Approximate query complexity - UNCG.edu

Approximate query complexity Clifford Smyth University of North Carolina Greensboro Greensboro, NC, USA [email protected] May 31, 2011

Abstract Let f : {0, 1}n → {0, 1}. Let µ be a product probability measure on {0, 1}n . For  ≥ 0 we define D (f ), the -approximate decision tree complexity of f , to be the minimum depth of a decision tree T with µ(T (x) 6= f (x)) ≤ . For j = 0 or 1 and for δ ≥ 0, we define Cj,δ (f ), the δ-approximate j-certificate complexity of f , to be the minimum certificate complexity of a set A ⊆ Ω with µ(A∆f −1 (j)) ≤ . Note that D0 (f ) = D(f ) and Cj,0 (f ) = Cj (f ) are the ordinary decision tree and j-certificate complexities of f , respectively. We extend the wellknown result, D(f ) ≤ C1 (f ) C0 (f ) [3, 6, 12], proving that for all  > 0 there exists a δ > 0 and a constant K = K(, δ) > 0 such that for all n, µ, f , D (f ) ≤ K C1,δ (f ) C0,δ (f ). We also answer a related question

1

on query complexity raised by Tardos [12]. We prove generalizations of these results to general product probability spaces.

1

Introduction

Let n ≥ 1 be an integer. For each i, 0 ≤ i ≤ n, let Ωi be a finite set with at Q least two elements. Let Ω = ni=1 Ωi and let f : Ω → Ω0 . Elements x ∈ Ω will be called assignments or inputs to f and elements of Ω0 will be called outputs. If f : {0, 1}n → {0, 1}, f is called a Boolean function. Consider a deterministic algorithm M which computes f , i.e. outputs M (x) = f (x) when given input x. The query complexity of M is the maximum, taken over all inputs x = (x1 , . . . , xn ), of the number of coordinates xi that M queries, i.e. reads, before returning f (x). The deterministic query complexity of f , written D(f ), is defined to be the minimum query complexity of a deterministic algorithm M computing f . Note that D(f ) ≤ n where n is the number of input coordinates. We will analyze our algorithms in the query complexity model: only the number of queries is counted and an algorithm otherwise is assumed to have unlimited computational power. Since only queries are relevant, any such algorithm can be presented in a standardized form known as a decision tree. A decision tree T with domain Ω and range Ω0 is a finite rooted tree such that every internal vertex v is labeled with a constant q(v) ∈ [n] := {1, 2, . . . , n} called the query coordinate of v and each leaf l is labeled with

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a constant o(l) ∈ Ω0 called the output of l. Each internal vertex v ∈ T with q(v) = i has |Ωi | children, one child c(v, j) for each j ∈ Ωi . For each input x, we define a path pT (x) := (v0 , . . . , vd ) in T as follows. pT (x) begins with the root v0 . For 0 ≤ i < d, vi+1 = c(vi , xq(vi ) ) and the path terminates upon reaching the leaf vd that is uniquely determined by T and x. We define T : Ω → Ω0 , the function computed by T , to be T (x) := o(vd ). We say T computes f , if T (x) = f (x) for all x. The query complexity of T on x is the length of pT (x), the number of queries T makes while computing T (x). D(T ), the query complexity of T , is the depth of T , the maximum length of a path in T from the root to a leaf. Clearly,

D(f ) = min{D(T ) : T is a decision tree computing f }.

Hence D(f ) is also known as the decision tree complexity of f . A certificate of Ω is a function c with domain dom(c) ⊆ [n] such that for each i ∈ dom(c), c(i) ∈ Ωi . The complexity or size of c is |c| := |dom(c)|. We say x satisfies c, written x |= c, if xi = c(i) for all i ∈ dom(c). Let [c] := {x ∈ Ω : x |= c} be the cylinder of assignments satisfying c. The empty certificate, λ, is the unique certificate with |λ| = 0. If A is a non-empty set of certificates, the certificate complexity of A is S C(A) := max{|a| : a ∈ A}. We define [A] := a∈A [a], the subset of Ω

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represented by A. If A ⊆ Ω, we define the certificate complexity of A,

C(A) := min{C(A) : [A] = A}.

We adopt the convention that C(A) = 0 if A is an empty set of certificates and that C(A) = 0 if A = ∅. Note that C(Ω) = 0 as [λ] = Ω. Let f : Ω → Ω0 . If j ∈ Ω0 and x ∈ f −1 (j), we say that c is a j-certificate for x with respect to f if x ∈ [c] ⊆ f −1 (j). The j-certificate complexity of f is Nj (f ) := C(f −1 (j)). We may think of Nj (f ) as the non-deterministic query complexity of the finite decision problem: “Given x ∈ Ω, is f (x) = j?”. If x ∈ f −1 (j), at most Nj (f ) coordinates of x need to be queried to prove this. The following was proven independently in [3], [6], and [12]. Theorem 1.1 If f : {0, 1}n → {0, 1} then

D(f ) ≤ N1 (f )N0 (f ).

(1)

For i ∈ [n], let µi be a discrete probability measure on Ωi . Let (Ω, µ) be the corresponding finite product probability space, with product measure Q µ = ni=1 µi . Without loss of generality, we may assume that for all i and for all ω ∈ Ωi , µi (ω) > 0. Given  ≥ 0, we define the following natural -approximate versions of decision tree and certificate complexity.

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For A ⊆ Ω, we define

C (A) := min{C(A) : µ([A]∆A) ≤ }

where A∆B := (A \ B) ∪ (B \ A). Note that C0 (A) = C(A). For f : Ω → Ω0 and a decision tree T , we say T -approximates f if µ({x ∈ Ω : T (x) 6= f (x)}) ≤ . We define the -approximate decision tree complexity of f ,

D (f ) := min{D(T ) : T is a decision tree that -approximates f }.

If j ∈ Ω0 then the -approximate j-certificate complexity of f is defined to be Nj, (f ) := C (f −1 (j)). Note that D0 (f ) = D(f ) and Nj,0 (f ) = Nj (f ). Our main result is the following generalization1 of Theorem 1.1 to the -approximate setting. Theorem 1.2 Given  > 0, let β0 = −0 where 0 is the unique solution of 0 +(0 /3)3 = . Then for all f : Ω → {0, 1}, with (Ω, µ) a finite product probability space and 1

A preliminary and weaker version of this result appeared in [11].

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for all α1 , α0 ≥ 0, if α = α1 + α0 < β0 then

D (f ) ≤ K N1,α1 (f ) N0,α0 (f ),

(2)

where K = (1 − γ/3)/((γ/3)3 − α) and γ =  − α. Note that K is independent of both the product probability space (Ω, µ) and the function f . Let A1 , . . . , Am be families of certificates. We define

U (A1 , . . . , Am ) := {x ∈ Ω : ∃!j ∈ [m] x ∈ [Aj ]},

the set of uniquely covered assignments and

amb(A1 , . . . , Am ) := Ω \ U (A1 , . . . , Am ),

the set of ambiguous assignments. We define [

δ(A1 , . . . , Am ) := µ(

[Ai ] ∩ [Aj ]).

1≤i 0 let δ0 = (/3)3 . Then for all finite product probability spaces (Ω, µ) and all sets of certificates A, B on Ω, if δ = δ(A, B) < δ0 then

G (A, B) ≤ K C(A)C(B).

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(3)

where K = (1 − /3)/((/3)3 − δ). Tardos made the following conjecture in [12]: Conjecture 1.6 There exists a positive constant c so that the following holds. For all  > 0 there is a δ > 0 such that for any n ≥ 1 and any sets of certificates A, B on {0, 1}n with |amb(A, B)| ≤ δ 2n , G (A, B) ≤ mc where m = max{C(A), C(B)}. Theorem 1.5 may be viewed as only a partial answer to this conjecture; when  → 0 the constant K of Theorem 1.5 goes to infinity. We define certificates a and b to be disjoint, denoted a ∼d b if dom(a) ∩ dom(b) = ∅. If A, B ⊆ Ω let

A ×d B := {(x, y) : ∃ certificates a ∼d b x ∈ [a] ⊆ A, y ∈ [b] ⊆ B}.

The crux of the proof Theorem 1.5 is Theorem 1.7 (The Dual Inequality [8]) Let (Ω, µ) be a finite product probability space and A, B ⊆ Ω. Then


(µ × µ) A ×d B ≤ µ(A ∩ B).

We will use the following version of Theorem 1.7 applied to sets of certificates A and B,  (µ × µ) 

 [

[a] × [b] ≤ µ([A] ∩ [B]).

a∈A,b∈B: a∼d b

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(4)

The dual inequality is so named due to its similarity to Reimer’s inequality (also known as the BKR inequality). If A, B ⊆ Ω let

A ∩d B := {x : ∃ certificates a ∼d b x ∈ [a] ⊆ A, x ∈ [b] ⊆ B}.

Theorem 1.8 (Reimer’s Inequality [10]) Let (Ω, µ) be a finite product probability space and A, B ⊆ Ω. Then

µ(A ∩d B) ≤ µ(A)µ(B) = (µ × µ) (A × B) .

Theorem 1.8 was conjectured in [2] and proved in [10]. Results extending Theorems 1.7 and 1.8 are given in [4]. We prove Theorems 1.2 and 1.5 in section 2. In the same section we show that Theorem 2.1, a generalization of Theorem 1.5 to the case in which there are m > 2 families of certificates, follows immediately as does Theorem 2.2, a generalization of Theorem 1.2 to functions taking on m > 2 values. In the final section we note some possible directions for future work.

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Proofs

If a, b are certificates, we say they are compatible, denoted a ∼c b, if a(i) = b(i) for all i ∈ dom(a)∩dom(b). If a and b are compatible let a\b be the certificate with dom(a\b) := dom(a)\dom(b) and (a\b)(i) := a(i) for all i ∈ dom(a\b). 9

The following proof of Theorem 1.4 (along the lines of [3], [6], and [12]) will form the basis of our proof of Theorem 1.5. Proof of Theorem 1.4.

The algorithm T proceeds in rounds. For each

round i ≥ 0, let qi = qi (x) be the certificate recording all the queries asked by T up to the beginning of round i when running on input x . Let dom(qi ) be the set of queried variables and let qi (j) = xj for each j ∈ dom(qi ). Let Qi = Qi (x) := [qi ]. Let Ai = Ai (x) := {a \ qi : a ∈ A, a ∼c qi } be the set of reduced certificates of A compatible with qi . We define Bi = Bi (x), the reduced certificates of B, analogously. Note that if a ∼c qi , then [a\qi ]∩[qi ] = [a] ∩ [qi ] and so [Ai ] ∩ Qi = [A] ∩ Qi and [Bi ] ∩ Qi = [B] ∩ Qi . We fix a total ordering of the certificates in A and in B and give Ai and Bi the respective induced orderings. Algorithm for the -compatibility game,  = 0.

Set i = 0, q0 = λ, Q0 = Ω, A0 = A and B0 = B. Repeatedly carry out the following steps until the algorithm halts. Begin round i: If Bi = ∅, halt and output T (x) = A. If Ai = ∅, halt and output T (x) = B. Select the first certificate a ∈ Ai and query xj for each j ∈ dom(a). Calculate qi+1 , Qi+1 = [qi+1 ], Ai+1 and Bi+1 accordingly. End round i and increment i to i + 1. 10

If T terminates on x, its output is correct according to the -compatibility game, see Definition 1.3. We now show that T always terminates in at most C(A)C(B) queries. Suppose that in round i neither halting condition holds. Let ai be the first certificate in Ai . Since [A] ∩ [B] = ∅, we have [Ai ] ∩ [Bi ] ∩ Qi = [A] ∩ [B] ∩ Qi = ∅. Since the conditions a ∼c b and [a] ∩ [b] 6= ∅ are equivalent, every b ∈ Bi must be incompatible with every a ∈ Ai . That is, we must have b(j) 6= a(j) for some j 6∈ dom(qi ). Thus querying xj for all j ∈ dom(ai ) yields C(Bi+1 ) ≤ C(Bi ) − 1. Thus we have C(Bi ) = 0 at the beginning of some round i ≤ C(B), i.e. we have Bi = ∅ or Bi = {λ}. In the second case, Ai = ∅ as no certificate can be incompatible with λ. Therefore the algorithm halts after at most C(B) rounds of at most C(A) queries each.

If x ∈ [A] we define

C(A, x) := min{|a| : x |= a ∈ A}.

Using our total ordering on A, we define

min(A, x) := min{a ∈ A : x |= a, |a| = C(A, x)}.

For the definition of min(A, x) recall that A is totally ordered and hence a minimum element may be selected from any non-empty subset. For a ∈ A 11

we define [a]0 := {x ∈ [a] : min(A, x) = a}. Note that {[a]0 : a ∈ A} is a partition of [A]. We define min(B, x) and [b]0 for b ∈ B similarly. Let qi , Qi , Ai , Bi be as in the proof of Theorem 1.4. Proof of Theorem 1.5.

Let A, B,  be given. Let δ = δ(A, B) = µ([A] ∩

[B]). Pick 1 , 2 , 3 > 0 so that 1 + (1 − 1 )(2 + 3 ) ≤ . Let K 0 = 2 (1 − 1 )/(21 2 − δ). If δ < 21 2 , we define a decision tree T , below, that plays the -compatibility game on A and B with D(T ) ≤ (1/3 )K 0 C(A)C(B). Taking 1 = 2 = 3 = /3 gives the statement of the theorem, with K = (1/3 )K 0 = (1 − /3)/((/3)3 − δ). Algorithm for the -compatibility game,  > 0.

Set i = 0, q0 = λ, Q0 = Ω, A0 = A and B0 = B. Repeatedly carry out the following steps until the algorithm halts. Begin round i: Test halting conditions 1: If µ([Ai ]|Qi ) > 1 − 1 or µ([Bi ]|Qi ) < 1 , halt and output A. If µ([Bi ]|Qi ) > 1 − 1 or µ([Ai ]|Qi ) < 1 , halt and output B. Test halting condition 2: If µ([Ai ] ∩ [Bi ]|Qi ) > (1/2 )δ, halt and output either A or B. Test halting Condition 3: If i > (1/3 )K 0 C(B), halt and output either A or B.

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Update: Select the first certificate a ∈ Ai such that  0  µ [b] Qi  ≥ 1/K 0 b∈Bi : b6∼d a 

[

(5)

and query all of its variables. Calculate qi+1 , Qi+1 = [qi+1 ], Ai+1 and Bi+1 accordingly. End round i and increment i to i + 1. The output in halting condition 2 or 3 can be arbitrary without affecting the analysis of the algorithm. We will soon show how the term satisfying (5) can be found in each update step. We only consider the nodes v of T that correspond to the beginning of some round i. Let r(v) = i be the round number of such a vertex v. Given v, let q(v), Q(v), A(v), and B(v), be the status of the variables qi , Qi , Ai , Bi , respectively, at the beginning of the round corresponding to v. We define a sequence {Tj }m j=0 of labeled subtrees of T . Vertices of Tj are either labeled unexpanded or expanded. Let T0 be the tree consisting only of the root u of T where u is labeled as unexpanded. For all j ≥ 0, Tj+1 is created from Tj as follows. Pick uj , an unexpanded leaf of Tj . Let e = Q(uj ), Ae = A(uj ), and Be = B(uj ). Let Tj+1 be the tree qe = q(uj ), Q that results from Tj by labeling uj as expanded and running one round of the e alone. If halting condition h applies, give uj algorithm on the inputs in Q an additional halting label h. Otherwise, let aj ∈ Ae be the certificate found in the updating step and add a new unexpanded child to uj , one for every outcome of querying the variables in aj . The process ends with Tm , a copy 13

of T in which all vertices are expanded and all leaves have halting lablels. We now show that the certificate aj ∈ Ae can be found in the update e > 0, we define the conditional probability measure, µ(X|Q) e := step. If µ(Q) e e for all X ⊆ Ω. This is a product measure; for all y ∈ Ω, µ(X ∩ Q)/µ( Q) e = ν(y) := Qn νk (yk ) where νk = µk for all k ∈ [n] \ dom(e µ(y|Q) q ) and k=1 νk = δxk for k ∈ dom(e q ). Here δz is the probability measure on Ωk with δz (z) = 1. e ν([B]) e ≤ 1 − 1 By halting conditions 1 and 2 on Q, we have 1 ≤ ν([A]), e ∩ [B]) e ≤ (1/2 )δ. Applying the dual inequality, (4), to Ae and Be on and ν([A] (Ω, ν) gives

S S e ∩ [B]) e ≤ (1/2 )δ. (ν × ν)( a∼d b [a]0 × [b]0 ) ≤ (ν × ν)( a∼d b [a] × [b]) ≤ ν([A] Here and in what follows, the indices a and b range over Ae and Be respectively. We also have

S S S (ν × ν)( a6∼d b [a]0 × [b]0 ) + (ν × ν)( a∼d b [a]0 × [b]0 ) = (ν × ν)( a,b [a]0 × [b]0 ) e × [B]) e = ν([A])ν([ e e ≥ 2 = (ν × ν)([A] B]) 1 These inequalities give

P

a

S S ν([a]0 )ν( b:b6∼d a [b]0 ) = (ν × ν)( a6∼d b [a]0 × [b]0 ) ≥ 21 − (1/2 )δ.

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Since λ :=

P

a

e ≤ 1 − 1 , we can divide this last inequality by ν([a]0 ) = ν([A])

λ to get P

a

where wa = ν([a]0 )/λ ≥ 0,

S wa ν( b:b6∼d a [b]0 ) ≥ 1/K 0

wa = 1, and K 0 = 2 (1 − 1 )/(21 2 − δ). Thus S e ≥ 1/K 0 . by convexity there is a term a ∈ Ae satisfying (5), µ( b:b6∼d a [b]0 |Q) P

a

Given x ∈ Ω, 0 ≤ j ≤ m let vj (x) be the leaf of Tj reached when T runs on x. Let rj (x) be the number of rounds taken to reach vj (x). Let qj (x), Qj (x), Aj (x), Bj (x) be the status of the the variables q, Q, A, B at vj (x). Let    C(Bj (x), x) if x ∈ [Bj (x)] `j (x) =   0 otherwise. Suppose vj (x) = uj . Let w = vj+1 (x). If `j (x) > 0, then `j (x) = |b| where b = min(B(x), x). Since x |= b we have x ∈ [b] ∩ Qj+1 (x) 6= ∅, and thus b is not removed from Bj (x) in the updating step. A copy of b, perhaps with some variables deleted, remains in Bj+1 (x). Thus if aj 6∼d b, we have e that `j+1 (x) ≤ `j (x) − 1. But by the choice of aj , the fraction of x ∈ Q S e ≥ 1/K 0 , and we have satisfy this condition is large, µ( b:b6∼d aj [b]0 |Q) e ≤ E[`j (x)|Q] e − 1/K 0 , E[`j+1 (x)|Q]

(6)

where the expectation is taken with respect to µ. Let E = {x ∈ U (A, B) : T makes an error on x}. We show µ(E) < .

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For 1 ≤ h ≤ 3 let

Hh := {x ∈ Ω : T halts on x due to halting condition h}, Eh := E ∩ Hh

We will show µ(E1 ) < 1 µ(H1 ), µ(H2 ) < 2 , and µ(H3 ) ≤ 3 so that

µ(E) < 1 µ(H1 ) + µ(H2 ) + µ(H3 ) ≤ 1 + (1 − 1 )(2 + 3 ) ≤ 

where the last inequality holds by the choice of the i . Note that T ’s output when halting conditions 2 or 3 apply is irrelevant to this analysis. For any Tj , {Qj (v) : v a leaf of T } is a partition of Ω. We have µ(E1 ) = P

Q

µ(E1 |Q(v))µ(Q(v)) where the sum is over all leaves v of Tm = T with

halting label 1. For these leaves we have µ(E1 |Q) < 1 and thus µ(E1 ) < 1 µ(H1 ). (If µ([Ai ]|Q) > 1 − 1 , output due to halting condition 1 is A and this will be in error iff x ∈ [Bi ] \ [Ai ], but µ([Bi ] \ [Ai ]|Q) ≤ 1 . The other cases follow similarly.) Clearly δ = µ([A] ∩ [B]) ≥

P

v

µ([A(v)] ∩ [B(v)]|Q(v))µ(Q(v)) where the

sum is over leaves v with halting label 2. Since µ([A] ∩ [B]|Q) > (1/2 )δ for such v, µ(H2 ) < 2 . It remains to show µ(H3 ) ≤ 3 . We return to the step passing from Tj to Tj+1 . Since we have `j+1 (x) ≤ e we may use (6) to get `j (x) for all x 6∈ Q,

E`j (x) =

X

e Q) e E[`j (x)|Q]µ(Q) + E[`j (x)|Q]µ(

e Q6=Q

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≥

X

e e + (1/K 0 )µ(Q) e E[`j+1 (x)|Q]µ(Q) + E[`j+1 (x)|Q)]µ( Q)

e Q6=Q

= E`j+1 (x) + (1/K 0 )µ(Q(v)) where the sums are over all Q = Q(v) where v is a leaf of Tj . Thus we have e Elj (x) − Elj+1 (x) ≥ (1/K 0 )µ(Q).

e and rj+1 (x) = rj (x) otherwise, we Since rj+1 (x) = rj (x) + 1 for x ∈ Q e Thus have Erj+1 (x) − Erj (x) = µ(Q). E`j (x) − E`j+1 (x) ≥ (1/K 0 )(Erj+1 (x) − Erj (x)).

Summing over 0 ≤ j ≤ m − 1 gives

E`0 (x) − E`m (x) ≥ (1/K 0 )(Erm (x) − Er0 (x)).

Since r0 (x) = 0 and `0 (x) ≤ C(B), K 0 C(B) ≥ Erm (x). By Markov’s inequality, µ(H3 ) ≤ µ({x : rm (x) > (1/3 )K 0 C(B)}) ≤ µ({x : rm (x) ≥ (1/3 )Erm (x)}) ≤ 3 . By halting condition 3, T performs ≤ (1/3 )K 0 C(B) rounds, each requiring at most C(A) queries. Thus D(T ) ≤ (1/3 )K 0 C(A)C(B) as claimed.

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Theorem 1.5 can be generalized to three or more families of certificates. Theorem 2.1 Given  > 0 and m ≥ 2, let δ0 = (/(3b))3 where b = dlog2 me. Then for all finite product probability spaces (Ω, µ) and all sets of certificates A0 , . . . , Am−1 on Ω, if δ = δ(A0 , . . . , Am−1 ) < δ0 then

G (A0 , . . . , Am−1 ) ≤ bK M1 M2

(7)

where K = (1 − /(3b))/((/(3b))3 − δ) and where M1 and M2 are the largest and second largest members of {C(Aj ) : 0 ≤ j < m}, respectively. Proof:

The case m = 2 is Theorem 1.5. Suppose m ≥ 3 and let b =

dlog2 me. For 0 ≤ i ≤ b − 1 and for j = 0 or 1 let Si,j be the set of all integers S k ∈ {0, . . . , m − 1} whose ith bit in binary is (k)i = j. Let Bi,j = k∈Si,j Ak . Since δ(Bi,0 , Bi,1 ) ≤ δ(A1 , . . . , Am ) = δ < δ0 = (/(3b))3 ,Theorem 1.5 implies we can construct a decision tree Ti with D(Ti ) ≤ KM1 M2 , playing the (/b)compatibility game on Bi,1 and Bi,0 . Let T be the decision tree formed by running each Ti and giving output T (x) = Ak if and only if Ti (x) = Bi,(k)i for all 0 ≤ i ≤ b − 1. Let E, E0 , . . . , Eb−1 be the error sets of T, T0 , . . . , Tb−1 respectively. To S prove the theorem, it suffices to show E ⊆ b−1 i=0 Ei . If x ∈ E, then x ∈ U (A0 , . . . , Am−1 ) and there exists k 6= k 0 such that x ∈ [Ak ] while T (x) = 18

Ak0 . Let 0 ≤ i ≤ b − 1 such that (k)i 6= (k 0 )i . If (k)i = 1, x ∈ [Bi,1 ] \ [Bi,0 ] and Ti (x) = Bi,0 , otherwise x ∈ [Bi,0 ] \ [Bi,1 ] and Ti (x) = Bi,1 . In either case, x ∈ Ei .

Theorem 1.2 is the case m = 2 of the following Theorem 2.2 Given  > 0 and m ≥ 2, let β0 =  − 0 where 0 is the unique solution of 0 + (0 /(3b))3 =  and b = dlog2 me. Then for all f : Ω → [m], with (Ω, µ) a P finite product probability space and for all α1 , . . . , αm ≥ 0, if α = αj < β0 then D (f ) ≤ bK M1 M2

(8)

where K = (1 − γ/(3b))/((γ/(3b))3 − α), γ =  − α, and where M1 and M2 are the largest and second largest values of {Nj,αj (f ) : j ∈ Ω0 }, respectively. Note the constant K in Theorem 2.2 is independent of (Ω, µ) and f . Let A1 , . . . , Am be sets of certificates with C(Ai ) = Ni,αi (f ) and P µ(EAi ) ≤ αi for all i ∈ [m] where EAi = [Ai ]∆f −1 (i). Let α = αi . We S claim amb(A1 , . . . , Am ) ⊆ i∈[m] EAi . Note that if x ∈ amb(A1 , . . . , Am )

Proof:

and f (x) = i then either x 6∈ [Ai ] and so x ∈ EAi or x ∈ [Ai ] ∩ [Aj ] for some j 6= i and so x ∈ EAj . Thus δ(A1 , . . . , Am ) ≤ µ(amb(A1 , . . . , Am )) ≤ α. By Theorem 2.1, if α < (γ/(3b))3 there is a decision tree T that plays the

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γ-approximate compatibility game on A1 , . . . , Am with query complexity

Gγ (A1 , . . . , Am ) ≤ bK M1 M2

(9)

where K = (1 − γ/(3b))/((γ/(3b))3 − α). We change the definition of T slightly so that T maps Ω to [m] instead of {A1 , . . . , Am } and T (x) = Ai is instead represented as T (x) = i. Let EU = {x ∈ U (A1 , . . . , Am ) : T makes an error on x}. By construction of T , µ(EU ) ≤ γ. It can be seen by a straightforward checking of cases, that {x ∈ Ω : T (x) 6= f (x)} ⊆ EU ∪

[

EAi .

i∈[m]

Since amb(A1 , . . . , Am ) ⊆

S

i∈[m]

EAi , it suffices to show the preceding inclu-

sion for all x ∈ U (A1 , . . . , Am ). Suppose T (x) = i and f (x) = j where i 6= j. If x ∈ [Ak ] with k 6= i then T (x) = i is an error and x ∈ EU . If x ∈ [Ai ] then x ∈ EAj . Thus µ(T (x) 6= f (x)) ≤ γ + α and so Dγ+α (f ) ≤ Gγ (A1 . . . , Am ). Since  − γ = α < β0 , γ >  − β0 = 0 , and thus γ + (γ/(3b))3 > . This implies α =  − γ < (γ/(3b))3 and

D (f ) = Dγ+α (f ) ≤ Gγ (A1 , . . . , Am ).

Combining this with (9) gives the result.

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3

Conclusion and Notes

It would be interesting to resolve Conjecture 1.6. Unfortunately the analysis of Theorem 1.5 is not sufficient as the constant K there goes to infinity as  → 0. It is natural to suppose Conjecture 3.1 Theorem 1.2 holds for all f and for all α <  and Theorem 1.5 holds for all A, B with δ(A, B) < . In [12] Tardos noted that the following result from communication complexity is analogous to Theorem 1.1. Theorem 3.2 [1, 5] For all X, Y finite and for all f : X × Y → {0, 1},

Dcc (f ) ≤ (N1cc (f ) + 1)(N0cc (f ) + 1).

Here, Dcc (f ) is the minimum depth of a two-party communication protocol for f .

For j ∈ {0, 1}, Njcc (f ) is the non-deterministic communi-

cation complexity of the decision problem, “Is f (x, y) = j?”. We have Njcc (f ) = dlog2 (Cjcc (f ))e where Cjcc (f ) is the minimum number of rectangles X 0 × Y 0 ⊆ f −1 (j) necessary to cover f −1 (j). See [9] for more information. It is natural to define approximate versions of these measures. Let Dcc (f ) be the minimum depth of a communication protocol for f that is correct on 21

all but an -fraction of inputs in X × Y . This measure is used in distribucc cc tional computational complexity. We define Nj,δ (f ) = dlog2 (Cj,δ (f ))e, where cc Cj,δ (f ) is the minimum number of rectangles needed so that their union forms

a set S with |S∆f −1 (j)| ≤ δ|X × Y |. It is natural to conjecture the following approximate version of Theorem 3.2: Conjecture 3.3 For all  > 0 there exists a β0 > 0 such that for all f : X × Y → {0, 1} with X, Y finite and for all α0 , α1 ≥ 0, if α = α0 + α1 < β0 then cc cc Dcc (f ) ≤ K N1,α (f ) N0,α (f ), 1 0

where K depends only on  and α. A simpler equivalent formulation is given by

cc cc ∀ > 0 ∃α > 0 ∃K > 0 ∀f Dcc (f ) ≤ K N1,α (f ) N0,α (f ).

Let g be -close to f if f 6= g on at most a -fraction of inputs. Let N cc (f ) = max{N1cc (f ), N0cc (f )}. If we define Ncc (f ) := min{N cc (g) : g is -close to f } then trivially, Dcc (f ) = O((Ncc (f ))2 ). However, the above question, in which the rectangle covers are decoupled (i.e. are viewed as not necessarily coming from a single function g) does not appear to have been considered in the literature. 22

Acknowledgments I would like to thank Mike Saks, Steven Rudich, and Russell Impagliazzo for many helpful discussions. In particular, the technique of “expansions” used in the proof of Theorem 1.5 was modeled after a technique developed by Rudich and Impagliazzo for related problems [7]. I would also like to thank the two referees for their insightful and helpful comments.

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