Approximating the Influence of a monotone Boolean function in O(\sqrt{n}) query complexity

0.16 MB·English

by  Dana Ron

Approximating the influence of a monotone Boolean function in
O n
(√ ) query complexity
DanaRon RonittRubinfeld MuliSafra OmriWeinstein
∗ † ‡ §
1 January 28,2011
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0
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Abstract
J

7 The Total Influence (Average Sensitivity) of a discrete function is one of its fundamental mea-
2 sures. We study the problem of approximating the total influence of a monotone Boolean function

f : 0,1 n 0,1 , which we denote by I[f]. We present a randomized algorithm that ap-
] { } → { }
S proximates the influence of such functions to within a multiplicative factor of (1 ǫ) by performing
±
D O √nlognpoly(1/ǫ) queries.WealsoprovealowerboundofΩ √n onthequerycomplexity
I[f] lognI[f]
. ·
s of(cid:16)anyconstant-factor(cid:17)approximationalgorithmforthisproblem(w(cid:16)hichhold(cid:17)sforI[f] = Ω(1)),hence
c
showingthatouralgorithmisalmostoptimalintermsofitsdependenceonn. Forgeneralfunctionswe
[
givealowerboundofΩ n ,whichmatchesthecomplexityofasimplesamplingalgorithm.
1 I[f]
v (cid:16) (cid:17)
5 Keywords: influenceofaBooleanfunction, sublinear approximation algorithms, random walks.
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5
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:
v
i
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r
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∗SchoolofElectricalEngineeringatTelAvivUniversity,[email protected]. ThisworkwassupportedbytheIsrael
ScienceFoundation(grantnumber246/08).
†CSAILat MIT, and theBlavatnik School of Computer Science at TelAviv University, [email protected]. This
work was supported by NSF grants 0732334 and 0728645, Marie Curie Reintegration grant PIRG03-GA-2008-231077 and the
IsraelScienceFoundationgrantnos.1147/09and1675/09
‡BlavatnikSchoolofComputerScienceatTelAvivUniversity,[email protected].
§ComputerScienceDepartment,PrincetonUniversity,[email protected].
1
1 Introduction
The influence of a function, first introduced by Ben-Or and Linial [2] in the context of “collective coin-
flipping”, captures the notion of the sensitivity of a multivariate function. More precisely, for a Boolean
function f : 0,1 n 0,1 , the individual influence of coordinate i on f is defined as I [f] d=ef
i
{ } → { }
Pr [f(x) = f(x( i))], where x is selected uniformly1 in 0,1 n and x( i) denotes x with the ith
x 0,1 n ⊕ ⊕
∈{ } 6 { }
bit flipped. The total influence of a Boolean function f (which we simply refer to as the influence of f)is
I[f]= I [f].
i i
The study of the influence of a function and its individual influences (distribution) has been the focus
P
of many papers ( [2, 21, 7, 16, 34, 8, 35, 14, 6, 15, 28, 11] to mention a few – for a survey see [17]).
The influence of functions has played a central role in several areas of computer science. In particular,
this is true for distributed computing (e.g., [2, 21]), hardness of approximation (e.g., [12, 22]), learning
theory(e.g.,[18,9,29,30,10])2 andproperty testing (e.g.,[13,4,5,26,31]). Thenotionofinfluence also
arises naturally in the context of probability theory (e.g., [32, 33, 3]), game theory (e.g., [24]), reliability
theory(e.g.,[23]),aswellastheoretical economicsandpolitical science(e.g.,[1,19,20]).
Giventhattheinfluenceissuchabasicmeasureoffunctionsanditplaysanimportantroleinmanyareas,
webelieve itisofinterest tostudy thealgorithmic question ofapproximating theinfluence ofafunction as
efficiently as possible, that is by querying the function on as few inputs as possible. Specifically, the need
for an efficient approximation for a function’s influence might arise in the design of sublinear algorithms,
andinparticular propertytestingalgorithms.
As we show, one cannot improve on a standard sampling argument for the problem of estimating the
influence of a general Boolean function, which requires Ω( n ) queries to the function, for any constant
I[f]
multiplicative estimation factor.3 This fact justifies the study of subclasses of Boolean functions, among
whichthefamily ofmonotone functions isaverynatural and central one. Indeed, weshow thatthe special
structure of monotone functions implies a useful behavior of their influence, and thus the computational
problem ofapproximating theinfluenceofsuchfunctions becomessignificantly easier.
1.1 Ourresults andtechniques
We present a randomized algorithm that approximates the influence of a monotone Boolean function to
within any multiplicative factor of (1 ǫ) in O √nlognpoly(1/ǫ) expected query complexity. We also
± I[f]
prove an almost matching lower bound of Ω (cid:16)√n on the qu(cid:17)ery complexity of any constant-factor
lognI[f]
·
approximation algorithm forthisproblem (whi(cid:16)chholdsf(cid:17)orI[f] = Ω(1)).
Asnoted above, the influence ofafunction can be approximated bysampling random edges (i.e., pairs
(x,x( i))thatdifferonasingle coordinate) fromthe 0,1 n lattice. Arandom edgehasprobability I[f] to
⊕ { } n
beinfluential (i.e, satisfy f(x) = f(x( i))), soastandard sampling argument implies that itsuffices toask
⊕
6
O( n poly(1/ǫ))queriesinordertoapproximate thisprobability towithin(1 ǫ).4
I[f] ±
1Theinfluencecanbedefinedwithrespecttootherprobabilityspaces(aswellasfornon-Booleanfunctions),butwefocuson
theabovedefinition.
2Herewereferencedseveralworksinwhichtheinfluenceappearsexplicitly.Theinfluenceofvariablesplaysanimplicitrolein
manylearningalgorithms,andinparticularthosethatbuildonFourieranalysis,beginningwith[25].
3Ifonewantsanadditiveerrorofǫ,thenΩ((n/ǫ)2)queriesarenecessary(whentheinfluenceislarge)[27].
4We also note that in the case of monotone functions, the total influence equals twice the sum of the Fourier coefficients
1
In order to achieve better query complexity, we would like to increase the probability of hitting an
influential edge in a single trial. The algorithm we present captures this intuition, by taking random walks
downthe 0,1 n lattice5,andthenaveraging thetotalnumber ofinfluential edges encountered inallwalks
{ }
over the number of walks taken. The crucial observation on which the algorithm relies, isthat amonotone
functioncanhaveatmostoneinfluentialedgeinasinglepath,andthusitissufficienttoqueryonlythestart
andendpointsofthewalktodetermine whetheranyinfluentialedgewastraversed.
Before continuing the technical discussion concerning the algorithm and its analysis, wemake the fol-
lowingmoreconceptual note. Random walkshave numerous applications inComputer Science astheyare
an important tool for mixing and sampling almost uniformly. In our context, where the walk is performed
onthedomainofanunknownfunction, itisusedforadifferentpurpose. Namely,byqueryingonlythetwo
endpoints ofarandomwalk(starting fromauniformlysampledelement)we(roughly)simulatetheprocess
oftakingamuchlargersampleofelements.
The main issue that remains is determining the length of the walk, which we denote by w. Let p (f)
w
denotetheprobabilitythatawalkoflengthw(downthelatticeandfromauniformlyselectedstartingpoint)
passesthrough someinfluential edge.6 Weareinterested inanalyzing howp (f)increases asafunction of
w
w. WeshowthatforwthatisO(ǫ n/logn),thevalueofp (f)increasesalmostlinearlywithw. Namely,
w
itis(1 ǫ) w I[f]. Thus,bytakingwtobeΘ(ǫ n/logn)wegetanimprovementbyafactorofroughly
± · n · p
√n on the basic sampling algorithm. We note though that by taking w to be larger we cannot ensure in
p
generalthesamebehavior ofp (f)asafunction ofw andI[f],sincethebehavior mightvarysignificantly
w
depending onf.
The way we prove the aforementioned dependence of p (f)on w is roughly as follows. For any edge
w
e in the Boolean lattice, let p (e) denote the probability that a walk of length w (as defined above) passes
w
through e. By the observation made previously, that a monotone function can have at most one influential
edgeinagivenpath,p (f)isthesumofp (e),takenoveralledgesethatareinfluentialwithrespecttof.
w w
For our purposes it is important that p (e) be roughly the same for almost all edges. Otherwise, different
w
functions that have the same number of influential edges, and hence the same influence I[f], but whose
influential edges are distributed differently in the Boolean lattice, would give different values for p (f).
w
We show that for w = O(ǫ n/logn), the value of p (e) increases almost linearly with w for all but a
w
negligible fraction of the influential edges (where ‘negligible’ is with respect to I[f]). This implies that
p
p (f)growsroughlylinearly inwforw = O(ǫ n/logn).
w
Todemonstrate the benefit of taking walks oflength O(√n), let us consider the classic example of the
p
Majority function on n variables. Here, all influential edges are concentrated in the exact middle levels of
the lattice (i.e, all of them are of the form (x,x( i)) where the Hamming weight of x is n/2 or n/2 ).
⊕
⌊ ⌋ ⌈ ⌉
Theprobability, p (e),ofawalkoflengthwpassingthroughaninfluentialedgeeissimplytheprobability
w
of starting the walk at distance at most w above the threshold n/2. Thus, taking longer walks allows us,
so to speak, to start our walk from a higher point in the lattice, and still hit an influential edge. Since the
probability of a uniformly chosen point to fall in each one of the the first √n levels above the middle is
roughly the same, the probability of hitting an influential edge in that case indeed grows roughly linearly
that correspond to singleton sets {i}, i ∈ {1,...,n}. Therefore, it is possible to approximate the influence of a function by
approximatingthissum,whichequals 21n·Pni=1(cid:16)Px∈{0,1}n:xi=1f(x)−Px∈{0,1}n:xi=0f(x)(cid:17).However,thedirectsampling
approachforsuchanapproximationagainrequiresΩ(n/I[f])samples.
5Thatis,startingfromarandomlyselectedpointin{0,1}n,ateachstep,ifthecurrentpointisx,weuniformlyselectanindex
isuchthatxi =1andcontinuethewalktox(⊕i).
6For technical reasons we actually consider a slightly different measure than p (f), but we ignore this technicality in the
w
introduction.
2
inthesizeofthewalk. Nevertheless, taking walksoflength whichsignificantly exceeds O(√n)(say, even
Ω( n log(n)))would add negligible contribution to that probability (as this contribution is equivalent to
·
theprobabilityofauniformlychosenpointtodeviateΩ( n log(n))levelsfromthemiddlelevel)andthus
p ·
thelineardependence onthelengthofthewalkisnolongerpreserved.
p
2 Preliminaries
Intheintroductionwedefinedtheinfluenceofafunctionasthesum,overallitsvariables,oftheirindividual
influence. Anequivalent definition is that the influence ofa function f isthe expected number ofsensitive
coordinates forarandominputx 0,1 n (thatis,thosecoordinates iforwhichf(x)= f(x( i))).
⊕
∈ { } 6
It will occasionally be convenient to view f as a 2-coloring of the Boolean lattice. Under this setting,
any “bi-chromatic” edge, i.e, an edge (x,x( i)) such that f(x) = f(x( i)), will be called an influential
⊕ ⊕
6
edge. ThenumberofinfluentialedgesofaBooleanfunction f is2n 1 I[f].7
−
·
We consider the standard partial order ‘ ’ over the (n-dimensional) Boolean lattice. Namely, for x =
≺
(x ,...,x ),y = (y ,...,y ) 0,1 n,weusethenotationx ytomeanthatx y forevery1 i n,
1 n 1 n i i
∈ { } ≺ ≤ ≤ ≤
and x < y for some 1 i n. A Boolean function f : 0,1 n 0,1 is said to be monotone
i i
≤ ≤ { } → { }
if f(x) f(y) for all x y. A well known isoperimetric inequality implies that any monotone Boolean
≤ ≺
functionsatisfiesI[f] = O(√n)(see[16]foraproof). ThisboundistightforthenotableMajorityfunction.
In this paper we deal mainly with monotone Boolean functions that have at least constant Influence
(i.e, I[f] c, for some c 0), since the computational problem we study arises more naturally when the
≥ ≥
function has some significant sensitivity. As shown in [21], the influence of a function is lower bounded
by 4 Pr[f = 1] Pr[f = 0], and so our analysis holds in particular for functions that are not too biased
· ·
(relatively balanced).
Notations. We use the notation f(n) = O˜(g(n)) if f(n) = O(g(n)polylog(g(n))). Similarly, f(n) =
Ω˜(g(n))iff(n)= Ω(g(n)/polylog(g(n))).
3 The Algorithm
Asnoted intheintroduction, wecaneasilygeta(1 ǫ)-factor estimateoftheinfluencewithhighconstant
±
probability by uniformly sampling Θ n ǫ 2 pairs (x,x( i)) (edges in the Boolean lattice), querying
I[f] · − ⊕
thefunctiononthesepairs,andconside(cid:16)ringthefra(cid:17)ctionofinfluentialedgesobservedinthesample. Werefer
tothisasthedirectsamplingapproach. However,sinceweareinterestedinanalgorithmwhosecomplexity
is √n poly(1/ǫ) we take a different approach. To be precise, the algorithm we describe works for ǫ that
I[f] ·
is above a certain threshold (of the order of logn/n). However, if ǫ is smaller, then n ǫ 2 is upper
I[f] · −
boundedby √n poly(1/ǫ),andwecantakepthedirectsamplingapproach. Thusweassumefromthispoint
I[f]·
onthatǫ = ω( logn/n).
Asdiscussed intheintroduction, instead ofconsidering neighboring pairs, (x,x( i)),weconsider pairs
p ⊕
(v,u)suchthatv uandthereisapathdownthelatticeoflengthroughlyǫ√nbetweenv andu. Observe
≻
thatsincethefunctionf ismonotone, ifthepath(downthelattice)fromv toucontains aninfluentialedge,
7Toverifythis,observethatwhenpartitioningtheBooleanlatticeintotwosetswithrespecttoacoordinatei,weendupwith
2n−1verticesineachset. Theindividualinfluenceofvariablei,Ii[f],isthefractionofthe“bi-chromatic”edgesamongalledges
crossingthecut.SinceI[f]=Pni=1Ii[f]wegetthatthetotalnumberofinfluentialedgesis2n−1·I[f].
3
thenf(v) = f(u),andfurthermore, anysuchpathcancontainatmostoneinfluentialedge. Theintuitionis
6
that since we “can’t afford” to detect influential edges directly, we raise our probability of detecting edges
byconsidering longerpaths.
In our analysis we show that this intuition can be formalized so as to establish the correctness of the
algorithm. We stress that when considering a path, the algorithm only queries its endpoints, so that it
“doesn’t pay” for the length of the path. The precise details of the algorithm are given in Figure 1. When
wesaythatwetakeawalkofacertain lengthw downtheBooleanlatticewithacut-off atacertainlevelℓ,
wemeanthatwestopthewalk(beforetakingallw steps)ifwereachapointinlevelℓ(i.e.,withHamming
weightℓ).
Note that m, the number of walks taken, is arandom variable. Namely, the algorithm continues taking
new walks until the number of “successful” walks (that is, walks that pass through an influential edge)
reaches a certain threshold, which is denoted by t. The reason for doing this, rather than deterministically
settingthenumberofwalksandconsideringtherandomvariablewhichisthenumberofsuccessfulwalks,is
thatthelatterapproach requires toknowalowerbound ontheinfluenceoff. Whileitispossible tosearch
for such a lower bound (by working iteratively in phases and decreasing the lower bound on the influence
betweenphases)ourapproach yieldsasomewhatsimpleralgorithm.
Algorithm1: ApproximatingtheInfluence(givenǫ,δ andoracleaccesstof)
1. Setǫ˜= ǫ/4,w = ˜ǫ√n ,s = 1 2nlog(2n),andt = 96ln(δ2).
16q2log(2ǫ˜n) ∗ 2q ˜ǫ ǫ2
2. Initialize α 0,m 0,andIˆ 0.
← ← ←
3. Repeatthefollowing untilα = t:
(a) Performarandomwalkoflengthwdownthe 0,1 n latticefromauniformlychosenpoint
{ }
v withacut-offatn/2 s 1,andletudenotetheendpoint ofthewalk.
∗
− −
(b) Iff(u)= f(v)thenα α+1.
6 ←−
(c) m m+1
←
4. Iˆ n t
← w · m
5. ReturnIˆ.
Figure1: Thealgorithmforapproximatingtheinfluenceofafunctionf.
In what follows we assume for simplicity that I[f] 1. As we discuss subsequently, this assumption
≥
can be easily replaced by I[f] c for any constant c > 0, or even I[f] n c, by performing a slight
−
≥ ≥
modificationinthesettingoftheparameters ofthealgorithm.
Theorem3.1 For every monotone function f : 0,1 n 0,1 such that I[f] 1, and for every δ > 0
{ } → { } ≥
andǫ = ω( logn/n),withprobability atleast1 δ,theoutput, Iˆ,ofAlgorithm1satisfies:
−
p (1 ǫ) I[f] Iˆ (1+ǫ) I[f].
− · ≤ ≤ ·
Furthermore, with probability at least 1 δ, the number of queries performed by the algorithm is
−
O log(1/δ) √nlog(n/ǫ) .
ǫ3 · I[f]
(cid:16) (cid:17)
4
We note that the (probabilistic) bound on the number of queries performed by the algorithm implies that
the expected query complexity of the algorithm is O log(1/δ) √nlog(n/ǫ) . Furthermore, the probability
ǫ3 · I[f]
that the algorithm performs a number of queries that(cid:16)is more than k time(cid:17)s the expected value decreases
exponentially withk.
Thenextdefinitioniscentraltoouranalysis.
Definition1 Fora(monotone)Booleanfunctionf andintegerswands ,letp (f)denotetheprobability
∗ w,s
∗
thatarandomwalkoflengthwdowntheBooleanlattice,fromauniformlyselectedpointandwithacut-off
atn/2 s 1,startsfromf(v) = 1andreachesf(u)= 0.
∗
− −
Given the definition of p (f), wenext state and prove the main lemma onwhich the proof of Theo-
w,s
∗
rem3.1isbased.
Lemma3.2 Let f satisfy I[f] 1, let ǫ > 0 satisfy ǫ > 8q2log(8ǫn), and denote ˜ǫ = ǫ/4. For any
≥ √n
w ˜ǫ√n andfors = 1√n 2log(2n)wehavethat
≤ 16 2log( 2n ) ∗ 2 · ˜ǫ
q ǫ˜I[f] q
w w
(1 ǫ/2) I[f] p (f) (1+ǫ/2) I[f].
w,s
− · n · ≤ ∗ ≤ · n ·
Proof: For a point y 0,1 n, let h(y) denote its Hamming weight (which we also refer to as the level
∈ { }
in the Boolean lattice that it belongs to). By the choice of s = 1√n 2log(2n), and since I[f] 1,
∗ 2 ˜ǫ ≥
the number of points y for which h(y) n/2+s or h(y) n/2 sq, is upper bounded by 2n ˜ǫI[f].
≥ ∗ ≤ − ∗ · n
Eachsuchpoint y isincident tonedges, and eachedgehastwoendpoints. Itfollowsthatthere areatmost
2n 1 ǫ˜I[f] edges (y,x) for which h(y) n/2+s or h(y) n/2 s . Recall that an influential edge
− ∗ ∗
· ≥ ≤ −
(y,x)forh(y) = h(x)+1,isanedgethatsatisfiesf(y) = 1andf(x) = 0. Lete (f)denote thenumber
s
∗
ofinfluentialedges(y,x)suchthatn/2 s h(x),h(y) n/2+s . Sincethetotalnumberofinfluential
∗ ∗
− ≤ ≤
edgesis2n 1I[f],wehavethat
−
(1 ˜ǫ) 2n 1I[f] e (f) 2n 1I[f]. (1)
− s −
− · ≤ ∗ ≤
Consider any influential edge (y,x) where h(y) = ℓ and ℓ n/2 s . Weare interested in obtaining
∗
≥ −
bounds on the probability that a random walk of length w (where w ǫ˜√n ) down the lattice,
≤ 16 2log( 2n )
q ǫ˜I[f]
starting from a uniformly selected point v 0,1 n, and with a cut-off at n/2 s 1, passes through
∗
∈ { } − −
(y,x). First,thereistheeventthatv = y andtheedge(y,x)wasselected inthefirststepofthewalk. This
event occurs with probability 2 n 1. Next there is the event that v is at distance 1 from y (and above it,
− · ℓ
thatis,h(v) = h(y)+1 = ℓ+1),andtheedges(v,y)and(y,x)areselected. Thisoccurswithprobability
2 n (n ℓ) 1 1. In general, for every 1 i w 1 we have (n ℓ) (n ℓ i +1) pairs
− · − · ℓ+1 · ℓ ≤ ≤ − − ··· − −
(v,P)wherev y and w(v) = ℓ+i,andwhereP isapathdownthelattice fromv toy. Theprobability
≻
of selecting v as the starting vertex is 2 n and the probability of taking the path P from v is 1 .
− (ℓ+i) (ℓ+1)
Therefore, theprobability thattherandom walkpassesthrough (y,x)is: ···
w 1 w 1i 1
1 − (n ℓ) (n ℓ i+1) 1 − − n ℓ j
2−n 1+ − ··· − − = 2−n 1+ − − . (2)
· ℓ · (ℓ+i) (ℓ+1) · ℓ  ℓ+i j 
!
i=1 ··· i=1 j=0 −
X XY
 
5
Letℓ = n/2+s(wheresmaybenegative),anddenote τ(ℓ,i,j) d=ef n−ℓ−j. Then
ℓ+i j
−
n/2 s j 2s+i
τ(ℓ,i,j) = − − = 1 . (3)
n/2+s+i j − n/2+s+i j
− −
Considerfirstthecasethatℓ n/2,i.eℓ = n/2+s(s 0). Inthatcaseitisclearthatτ(ℓ,i,j) 1(since
j ≤ i),so ij−=10τ(ℓ,i,j)isu≥pperbounded by1. Inord≥ertolowerbound ij−=10τ(ℓ,i,j), wenote≤that
Q Q
2s+w 2(2s+w)
τ(ℓ,i,j) 1 = 1 . (4)
≥ − n/2 − n
Thus,fors s wehave
∗
≤
i 1 i 1
− − 2(2s+w)
τ(ℓ,i,j) 1
≥ − n
j=0 j=0(cid:18) (cid:19)
Y Y
w
2(2s+w)
1 (sincei w)
≥ − n ≤
(cid:18) (cid:19)
2(2s+w)w
1
≥ − n
6s w
∗
1 (2s+w 3s sinces s andw s )
∗ ∗ ∗
≥ − n ≥ ≤ ≤
3ǫ˜
= 1 (bythedefinitionsofs andw)
∗
− 16
1 ˜ǫ/2 . (5)
≥ −
Therefore, wehavethatforn/2 ℓ n/2+s ,
∗
≤ ≤
i 1
− n ℓ j
1 ǫ˜/2 − − 1 , (6)
− ≤ ℓ+i j ≤
j=0 −
Y
andforℓ > n/2+s itholdsthat
∗
i 1
− n ℓ j
− − 1. (7)
ℓ+i j ≤
j=0 −
Y
Weturntothecasewheren/2 s ℓ < n/2. Herewehave
∗
− ≤
2s i 2w 4w
τ(ℓ,i,j) = 1+ − 1 1 (8)
n/2 s+i j ≥ − n 2w ≥ − n
− − −
wherethelastinequality followsfromthefactthatw < n/4. Thus,
2
i−1 4w w 4w2 4 ǫ˜√n
τ(ℓ,i,j) 1 1 = 1 > 1 ǫ˜2/2 > 1 ǫ˜/2 . (9)
jY=0 ≥ (cid:18) − n (cid:19) ≥ − n −n·16 2log(ǫ˜I2[nf]) − −
 q 
6
Ontheotherhand,
2s i 2s 8s
∗
τ(ℓ,i,j) = 1+ − 1+ 1+ , (10)
n/2 s+i j ≤ n/2 s ≤ n
− − −
wherethelastinequality holdssincen 2s. Thus,wehave
≥
i 1 w
− 8s∗ 16s∗w
τ(ℓ,i,j) 1+ 1+ = 1+ǫ˜/2 . (11)
≤ n ≤ n
j=0 (cid:18) (cid:19)
Y
wherethesecond inequality followsfromtheinequality (1+α)k 1+2αk whichholdsforα < 1/(2k);
≤
Indeed, inourcase8s /n 1/(2w) (thisisequivalent tow n/16s whichholds givenoursetting ofs
∗ ∗ ∗
≤ ≤
andtheupperboundonw).
Wethereforehavethatforn/2 s ℓ < n/2,
∗
− ≤
i 1
− n ℓ j
1 ǫ˜/2 − − 1+ǫ˜/2 . (12)
− ≤ ℓ+i j ≤
j=0 −
Y
CombiningEquations(6)and(12),wehavethatforn/2 s ℓ n/2+s ,
∗ ∗
− ≤ ≤
i 1
− n ℓ j
1 ǫ˜/2 − − 1+ǫ˜/2 . (13)
− ≤ ℓ+i j ≤
j=0 −
Y
Now,weareinterestedinsumminguptheprobability,overallrandomwalks,thatthewalkpassesthroughan
influential edge. Sincethefunction ismonotone, everyrandom walkpasses through atmostoneinfluential
edge,sothesetsofrandomwalksthatcorrespondtodifferentinfluentialedgesaredisjoint(thatis,theevent
thatawalkpassesthroughaninfluentialedge(y,x)isdisjointfromtheeventthatitpassesthroughanother
influential edge (y ,x)). Since the edges that contribute top (f)are all from levels ℓ n/2 s (and
′ ′ w,s ∗
∗ ≥ −
sincethereare2n 1I[f]influentialedgesintotal),byEquations(2),(7)and(13)wehave
−
w 1
1 −
p (f) 2n 1I[f]2 n 1+ (1+ǫ˜/2) (14)
w,s − −
∗ ≤ · n/2 s
∗ !
− i=1
X
1 1
I[f] w(1+ǫ˜/2) (15)
≤ 2 · n/2 s ·
∗
−
1 2
I[f] (1+ǫ˜) w(1+˜ǫ/2) (16)
≤ 2 · n ·
I[f] w
· (1+2ǫ˜) (17)
≤ n ·
I[f] w
= · (1+ǫ/2) , (18)
n
where Equation (16) follows from the definition of s , the premise of the lemma that ǫ > 8q2log(8ǫn) and
∗ √n
ǫ˜= ǫ/4.
7
Forlower bounding p (f), wewillconsider only the contribution of the influential edges that belong to
w,s
∗
levelsℓ n/2+s . Consequently, Equations (1),(2)and(13)giveintotal
∗
≤
w 1
1 −
p (f) 2n 1(1 ǫ˜)I[f]2 n 1+ (1 ǫ˜/2) (19)
w,s − −
∗ ≥ − · n/2+s −
∗ !
i=1
X
1 1
˜
I[f](1 ǫ˜)w(1 ǫ/2) (20)
≥ 2 − − · n/2+s
∗
1 2
˜
I[f] w(1 ǫ˜)(1 ǫ/2) (1 ˜ǫ) (21)
≥ 2 · − − · n −
I[f] w
· (1 2ǫ˜) (22)
≥ n −
I[f] w
= · (1 ǫ/2) , (23)
n −
where Equation (21) follows from the definition of s , the premise of the lemma that ǫ > 8q2log(8ǫn) and
∗ √n
ǫ˜= ǫ/4.
Equations(18)and(23)give
w w
(1 ǫ/2) I[f] p (f) (1+ǫ/2) I[f], (24)
w,s
− · n · ≤ ∗ ≤ · n ·
asclaimedintheLemma.
ProofofTheorem 3.1: Forw ands assetbythealgorithm, letp (f)beasinDefinition1,wherewe
∗ w,s
∗
shallusetheshorthandp(f). Recallthatmisarandomvariabledenotingthenumberofiterationsperformed
by the algorithm until it stops (once α = t). Let m˜ = t , m˜ = m˜ , and m˜ = m˜ . We say
p(f) 1 (1+ǫ/4) 2 (1 ǫ/4)
thataniterationofthealgorithm issuccessfulifthewalktakeninthatiterationpassesthrough−aninfluential
edge(sothatthevalueofαisincreased by1). Letpˆ(f) = t denotethefractionofsuccessful iterations.
m
Supposethatm˜ m m˜ . Insuchacase,
1 2
≤ ≤
(1 ǫ/4) p(f) pˆ(f) (1+ǫ/4)p(f) (25)
− · ≤ ≤
since pˆ(f) = t = p(f)·m˜. By the definition of the algorithm, Iˆ = n t = n pˆ(f) so by Lemma 3.2
m m w · M w ·
(recallthatbythepremiseofthetheorem, ǫ = ω( logn/n))wehave
(1 ǫ)I[f] (1 ǫ/2)(1 ǫ/4)I[f] p Iˆ (1+ǫ/4)(1+ǫ/2)I[f] (1+ǫ)I[f] (26)
− ≤ − − ≤ ≤ ≤
andthus(assumingm˜ m m˜ ),theoutputofthealgorithm providestheestimationwearelookingfor.
1 2
≤ ≤
It remains to prove that m˜ m m˜ with probability at least 1 δ. Let X denote the indicator
1 2 i
≤ ≤ −
random variable whose value is 1 if and only if the ith iteration of the algorithm was successful, and let
X = m˜1 X . BythedefinitionofX ,wehavethatE[X ]= p(f),andso(bythedefinitionofm˜ andm˜)
i=1 i i i 1
wehavethatE[X] = m˜ p(f)= t Hence,byapplying themultiplicative Chernoffbound,
P 1· 1+ǫ/4
1 ǫ 2 t ǫ2t
Pr[m < m˜ ]= Pr[X >t] = Pr[X > (1+ǫ/4)E[X]] exp exp
1
≤ −3 4 1+ǫ/4 ≤ −96
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17) (27)
8
Thus, for t = 96ln(2δ) we have that Pr[m < m˜ ] δ. By an analogous argument we get that Pr[m >
ǫ2 1 ≤ 2
m˜ ] δ,andsom˜ m m˜ withprobability atleast1 δ,asdesired.
2 ≤ 2 1 ≤ ≤ 2 −
Since we have shown that m m˜ with probability at least 1 δ, and the query complexity of the
2
≤ −
algorithm isO(m),wehavethat,withprobability atleast1 δ,thequerycomplexity isupperbounded by
−
t t n log(1/δ) √nlog(n/ǫ)
O(m˜ ) = O = O · = O , (28)
2 p(f) w I[f] ǫ3 · I[f]
(cid:18) (cid:19) (cid:18) · (cid:19) (cid:18) (cid:19)
asrequired.
Remark. We assumed that I[f] 1 only for the sake of technical simplicity. This assumption can be
≥
replaced with I[f] 1 for any constant c 0, and the only modifications needed in the algorithm and
≥ nc ≥
its analysis are the following. The level of the cutoff s should be set to s = n/2 log( 2n ) =
∗ ∗ · ǫ˜n−c
1√n 2clog(2n)+log(1/ǫ˜) (which is a constant factor larger than the current septting), anqd the length w
2
of the walks in the algorithm should be set to w = ǫ˜√n (which is a constant factor smaller than
p
16 2log( 2n )
q ǫ˜n−c
thecurrentsetting).
Thefirstmodificationfollowsfromthefactthatthenumberofpointsy whoseHammingweighth(y)is
at least n/2+r n/2 or atmost n/2 r n/2 isupper bounded by 2n 2e r2. This implies that the
−
· − · ·
numberofedges(y,x)(whereh(y) = h(x)+1)suchthath(y) n/2+r n/2orh(y) n/2 r n/2
p p ≥ · ≤ − ·
isupperbounded byn 2n 2 r2. Requiring thatthelatterisnomorethan˜ǫ I[f]2n 1 ˜ǫ n c2n 1 (i.e,
· · − p· − ≥ · − −p
ǫ˜-fraction of the total number of influential edges), yields the desired r, where s = r n/2. The second
∗
modification, i.e, in the length of the walk, is governed by the choice of s , since, by the analysis, their
∗ p
product should be bounded by O(ǫ˜n). Since in both expressions 1/I[f] = nc appears only inside a log
term,thistranslates onlytoconstant factorincrease.
We note that the lower bound we give in Section 4 applies only to functions with (at least) constant
influence, and so in the above case where I[f] = 1/poly(n), the tightness of the algorithm (in terms of
querycomplexity)isnotguaranteed.
4 A Lower Bound
In this section we prove a lower bound of Ω √n on the query complexity of approximating the
I[f]logn
·
influence of monotone functions. Following it(cid:16)we explai(cid:17)n how a related construction gives a lower bound
of Ω n on approximating the influence of general functions. The idea for the first lower bound is the
I[f]
follow(cid:16)ing.(cid:17)Weshowthatanyalgorithm thatperformso √n queriescannotdistinguish withconstant
I[f]logn
·
successprobability betweenthatfollowing: (1)Acertai(cid:16)nthreshol(cid:17)dfunction(overarelativelysmallnumber
ofvariables), and (2) Afunction selected uniformly atrandom from acertain family offunctions that have
significantly higher influence than the threshold function. The functions in this family can be viewed as
“hidingtheirinfluencebehindthethreshold function”. Moreprecisedetailsfollow.
We first introduce one more notation. For any integer 1 k n and 0 t k, let τt : 0,1 n
≤ ≤ ≤ ≤ k { } →
0,1 bethet-threshold function over x ,...,x . Thatis, τt(x) = 1ifand only if k x t. Observe
{ } 1 k k i=1 i ≥
that(sinceforevery1 i kwehavethatI [τt] = 2 k 2 k 1 whilefori> kwehavethatI [τt] = 0),
≤ ≤ i k − · · t−1 P i k
I[τt]= k 2 (k 1) k 1 . −
k · − − · t−1 (cid:0) (cid:1)
−
(cid:0) (cid:1)
9