii) decimal ( b i ∗ ) ≤ decimal ( a i ∗ ) if and only if x j = 0.We will use the above observation to show the correctness of the protocol P ′ .First consider the case Disj X | R×R ( A, B ) = 0. Then, by Claim 10, there exists an i ∈ [ d ] suchthat R i and R ′ i intersects at a point in J p i . From Observation 11 (i), we can say R i ∗ ∩ R ′ i ∗ = ∅ .Combining R i ∗ ∩ R ′ i ∗ = ∅ with Observations 11 (ii) and (iii), we have x j = 0. Hence, Disj X | R×R ( A, B ) = 0 implies x j = 0. The converse part, i.e., x j = 0 implies Disj X | R×R ( A, B ) = 0, can beshown in the similar fashion.The one-way communication complexity of protocol P ′ for AugIndex d log m is the same as thatof P for Disj X | R×R , that is, o ( d log m ). However, this is impossible as the one-way communicationcomplexity of Augmented Indexing , over d log m bits, is Ω( d log m ) = Ω (cid:0) d log nd (cid:1) bits. Thiscompletes the proof of Theorem 7. In this section, we prove the following theorems.
Theorem 12.
For all n ≥ d , there exists a G ⊂ Z with | G | = n and T ⊆ G with VC-dim ( T ) = 2 d ,such that T ⊆ G ∩ [ ≤ j ≤ d ℓ j | { ℓ , . . . , ℓ d } ∈ L and R ( Disj G | T ×T ) = Ω (cid:18) d log( n/d )log log( n/d ) (cid:19) . The set L is as defined in Problem 1. Theorem 13.
For all n ≥ d , there exists a G ⊂ Z with | G | = n and T ⊆ G with VC-dim ( T ) = 2 d ,such that T ⊆ G ∩ [ ≤ j ≤ d ℓ j | { ℓ , . . . , ℓ d } ∈ L and R ( Int G | T ×T ) = Ω (cid:16) d log nd (cid:17) . The set L is as defined in problem 1. Remark 2.
Theorem 12 takes care of Theorem 4 and 6(2). Theorem 13 takes care of Theorem 5and 6(3).Note that the same set system will be used for the proofs of the above theorems. The hard instance, for the proof of the above theorems, is inspired by point line incidence set systems incomputational geometry and is constructed in Section 3.1. We prove Theorems 12 and 13 inSections 3.2 and 3.3, respectively, using reductions.7 .1 The hard instance for the proofs of Theorems 12 and 13
In this subsection, we give the description of G ⊂ Z with | G | = n and T ⊆ G , withVC-Dim( T ) = 2 d . The same G and T will be our hard instance for the proofs of Theorems 12and 13. In this subsection, without loss of generality, we can assume that d divides n and n/d is aperfect square.Informally, G is the set of points present in the union of d many pairwise disjoint square gridseach containing nd points and the grids are taken in such a way that any straight line of non-negativecan intersects with at most one grid. Also, each set in T is the union of the set of points presentin d many lines of non-negative slope such that one line intersects with exactly one grid. Moreover,all of the d lines have slopes either zero or positive. Formally, the description of G and T are givenbelow along with some of its properties that are desired to show Theorems 12 and 13. The ground set G : Let m = p nd , and G (0 , = (cid:8) ( x, y ) ∈ Z : 0 ≤ x, y ≤ m − (cid:9) be the gridof size m × m anchored at the origin (0 , p, q ∈ Z , the m × m grid anchored at( p, q ) will be denoted by G ( p,q ) , i.e., G ( p,q ) = (cid:8) ( i + p, j + q ) : ( i, j ) ∈ G (0 , (cid:9) . For d ∈ N , consider G ( p ,q ) , . . . , G ( p d ,q d ) satisfying the following property: Property
For any i, j ∈ [ d ] , with i = j , let L and L be lines of non-negative slopes that passthrough at least two points of G ( p i ,q i ) and G ( p j ,q j ) , respectively. Then L and L does not intersectat any point inside S dℓ =1 G ( p ℓ ,q ℓ ) . Observe that there exists G ( p ,q ) , . . . , G ( p d ,q d ) satisfying Property . We will take G = S dℓ =1 G ( p ℓ ,q ℓ ) as the ground set. Without loss of generality, we can assume that ( p , q ) = (0 , G ⊂ Z and | G | = dm = n . The subsets of G in T : T contains two types of subsets T and T of G , and they are generatedby the following ways: • Take any d lines L , . . . , L d of non negative slope such that, ∀ i ∈ [ d ], L i passes through( p i , q i ) ∈ G ( p i ,q i ) and (at least) another point in G ( p i ,q i ) . Note that L i does not contain anypoint from G \ G ( p i ,q i ) . The set A = S di =1 (cid:0) L i ∩ G ( p i ,q i ) (cid:1) is in T , and we say A is generated by the lines L , . . . , L d . • Take any d vertical lines L ′ , . . . , L ′ d such that, ∀ i ∈ [ d ], L ′ i contains at least one point from G ( p i ,q i ) . Note that L ′ i does not contain any point from G \ G ( p i ,q i ) . The set B = S di =1 ( L ′ i ∩ G ( p i ,q i ) ) is in T , and we say B is generated by the lines L ′ , . . . , L ′ d .The following claim bounds the VC dimension of T , which as described above. Claim 14.
For G ⊂ Z and T ⊆ G as described above, VC-dim ( T ) = 2 d .Proof. The proof follows from the fact that any subset of X containing 2 d + 1 points will containat least three points from some G ( p j ,q j ) , j ∈ [ d ]. These points in G ( p j ,q j ) can not be shattered bythe sets in T . Also, observe that there exists 2 d points two from each G ( p j ,q j ) that can be shatteredby the sets in T . 8ow, we give two claims about G and T , constructed above, that follow directly from ourconstruction of G ⊂ Z and T ⊆ G . Claim 15.
Let A ∈ T and B ∈ T such that A is generated by lines L , . . . , L d and A is generatedby lines L ′ , . . . , L ′ d . Then A and B intersect if and only if there exists i ∈ [ d ] such that L i and L ′ i intersect at a point in G ( p i ,q i ) . Claim 16.
Let A ∈ T and B ∈ T such that A is generated by lines L , . . . , L d and B is generatedby lines L ′ , . . . , L ′ d . Also let | A ∩ B | = d . Then for each i ∈ [ d ] , L i and L ′ i intersect at a point in G ( p i ,q i ) . Moreover, A ( B ) can be determined if we know B ( A ) and A ∩ B . The above claims will be used in the proofs of Theorems 12 and 13.
Let us consider a problem in communication complexity denoted by
Or-Disj t { , } ℓ that willbe used in our proof. In Or-Disj t { , } ℓ , Alice gets t strings x , . . . , x t ∈ { , } ℓ and Bob alsogets t strings y , . . . , y t ∈ { , } ℓ . The objective is to compute t W i =1 Disj { , } ℓ ( x i , y i ). Note that Disj { , } ℓ ( x i , y i ) is a binary variable that takes value 1 if and only if x i ∩ y i = ∅ . Proposition 17 (Jayram et al. [JKS03]) . R (cid:16) Or-Disj t { , } ℓ (cid:17) = Ω( ℓt ) . Note that Proposition 17 directly implies the following result.
Proposition 18. R (cid:16) Or-Disj t { , } ℓ | S ℓ × S ℓ (cid:17) = Ω( ℓt ) , where S ℓ = { , } ℓ \ { ℓ } . Let k ∈ N be the largest integer such that first k consecutive primes p , . . . , p k satisfy thefollowing inequalty: Π ki =1 p i ≤ r nd . (4)Using the fact that Π ki =1 p i = e (1+ o (1)) k log k , we get k = Θ (cid:16) log( n/d )log log( n/d ) (cid:17) .We prove the theorem by a reduction from Or-Disj d { , } k | S k × S k to Disj G | T ×T , where S k := { , } k \ { k } . Note that G ⊂ Z with | G | = n , and T ⊆ G , with VC-dim( T ) = 2 d , are the same as that weconstructed in Section 3.1. To reach a contradiction, assume that there exists a two way protocol P that solves Disj G | T ×T with communication cost of o (cid:16) d log m log log m (cid:17) = o (cid:16) d log( n/d )log log( n/d ) (cid:17) bits. Now,we give protocol P ′ that solves Or-Disj d { , } k | S k × S k , as described below. Protocol P ′ for Or-Disj d { , } k | S k × S k Step-1
Let A = ( x , . . . , x d ) ∈ [ S k ] d and B = ( y , . . . , y d ) ∈ [ S k ] d be the inputs of Alice andBob for Or-Disj d { , } k | S k × S k . Recall that S k = { , } k \ { k } . Bob finds ¯ B = (¯ y , . . . , ¯ y d ) ∈ (cid:2) { , } k (cid:3) d , where ¯ y i is obtained by complementing each bit of y i . For a set W , [ W ] d = W × . . . × W ( d times). tep-2 Both Alice and Bob privately determine first k prime numbers p , . . . , p k without anycommunication. Step-3
Let Φ : { , } k → { , }⌈ log ( √ nd ) ⌉ be the function such that φ ( x ) is the bit representationof the number Q ki =1 p x i i , where x = ( x , . . . , x k ) ∈ { , } k . Alice finds A ′ = ( a , . . . , a d ) ∈ h { , }⌈ log ( √ nd ) ⌉ i d and Bob finds B ′ = ( b , . . . , b ∈ h { , }⌈ log ( √ nd ) ⌉ i d privately withoutany communication, where a i = φ ( x i ) and b i = φ (¯ y i ) for each i ∈ [ d ]. Step-4
For each i ∈ [ d ], let L i and L ′ i be the lines having equation y − q i = decimal ( a i ) − decimal ( a i ) ( x − p i ) and x − p i = decimal ( b i ) respectively. Alice finds A ′′ ∈ T that is generated by the lines L , . . . , L d ,and Bob finds B ′′ ∈ T which is generated by the lines L ′ , . . . , L ′ d , i.e., A ′′ = S i ∈ [ d ] ( L i ∩ G ( p i ,q i ) )and B ′′ = S i ∈ [ d ] ( L ′ i ∩ G ( p i ,q i ) ). Step-5
Then Alice and Bob solve
Disj G | T ×T ( A ′′ , B ′′ ), and report d W i =1 Disj { , } k ( x i , y i ) = 1 if andonly if Disj G | T ×T ( A ′′ , B ′′ ) = 0.Now we argue for the correctness of the protocol P ′ . Let Disj G | T ×T ( A ′′ , B ′′ ) = 0, that is, A ′′ ∩ B ′′ = ∅ . By Claim 15 and from the description of P ′ , there exists i ∈ [ d ] such that the lines L i : y − q i = decimal ( a i ) − decimal ( a i ) ( x − p i ) and L ′ i : x − p i = decimal ( b i ) intersect at a point in G ( p i ,q i ) , thatis, the lines y = decimal ( a i ) − decimal ( a i ) x and x = decimal ( b i ) intersect at a point in G (0 , . Now, we can saythat, there exists i ∈ [ d ] such that decimal ( a i ) divides decimal ( b i ), equivalently, φ ( x i ) divides φ (¯ y i ).This implies x i is a subset of ¯ y i ( or x i ∩ y i = ∅ ) for some i ∈ [ d ]. Hence, d W i =1 Disj { , } k ( x i , y i ) = 1.The converse part, that is, d W i =1 Disj { , } k ( x i , y i ) = 1 implies Disj G | T ×T ( A ′′ , B ′′ ) = 0 can be shownin the similar fashion.Observe that the communication cost of protocol P ′ for Or-Disj d { , } k | S × S is same as that ofprotocol P for Disj G | T ×T , which is o (cid:16) d log m log log m (cid:17) = o (cid:16) d log( n/d )log log( n/d ) (cid:17) = o ( dk ) as m = p nd and k =Θ (cid:16) log( n/d )log log( n/d ) (cid:17) . This contradicts Proposition 18 which says that R (cid:16) Or-Disj d { , } k | S × S (cid:17) = Ω( dk ). With out loss of generality, we also assume that d divides n and, more over, n/d is a perfectsquare.First, consider the problem Learn G | T ×T , where the objective of Alice and Bob is to learn eachother’s set. Note that G ⊂ Z with | G | = n and T ⊆ G with VC-Dim( T ) = 2 d are same as thatconstructed in Section 3.1. In Learn G | T ×T , Alice and Bob get two sets A and B , respectively,from T with a promise | A ∩ B | = d . The objective of Alice (Bob) is to learn B ( A ). Observethat R ( Learn G | T ×T ) = Ω( d log n ) as there are Ω( m d ) = Ω (cid:16)(cid:0)p nd (cid:1) d (cid:17) many candidate sets for theinputs of Alice and Bob. We prove the theorem by a reduction from Learn G | T ×T to Int G | T ×T .Let by contradiction consider a protocol P that solves Int G | T ×T by using o ( d log n ) bits ofcommunication. To solve Learn G | T ×T , Alice and Bob first run a protocol P and finds A ∩ . Now by Claim 15, it is possible for Alice (Bob) to determine B ( A ) by combining A ( B )along with A ∩ B , with out any communication with Bob (Alice). Now, we have a protocol P ′ that solves Learn G | T ×T with o ( d log n ) bits of communication. However, this is impossible as R ( Learn G | T ×T ) = Ω( d log n ). Hence, we are done with the proof of Theorem 13. In this paper, we studied
Disj n | S×S and
Int n | S×S when S is a subset of 2 [ n ] and VC-dim( S ) ≤ d . One of the main contributions of our work is the result (Theorem 6) showing that unlike in thecase of d -SparseDisj n and d -SparseDisj n functions, there is no separation between random-ized and deterministic communication complexity of Disj n | S×S and
Int n | S×S functions whenVC-dim( S ) ≤ d . Note that we have settled both the one-way and two-way (randomized) commu-nication complexities of Int n | S×S when VC-dim( S ) ≤ d (Theorem 6 (1) and (3)). In the contextof Disj n | S×S , we have settled the one-way (randomized) communication complexity. The two-waycommunication complexity for
Disj n | S×S is tight up to factor log log nd (See Theorem 6 (2)). How-ever, we believe that the factor of log log nd should not be present in the statement of Theorem 6(2). Conjecture 1.
There exists
S ⊆ [ n ] with VC-dim ( S ) ≤ d and R ( Disj n | S×S ) = Ω (cid:0) d log nd (cid:1) . Recall G ⊂ Z with | G | = n and T ⊆ G with VC-Dim( T ) = 2 d construction from Section 3.1,that served as the hard instance for the proof of Theorem 12 and Theorem 13. The same G and T can not be the hard instance for the proof of Conjecture 1 because of the following result. Theorem 19.
Let us consider G ⊂ Z with | G | = n and T ⊆ G with VC-Dim ( T ) = 2 d as definedin Section 3.1. Also, recall the definition of T and T . There exists a randomized communicationprotocol that can, ∀ A ∈ T and ∀ B ∈ T , can compute Disj G | T ×T ( A, B ) , with probability at least / , and uses O (cid:16) d log d log nd log log nd · log log log nd (cid:17) bits of communication. We use the following observation to prove the above theorem.
Observation 20.
Let us consider the communication problem
Gcd k ( a, b ), where Alice and Bob get a and b respectively from { , . . . , k } , and the objective is for both the players to compute gcd( a, d ).Then there exists a randomized protocol, with success probability at least 1 − δ , for Gcd k thatuses O (cid:16) log k log log k · log log log k · log δ (cid:17) bits of communication. Proof.
We will give a protocol P for the case when δ = 1 / O (cid:16) log k log log k · log log log k (cid:17) bits of communication. By repeating O (cid:0) log δ (cid:1) times protocol P and reporting the majority ofthe outcomes as the output, we will get the correct answer with probability at least 1 − δ . BothAlice and Bob generate all the prime numbers p , . . . , p t between 1 and k . From the Prime NumberTheorem, we known that t = Θ (cid:16) k log k (cid:17) . Alice and Bob separately, construct the sets S a and S b thatcontain the prime numbers that divides a and b respectively. Note that | S a | and | S b | is bounded by O (cid:16) log k log log k (cid:17) . Alice and Bob compute S a ∩ S b by solving Sparse Set Intersection problem on input The product of first t prime numbers is e (1+ o (1)) t log t . a and S b using O (cid:16) log k log log k (cid:17) bits of communication [BCK + p ∈ S a ∩ S b , let α p,a and α p,b denote the exponent of p in a and b , respectively. Observe thatgcd( a, b ) = Y p ∈ S a ∩ S b p min { α p,a ,α p,b } . For each p ∈ S a , Alice sends α p,a to Bob. Number of bits of communication required to send theexponents of all the primes in S a ∩ S b , is | S a ∩ S b | + X p ∈ S a ∩ S b log( α p,a ) ≤ O (cid:18) log k log log k (cid:19) + | S a ∩ S b | log P p ∈ S a ∩ S b α p,a | S a ∩ S b | ≤ O (cid:18) log k log log k (cid:19) + | S a ∩ S b | log (cid:18) log k | S a ∩ S b | (cid:19) ≤ O (cid:18) log k log log k · log log log k (cid:19) In the above inequalities, we used the facts that | S a ∩ S b | = O (cid:16) log k log log k (cid:17) , P p ∈ S a ∩ S b α p,a ≤ log k andlog x is a concave function. After getting the exponents α p,a of the primes p ∈ S a ∩ S b from Alice,Bob also sends the exponents α p,b of the primes p ∈ S a ∩ S b to Alice using O (cid:16) log k log log k log log log k (cid:17) bits of communication to Alice. Since both Alice and Bob now know the set S a ∩ S b , and theexponents α p,a and α p,b for all p ∈ S a ∩ S b , both of them can compute gcd( a, b ). Total number ofbits communicated in this protocol is O (cid:16) log k log log k log log log k (cid:17) .We will now give the proof of Theorem 19. Proof of the Theorem 19.
Consider the case when d = 1. From the description of G and T inSection 3.1, we can say that G = G (0 , , where G (0 , = { ( x, y ) ∈ Z : 0 ≤ x, y ≤ √ n } . Moreover,each set in T is a set of points present in a straight line of non-negative slope that passes throughtwo points of G (0 , with one point being (0 ,
0) and each set in T is a set of points present in avertical straight line that passes through exactly √ n many grid points. Keeping Claims 15 and 16in mind, we will be done if we can show the existence of a randomized communication protocol forcomputing the function Disj G | T ×T , with probability of success at least 1 − δ and number of bitscommunicated by the protocol being bounded by O (cid:16) log n log log n · log log log n · log δ (cid:17) , for the specialcase when d = 1. This is because for general d , we will be solving d instances of the above problem,with the number of points in each grid being nd and setting δ = d for each of the d instances. Protocol for d = 1 . Alice and Bob get A and B from T and T , respectively. Let A is generatedby the straight line L A and B is generated by L B , where L A is a straight line with non-negativeslope and L B is a vertical line. If L A is a horizontal one : Alice just sends this information to Boband then both report that A ∩ B = ∅ . If L A is a vertical line : Alice sends this information to Boband he reports A ∩ B = ∅ if and only if L B passes through origin. Now assume that L A is neither a With out loss of generality assume that √ n is an integer Recall that we have assumed, without loss of generality, that d divides n . L A be y = pq x , where 1 ≤ p, q ≤ √ n , and p and q are relatively prime to each other. Also, let equation of Bob’s line L B be x = r , where 0 ≤ r ≤ √ n .Observe that A ∩ B = ∅ if and only if L A and L B intersects at a point of G (0 , . Moreover, L A and L B intersects at a grid point if and only if q divides r and 1 ≤ prq ≤ √ n . So, Alice and Bobrun the communication protocol for Gcd √ n ( q, r ) to decide whether q = gcd( q, r ). If q = gcd( q, r )and 1 ≤ prq ≤ √ n (again Alice and Bob can decide this using O (1) bits of communications) then A ∩ B = ∅ , otherwise A ∩ B = ∅ . Alice and Bob can decide if q = gcd( q, r ) and 1 ≤ prq ≤ √ n using O (1) bits of communication.The communication cost of our protocol is dominated by the communication complexity of Gcd √ n ( q, r ), which is equal to O (cid:16) log n log log n log log log n log δ (cid:17) by Observation 20. Acknowledgments
Anup Bhattacharya is supported by SERB-National Post Doctoral Fellowship, India. ArijitGhosh is supported by Ramanujan Fellowship (No. SB/S2/RJN-064/2015), India. The authorswould like to thank Sudeshna Kolay and Arijit Bishnu for the many discussions in the early stagesof this work.
A VC dimension, and Problems 1 and 2
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