On polynomial approximations over $\mathbb{Z}/2^k\mathbb{Z}$

0.26 MB·

by  Abhishek Bhrushundi

On polynomial approximations over Z/2kZ
∗
AbhishekBhrushundi† Prahladh Harsha‡ Srikanth Srinivasan§
7
1
Abstract
0
2
We study approximation of Boolean functions by low-degree polynomials over the ring

n Z/2kZ. More precisely, given a Boolean function F : 0,1 n 0,1 , define its k-lift to be
{ } → { }
a F : 0,1 n 0,2k 1 by F (x) = 2k F(x) (mod 2k). We consider the fractionalagreement
J k { } → { − } k −
3 (whichwerefertoasγd,k(F))ofFk withdegreedpolynomialsfromZ/2kZ[x1,...,xn].
Ourresultsarethefollowing:
2
] • Increasing k can help: We observe that as k increases, γd,k(F) cannot decrease. We give
C twokindsofexampleswhereγd,k(F)actuallyincreases. Thefirstisaninfinite familyof
functions F such that γ (F) γ (F) Ω(1). The second is an infinite family of
C functionsFsuchthatγ 2d(,2F) −1+3od−(11),1—as≥smallaspossible—butγ (F) 1+Ω(1).
s. d,1 ≤ 2 d,3 ≥ 2
c Increasingkdoesn’talwayshelp:AdaptingaproofofGreen[Comput.Complexity,9(1):16–
[ • 38,2000],weshowthatirrespectiveofthevalueofk,theMajorityfunctionMaj satisfies
n

1
v 1 O(d)
γ (Maj ) + .
8 d,k n ≤ 2 √n
6
2 In other words, polynomials over Z/2kZ for large k do not approximate the majority
6
functionanybetterthanpolynomialsoverZ/2Z.
0
1. Weobservethatthemodelwestudysubsumesthemodelofnon-classicalpolynomialsinthe
0 sense that proving bounds in our model implies bounds on the agreement of non-classical
7 polynomials with Boolean functions. In particular, our results answer questions raised by
1
BhowmickandLovett[InProc.30thComputationalComplexityConf.,pages72—87,2015]thatask
:
v whether non-classical polynomials approximate Boolean functions better than classical poly-
i nomialsofthesamedegree.
X
r
a
∗A preliminary version of this paper appeared in Proc. 34th Annual Symp. on Theoretical Aspects of Comp. Science
(STACS),2017[BHS17].
†Department of Computer Science, Rutgers University, USA. [email protected]. Work done while
the author was visiting the Tata Institute of Fundamental Research. Research supported in part by UGC-ISF grant
6-2/2014(IC).
‡TataInstitute ofFundamentalResearch,India. [email protected]. ResearchsupportedinpartbyUGC-ISF
grant6-2/2014(IC).
§DepartmentofMathematics,IndianInstituteofTechnology,Bombay,[email protected].
1 Introduction
Manylowerboundresultsincircuitcomplexityareprovedbyshowingthatanysmallsizedcircuit
in a given circuit class can be approximated by a function from a simple computational model
(e.g., small depthcircuits by low-degreepolynomials) and subsequentlyshowing that this is not
possibleforsomesuitable“hardfunction”.
AclassiccaseinpointistheworkofRazborov[Raz87]whichshowslowerboundsforAC0[ ],
⊕
the class of constant depth circuits made up of AND, OR and gates. Razborov shows that
⊕
any small AC0[ ] circuit C can be well approximated by a low-degree multivariate polynomial
⊕
Q(x ,...,x ) F [x ,...,x ]inthesensethat
1 n 2 1 n
∈
Pr [Q(x) = C(x)] = o(1).
x 0,1 n 6
∼{ }
Thenextstepintheproofistoshowthatthehardfunction,ontheotherhand,doesnothaveany
such approximation. Razborov does this for a suitable symmetric function, Smolensky [Smo87]
fortheMOD function (forconstantoddq), andSzegedy[Sze89]andSmolensky[Smo93]forthe
q
MajorityfunctionMaj onnbits.
n
Giventheimportanceoftheabovelowerbound,polynomialapproximationsinotherdomains
and metricshave beenintenselyinvestigatedand have resultedin interestingcombinatorial con-
structionsanderror-correctingcodes[Gro00,Efr12],learningalgorithms[LMN93,KS04]andmore
recentlyinthedesignofalgorithmsforcombinatorialproblems[Wil14,AWY15]aswell.
Todescribethemodelofpolynomialapproximationconsideredinthispaper,wefirstrecallthe
Razborov [Raz87] model of polynomial approximation. Given a Boolean function F : 0,1 n
{ } →
0,1 anddegreed n,Razborovconsidersthelargestγsuchthatthereisadegreedpolynomial
{ } ≤
Q F [x ,...,x ] that has agreement at least γ with F (i.e., Pr [Q(x) = F(x)] γ). Call this
2 1 n x
∈ ≥
γ (F). Inthisnotation,Szegedy[Sze89]andSmolensky’s[Smo93]resultsfortheMajorityfunction
d
canbesuccinctlystatedas
1 O(d)
γ (Maj ) + .
d n ≤ 2 √n
WeconsiderageneralizationoftheabovemodeltoringsZ/2kZ inthefollowingsimplemanner.
Tobeginwith,weconsidertheringZ/4Z. GivenaBooleanfunctionF,letF : 0,1 n 0,2
2
{ } → { } ⊆
Z/4Z be the 2-lift of F defined as F (x) := 22 F(x) (i.e., F (x) := 0 if F(x) = 0 and F (x) := 2
2 − 2 2
otherwise). Once again, we can define γ (F) to be the largest γ such that there exists a degree
d,2
d polynomial Q Z/4Z[x ,...,x ] that has agreement γ with F . Note that γ (F) γ (F)
2 1 n 2 d,2 d
∈ ≥
since if, for instance, Q(x) = x x + x F [x ,...,x ] has agreement γ with F, then Q :=
1 2 3 2 1 n 2
∈
2(x x +x ) Z/4Z[x ,...,x ] also has the same agreement γ with F . Hence, proving upper
1 2 3 1 n 2
∈
boundsforγ (F)isatleastashardasprovingupperboundsforγ (F).
d,2 d
More generally,we can extendthesedefinitionsto γ (F), theagreementof F , the k-lift of F,
d,k k
defined as F (x) = 2k F(x) mod 2k, with degree d polynomials from Z/2kZ[x ,...,x ]. It is not
k − 1 n
hard to show that γ (F) γ (F) and hence as k increases, the problem of proving upper
d,k+1 d,k
≥
boundsonγ (F)canonlygetharder.
d,k
OurmotivationforthismodelcomesfromarecentworkofBhowmickandLovett[BL15],who
studythemaximum agreementbetweennon-classical polynomials ofdegreed anda Booleanfunc-
tion F,whichissimilar toγ (F)(seeSection5foranexacttranslationbetweentheabove model
d,d
1
and non-classical polynomials). In particular, non-classical polynomials of degree d can be con-
sideredasasubsetofthedegreedpolynomialsinZ/2dZ[x ,...,x ]. Withrespecttocorrelation1,
1 n
BhowmickandLovettshowedthatthereexistnon-classical polynomials(andhencepolynomials
in Z/2dZ[x ,...,x ]) of logarithmic degree that have very good correlation with the Maj func-
1 n n
tion. With respect to agreement, they show that low-degree non-classical polynomials can only
havesmallagreementwiththeMajorityfunction. Theirresultsstatedinourlanguage,implythat
1 O(d 2d)
γ (Maj ) + · .
d,d n ≤ 2 √n
In particular, if d = Ω(logn), this result unfortunately does not give any non-trivial bound on
the maximum agreement between non-classical polynomials of degree d and the Maj function.
n
Bhowmick and Lovett, however, conjectured that this result could be improved and left open
the question of whether non-classical polynomials of degree d can do any better than classical
polynomials of the same degree in approximating the Majority function. More generally, they
informally conjectured that although non-classical polynomials achieve better correlation with
Boolean functions than their classical counterparts, they possibly do not approximate Boolean
functions any better than classical polynomials. Our work stems from trying to answer these
questions.
1.1 Our results
We prove the following results about agreement of Boolean functions with polynomials over the
ringZ/2kZ:
1. We explore whether there exist Boolean functions for which agreement can increase by in-
creasingk. Inparticular,dothereexistBooleanF suchthatγ (F) > γ (F)?
d,k d,1
It is not hard to show that this is impossible for d = 1. Further, it can be shown that if
γ (F) > 1 1 , thenγ F) = γ (F). Keepingthis in mind, thefirstplace wherewecan
d,k − 2d d,k( d,1
expectlargerk toshowbetteragreementisγ vs. γ . Ourfirstresultshowsthatthereare
2,2 2,1
indeedseparatingexamplesintheregime.
(a) Fix d N tobe any powerof2. Forinfinitely many n, thereexistsa Boolean function
∈
F : 0,1 n 0,1 suchthatγ (F) 5/8+o(1)butγ (F) 3/4 o(1).
3d 1,1 2d,2
{ } → { } − ≤ ≥ −
Note that since F is Boolean, γ (F) 1/2 for any d,k. We then ask if there exist Boolean
d,k
≥
functions Fsuchthatγ (F)ismoreorlessthetrivialboundof1/2,whileγ (F)issignif-
d,1 d,k
′
>
icantlylargerford dandsomek 1. Inthiscontext,weshowthefollowingresult.
′
≤
(b) Fixanyℓ 2. Forlargeenoughn,thereisaBooleanfunction F : 0,1 n 0,1 such
thatγ2ℓ 1,≥1(F) ≤ 1/2+o(1) butγd,3(F) ≥ 9/16−o(1),ford = 2ℓ{−1+}2ℓ−→2 ≤{ 2ℓ}−1.
−
2. WeshowthatforMaj ,themajorityfunctiononnbits,andanyd,k Z+,
n ∈
1 O(d)
γ (Maj ) + ,
d,k n ≤ 2 √n
1ThecorrelationbetweenF,G: 0,1 n Z/2kZisdefinedtobeEx[ωF(x)−G(x)]whereωistheprimitive2kthroot
{ } →
ofunityinC.IfF,Gare 0,2k−1 -valued,thenthisquantityisexactly2γ 1whereγistheagreementbetweenFand
{ } −
G.Otherwise,however,itdoesnotmeasureagreement.
2
2 by adapting a proof due to Green [Gre00] of a result on the approximability of the parity
functionbylow-degreepolynomialsovertheringZ/pkZ forprime p = 2.
6
CoupledwiththeobservationthattheclassofpolynomialsoverringsZ/2kZ subsumestheclass
ofnon-classicalpolynomials,part(b) ofthefirstresultprovidesacounterexampletoaninformal
conjectureofBhowmickandLovett[BL15]that,foranyBooleanfunction F,non-classicalpolyno-
mials ofdegreeddonotapproximate F anybetterthanclassical polynomialsofthesamedegree,
andthesecondresultconfirmstheirconjecturethatnon-classicalpolynomialsdonotapproximate
theMajorityfunctionanybetterthanclassicalpolynomials.
1.2 Organisation
We start with some preliminaries in Section2. In Section3, we show some separation results.
Next, in Section4, we prove upper bounds for γ (Maj ). Finally, in Section5, we discuss how
d,k n
our model relates to non-classical polynomials, answering questions raised by Bhowmick and
Lovett.
2 Preliminaries
For x 0,1 n, x denotes the Hamming weight of x, and for i 0, x is the (i+1)th least
i
∈ { } | | ≥ | |
significant bit of x in base 2. For d N, we use 0,1 n (resp. 0,1 n ) to denote the set of
| | ∈ { } d { }=d
elements in 0,1 n of Hamming weight at most d (resp.≤exactly d). We use to denote the
n
{ } F
collectionofallBooleanfunctionsdefinedon 0,1 n.
{ }
2.1 Elementarysymmetricpolynomials
Recall that for t 1, the elementary symmetricpolynomial of degree t over F , S (x ,...,x ), is
2 t 1 n
≥
defined as St(x1,...,xn) = 1≤a1<...<at≤nxa1 ...xat. Here ⊕ denotes addition modulo two. This
maybeinterpretedas
L
x
S (x ,...,x ) = | | mod2. (2.1)
t 1 n
t
(cid:18) (cid:19)
AdirectconsequenceofLucas’stheorem(see,e.g.,[Knu97,Section1.2.6,Ex. 10])andEq.(2.1)
isthefollowing:
Lemma 2.1. For every ℓ ≥ 0, S2ℓ(x) = |x|ℓ. More generally, St(x) = ∏i|x|i where the product runs
th
overalli 0suchthatthe(i+1) leastsignificantbitofthebinaryexpansion oftis1.
≥
ThefollowingresultfollowsfromtheworkofGreenandTao[GT09,Theorem11.3],whobuild
upontheideasofAlonandBeigel [AB01].
Theorem2.2(Green-Tao[GT09],Alon-Beigel[AB01]). Fixℓ 0. Then,foreverymultilinearpolyno-
mial P ∈ F2[x1,...,xn]ofdegreeatmost2ℓ−1,wehavePrx∼{0,≥1}n[S2ℓ(x) = P(x)] ≤ 1/2+o(1).
Theorem2.2hasanicecorollary:
2TheconstantintheO( )isanabsoluteconstant.
·
3
Corollary2.3. Forevery fixed ℓ ≥ 0,thefunctions{S2i(x)}0≤i≤ℓ arealmostbalancedandalmostuncor-
related, i.e.
0 i ℓ, Pr[S (x) = 0] Pr[S (x) = 1] = o(1)
• ∀ ≤ ≤ | 2i − 2i |
• ∀ a0,...,aℓ ∈ {0,1}, |Pr 0≤i≤ℓ(S2i(x) = ai) − 2ℓ1+1| = o(1).
CombiningCorollary2.3w(cid:2)iVthLemma2.1,weg(cid:3)etanotherusefulfact:
Lemma2.4. Letxbeuniformlydistributedover 0,1 n. Then,foreveryfixedr 1,therandomvariables
{ } ≥
x arealmostuniformandalmostr-wiseindependenti.e.
i 0 i r 1
{| | } ≤≤ −
0 i r 1, Pr[ x = 0] Pr[ x = 1] = o(1).
i i
• ∀ ≤ ≤ − | | | − | | |
(a ,...,a ) 0,1 r, Pr[( x ,..., x ) = (a ,...,a )] 1 = o(1).
• ∀ 0 r−1 ∈ { } | | |0 | |r−1 0 r−1 − 2r|
2.2 Boolean functionsand polynomials over Z/2kZ
Given an F and k 1, we define the k-lift of F to be the function F : 0,1 n Z/2kZ
n k
∈ F ≥ { } →
definedasfollows. Forany x 0,1 n,
∈ { }
0 if F(x) = 0,
F (x) =
k 2k 1 otherwise.
−
(cid:26)
For d N and k 1, will denotethe set of multilinear polynomials of degree at most d
d,k
∈ ≥ P
overtheringZ/2kZ.
For functions F,G : D R for some finite domain D and range R, the agreement between F
→
and G,denotedbyagr(F,G),isdefinedtobethefractionofinputswheretheyagree,i.e.,
agr(F,G) = Pr [F(x) = G(x)].
x D
∼
WewillconsiderhowwellmultilinearpolynomialsofdegreedcanapproximateBooleanfunc-
tionsintheabovesense. Moreprecisely,foranyBooleanfunction F ,wedefine
n
∈ F
γ (F) = max agr(F ,Q).
d,k k
Q
∈Pd,k
Following[Gop08],wecallaset I 0,1 n aninterpolatingset3 for iftheonlypolynomial
d,k
⊆ { } P
P thatvanishesatallpointsin I iszeroeverywhere. Formally,forany P ,
d,k d,k
∈ P ∈ P
( x I P(x) = 0) ( y 0,1 n P(y) = 0).
∀ ∈ ⇒ ∀ ∈ { }
WenowstateanumberofstandardfactsregardingBooleanfunctionsandmultilinearpolyno-
mialsoverZ/2kZ. Theomittedproofsareeithereasyorwell-known.
Unlessmentionedotherwise,letn,d,kbeanyintegerssatisfyingn 1,d 0,k 1.
≥ ≥ ≥
Lemma2.5. Anypolynomial Q satisfiesthefollowing:
d,k
∈ P
1. (Schwartz-Zippel) If Qisnon-zero,thenPr [Q(x) = 0] 1 .
x∼{0,1}n 6 ≥ 2d
3Thisisalsocalledahittingsetintheliterature.
4
2. Qisthezeropolynomialiff Q(x) = 0forall x 0,1 n.
∈ { }
3. (Mo¨bius Inversion) Say Q(x) = ∑ c x , where c Z/2kZ and x denotes ∏ x . Then,
|S|≤d S S S ∈ S i∈S i
c = ∑ ( 1)S T Q(1 )where1 0,1 n isthecharacteristic vectorof T.
S T⊆S − | |−| | T T ∈ { }
4. ( 0,1 n is an interpolating set) Q vanishes at all points in 0,1 n iff Q vanishes at all points of
{ } d { }
0,1 n≤. By shifting the origin to any point of 0,1 n, the same is true of any Hamming ball of
{ } d { }
radius≤din 0,1 n.
{ }
Proof. Point1: WriteQas Q(x) = 2ℓ Q (x),whereℓ < kisthelargestpowerof2thatdividesthe
′
·
GCDofthecoefficientsofQ. ProjectingQ toanon-zeropolynomialoverZ/2Zbydroppingallits
′
coefficients modulo 2 and applying the standard Schwartz-Zippel lemma over Z/2Z completes
theproof.
Point2followsfrompoint1,andpoint4frompoint3.
Lemma2.6. Fixany F .
n
∈ F
1. γ (F) 1.
d,k ≥ 2
2. γ (F) γ (F).
d,k+1 d,k
≥
3. γ (F) > 1 1 γ (F) = γ (F).
d,k − 2d ⇒ d,k d,1
4. γ (F) = γ (F).
1,k 1,1
Proof. Point1istrivialsincethereisaconstantpolynomialthathasagreementatleast 1 with F .
2 k
Point 2: Say P has agreement α with F . Then, 2 P (interpretednaturally as a polyno-
d,k k
∈ P ·
mialin )hasagreementαwith F .
d,k+1 k+1
P
Forpoint3,considerapolynomialQ thatachievesthemaximumagreementα > 1 1
∈ Pd,k − 2d
with F . LetQ bethepolynomialobtainedfromQbydroppingallitsco-efficientsmodulo
k ′ d,1
∈ P
2k 1. Notethatfor any x, Q(x) 0,2k 1 implies that Q (x) = 0(in thering Z/2k 1Z). Hence,
− − ′ −
∈ { }
the probability that Q is zero is at least α > 1 1 . Lemma2.5 point 1 implies that Q must be
′ − 2d ′
the zero polynomial. Equivalently, all of the coefficients of Q are divisible by 2k 1 and hence Q
−
canbenaturallyidentifiedwith2k 1 Q forsomeQ Z/2Z[x ,...,x ]. Itiseasytocheckthat
− ′′ ′′ 1 n
· ∈
agr(Q ,F ) = αandhencewehaveγ (F) γ (F). Ontheotherhand,frompoint2,wealready
′′ 1 d,1 d,k
≥
knowthatγ (F) γ (F). Hencewearedone.
d,k d,1
≤
Point4followsfrompoints1and3.
3 Some separation results
3.1 Symmetricfunctionsasseparatingexamples
We know from Theorem2.2 that, for every fixed ℓ ≥ 2, γ2ℓ 1,1(S2ℓ) ≤ 12 +o(1). In contrast, the
−
mainresultofthissectionshowsthat
Theorem3.1. Foreveryfixedℓ ≥ 2,γd,3(S2ℓ) ≥ 196 −o(1),whered = 2ℓ−1+2ℓ−2.
5
Noticethat2ℓ−1+2ℓ−2 ≤ 2ℓ−1forℓ ≥ 2. Thisimpliesthat,forℓ ≥ 2, S2ℓ(x) isanexampleof
a function F for which there exist k,d N such that γ (F) 1 +o(1) but γ (F) 1 +Ω(1)
∈ d,1 ≤ 2 d′,k ≥ 2
forsomed d.
′
≤
ProofofTheorem 3.1. Lemma2.1fromSection2tellsusthatS2ℓ(x) = |x|ℓ. Thus,S2ℓ,3(x) ∈ Z/8Z[x1,...,xn],
the3-liftofS2ℓ(x),isgivenby
4 if x ℓ = 1
S2ℓ,3(x) = | | (3.1)
(0 otherwise
Fix d to be 2ℓ 1+2ℓ 2. Consider the polynomial P(x) = ∑ ∏ x in Z/8Z[x ,...,x ].
− − T ([n]) i T i 1 n
∈ d ∈
Toprovethetheorem,itsufficestoshowthat
1 1
x P0r,1 n[P(x) = S2ℓ,3(x)] ≥ 2 + 16 −o(1).
∼{ }
x
Clearly, P(x) = | | mod8,and
d
(cid:18) (cid:19)
x
0 if8 | |
| d
P(x) =  (cid:18) (cid:19) (3.2)
x x
4 if4 | | but8∤ | |

| d d
(cid:18) (cid:19) (cid:18) (cid:19)


The following theorem due to Kummer (see, e.g., [Knu97, Section 1.2.6, Ex. 11]) determines
thelargestpowerofaprimethatdividesabinomialcoefficient.
Theorem 3.2 (Kummer). Let p be a prime and N,M N such that N M. Suppose r is the largest
∈ ≥
integersuchthat pr (N). Thenr isequaltothenumberofborrowsrequired whensubtracting M from N
| M
inbase p.
Let B(x) be thenumber ofborrows requiredwhen subtracting d from x . Rewriting Eq.(3.2)
| |
intermsofB(x)usingKummer’stheorem,weget
4 if B(x) = 2
P(x) = (3.3)
(0 if B(x) 3
≥
Wewillneedthefollowinglemma.
Lemma3.3. P(x) = S2ℓ,3(x)ifeither
1. x ℓ 2 = 0,or
| | −
2. ( x ℓ 2, x ℓ 1, x ℓ, x ℓ+1) = (1,0,0,0).
| | − | | − | | | |
Proof. Sinced = 2ℓ−1+2ℓ−2,allthebitsofdexceptdℓ 1anddℓ 2arezero. Thus,whensubtracting
− ℓ −
dfrom x ,noborrowsarerequiredbythebits x ,0 i 3.
i
| | | | ≤ ≤ −
Using the above observation, the reader can verify that when ( x ℓ 2, x ℓ 1, x ℓ, x ℓ+1) =
| | − | | − | | | |
(1,0,0,0) the number of borrows required is at least 3 i.e. B(x) 3, which in turn implies that
≥
P(x) = 0. Since|x|ℓ = 0,S2ℓ,3(x) = 0. Thisprovesthesecondpartofthelemma.
6
To prove the first part, suppose x ℓ 2 = 0. Since dℓ 1 = dℓ 2 = 1, it follows that both x ℓ 2
| | − − − | | −
and x ℓ 1 will need to borrow when subtracting d from x . As argued before, no borrows are
| | − | |
requiredbythebitsbefore(i.e. lesssignificantthan) x ℓ 2,andthusthetotalnumberofborrows
ℓ | | −
requiredbythebits x ,0 i 1,is2.
i
| | ≤ ≤ −
Notethatthebit x ℓ 1 borrowsfrom x ℓ. Considerthefollowingcaseanalysis:
| | − | |
Case x ℓ = 1: x ℓ will not need to borrow since dℓ = 0. In fact, none of the bits after
• | | | |
(i.e. more significant than) x ℓ will need to borrow, and thus B(x) = 2. This implies that
| |
P(x) = 4. WealsohaveS2ℓ,3(x) = 4andhence P(x) = S2ℓ,3(x).
Case x ℓ = 0: x ℓwillrequireaborrowandthismeansB(x) 3. ThisimpliesthatP(x) = 0.
• | | | | ≥
Since|x|ℓ = 0,itfollowsthat P(x) = S2ℓ,3(x).
Thiscompletestheproof.
ByLemma3.3,wehave
Pr[P(x) = S2ℓ,3(x)] ≥ Pr[|x|ℓ−2 = 0]+Pr[(|x|ℓ−2,|x|ℓ−1,|x|ℓ,|x|ℓ+1) = (1,0,0,0)] (3.4)
UsingLemma2.4fromSection2,wehave
1
Pr[ x ℓ 2 = 0] o(1)
| | − ≥ 2 −
1
Pr[( x ℓ 2, x ℓ 1, x ℓ, x ℓ+1) = (1,0,0,0)] o(1)
| | − | | − | | | | ≥ 16 −
which,togetherwithEq.(3.4),implies
1 1
Pr[P(x) = S2ℓ,3(x)] ≥ 2 + 16 −o(1).
3.2 A separationat k = 2
Let d N be any power of 2. In this section, we show that there are functions F for which
∈
γ (F) > γ (F).
2d,2 3d 1,1
−
Theorem 3.4. For large enough n, there exists a function F such that γ (F) 3 o(1) but
∈ F2n 2d,2 ≥ 4 −
γ (F) 5 +o(1).
3d−1,1 ≤ 8
Inparticular, weseethatγ (F) > γ (F). Thisresultisnotable,sinceitshowsthatthereisa
2,2 2,1
separationat thefirst place where it is possible tohave one(Recall that γ (F) = γ (F) for any
1,k 1,1
F byLemma2.6).
n
∈ F
LetusbegintheproofofTheorem3.4. WefirstdefineafamilyofBooleanfunctionson 0,1 2n.
{ }
Wedenotethe2n variables by x1,...,xn and y1,...,yn. Weuse (|dx|) todenotethe dth elementary
symmetricpolynomialfromtheringZ/4Z[x1,...,xn],i.e.,(|dx|) = ∑S ([n])∏i Sxi.4
∈ d ∈
WewillneedthefollowingeasycorollaryofTheorem3.2.
4We distinguish between (|dx|) and Sd(x) since the former is from Z/4Z[x1,...,xn] and latter a polynomial in
F2[x1,...,xn].
7
Corollary3.5. Letdbeapowerof2. Then,for N d, thehighest powerof2dividing (N) isequaltothe
≥ d
highestpowerof2dividing N .
⌊d⌋
LetS = (x,y) (|x|),(|y|) 1 (mod 2) . Given anyfunction H : 0,1 2n 0,1 , wedefine
{ | d d ≡ } { } → { }
theBooleanfunction F (x ,...,x ,y ,...,y )asfollows:
H 1 n 1 n
0 if(|x|) (|y|) 0 (mod 4),
d · d ≡
FH(x,y) =  1 if(|x|) (|y|) 2 (mod 4),
d · d ≡
 H(x,y) otherwise.

Define P(x,y) = (|xd|)·(|yd|) ∈ Z/4Z[x1,...,xn,y1,...,yn]. Notethat FH(x,y)isdefinedsothat
its 2-lift agrees with P(x,y) on points (x,y) where P(x,y) 0,2 . Also Corollary3.5 implies
∈ { }
that the following is an alternate equivalent definition of F in terms of elementary symmetric
H
polynomialsmodulo2.
0 ifS (x) = S (y) = 0,
d d
S (y) ifS (x) = 1andS (y) = 0,
F (x,y) =  2d d d (3.5)
H S (x) ifS (x) = 0andS (y) = 1,
 2d d d
 H(x,y) otherwise.


We now begin the proof of Theorem3.4. First of all, let us note that for any choice of H, we
have:
Lemma3.6. γ (F ) 3 o(1).
2d,2 H ≥ 4 −
Proof. Consider the polynomial P(x,y) defined above. From Eq.(3.5), it follows that the
2d,2
∈ P
probabilitythatP(x,y) = F (x,y)5islessthanorequaltotheprobabilitythatS (x) = S (y) = 1,
H,2 d d
6
whichis 1 +o(1) byCorollary2.3. Thisgivestheclaim.
4
Themainlemmaisthefollowing.
Lemma3.7. Say H : 0,1 2n 0,1 ischosenuniformlyatrandom. Then,
{ } → { }
5
Pr[γ (F ) > +o(1)] = o(1).
3d 1,1 H
H − 8
ThiswillproveTheorem3.4. Wewillprovetheabovelemmainthefollowingsubsection.
3.3 Proof of Lemma3.7
The outline ofthe proofis as follows. Fix any polynomial Q F [x ,...,x ,y ,...,y ] ofdegree
2 1 n 1 n
∈
at most3d 1. Weneed toshow that agr(F ,Q) 5 +o(1) for a random H : 0,1 2n 0,1 .
− H ≤ 8 { } → { }
The fact that H is random ensures that any Q cannot agree with H on significantly more than
halftheinputsinS. ForinputsoutsideS,weneedamoreinvolvedargument,followingAlonand
Beigel[AB01]. WeshowthatforanyQwecanfindsomewhatlargesetsIand Jofxandyvariables
respectively such that when we set the variables outside I J, we obtain a polynomial that is
∪
symmetricinthevariablesof I J. ThisisaRamseytheoreticargumenta´laAlon-Beigel[AB01].
∪
5F denotesthe2-liftofF .
H,2 H
8
Following this argument, we only need to prove the agreement upper bound for Q that is
symmetricin xandyvariables. Thiscanbedonebyreductiontoaconstant-sizedproblem,aswe
showbelow. Acarefulcomputationtosolvetheconstant-sizedproblemwillfinishtheproof.
We begin with some notation that will be useful in the proof. Throughout, we work with
disjointsetsofx-variablesandy-variablesofequalsizeandconsiderpolynomialsoverthesevari-
ables. Letthe x-variables be x ,...,x andthey-variables be y ,...,y . For I,J [n],theset
1 n 1 n
{ } { } ⊆
ofF -polynomialsQoverthevariables x i I and y j J isdenotedF [x ,y ]. Similarly,
2 i j 2 I J
{ | ∈ } { | ∈ }
weuseQ F [x ]todenotethefactthat Qisapolynomialonlyoverthevariables x i I .
2 I i
∈ { | ∈ }
We use to denote Boolean assignments σ : x i I y j J 0,1 . Given
I,J i j
A { | 6∈ }∪{ | 6∈ } → { }
F : F2n F andσ ,weuse F F [x ,y ]todenoteitsnaturalrestrictiontothevariables
2 → 2 ∈ AI,J |σ ∈ 2 I J
indexedby I J.
∪
Wesaythat Q F [x ,y ]is(x,y)-symmetric ifitisalinearcombinationofthepolynomialsin
2 I J
∈
the set S (x ) S (y ) d ,d N . We note that being (x,y)-symmetric depends on the sets
{ d1 I · d2 J | 1 2 ∈ }
I,J underconsideration. Thiswillbeimplicitwhenused.
Given a multilinear monomial m over the x and y-variables, its multidegree is defined to be
(i,j) if m multiplies i x-variables and j y-variables. Let = (i,j) i+j 3d 1 be the set of
D { | ≤ − }
multidegreesofmonomialsofdegreeatmost3d 1. Weorder inascendingorderaccordingto
− D
i+j,i.e.,fixatotalordering of suchthatifi +j < i +j ,then(i ,j ) (i ,j )6. Let(i ,j )
1 1 2 2 1 1 2 2 0 0
(cid:22) D (cid:22)
be the largest element in the ordering . We will define multdeg(Q) to be the largest (w.r.t. )
(cid:22) (cid:22)
multidegreeofamonomialthathasanon-zerocoefficientin Q. For(i,j) suchthati+j 3d 1,
≤ −
wesaythatapolynomialQ F [x ,y ]is(x,y;i,j)-symmetricifwecanwriteQas
2 I J
∈
Q = Q Q (3.6)
1 2
⊕
where multdeg(Q ) (i,j) and Q is (x,y)-symmetric. Note that if (i,j) = (i ,j ), then any
1 2 0 0
(cid:22)
polynomial of degree at most 3d 1 is (x,y;i ,j )-symmetric, since we can take Q = Q and
0 0 1
−
Q = 0.
2
We also need the following variant of the function F defined above. Call a function Φ
H
∈
F [x ,...,x ] (resp. Ψ F [y ,...,y ]) d-simple w.r.t. x (resp.w.r.t.y)if itis alinear combination
2 1 n 2 1 n
∈
of symmetricpolynomials in x (resp.in y) of degreestrictly lessthan d. Equivalently, we can say
thatΦ(x) onlydependsupon x ,..., x ,andsimilarlyforΨ(y) w.r.t. y.
0 lgd 1
Givenpairsofpolynomials|Φ| = (Φ|,|Φ )− F [x ,...,x ] F [x ,...,x ],andΨ = (Ψ ,Ψ )
1 2 2 1 n 2 1 n 1 2
∈ × ∈
F [y ,...,y ] F [y ,...,y ],suchthatΦ ,Φ andΨ ,Ψ are d-simplew.r.t. xandyrespectively,
2 1 n 2 1 n 1 2 1 2
×
define
0 ifS (x) = Φ (x),S (y) = Ψ (y),
d 1 d 1
S (y) Ψ (y) ifS (x) = 1 Φ (x)andS (y) = Ψ (y),
FH,Φ,Ψ(x,y) =  S2d(x)⊕Φ2(x) ifSd(x) = Φ⊕(x)1andS (y)d= 1 Ψ1(y), (3.7)
 H2(dx,y)⊕ 2 othderwise. 1 d ⊕ 1


Also,defineSΦ,Ψ = (x,y) Sd(x) = 1 Φ1(x),Sd(y) = 1 Ψ1(y) .
{ | ⊕ ⊕ }
Withthenotationabove,wearereadytostateaclaimthatgeneralizesLemma3.7.
Lemma3.8. Fixany d N. For (i,j) and ε (0,1), there is an n(i,j,ε) suchthat given any n
∈ ∈ D ∈ ≥
n(i,j,ε), for uniformlyrandom H : 0,1 2n 0,1 , wehaveforany Φ = (Φ ,Φ ) F [x ,...,x ]2
1 2 2 1 n
{ } → { } ∈
6If(i ,j )=(i ,j ),buti +j =i +j ,thentherelationbetween(i ,j )and(i ,j )isfixedinanarbitrarymanner.
1 1 6 2 2 1 1 2 2 1 1 2 2
9