Multiparty Karchmer-Wigderson Games and Threshold Circuits
aa r X i v : . [ c s . CC ] F e b Multiparty Karchmer – Wigderson Games andThreshold Circuits
Alexander Kozachinskiy ∗ and Vladimir Podolskii † University of Warwick, Coventry, UK Steklov Mathematical Institute, Russian Academy of Sciences, Moscow, Russia National Research University Higher School of Economics, Moscow, Russia
Abstract
We suggest a generalization of Karchmer – Wigderson communication gamesto the multiparty setting. Our generalization turns out to be tightly connectedto circuits consisting of threshold gates. This allows us to obtain new explicitconstructions of such circuits for several functions. In particular, we provide anexplicit (polynomial-time computable) log-depth monotone formula for Majorityfunction, consisting only of 3-bit majority gates and variables. This resolves aconjecture of Cohen et al. (CRYPTO 2013).
Contents Q k ( R k )-hypotheses games . . . . . . . . . . . . . . . . . 61.6 Organization of the paper . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Q k ( R k )-hypotheses games 114 Results for Majority 13 ∗ [email protected]. Supported by the EPSRC grant EP/P020992/1 (SolvingParity Games in Theory and Practice). † [email protected] Proof of the main theorem 17
Karchmer and Wigderson established tight connection between circuit depth and com-munication complexity [11] (see also [12, Chapter 9]). Namely, they showed that foreach Boolean function f one can define a communication game which communicationcomplexity exactly equals the depth of f in the standard De Morgan basis. This discov-ery turned out to be very influential in Complexity Theory. A lot of circuit depth lowerbounds as well as formula size lower bounds rely on this discovery [10, 13, 5, 7, 4]. Karch-mer – Wigderson games have been used also in adjacent areas like Proof Complexity (see,e.g., [14]).Karchmer – Wigderson games represent a deep connection of two-party communica-tion protocols with De Morgan circuits. Loosely speaking, in this connection one partyis responsible for ∧ gates and the other party is responsible for ∨ gates. In this paper weaddress the question of what would be a natural generalization of Karchmer – Wigdersongames to the multiparty setting. Is it possible to obtain in this way a connection withother types of circuits?We answer positively to this question: we suggest such a generalization and showits connection to circuits consisting of threshold gates . To motivate our results we firstpresent applications we get from this new connection. There are two classical constructions of O (log n )-depth monotone formulas for the Ma-jority function, MAJ n +1 . The one was given by Valiant [15]. Valiant used probabilisticmethod which does not give an explicit construction. The other construction is theAKS sorting network [1]. This construction actually gives polynomial-time computable O (log n )-depth O ( n log n )-size monotone circuit for MAJ n .Several authors (see, e.g., [6, 3]) noticed that the Valiant’s probabilistic argumentactually gives a O (log n )-depth formula for MAJ n , consisting only of MAJ gates andvariables. Is it possible to construct a O (log n )-depth circuit for MAJ n +1 , consistingonly of MAJ gates and variables, deterministically in polynomial time ? Note that AKS sorting network does not provide a solution because it consists of ∧ and ∨ gates. O (log n )-depth circuit, con-sisting only of MAJ gates and variables, which coincides with MAJ n for all inputs inwhich the fraction of ones is bounded away from 1 / − Θ( √ log n ) .We show that the conjecture of Cohen et al. is true (unconditionally). Theorem 1.
There exists polynomial-time computable O (log n ) -depth formula for MAJ n +1 , consisting only of MAJ gates and variables. In the proof we use the AKS sorting network. In fact, one can use any constructionof polynomial-time computable O (log n )-depth monotone circuit for MAJ n +1 . We alsoobtain the following general result: Theorem 2.
If there is a monotone formula (i.e., formula, consisting of ∧ , ∨ gatesand variables) for MAJ n +1 of size s , then there is a formula for MAJ n +1 of size O ( s · n log (3) ) = O ( s · n . ... ) , consisting only of MAJ gates and variables. Transformation from the last theorem, however, is not efficient. We can make thistransformation polynomial-time computable, provided log (3) is replaced by 1 / (1 − log (2)) ≈ .
71. In turn, we view Theorem 2 as a potential approach to obtain super-quadratic lower bounds on monotone formula size for MAJ n +1 . However, this approachrequires better than n (3) lower bound on formula size of MAJ n +1 in the { MAJ } basis. Arguably, this basis may be easier to analyze than the standard monotone basis.The best known size upper bounds in the {∧ , ∨} basis and the { MAJ } basis are, re-spectively, O ( n . ) and O ( n . ) [8]. Both bounds are due to Valiant’s method (see [8]also for the limitations of Valiant’s method).We also study a generalization of the conjecture of Cohen et al. to threshold functions.By THR ba we denote the following Boolean function:THR ba : { , } b → { , } , THR ba ( x ) = x contains at least a ones,0 otherwise.For some reasons (to be discussed below) a natural generalization would be a questionof whether THR kn +1 n +1 can by computed by a O (log n )-depth circuit, consisting only ofTHR k +12 gates and variables (initial conjecture can be obtained by setting k = 2). Thisquestion was also addressed by Cohen et al. in [3]. First, they observed that there isa construction of depth O ( n ) (and exponential size). Secondly, they gave an explicitconstruction of depth O (log n ), which coincides with THR kn +1 n +1 for all inputs in whichthe fraction of ones is bounded away from 1 /k by Θ(1 / √ log n ).However, no exact (even non-explicit) construction with sub-linear depth or sub-exponential size was known. In particular, Valiant’s probabilistic construction does notwork for k >
3. Nevertheless, in this paper we improve depth O ( n ) to O (log n ) and sizefrom exp { O ( n ) } to n O (1) for this problem: Theorem 3.
For any constant k > there exists polynomial-time computable O (log n ) -depth polynomial-size circuit for THR kn +1 n +1 , consisting only of THR k +12 gates and vari-ables. .2 Applications to Multiparty Secure Computations The conjecture stated in [3] was motivated by applications to Secure Multiparty Compu-tations. The paper [3] establishes an approach to construct efficient multiparty protocolsbased on protocols for small number of players. More specifically, in their framework onestarts with a protocol for small number of players and a formula F computing certainboolean function. Then one combines a protocol for a small number of players with itselfrecursively, where the recursion mimics the formula F .It is shown in [3] that from our result it follows that for any n there is an explicitpolynomial size protocol for n players secure against a passive adversary that controlsany t < n players. It is also implicit in [3] that from Theorem 3 for k = 3 it followsthat for any n there is a protocol of size 2 O (log n ) for n players secure against an activeadversary that controls any t < n players. An improvement of the depth of the formulain Theorem 3 to O (log n ) would result in a polynomial size protocol. We refer to [3] formore details on the secure multiparty computations. We now reveal a bigger picture to which the above results belong to. Namely, they canbe put into framework of multiparty Karchmer – Wigderson games.Before specifying how we define these games let us give an instructive example.Consider ordinary monotone Karchmer – Wigderson game for MAJ n +1 . In this gameAlice receives a string x ∈ MAJ − n +1 (0) and Bob receives a string y ∈ MAJ − n +1 (1). Inother words, the number of ones in x is at most n and the number of ones in y is atleast n + 1. The goal of Alice and Bob is to find some coordinate i such that x i = 0 and y i = 1. Next, imagine that Bob flips each of his input bits. After that parties have twovectors in both of which the number of ones is at most n . Now Alice and Bob have tofind any coordinate in which both vectors are 0.In this form this problem can be naturally generalized to the multiparty setting.Namely, assume that there are k parties, and each receives a Boolean vector of length kn + 1 with at most n ones. Let the task of parties be to find a coordinate in which all k input vectors are 0. How many bits of communication are needed for that?For k = 2 the answer is O (log n ), because there exists a O (log n )-depth monotonecircuit for MAJ n +1 and hence the monotone Karchmer – Wigderson game for MAJ n +1 can be solved in O (log n ) bits of communication. For k > O (log n )-bit solution based on the binary search.Now, let us look at the case k > kn +1 n +1 equals 0. The goal is to find a common zero. Note that we can consider a similar problemfor any function f satisfying so-called Q k -property : any k vectors from f − (0) have acommon zero. In the next definition we define Q k -property formally and also introducerelated R k -property. Definition 1.
Let Q k be the set of all Boolean functions f satisfying the following roperty: for all x , x , . . . , x k ∈ f − (0) there is a coordinate i such that x i = x i = . . . = x ki = 0 .Further, let R k be the set of all Boolean functions f satisfying the following property:for all x , x , . . . , x k ∈ f − (0) there is a coordinate i such that x i = x i = . . . = x ki . For f ∈ Q k let Q k -communication game for f be the following communication prob-lem. In this problem there are k parties. The j th party receives a Boolean vector x j ∈ f − (0). The goal of players is to find any coordinate i such that x i = x i = . . . = x ki = 0.Similarly we can define R k -communication games for functions from R k . In the R k -communication games the objective of parties is slightly different: their goal is to findany coordinate i and a bit b such that x i = x i = . . . = x ki = b .Self-dual functions belong to R and monotone self-dual functions belong to Q . Itis easy to see that R -communication games are equivalent to Karchmer – Wigdersongames for self-dual functions (one party should flip all the input bits). Moreover, Q -communication games are equivalent to monotone Karchmer – Widgerson games formonotone self-dual functions.In this paper we consider R k -communication games as a multiparty generalizationof Karchmer – Wigderson games. In turn, Q k -communication games are considered asa generalization of monotone Karchmer – Wigderson games. To justify this choice oneshould relate them to some type of circuit complexity.
Every function from Q k can be lower bounded by a circuit, consisting only of THR k +12 gates and variables. More precisely, let us write C f for a Boolean circuit C anda Boolean function f if for all x ∈ f − (0) we have C ( x ) = 0. Then the followingproposition holds: Proposition 4 ([3]) . The set Q k is equal to the set of all Boolean functions f for whichthere exists a circuit C f , consisting only of THR k +12 gates and variables. There is a similar characterization of the set R k . Proposition 5.
The set R k is equal to the set of all Boolean functions f for which thereexists a circuit C f , consisting only of THR k +12 gates and literals . The proof from [3] of Proposition 4 with obvious modifications also works for Propo-sition 5.Given f ∈ Q k , what is the minimal depth of a circuit C f , consisting only ofTHR k +12 gates and variables? We show that this quantity is equal (up to constantfactors) the communication complexity of Q k -communication game for f . Theorem 6.
Let k > be any constant. Then for any f ∈ Q k the following twoquantities are equal up to constant factors: We stress that negations can only be applied to variables but not to THR k +12 gates. the communication complexity of Q k -communication game for f ; • minimal d for which there exists a d -depth circuit C f , consisting only of THR k +12 gates and variables. Similar result can be obtained for R k -communication games. Theorem 7.
Let k > be any constant. Then for any f ∈ R k the following twoquantities are equal up to constant factors: • the communication complexity of R k -communication game for f ; • minimal d for which there exists a d -depth circuit C f , consisting only of THR k +12 gates and literals. Proofs of both theorems are divided into two parts:(a) transformation of a d -depth circuit C f , consisting only of THR k +12 gates andvariables (literals), into a O ( d )-bit protocol computing Q k ( R k )-communicationgame for f ;(b) transformation of a d -bit protocol computing Q k ( R k )-communication game for f into a d -depth circuit C f , consisting only of THR k +12 gates and variables (liter-als).The first part is simple and the main challenge is the second part. Later in thispaper (Section 6) we also formulate refined versions of Theorems 6 and 7. Namely, werefine these theorems in the following two directions. Firstly, we take into account circuitsize and for this we consider dag-like communication protocols. Secondly, we show thattransformations (a-b) can be done in polynomial time (under some mild assumptions).We derive our upper bounds on the depth of MAJ n +1 and THR kn +1 n +1 (Theorems 1and 3) from Theorem 6. We first solve the corresponding Q k -communication games withsmall number of bits of communication. Namely, for the case of MAJ n +1 we use AKSsorting network to solve the corresponding Q -communication game with O (log n ) bitsof communication. For the case of THR kn +1 n +1 with k > Q k -communication game by a simple binary search protocol with O (log n ) bits of com-munication. This is where we get depth O (log n ) for Theorem 1 and depth O (log n ) forTheorem 3. Again, some special measures should be taken to make the resulting circuitspolynomial-time computable and to control their size . Q k ( R k )-hypotheses games As we already mentioned, the hard part of our main result is to transform a protocolinto a circuit. We should only care about the size in case of Theorem 3, because depth O (log n ) immediately givespolynomial size. f in Q k ( R k ) we introduce the corresponding Q k ( R k )-hypotheses game for f . We show that strategies in these games exactly capture depth and size of circuits,consisting only of THR k +12 gates and variables (literals). It turns out that strategiesare more convenient than circuits to simulate protocols, since they operate in the sametop-bottom manner.Once we establish the equivalence of circuits and hypotheses games, it remains forus to transform a communication protocol into a strategy in a hypotheses games. Thisis an elaborate construction that is presented in Propositions 16 and 19. Below in thissection we introduce hypotheses games and as an illustration sketch the construction ofa strategy in a hypothesis game that is used in the proof of Theorem 1.Here is how we define these games. Fix f : { , } n → { , } . There are two players,Nature and Learner. Before the game starts, Nature privately chooses z ∈ f − (0),which is then can not be changed. The goal of Learner is to find some i ∈ [ n ] suchthat z i = 0. The game proceeds in rounds. At each round Learner specifies k + 1families H , H , . . . , H k ⊂ f − (0) to Nature. We understand this as if Learner makesthe following k + 1 hypotheses about z :“ z ∈ H ” , “ z ∈ H ” , ...“ z ∈ H k ” . Learner looses immediately if less than k hypotheses are true, i.e., if the number of j ∈ { , , . . . , k } satisfying z ∈ H j is less than k . Otherwise Nature points out tosome hypothesis which is true. In other words, Nature specifies to Learner some j ∈{ , , . . . , k } such that z ∈ H j . The game then proceeds in the same manner for somefinite number of rounds. At the end Learner outputs an integer i ∈ [ n ]. We say thatLearner wins if z i = 0.It is not hard to show that Learner has a winning strategy in Q k -hypotheses gamefor f if and only if f ∈ Q k . Since we will use similar arguments in the paper, let us gothrough the “if” part: if f ∈ Q k , then Learner has a winning strategy. Denote by Z bethe set of all z ’s which are compatible with Nature’s answers so far. At the beginning Z = f − (0). If |Z| > k + 1, Learner takes any distinct z , z , . . . , z k +1 ∈ Z and makesthe following hypotheses: “ z = z ” , “ z = z ” , ...“ z = z k +1 ” . At least k hypotheses are true, and the Nature’s response strictly reduces the size of Z .When the size of Z becomes k , Learner is ready to give an answer due to Q k -propertyof f . 7his strategy requires exponential in n number of rounds. This can be easily improvedto O ( n ) rounds. Indeed, instead of choosing k + 1 distinct elements of Z split Z into k + 1 disjoint almost equal parts. Then let the i th hypotheses be “ z is not in the i thpart”. Nature’s response to this reduces the size of Z by a constant factor, until the sizeof Z is k .For f ∈ Q k we can now ask what is the minimal number of rounds on in a Learner’swinning strategy. The following proposition gives an exact answer: Proposition 8.
For any f ∈ Q k the following holds. Learner has a d -round winningstrategy in Q k -hypotheses game for f if and only if there exists a d -depth circuit C f ,consisting only of THR k +12 gates and variables. Proposition 8 is the core result for our applications. For instance, we prove Theorem1 by giving an explicit O (log n )-round winning strategy of Learner in Q -hypothesesgame for MAJ n +1 . Let us now sketch our argument (the complete proof can be foundin Section 4).Assume that Nature’s input vector is z . We notice that in O (log n ) rounds one caneasily find two integers i, j ∈ [2 n + 1] such that either z i = 0 or z j = 0. However, weneed to know for sure. For that we take any polynomial time computable O (log n )-depthmonotone formula F for MAJ n +1 (for instance one that can be obtained from the AKSsorting network). We start to descend from the output gate of F to one of F ’s inputs.Throughout this descending we maintain the following invariant. If g is the current gate,then either g ( z ) = 0 ∧ z i = 0 or g ( ¬ z ) = 1 ∧ z j = 0 (here ¬ denotes bit-wise negation).It can be shown that in one round one can either exclude i or j (which will already giveus an answer) or replace g by some gate which is fed to g . If we reach an input to F , weoutput the index of the corresponding variable.Similarly one can define R k -hypotheses game for any f : { , } n → { , } . In R k -hypotheses game Nature and Learner play in the same way except that now Learner’sobjective is to find some pair ( i, b ) ∈ [ n ] × { , } such that z i = b . The following analogof Proposition 8 holds: Proposition 9.
For any f ∈ R k the following holds. Learner has a d -round winningstrategy in R k -hypotheses game for f if and only if there exists a d -depth circuit C f ,consisting only of THR k +12 gates and literals. In Section 2 we give Preliminaries. In Section 3 we define Q k ( R k )-hypotheses gamesformally and derive Proposition 8 and 9. In Section 4 we obtain our results for Majorityfunction (Theorems 1 and 2) using simpler arguments than in our general results. Thenin Section 5 we prove these general results (Theorems 6 and 7). In Section 6 we refineTheorems 6 and 7 in order to take into account the circuit size and computational aspects(Theorems 21 and 23 below). In Section 7 we derive Theorem 3 and provide anotherproof for Theorem 1. Finally, in Section 8 we formulate some open problems.8 Preliminaries
Let [ n ] denote the set { , , . . . , n } for n ∈ N . For a set W we denote the set of allsubsets of W by 2 W . For two sets A and B by A B we mean the set of all functions ofthe form f : B → A .We usually use subscripts to denote coordinates of vectors. In turn, we usually usesuperscripts to numerate vectors.We use standard terminology for Boolean formulas and circuits [9]. We denote the sizeof a circuit C by size( C ) and the depth by depth( C ). By De Morgan formulas/circuitswe mean formulas/circuits consisting of ∧ , ∨ gates of fan-in 2 and literals (i.e., we assumethat negations are applied only to variables). By monotone formulas/circuits we meanformulas/circuits consisting of ∧ , ∨ gates of fan-in 2 and variables. We also considerformulas/circuits consisting only of THR k +12 gates and variables (literals). We stressthat in such circuits we do not use constants. Allowing literals as inputs we allow toapply negations only to variables. We also assume that negations in literals do notcontribute to the depth of a circuit.We use the notion of deterministic communication protocols in the multiparty number-in-hand model. However, to capture the circuit size in our results we considernot only standard tree-like protocols, but also dag-like protocols. This notion was con-sidered by Sokolov in [14]. We use slightly different variant of this notion, arguably moreintuitive one. In the next subsection we provide all necessary definitions. To obtain adefinition of a standard protocol one should replace dags by binary trees. We use the following terminology for directed acyclic graphs (dags). Firstly, we allowmore than one directed edge from one node to another. A terminal node of a dag G isa node with no out-going edges. Given a dag G , let • V ( G ) denote the set of nodes of G ; • T ( G ) denote the set of terminal nodes of G .For v ∈ V ( G ) let Out G ( v ) be the set of all edges of G that start at v . A dag G is called t -ary if every non-terminal node v of G we have | Out G ( v ) | = t . An ordered t -ary dag isa t -ary dag G equipped with a mapping from the set of edges of G to { , , . . . , t − } .This mapping restricted to Out G ( v ) should be injective for every v ∈ V ( G ) \ T ( G ). Thevalue of this mapping on an edge e will be called the label of e . In terms of labels werequire for ordered t -ary dags that any t edges, starting at the same node, have differentlabels.By a path in G we mean a sequence of edges h e , e , . . . , e m i such that for every j ∈ [ m −
1] edge e i ends in the same node in which e j +1 starts. Note that there maybe two distinct paths visiting same nodes (for instance, there may be two parallel edgesfrom one node to another). 9e say that a node w is a descendant of a node v if there is a path from v to w . Wecall w a successor of v if there is an edge from v to w . A node s is called starting node ifany other node is a descendant of s . Note that any dag has at most one starting node.If a dag G has the starting node s , then by depth of v ∈ V ( G ) we mean the maximallength of a path from s to v . The depth of G then is the maximal depth of its nodes.Assume that X , X , . . . , X k , Y are some finite sets. Definition 2. A k -party dag-like communication protocol π with inputs from X × X × . . . X k and with outputs from Y is a tuple h G, P , P , . . . , P k , φ , φ , . . . , φ k , l i , where • G is an ordered -ary dag with the starting node s ; • P , P , . . . , P k is a partition of V ( G ) \ T ( G ) into k disjoint subsets; • φ i is a function from P i × X i to { , } ; • l is a function from T ( G ) to Y . The depth of π (denoted by depth( π )) is the depth of G . The size of π (denoted bysize( π )) is | V ( G ) | .The underlying mechanics of the protocol is as follows. Parties descend from s toone of the terminals of G . If the current node v is not a terminal and v ∈ P i , thenat v the i th party communicates a bit to all the other parties. Namely, the i th partycommunicates the bit b = φ i ( v, x ), where x ∈ X i is the input of the i th party. Amongthe two edges, starting at v , parties choose one labeled by b and descend to one of thesuccessors of v along this edge. Finally, when parties reach a terminal t , they output l ( t ).We say that x ∈ X i is i -compatible with an edge e from v to w if one of the followingtwo condition holds: • v / ∈ P i ; • v ∈ P i and e is labeled by φ i ( v, x ).We say that x ∈ X i is i -compatible with a path p = ( e , e , . . . , e m ) of G if for every j ∈ [ m ] it holds that x is i -compatible with e j . Finally, we say that x ∈ X i is i -compatiblewith a node v ∈ V ( G ) if there is a path p from s to v such that x is i -compatible with v .We say that an input ( x , x , . . . , x k ) ∈ X × X × . . . X k visits a node v ∈ V ( G ) ifthere is a path p from s to v such that for every i ∈ [ k ] it holds that x i is i -compatiblewith p . Note that there is unique t ∈ T ( G ) such that ( x , x , . . . , x k ) visits t .To formulate an effective version of Theorems 6 and Theorem 7 we need the followingdefinition. Definition 3.
The light form of a k -party dag-like communication protocol π = h G, P , P , . . . , P k , φ , φ , . . . , φ k , l i is a tuple h G, P , P , . . . , P k , l i . π we just forget about φ , φ , . . . , φ k . In other words,the light form only contains the underlying graph of π , the partition of non-terminalnodes between parties and the labels of terminals. On the other hand, in the light formthere is no information at all how parties communicate at the non-terminal nodes.Protocol π computes a relation S ⊂ X × X × . . . × X k × Y if the following holds.For every ( x , x , . . . , x k ) ∈ X × X × . . . × X k there exist y ∈ Y and t ∈ T ( G ) such that( x , . . . , x k ) visits t , l ( t ) = y and ( x , x , . . . , x k , y ) ∈ S .Using language of relations, we can formally define Q k - and R k -communication games.Namely, given f : { , } n → { , } , f ∈ Q k , we define Q k -communication game for f asthe following relation: S ⊂ f − (0) × . . . × f − (0) | {z } k × [ n ] ,S = n ( x , . . . , x k , j ) | x j = . . . = x kj = 0 o . Similarly, given f : { , } n → { , } , f ∈ R k , we define R k -communication game for f asthe following relation: S ⊂ f − (0) × . . . × f − (0) | {z } k × ([ n ] × { , } ) ,S = n ( x , . . . , x k , ( j, b )) | x j = . . . = x kj = b o . It is easy to see that a dag-like protocol for S can be transformed into a tree-likeprotocol of the same depth, but this transformation can drastically increase the size. Q k ( R k )-hypotheses games Fix f ∈ Q k , f : { , } n → { , } . Here we define Learner’s strategies in Q k -hypothesesgame for f formally. We consider not only tree-like strategies but also dag-like. Tospecify a Learner’s strategy S in Q k -hypotheses game we have to specify: • An ordered ( k + 1)-ary dag G with the starting node s ; • a subset H j ( p ) for every j ∈ { , , . . . , k } and for every path p in G from s to somenode in V ( G ) \ T ( G ); • a number i t ∈ [ n ] for every terminal t .The underlying mechanics of the game is as follows. Let Nature’s vector be z ∈ f − (0).Learner and Nature descend from s to one of the terminals of G . More precisely, aposition in the game is determined by a path p , starting at s . If the endpoint of p is nota terminal, then Learner specifies some sets H ( p ) , H ( p ) , . . . , H k ( p ) as his hypotheses.If less than k of these sets contain z , then Nature wins. Otherwise Nature specifies some j ∈ { , , . . . , k } such that z ∈ H j ( p ). Among k + 1 edges that start at the endpoint of11 players choose one which is labeled by j . After that they extend p by this edge. Atsome point parties reach some terminal t (i.e., the endpoint of p becomes equal t ). Thenthe game ends and Learner output i t .We stress that Learner’s output depends only on t but not on a path to t (unlikeLearner’s hypotheses). This property will be crucial in establishing connection of Q k -hypotheses games to circuits.We now proceed to a formal definition of what does it mean that S is winning forLearner.We say that z ∈ f − (0) is compatible with a path p = h e , . . . , e m i , starting in s ,if the following holds. If p is of length 0, then every z ∈ f − (0) is compatible with p .Otherwise for every i ∈ { , . . . , e m } it should hold that z ∈ H j ( h e , . . . , e i − i ), where j is the label of edge e i . Informally this means that Nature, having z on input, can reacha position in the game which corresponds to a path p .We say that strategy S is winning for Learner in Q k -hypotheses game for f if forevery path p , starting at s , and for every z ∈ f − (0), compatible with p , the followingholds: • if the endpoint of p is not a terminal, then the number of j ∈ { , , . . . , k } suchthat z ∈ H j ( p ) is at least k ; • if the endpoint of p is t ∈ T ( G ), then z i t = 0.We will formulate a stronger version of Proposition 8. For that we need the notion ofthe light form of the strategy S . Namely, the light form of S is its underlying dag G equipped with a mapping which to every t ∈ T ( G ) assigns i t . In other words, the lightform contains a “skeleton” of S and Learner’s outputs in terminals (and no informationabout Learner’s hypotheses).We can identify the light form of any strategy S with a circuit, consisting only ofTHR k +12 gates and variables. Namely, place THR k +12 gate in every v ∈ V ( G ) \ T ( G ) andfor every t ∈ T ( G ) place a variable x i t in t . Set s to be the output gate. Proposition 10.
For all f ∈ Q k , f : { , } n → { , } the following holds: (a) if S if a Learner’s winning strategy in Q k -hypotheses game for f , then its lightform, considered as a circuit C consisting only of THR k +12 gates and variables,satisfies C f . (b) Assume that C f is a circuit, consisting only of THR k +12 gates and variables.Then there exists a Learner’s winning strategy S in Q k -hypotheses game for f suchthat the light form of S coincides with C . We omit the proof of (b) as in the paper we only use (a) . Proof of (a) of Proposition 10.
For a node v ∈ V ( G ) let f v : { , } n → { , } be thefunction, computed by the circuit C at the gate, corresponding to v .We shall prove the following statement. For any path p , starting in s , and for any z which is compatible with p it holds that f v ( z ) = 0, where v is the endpoint of p . To12ee why this implies C f take any z ∈ f − (0) and note that z is compatible with thepath of length 0. The endpoint of such path is s and hence 0 = f s ( z ) = C ( z ).We will prove the above statement by the backward induction on the length of p .The longest path p ends in some t ∈ T ( G ). By definition f t = x i t . On the other hand,since S is winning, z i t = 0 for any z compatible with p . In other words, f t ( z ) = 0 forany z compatible with p . The base is proved.Induction step is the same if p ends in some other terminal. Now assume that p endsin v ∈ V ( G ) \ T ( G ). Take any z ∈ f − (0) compatible with p . Let p j be the extension of p by the edge which starts at v and is labeled by j ∈ { , , . . . , k } . Next, let v j be theendpoint of p j (nodes v , v , . . . , v k are successors of v ). Since S is winning, the numberof j ∈ { , , . . . , k } such that z ∈ H j ( p ) is at least k . Hence by definition the number of j ∈ { , , . . . , k } such that z is compatible with p j is at least k . Finally, by the inductionhypothesis this means that the number of j ∈ { , , . . . , k } such that f v j ( z ) = 0 is atleast k . On the other hand: f v = THR k +12 ( f v , f v , . . . , f v k ) . Therefore f v ( z ) = 0, as required.One can formally define analogues notions for R k -hypotheses games. We skip this asmodifications are straightforwards and only formulate an analog of Proposition 10. Proposition 11.
For all f ∈ R k , f : { , } n → { , } the following holds: (a) if S if a Learner’s winning strategy in R k -hypotheses game for f , then its light form,considered as a circuit C consisting only of THR k +12 gates and literals, satisfies C f . (b) Assume that C f is a circuit, consisting only of THR k +12 gates and literals. Thenthere exists a Learner’s winning strategy S in R k -hypotheses game for f such thatthe light form of S coincides with C . Remark.
It might be unclear why we prefer to construct strategies instead of constructingcircuits directly, because beside the circuit itself we should also specify Learner’s hypothe-ses. The reason is that strategies can be seen as proofs that the circuit we construct iscorrect.
Proof of Theorem 1.
There exists an algorithm which in n O (1) -time produces a monotoneformula F of depth d = O (log n ) computing MAJ n +1 . Below we will define a strategy S F in the Q -hypotheses game for MAJ n +1 . Strategy S F will be winning for Learner.Moreover, its depth will be d + O (log n ). In the end of the proof we will refer toProposition 10 to show that S F yields a O (log n )-depth polynomial-time computableformula for MAJ n +1 , consisting only of MAJ gates and variables.13trategy S F has two phases. The first phase does not uses F at all, only the secondphase does. The objective of the first phase is to find some distinct i, j ∈ [2 n + 1] suchthat either z i = 0 ∧ z j = 1 or z i = 1 ∧ z j = 0, where z is the Nature’s vector. This canbe done as follows. Lemma 12.
One can compute in polynomial time a 3-ary tree T of depth O (log n ) withthe set of nodes v ( T ) and a mapping w : v ( T ) → [2 n +1] such that the following holds: • if r is the root of T , then w ( r ) = [2 n + 1] ; • if v is not a leaf of T and v , v , v are children of v , then every element of w ( v ) is covered at least twice by w ( v ) , w ( v ) , w ( v ) ; • if l is a leaf of T , then w ( r ) is of size .Proof. We start with a trivial tree, consisting only of the root, to which we assign [2 n +1].Then at each iteration we do the following. We have a 3-ary tree in which nodes areassigned to some subsets of [2 n + 1]. If every leaf is assigned to a set of size 2, weterminate. Otherwise we pick any leaf l of the current tree which is assigned to a subset A ⊂ [2 n + 1] of size at least 3. We split A into 3 disjoint subsets A , A , A of sizes ⌊| A | / ⌋ , ⌊| A | / ⌋ and | A | − ⌊| A | / ⌋ . We add 3 children to l (which become new leafs)and assign A ∪ A , A ∪ A , A ∪ A to them.It is easy to verify that the sizes of A ∪ A , A ∪ A , A ∪ A are at least 2 and at most · | A | . Hence the size of the set assigned to a node of depth h is at most (cid:16) (cid:17) h · (2 n + 1).This means that the depth of the tree is at any moment at most log / (2 n +1) = O (log n ).Therefore we terminate in 3 O (log n ) = n O (1) iterations, as at each iterations we add 3 newnodes. Each iteration obviously takes polynomial time.We use T to find two i, j ∈ [2 n + 1] such that either z i = 0 or z j = 0. Namely, wedescend from the root of T to one of its leafs. Learner maintains an invariant that theleftmost 0-coordinate of z is in w ( v ), where v is the current node of T . Let v , v , v be 3children of v . Learner for every i ∈ [3] makes a hypothesis that the leftmost 0-coordinateof z is in w ( v i ). Due to the properties of w at least two hypotheses are true. Natureindicates some v i for which this is true, and Learner descends to v i . When Learnerreaches a leaf, he knows a set of size two containing the leftmost 0-coordinate of z . Letthis set be { i, j } .We know that either z i or z j is 0. Thus z i z j ∈ { , , } . At the cost of one roundwe can ask Nature to identify an element of { , , } which differs from z i z j . If 10 isidentified, then z i z j ∈ { , } , and hence z i = 0, i.e., we can already output i . Similarthing happens when 01 is identified. Finally, if 00 is identified, then the objective of thefirst phase is fulfilled and we can proceed to the second phase.The second phase takes at most d rounds. In this phase Learner produces a sequence g , g , . . . , g d ′ , d ′ d of gates of F , where the depth of g i is i , the last gate g d ′ is an inputvariable (i.e., a leaf of F ) and each g ∈ { g , g , . . . , g d ′ } satisfies:( g ( z ) = 0 ∧ z i z j = 01) ∨ ( g ( ¬ z ) = 1 ∧ z i z j = 10) . (1)14ere ¬ z denotes the bit-wise negation of z .At the beginning Learner sets g = g out to be the output gate of F . Let us explainwhy (1) holds for g out . Nature’s vector is an element of MAJ − n +1 (0). I.e., the number ofones in z is at most n . In turn, in ¬ z there are at least n + 1 ones. Since g out computesMAJ n +1 , we have that g out ( z ) = 0 and g out ( ¬ z ) = 1. In turn, by the first phase it isguarantied that z i z j = 01 ∨ z i z j = 10.Assume now that the second phase is finished, i.e., Learner has produced some g d ′ = x k satisfying (1). Then by (1) either g d ′ ( z ) = z k = 0 or g d ′ ( ¬ z ) = ( ¬ z ) k = 1. In bothcases z k = 0, i.e., Learner can output k .It remains to explain how to fulfill the second phase. It is enough to show thefollowing. Assume that Learner knows a gate g l of F of depth l satisfying (1) and that g l is not an input variable. Then in one round he can either find a gate g l +1 of depth l + 1 satisfying (1) or give a correct answer to the game.The gate g l +1 will be one of the two gates which are fed to g l . Assume first that g l is an ∧ -gate and g l = u ∧ v . From (1) we conclude that from the following 3 statementsexactly 1 is true for z : u ( z ) = 0 and z i z j = 01 , (2) u ( z ) = 1 , v ( z ) = 0 and z i z j = 01 , (3) u ( ¬ z ) = v ( ¬ z ) = 1 and z i z j = 10 . (4)At the cost of one round Learner can ask Nature to indicate one statement which is falsefor x . If Nature says that (2) is false for z , then (1) holds for g l +1 = v . Next, if Naturesays that (3) is false for z , then (1) holds for g l +1 = u . Finally, if Nature says that (4)is false for z , then we know that z i z j = 01, i.e., Learner can already output i .In the same way we can deal with the case when g l is an ∨ -gate and g l = u ∨ v . By(1) exactly 1 of the following 3 statements is true for z : u ( z ) = v ( z ) = 0 and z i z j = 01 , (5) u ( ¬ z ) = 1 and z i z j = 10 , (6) u ( ¬ z ) = 0 , v ( ¬ z ) = 1 and z i z j = 10 . (7)Similarly, Learner asks Nature to indicate one statement which is false for z . If Naturesays that (5) is false for z , then z i z j = 10, i.e., Learner can output j . Next, if Naturesays that (6) is false for z , then (1) holds for g l +1 = v . Finally, if Nature says that (7) isfalse for z , then (1) holds for g l +1 = u .Thus S F is a O (log n )-depth winning strategy of Learner. Apply Proposition 10 to S F . We get a O (log n )-depth formula F ′ MAJ n +1 , consisting only of MAJ gatesand variables. From the self-duality of MAJ n +1 and MAJ it follows that F ′ computesMAJ n +1 . Finally, let us explain how to compute F ′ in polynomial time. To do so wehave to compute in polynomial time the light form of S F , i.e., the underlying tree of S F and the outputs of Learner in the leafs. It is easy to see that one can do this as follows.First, compute F and compute T from Lemma 12. For each leaf l of T do thefollowing. Let w ( l ) = { i, j } . Add 3 children to l . Two of them will be leafs of S F , in one15earner outputs i and in the other Learner outputs j . Attach a tree of F to the thirdchild. Then add to each non-leaf node of F one more child so that now the tree of F is3-ary. Each added child is a leaf of S F . If a child was added to an ∧ -gate, then Learneroutputs i in this child. In turn, if a child was added to an ∨ gate, then Learner outputs j in it. Finally, there are leafs that were in F initially, each labeled by some input variable.In these nodes Learner outputs the index of the corresponding input variable. Proof of Theorem 2.
How many rounds takes the first phase of the strategy S F from theprevious proof? Initially the left-most 0-coordinate of z takes O ( n ) values. At the costof one round we can shrink the number of possible values almost by a factor of 3 / / ( n ) + O (1). The size ofthat tree is hence 3 log / ( n )+ O (1) = O ( n / (1 − log (2)) ) = O ( n . ... ). To some of its leafs weattach a tree of the same size as the initial formula F . As a result we obtain a formula F ′ of size O ( n . ... · s ) for MAJ n +1 , consisting of MAJ gates and variables (here s isthe size of the initial formula F ).Let us show that we can perform the first phase in log ( n ) + O (1) rounds. Thiswill improve the size of the previous construction to O (3 log ( n )+ O (1) · s ) = O ( n log (3) · s ).However, the construction with log ( n ) + O (1) rounds will not be explicit. We need thefollowing Lemma: Lemma 13.
There exists a formula D with the following properties: • formula D is a complete ternary tree of depth ⌈ log ( n ) ⌉ + 10 ; • every non-leaf node of D contains a MAJ gate and every leaf of D contains aconjunction of 2 variables; • D ( x ) = 0 for every x ∈ { , } n +1 with at most n ones. Let us at first explain how to use formula D from Lemma 13 to fulfill the first phase.Recall that our goal is to find two indices i, j ∈ [2 n + 1] such that either z i = 0 or z j = 0.To do so Learner descends from the output gate of D to some of its leafs. He maintainsan invariant that for his current gate g of D it holds that g ( z ) = 0. For the output gatethe invariant is true because by Lemma 13 D is 0 on all Nature’s possible vectors. If wereached a leaf so that g is a conjuction of two variables z i and z j , then the first phase isfulfilled (by the invariant z i ∧ z j = 0). Finally, if g is a non-leaf node of D , i.e., a MAJ gate, then we can descend to one of the children of g at the cost of one round withoutviolating the invariant. Indeed, as g ( z ) = 0, then the same is true for at least 2 childrenof g . For each child g i of g Learner makes a hypotheses that g i ( z ) = 0. Any Nature’sresponse allows us to replace g by some g i . Proof of Lemma 13.
We will show existence of such D via probabilistic method. Namely,independently for each leaf l of D choose ( i, j ) ∈ [2 n + 1] uniformly at random and putthe conjuction z i ∧ z j into l . It is enough to demonstrate that for any x ∈ { , } n +1 with at most n ones it hols that Pr [ D ( x ) = 1] < − n − .16o do so we use the modification of the standard Valiant’s argument. For any fixed x let p be the probability that a leaf l of D equals 1 on x . This probability is the same forall the leafs and is at most 1 /
4. Now, Pr [ D ( x ) = 1] can be expressed exactly in termsof p as follows: Pr [ D ( x ) = 1] = f ( f ( f ( . . . f | {z } ⌈ log ( n ) ⌉ + 10 ( p ))) . . . ) , where f ( t ) = t + 3 t (1 − t ) = 3 t − t . Observe that 3 f ( t ) (3 t ) . Hence3 Pr [ D ( x ) = 1] (3 p ) ⌈ log2( n ) ⌉ +10 (3 / n < (1 / − n − . Theorem 6 follows from Proposition 14 (Subsection 5.1) and Proposition 16 (Subsec-tion 5.2). In turn, Theorem 7 follows from Proposition 15 (Subsection 5.1) and Propo-sition 19 (Subsection 5.2).
Proposition 14.
For any constant k > the following holds. Assume that f ∈ Q k and C f is a circuit, consisting only of THR k +12 gates and variables. Then there is a pro-tocol π , computing Q k -communication game for f , such that depth( π ) = O (depth( C )) .Proof. Let the inputs to parties be z , . . . , z k ∈ f − (0). Parties descend from the outputgate of C to one of the inputs. They maintain the invariant that for the current gate g of C it holds that g ( z ) = g ( z ) = . . . = g ( z k ) = 0. If g is not yet an input, then g is a THR k +12 gate and g = THR k +12 ( g , . . . , g k +1 ) for some gates g , . . . , g k +1 . For each z i we have g ( z i ) = THR k +12 ( g ( z i ) , . . . , g k +1 ( z i )) = 0. Hence for each z i there is at mostone gate out of g , . . . , g k +1 satisfying g j ( z i ) = 0. Hence in O (1) bits of communicationparties can agree on the index j ∈ [ k + 1] satisfying g j ( z ) = g j ( z ) = . . . g j ( z k ) = 0.Thus in O (depth( π )) bits of communication they reach some input of C . If this inputcontains the variable x l , then by the invariant z l = z l = . . . = z kl = 0, as required.Exactly the same argument can be applied to the following proposition. Proposition 15.
For any constant k > the following holds. Assume that f ∈ R k and C f is a circuit, consisting only of THR k +12 gates and literals. Then there is a protocol π , computing R k -communication game for f , such that depth( π ) = O (depth( C )) . .2 From protocols to circuits Proposition 16.
For every constant k > the following holds. Let f ∈ Q k . Assumethat π is a communication protocol computing Q k -communication game for f . Thenthere is a circuit C f , consisting of THR k +12 gates and variables, such that depth( C ) = O (depth( π )) .Proof. In the proof we will use the following terminology for strategies in Q k -hypothesesgame. Fix some strategy S . A current play is a finite sequence r , r , r , . . . r j of integersfrom 0 to k . By r i we mean Nature’s response in the i th round. Given a current play, let H i , . . . , H ik ⊂ f − (0) be k + 1 hypotheses Learner makes in the i th round according to S if Nature’s responses in the first i − r , . . . , r i − . If after that Nature’sresponse is r i , then Nature’s input vector z satisfies z ∈ H ir i . We say that z ∈ f − (0)is compatible with the current play r , . . . , r j if z ∈ H r , . . . , z ∈ H jr j . Informally, thismeans that Nature, having z on input, can produce responses r , . . . , r j by playing againststrategy S .Set d = depth( π ). By Proposition 10 it is enough to give a O ( d )-round winningstrategy of Learner in the Q k -hypotheses game for f . Strategy proceeds in d iterations,each iteration takes O (1) rounds.As the game goes on, a sequence of Nature’s responses r , r , r . . . is produced.Assume that r , . . . , r h ′ are Nature’s responses in the first h iteration (here h ′ is thenumber of rounds in the first h iterations). Given any r , r , r . . . , by Z h we denote theset of all z ∈ f − (0) which are compatible with r , . . . r h ′ , . We also say that elements of Z h are compatible with the current play after h iterations .Let V be the set of all nodes of the protocol π and let T be the set of all terminalsof the protocol π .Consider a set Z ⊂ f − (0), a set of nodes U ⊂ V and a function g : Z → C , where | C | = k . A g -profile of a tuple ( z , . . . , z k ) ∈ Z is a vector ( g ( z ) , . . . , g ( z k )) ∈ C k .We say that g : Z → C is complete for Z with respect to the set of nodes U if thefollowing holds. For every vector ¯ c ∈ C k there exists a node v ∈ U such that all tuplesfrom Z k with g -profile ¯ c visit v in the protocol π .We say that a set of nodes U ⊂ T is complete for Z if there exists g : Z → C , | C | = k which is complete for Z with respect to U .Note that we can consider only complete sets of size at most k k . Formally, if U iscomplete for Z , then there is a subset U ′ ⊂ U of size at most k k which is also completefor Z . Indeed, there are k k possible g -profiles and for each we need only one node in U . Lemma 17.
Assume that U ⊂ T is complete for Z ⊂ f − (0) . Then there exists i ∈ [ n ] such that z i = 0 for every z ∈ Z .Proof. If Z is empty, then there is nothing to prove. Otherwise let g : Z → C , | C | = k be complete for Z with respect to U . Take any vector ¯ c = ( c , . . . , c k ) ∈ C k such that { c i | i ∈ [ k ] } = g ( Z ). There exists a node v ∈ U such that any tuple from Z k with g -profile ¯ c visits v . Note that v is a terminal of π and let i be the output of π in g .Let us show that for any z ∈ Z it holds that z i = 0. Indeed, note that there exists a18uple ¯ z ∈ Z k which includes z and which has g -profile ¯ c . This tuple visits v . Since π computes Q k -communication game for f , every element of the tuple ¯ z should have 0 atthe i th coordinate. In particular, this holds for z .After d iterations Learner should be able to produce an output. For that there shouldexist i ∈ [ n ] such that for any z ∈ Z d it holds that z i = 0. We will use Lemma 17 toensure that. Namely, we will ensure that there exists U ⊂ T which is complete for Z d .Learner achieves this by maintaining the following invariant.Let us say that a set of nodes U is h -low if every element of U is either a terminal ora node of depth at least h . Invariant 1
There is a h -low set U which is complete for Z h .This invariant implies that Learner wins in the end, as any d -low set consists only ofterminals.A 0-low set which is complete for Z = f − (0) is a set consisting only of the startingnode of π .Assume that Invariant 1 holds after h iteration. Let us show how to perform thenext iteration to maintain the invariant. For that we need a notion of communicationprofile .A communication profile of z ∈ f − (0) with respect to a set of nodes U ⊂ V is afunction p z : U → { , } . For v ∈ U the value of p z ( v ) is defined as follows. If v is aterminal, set p z ( v ) = 0. Otherwise let i ∈ [ k ] be the index of the party communicatingat v . Set p z ( v ) to be the bit transmitted by the i th party at v on input z . I.e., p z forevery v ∈ U contains information where the protocol goes from the node v if the party,communicating at v , has z on input.We also define a communication profile of the tuple ( z , . . . , z k ) ∈ ( f − (0)) k as( p z , . . . , p z k ). Lemma 18.
Let ( z , . . . , z k ) , ( y , . . . , y k ) ∈ ( f − (0)) k be two inputs visiting the samenode v ∈ V \ T . Assume that their communication profiles with respect to { v } coincide.Then these two inputs visit the same successor of v .Proof. Let their common communication profile with respect to { v } be ( p , . . . , p k ).Next, assume that i is the index of the party communicating at v . Then the informationwhere these inputs descend from v is contained in p i .Here is what Learner does during the ( h + 1)st iteration. He takes any h -low U of sizeat most k k which is complete for Z h . Then he takes any g : Z h → C , | C | = k which iscomplete for Z h with respect to U . He now devises a new function g ′ taking elements ofthe set Z h on input. The value of g ′ ( z ) is a pair ( p z , g ( z )), where p z is a communicationprofile of z with respect to U . There are at most 2 | U | k k different communicationprofiles with respect to U . Hence g ′ ( z ) takes at most 2 k k · k = O (1) values.19t each round of the ( h + 1)st iteration Learner asks Nature to identify some pair( p, c ), where p : U → { , } and c ∈ C , such that g ′ ( z ) = ( p, c ) for the Nature’s vector z . Namely, we take any k + 1 values of g ′ which are not yet rejected by Nature andask Nature to reject one of them. We do so until there are only k possible values( p , c ) , . . . ( p k , c k ) left. This takes O (1) rounds and the ( h + 1)st iteration is finished.Any z ∈ f − (0) which is compatible with the responses Nature’ gave during the ( h + 1)stiteration in the current play satisfies g ′ ( z ) ∈ C ′ = { ( p , c ) , . . . ( p k , c k ) } . In particular,any z ∈ Z h +1 satisfies g ′ ( z ) ∈ C ′ . I.e., the restriction of g ′ to Z h +1 is a function of theform g ′ : Z h +1 → C ′ . Let us show that g ′ : Z h +1 → C ′ is complete for Z h +1 with respectto some ( h + 1)-low set U ′ . This will ensure that Invariant 1 is maintained after h + 1iterations.We define U ′ is follows. Take any vector ¯ c ∈ ( C ′ ) k . It is enough to show that all theinputs from ( Z h +1 ) k with g ′ -profile ¯ c visit the same node v ′ which is either a terminalor of depth at least h + 1. Then we just set U ′ to be the union of all such v ′ over allpossible g ′ -profiles.All the tuples from ( Z h +1 ) k with the same g ′ -profile visit the same node v ∈ U . Thisis because g ′ -profile of a tuple determines its g -profile (the value of g ′ determines thevalue of g ) , and hence we can use Invariant 1 for Z h − here. If v is a terminal, thereis nothing left to prove. Otherwise, note that g ′ -profile of a tuple also determines itscommunication profile with respect to U and hence with respect to { v } ⊂ U . Thereforeall the tuples with the same g ′ -profile by Lemma 18 visit the same successor of v .With straightforward modifications one can obtain a proof of the following: Proposition 19.
For every constant k > the following holds. Let f ∈ R k . Assume that π is a dag-like protocol computing R k -communication game for f . Then there is a circuit C f , consisting of THR k +12 gates and literals, satisfying depth( C ) = O (depth( π )) . Corollary 20 (Weak version of Theorem 3) . For any constant k > there exists O (log n ) -depth formula for THR kn +1 n +1 , consisting only of THR k +12 gates and variables.Proof. We will show that there exists O (log n )-depth protocol π computing Q k -communication game for THR kn +1 n +1 . By Proposition 16 this means that there is a O (log n )-depth formula F THR kn +1 n +1 , consisting only of THR k +12 gates and variables.It is easy to see that F actually coincides with THR kn +1 n +1 . Indeed, assume that F ( x ) = 0for some x with at least n + 1 ones. Then it is easy to construct x , . . . , x k , each with n ones, such that there is no common 0-coordinate for x, x , . . . , x k . On all of these vectors F takes value 0. However, the function computed by F should belong to Q k (Proposition4). Let π be the following protocol. Assume that the inputs to parties are x , x , . . . , x k ∈{ , } kn +1 , without loss of generality we can assume that in each x r there are exactly n ones. For x ∈ { , } kn +1 define supp( x ) = { i ∈ [ kn + 1] | x i = 1 } . Let T be a binaryrooted tree of depth d = log ( n ) + O (1) with kn + 1 leafs. Identify leafs of T withelements of [ kn + 1]. For a node v of T let T v be the set of all leafs of T which aredescendants of v . Once again, we view T v as a subset of [ kn + 1].20he protocol proceeds in at most d iterations. After i iterations, i = 0 , , , . . . , d ,parties agree on a node v of T of depth i , satisfying the following invariant: k X r =1 | supp( x r ) ∩ T v | < | T v | . (8)At the beginning Invariant (8) holds just because v is the root, T v = [ kn + 1] and eachsupp( x r ) is of size n .After d iterations v = l is a leaf of T . Parties output l . This is correct because by(8) we have | T l | = 1 = ⇒ | supp( x r ) ∩ T l | = 0 = ⇒ x l = 0 for every r ∈ [ k ].Let us now explain what parties do at each iteration. If the current v is not a leaf,let v , v be two children of v . Each party sends | supp( x r ) ∩ T v | and | supp( x r ) ∩ T v | ,using O (log n ) bits. Since T v and T v is a partition of T v , we have: X b =0 k X r =1 | supp( x r ) ∩ T v b | = k X r =1 | supp( x r ) ∩ T v | < | T v | = X b =0 | T v b | . Thus the inequality: k X r =1 | supp( x r ) ∩ T v b | < | T v b | (9)is true either for b = 0 or for b = 1. Let b ∗ be the smallest b ∈ { , } for which (9) istrue. Parties proceed to the next iteration with v being replaced by v b ∗ .There are d = O (log n ) iterations, at each parties communicate O (log n ) bits. Hence π is O (log n )-depth, as required. Remark.
Strategy from the proof of Proposition 16 is efficient only in terms of thenumber of rounds. In the next section we give another version of this strategy. Thisversion will ensure that circuits we obtain from protocols for Q k -communication gamesare not only low-depth, but also polynomial-size and explicit. For that, however, werequire a bit more from the protocol π . Fix f ∈ Q k . We say that a dag-like communication protocol π strongly computes Q k -communication game for f if for every terminal t of π , for every x ∈ f − (0) and for every i ∈ [ k ] the following holds. If x is i -compatible with t , then x j = 0, where j = l ( t ) is thelabel of terminal t in the protocol π .Similarly, fix f ∈ R k . We say that a dag-like communication protocol π strongly computes R k -communication game for f if for every terminal t of π , for every x ∈ f − (0)and for every i ∈ [ k ] the following holds. If x is i -compatible with t , then x j = b , where( j, b ) = l ( t ) is the label of terminal t in the protocol π .Strong computability essentially (but not completely) coincides with the notion ofcomputability that Sokolov gave in [14] for general relations. Strong computability im-plies more intuitive notion of computability that we gave in the Preliminaries. Theopposite direction is false in general. 21ext we prove an effective version of Proposition 16. Theorem 21.
For every constant k > there exists a polynomial-time algorithm A suchthat the following holds. Assume that f ∈ Q k and π is a dag-like protocol which stronglycomputes Q k -communication game for f . Then, given the light form of π , the algorithm A outputs a circuit C f , consisting only of THR k +12 gates and variables, such that depth( C ) = O (depth( π )) , size( C ) = O (cid:16) size( π ) O (1) (cid:17) .Proof. We will again give a O ( d )-round winning strategy of Learner in the Q k -hypothesesgame for f . Now, however, we should ensure that the light form of our strategy is ofsize O (cid:16) size( π ) O (1) (cid:17) and can be computed in time O (cid:16) size( π ) O (1) (cid:17) from the light form of π . Instead of specifying the light form of our strategy directly we will use the followingtrick. Assume that Learner has a working tape consisting of O (log size( π )) cells, whereeach cell can store one bit. Learner memorizes all the Nature’s responses so that heknows the current position of the game. But he does not store the sequence of Nature’sresponses on the working tape (there is no space for it). Instead, he first makes hishypotheses which depend on the current position. Then he receives a Nature’s response r ∈ { , , . . . , k } . And then he modifies the working tape, but the result should dependonly on the current content of the working tape and on r (and not on the currentposition in a game). Moreover, we will ensure that modifying the working tape takes O (cid:16) size( π ) O (1) (cid:17) time, given the light form of π .The main purpose of the working tape manifests itself in the end. Namely, at somepoint Learner decides to stop making hypotheses. This should be indicated on theworking tape. More importantly, Learner’s output should depend only on the contentof working tape in the end (and not on the whole sequence of Nature’s responses).Moreover, this should take O (cid:16) size( π ) O (1) (cid:17) time to compute that output, given the lightform of π .If a strategy satisfies these restrictions, then its light form is computable in O (cid:16) size( π ) O (1) (cid:17) time given the light form of π . Indeed, the underlying dag will con-sist of all possible configurations of the working tape. There are O (cid:16) size( π ) O (1) (cid:17) of them,as working tape uses O (log size( π )) bits. For all non-terminal configurations c we gothrough all r ∈ { , , . . . , k } . We compute what would be a configuration c r of theworking tape if the current configuration is c and Nature’s response is r . After that weconnect c to c , c , . . . , c k . Finally, in all terminal configurations we compute the outputsof Learner. This gives a light form of our strategy in O (cid:16) size( π ) O (1) (cid:17) time.Let V be the set of nodes of π and T be the set of terminals of π . Strategy proceedsin d iterations, each taking O (1) rounds. We define sets Z h exactly as in the proofof Proposition 16. We also use the same notion of communication profile. However,we define completeness in a different way. First of all, instead of working with sets ofnodes with no additional structure we will work with multidimensional arrays of nodes.Namely, we will consider k -dimensional arrays in which every dimension is indexed byintegers from [ k ]. Formally, such arrays are functions of the form M : [ k ] k → V . We willuse notation M [ c , . . . , c k ] for the value of M on ( c , . . . , c k ) ∈ [ k ] k .22onsider any Z ⊂ f − (0). We say that g : Z → [ k ] is complete for Z with respect toa multidimensional array M : [ k ] k → V if for every ( c , . . . , c k ) ∈ [ k ] k , for every i ∈ [ k ]and for every z ∈ Z the following holds. If c i = g ( z ), then z is i -compatible with M [ c , . . . , c k ].We say that a multidimensional array M : [ k ] k → V is complete for Z if there exists g : Z → [ k ] which is complete with respect to M .To digest the notion of completeness it is instructive to consider the case k = 2. Inthis case M is a 2 × π . The function g : Z → [2] iscomplete for Z with respect to M if the following holds. First, for every z ∈ Z twonodes in the g ( z )th row of M should be 1 -compatible with z . Second, for every z ∈ Z two nodes in the g ( z )th column of M should be 2 -compatible with z .Let us now establish an analog of Lemma 17. Lemma 22.
Assume that M : [ k ] k → T is complete for Z ⊂ f − (0) . Let l be the outputof π in the terminal M [1 , , . . . , k ] . Then z l = 0 for every Z .Proof. Since π strongly computes Q k -communication game for f , it is enough to showthat every z ∈ Z is i -compatible with M [1 , , . . . , k ] for some i . Take g : Z → [ k ]which is complete for Z with respect to M . By definition z is g ( z )-compatible with M [1 , , . . . , k ].We now proceed to the description of the Learner’s strategy. The working tape ofLearner consists of: • an integer iter ; • a multidimensional array M : [ k ] k → V ; • O (1) additional bits of memory.Integer iter will be at most d size( π ) so to store all this information we need O (log(size( π ))) bits, as required. Integer iter always equals the number of iterationsperformed so far (at the beginning iter = 0). The array M changes only at the momentswhen iter is incremented by 1. So let M h denote the content of the array M when iter = h .We call an array of nodes h -low if every node in it is either terminal or of depth atleast h . Learner maintains the following invariant. Invariant 2 M h is h -low and M h is complete for Z h .At the beginning Learner sets every element of M to be the starting node of π sothat Invariant 2 trivially holds.Note that every node in M d is a terminal of π . After d iterations Learner outputsthe label of terminal M d [1 , , . . . , k ] in the protocol π . As M d is complete for Z d due to Invariant 2 , this by Lemma 22 will be a correct output in the Q k -hypotheses game for23 . Obviously producing the output takes polynomial time given the light form of π andthe content of Learner’s working tape in the end.Now we need to perform an iteration. Assume that h iterations passed and Invariant2 still holds. Let U h be the set of all nodes appearing in M h . Take any function g : Z h → [ k ] which is complete for Z h with respect to M h .At each round of the ( h + 1)st iteration Learner asks Nature to specify some pair( p, c ) ∈ { , } U h × [ k ] such that ( p z , g ( z )) = ( p, c ), where z is the Nature’s vector and p z is a communication profile of z with respect to U h . Learner stores each ( p, c ) usinghis O (1) additional bits on the working tape. Learner can do this until there are only k pairs from ( p , c ) , . . . , ( p k , c k ) ∈ { , } U h × [ k ] left which are not rejected by Nature.When this moment is reached, the ( h + 1)st iteration is finished. The iteration takes2 | U h | · k − k = O (1) rounds, as required. For any z compatible with the current play after h + 1 iterations we know that ( p z , g ( z )) is among ( p , c ) , . . . , ( p k , c k ), i.e,( p z , g ( z )) ∈ { ( p , c ) , . . . , ( p k , c k ) } for all z ∈ Z h +1 . (10)Learner writes ( p , c ) , . . . , ( p k , c k ) on the working tape (all the pairs that were ex-cluded are on the working tape and hence he can compute the remaining ones). Learnerthen computes a ( h + 1)-low array M h +1 which will be complete for Z h +1 . To compute M h +1 he will only need to know M h , ( p , c ) , . . . , ( p k , c k ) (this information is on theworking tape) and the light form of π .Namely, Learner determines M h +1 [ d , . . . , d k ] for ( d , . . . , d k ) ∈ [ k ] k as follows. Con-sider the node v = M h [ c d , . . . , c d k ]. If v is a terminal, then set M h +1 [ d , . . . , d k ] = v .Otherwise let i ∈ [ k ] be the index of the party communicating at v . Look at p d i , whichcan be considered as a function of the form p d i : U h → { , } . Define r = p d i ( v ). Amongtwo edges, starting at v , choose one which is labeled by r . Descend along this edge from v and let the resulting successor of v be M h +1 [ d , . . . , d k ].Obviously, computing M h +1 takes O (cid:16) size( π ) O (1) (cid:17) . To show that Invariant 2 ismaintained we have to show that (a) M h +1 is ( h + 1)-low and (b) M h +1 is complete for Z h +1 .The first part, (a) , holds because each M h +1 [ d , . . . , d k ] is either a terminal or asuccessor of a node of depth at least h . For (b) we define the following function: g ′ : Z h +1 → [ k ] , g ′ ( z ) = i, where i is such that ( p z , g ( z )) = ( p i , c i ) . By (10) this definition is correct. We will show that g ′ is complete for Z h +1 with respectto M h +1 .For that take any ( d , . . . , d k ) ∈ [ k ] k , z ∈ Z h +1 and i ∈ [ k ] such that d i = g ′ ( z ). Weshall show that z is i -compatible with a node M h +1 [ d , . . . , d k ]. By definition of g ′ wehave that g ( z ) = c d i . As by Invariant 2 function g is complete for Z h with respect to M h , this means that z is i -compatible with v = M [ c d , . . . , c d k ]. If v is a terminal, then M h +1 [ d , . . . , d k ] = v and there is nothing left to proof.Otherwise v ∈ V \ T . Let j be the index of the party communicating at v . Bydefinition M h +1 [ d , . . . , d k ] is a successor of v . If j = i , i.e., not the i th party communi-cates at v , then any successor of v is i -compatible with z . Finally, assume that j = i .24ode M h +1 [ d , . . . , d k ] is obtained from v by descending along the edge which is labeledby r = p d i ( v ). Hence to show that z is i -compatible with M h +1 [ d , . . . , d k ] we shouldverify that at v on input z the i th party transmits the bit r . For that again recall that g ′ ( z ) = d i , which means by definition of g ′ that p z = p d i . I.e., p d i is the communicationprofile of z with respect to U h . In particular, the value r = p d i ( v ) is the bit transmittedby the i th party on input z at v , as required.In the same way one can obtain an analog of the previous theorem for the R k -case. Theorem 23.
For every constant k > there exists a polynomial-time algorithm A such that the following holds. Assume that f ∈ R k and π is a dag-like protocol whichstrongly computes R k -communication game for f . Then, given the light form of π , thealgorithm A outputs a circuit C f , consisting only of THR k +12 gates and literals, suchthat depth( C ) = O (depth( π )) , size( C ) = O (cid:16) size( π ) O (1) (cid:17) . In this section we obtain Theorems 1 and 3 by devising protocols strongly computingthe corresponding Q k -communication games. Unfortunately, establishing strong com-putability requires diving into straightforward but tedious technical details, even forsimple protocols. Alternative proof of Theorem 1.
We will show that there exists O (log n )-depth protocol π with polynomial-time computable light form, strongly computing Q -communicationgame for MAJ n +1 . By Theorem 21 this means that there is a polynomial-time com-putable O (log n )-depth formula F MAJ n +1 , consisting only of MAJ gates and vari-ables. From self-duality of MAJ n +1 and MAJ it follows that F computes MAJ n +1 .Take a polynomial-time computable O (log n )-depth monotone formula F ′ forMAJ n +1 . Consider the following communication protocol π . The tree of π coincideswith the tree of F ′ . Inputs to F ′ will be leafs of π . In a leaf containing input variable x i the output of the protocol π is i . Remaining nodes of π are ∧ and ∨ gates. In the ∧ gates communicates the first party, while in the ∨ gates communicates the second party.Fix an ∧ gate g (which belongs to the first party). Let g , g be gates which are fedto g , i.e., g = g ∧ g . There are two edges, starting at g , one leads to g (and is labeledby 0) and the other leads to g (and is labeled by 1). Take an input a ∈ MAJ − n +1 (0) tothe first party. On input a at the gate g the first party transmits the bit r = min { c ∈{ , } | g c ( a ) = 0 } . If the minimum is over the empty set, then we set r = 0.Take now an ∨ gate h belonging to the second party. Similarly, there are two edges,starting at h , one leads to h (and is labeled by 0) and the other leads to h (and islabeled by 1). Here h , h are two gates which are fed to h , i.e., h = h ∨ h . Take aninput b ∈ MAJ − n +1 (0) to the second party. On input b at the gate h the second partytransmits the bit r = min { c ∈ { , } | h c ( ¬ b ) = 1 } . If the minimum is over the emptyset, then we set r = 0. Here ¬ denotes the bit-wise negation. Description of the protocol π is finished. 25learly, the protocol π is of depth O (log n ) and its light form is polynomial-timecomputable. It remains to argue that the protocol strongly computes Q -communicationgame for MAJ n +1 . Nodes of the protocol may be identified with the gates of F ′ . Con-sider any path p = h e , . . . , e m i in the protocol π . Assume that e j is an edge from g j − to g j and g is the output gate of F ′ . We shall show that the following: if a ∈ MAJ − n +1 (0)is 1-compatible with p , then g ( a ) = g ( a ) = . . . = g m ( a ) = 0. Indeed, g ( a ) = 0 holdsbecause F ′ computes MAJ n +1 . Now, assume that g j ( a ) = 0 is already proved. If g j isan ∨ gate, then g j +1 ( a ) = 0 just because g j +1 feds to g j . Otherwise g j is an ∧ gate whichtherefore belongs to the first party. Let r ∈ { , } is the label of the edge e j +1 . Notethat g j +1 = g jr , where g j , g j are two gates which are fed to g j . . Since a is 1-compatiblewith p , it holds that r coincides with the bit that the first party transmits at g j on input a , i.e., with min { c ∈ { , } | g jc ( a ) = 0 } . The set over which the minimum is takenis non-empty because g j ( a ) = 0. In particular r belongs to this set, which means that g j +1 ( a ) = g jr ( a ) = 0, as required.Similarly one can verify that if b ∈ MAJ − n +1 (0) is 2-compatible with p , then g ( ¬ b ) = g ( ¬ b ) = . . . = g m ( ¬ b ) = 0. Hence we get that if a leaf l is 1-compatible (2-compatible)with a ( b ) and l contains a variable x i , then a i = 0 ( ¬ b i = 1). Hence the protocolstrongly computes the Q -communication game for MAJ n +1 . Proof of Theorem 3.
We will realize the protocol from the proof of Corollary 20 insuch a way that it will give us O (log n )-depth polynomial-size dag-like protocol withpolynomial-time computable light form, strongly computing Q k -communication gamefor THR kn +1 n +1 . By Theorem 21 this means that there is a polynomial-time computable O (log n )-depth polynomial-size circuit C THR kn +1 n +1 , consisting only of THR k +12 gatesand variables. With the same argument as in Corollary 20 one can show that C coincideswith THR kn +1 n +1 .We will use the same tree T as in the proof of Corollary 20. Let us specify theunderlying dag G of our protocol π . For a node v of T let S v be the set of all tuples( s , s , . . . , s k ) ∈ { , , . . . , kn + 1 } k such that s + s + . . . + s k < | T v | . For every node v of T and for every ( s , s , . . . , s k ) ∈ S v the dag G will contain a node identified witha tuple ( v, s , s , . . . , s k ). These nodes of G will be called the main nodes (there will besome other nodes too). The starting node of G will be ( r, n, . . . , n ), where r is the rootof T . Note that if l is a leaf of T , then | T l | = 1. Hence the only main node having l as the first coordinate is ( l, , . . . , π will coincide with theset of all main nodes of the form ( l, , . . . , l is a leaf of T . The output of π in( l, , . . . ,
0) is l .For an integer s kn + 1 let W ( s ) be a binary tree of depth O (log n ) with |{ ( a, b ) | a, b ∈ { , , . . . , s } , a + b = s }| leaves. We assume that leaves of W ( s ) are iden-tified with elements of { ( a, b ) | a, b ∈ { , , . . . , s } , a + b = s } . We use W ( s ) in theconstruction of G . Namely, take any main node ( v, s , s , . . . , s k ) with a non-leaf v .Attach W ( s ) to it. Then attach to every leaf of W ( s ) a copy of W ( s ). Next, toevery leaf of the resulting tree attach a copy of W ( s ) and so on. In this way we ob-tain a binary tree W ( v, s , . . . , s k ) of depth O (log n ) growing at ( v, s , . . . , s k ). Its leavescan be identified with tuples of integers ( a , b , . . . , a k , b k ) satisfying a , b , . . . , a k , b k > , a + b = s , . . . , a k + b k = s k . We will merge every leaf of W ( v, s , . . . , s k ) with somemain node. Namely, take a leaf ( a , b , . . . , a k , b k ). If a + . . . + a k < | T v | , then wemerge ( a , b , . . . , a k , b k ) with the main node ( v , a , . . . , a k ). Otherwise it should holdthat b + . . . + b k < | T v | . In this case we merge ( a , b , . . . , a k , b k ) with the main node( v , b , . . . , b k ).Description of the dag of π is finished. Since k is constant, there are n O (1) mainnodes and to each we attach a tree of depth O (log n ). Hence π is O (log n )-depth and n O (1) -size. Let us define a partition of non-terminal nodes between parties. Take a mainnode ( v, s , . . . , s k ), where v is not a leaf of T . The tree W ( v, s , . . . , s k ), growing from( v, s , . . . , s k ) consists of copies of W ( s ) , . . . , W ( s k ). We simply say that the i th partycommunicates in copies of W ( s i ). After that we conclude that the light form of π ispolynomial-time computable.Now let us specify how the i th party communicates inside W ( s i ). Assume that x ∈ { , } kn +1 is the input to the i th party. If | T v ∩ supp( x ) | 6 = s i , then the i th partycommunicates arbitrarily. Now, assume that | T v ∩ supp( x ) | = s i . Then the i th partycommunicates in such a way that the resulting path descends from the root of W ( s i ) tothe leaf identified with a pair of integers ( | T v ∩ supp( x ) | , | T v ∩ supp( x ) | ).From this we immediately get the following observation. Let p be a path from the rootof W ( v, s , . . . , s k ) to a leaf identified with a tuple ( a , b , . . . , a k , b k ). Further, assumethat x ∈ (THR kn +1 n +1 ) − (0), satisfying | T v ∩ supp( x ) | = s i , is i -compatible with p . Then a i = | T v ∩ supp( x ) | and b i = | T v ∩ supp( x ) | . Indeed, any such p passes though a copy W ( s i ) and leaves W ( s i ) in a leaf identified with ( | T v ∩ supp( x ) | , | T v ∩ supp( x ) | ).From this observation one can easily deduce that if x ∈ (THR kn +1 n +1 ) − (0) is i -compatible with a main node ( v, s , . . . , s k ), then | T v ∩ supp( x ) | = s i . Indeed, we canobtain this by induction on the depth of v . Induction step easily follows from the pre-vious paragraph. As for induction base we notice that | T r ∩ supp( x ) | = n for the root r of T (as in the proof of Corollary 20 we assume that | supp( x ) | = n as party can alwaysadd missing 1’s).In particular, this means that π strongly computes Q k -communication game forTHR kn +1 n +1 . Indeed, any terminal of π is of the form ( l, , . . . , l is a leaf of T . If x ∈ (THR kn +1 n +1 ) − (0) is i -compatible with ( l, , . . . , | T l ∩ supp( x ) | = |{ l } ∩ supp( x ) | = 0. This means that x l = 0 and hencethe output of the protocol is correct. • Can Q k -communication game for THR kn +1 n +1 be solved in O (log n ) bits of commu-nication for k >
3? Equivalently, can THR kn +1 n +1 be computed by O (log n )-depthcircuit, consisting only of THR k +12 and variables? Can a deeper look into the con-struction of AKS sorting network help here (note that we only use this sortingnetwork as a black-box)? 27 Can at least R k -communication game for THR kn +1 n +1 be solved in O (log n ) bits ofcommunication for k >
3? Again, this is equivalent to asking whether THR kn +1 n +1 can be computed by O (log n )-depth circuit, consisting only of THR k +12 and lit-erals . Note that if we allow literals (along with ∧ and ∨ gates), then there aremuch simpler constructions of a O (log n )-depth formula for MAJ n and, in fact, forevery symmetric Boolean function [16]. Moreover, this can be done in terms ofcommunication complexity [2]. A natural approach would be to apply ideas of [2]to R k -communication games. • Are there any other interesting functions in Q k and R k which can be analyzed withour technique? Acknowledgments.
The authors are grateful to Alexander Shen for suggesting togeneralize our initial results.
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