Correction Baccalauréat ES Pondichéry 17 avril 2012
17 Apr 2012 17 avril 2012. EXERCICE 1. 4 points. Commun à tous les candidats. 1. Réponse a. [En effet la droite (AB) est tangente à Cg au point A ...
Corrigé du brevet des collèges Pondichéry avril 2012
2 Apr 2012 Corrigé du brevet des collèges Pondichéry avril 2012. Activités numériques. 12 points. EXERCICE 1. 1. Non ! Car 88 = 10×8+8 : on perdra en ...
Corrigé du baccalauréat S Pondichéry 18 avril 2012
Corrigé du baccalauréat S Pondichéry. 18 avril 2012. EXERCICE 1. 6 points. Commun à tous les candidats. Les deux parties sont indépendantes. Partie A.
Correction Pondichéry avril 2012 – BREVET
Correction Pondichéry avril 2012 http ://exos2math.free.fr/. Activités numériques. 12 points. EXERCICE 1. La surface totale vaut donc S = 88×100 = 9680 cm2
Brevet 2012 Lintégrale davril à décembre 2012
Brevet des collèges Pondichéry avril 2012. Activités numériques À l'entrée du parc d'ani-math-ion figurent les informations suivantes : Tarifs. Horaires.
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Mai 2012. Pondichéry – Avril 2012 – Série S – Exercice Mai 2012. Analyse. Une jolie application de la congruence (chiffrement de Hill).
The cofactor preference of glucose?•6?•phosphate dehydrogenase
April 2012 accepted 19 April 2012) doi:10.1111/j.1742-4658.2012.08610.x. In Escherichia coli
Open Problems in Analysis of Boolean Functions
29 Apr 2012 DM] 29 Apr 2012. Open Problems in ... Compiled for the Simons Symposium February 5–11
PGE PGO
(www.passerelle-esc.com) du 30 novembre 2012 jusqu'au 2 avril 2013. Paiement Pour les classes préparatoires scientifiques (Math Spé ENS Cachan.
Maria J. ESTEBAN Née `a Alonsotegi (Pays Basque) le 6 Avril 1956
2013-2016 : Membre fondateur et membre du Board d'EU-MATHS-IN. 2012-2016 : Présidente du Comité Scientifique de l'IFCAM l'unité mixte franco-indienne en
Open Problems in
Analysis of Boolean Functions
Compiled for the Simons Symposium, February 5-11, 2012For notation and definitions, see e.g.
http://analysisofbooleanfunctions.org 2Correlation Bounds for Polynomials
Statement:Find an explicit (i.e., inNP) functionf: ?n2→
?2such that we ?2-polynomial p: ?n2→
?2of degree at most log2n.Source:Folklore dating back to [
Raz87,Smo87]
Remarks:
• The problem appears to be open even with correlation bound 1/? nre- placing 1/n. • Define the mod3function to be 1 if and only if the number of 1"s in
its input is congruent to 1 modulo 3. Smolensky [Smo87] showed that
mod3has correlation at most 2/3 with every
?2-polynomial of degree at mostc? n(wherec>0 is an absolute constant). For related bounds us- ing his techniques, there seems to be a barrier to obtaining correlation o(1/? n). • Babai, Nisan, and Szegedy [BNS92] implicitly showed a function inP
which has correlation at most exp(-nΘ(1)) with any ?2-polynomial of degree at most.99log2n; see also [VW08]. Bourgain [Bou05] (see
also [ GRS05]) showed a similar (slightly worse) result for the mod3 function.Tomaszewski"s Conjecture
Statement:Leta?
Source:Question attributed to Tomaszewski in [
Guy89]
Remarks:
• The bound of 1/2 would be sharp in light ofa=(1/?2,1/?2).
• Holman and Kleitman [HK92] proved the lower bound 3/8. In fact they
provedPrx≂{-1,1}n[|?a,x?| <1]≥3/8 (assumingai?= ±1 for alli), which is sharp in light ofa=(1/2,1/2,1/2,1/2). Talagrand"s Convolution with a Biased Coin" ConjectureStatement:Letf:{-1,1}n→
?≥0haveE[f]=1. Fix any 0<ρ<1. ThenPr[Tρf≥t] Source:[
Tal89]
Remarks:
• Talagrand in fact suggests the boundO(1 t?logt). • Talagrand offers a $1000 prize for proving this. • Even the special case" whenf"s domain is ?nwith Gaussian measure is open. In this Gaussian setting, Ball, Barthe, Bednorz, Oleszkiewicz, 3 and Wolff [ BBB+10] have shown the upper boundO(1t?logt) forn=1 and the boundO(loglogt t?logt) for any fixed constant dimension. Sensitivity versus Block Sensitivity
where sens[f] is the (maximum) sensitivity, maxx|{i?[n]:f(x)?=f(x?i)}|. Source:[
CFGS88,Sze89,GL92,NS94]
Remarks:
where bs[f] is the block sensitivity". However the version with degree is equally old, and in any case the problems are equivalent since it is known that bs[f] and deg(f) are polynomially related. • The best known gap is quadratic ([ CFGS88,GL92]) and it is suggested
GL92]) that this may be the worst possible.
Gotsman-Linial Conjecture
Statement:Among degree-kpolynomial threshold functionsf:{-1,1}n→ {-1,1}, the one with maximal total influence is the symmetric onef(x)= sgn(p(x1+···+xn)), wherepis a degree-kunivariate polynomial which al- ternates sign on thek+1 values ofx1+···+xnclosest to 0. Source:[
GL94] Remarks:
• The casek=1 is easy. • Slightly weaker version: degree-kPTFs have total influenceO(k)·? n. • Even weaker version: degree-kPTFs have total influenceOk(1)·? n. • The weaker versions are open even in the casek=2. Thek=2 case may be related to the following old conjecture of Holzman: Ifg:{-1,1}n→ ?has degree 2 (forneven), thenghas at most?n n/2?local strict minima. • It is known that bounding total influence byc(k)·? nis equivalent to a boundingδ-noise sensitivity byO(c(k))·? • The Gaussian special case" was solved by Kane [ Kan09].
• The best upper boundsknown are 2n1-1/2kand 2O(k)·n1-1/O(k)[ DHK+10].
Polynomial Freiman-Ruzsa Conjecture (in the
?n 2setting)
Statement:Suppose??=A?
?n ered by the union of poly(C) affine subspaces, each of cardinality at most|A|. Source:Attributed to Marton in [
Ruz93]; for the?n
2version, see e.g. [Gre05b]
Remarks:
4 • The following conjecture is known to be equivalent: Supposef: ?n 2→
?n 2satisfiesPrx,y[f(x)+f(y)=f(x+y)]≥?, wherexandyare in-
dependent and uniform on ?n 2. Then there exists a linear function
f: ?n 2→
?n 2such thatPr[f(x)=?(x)]≥poly(?).
• The PFR Conjecture is known to follow from thePolynomial Bo- golyubov Conjecture[ GT09]: LetA??n
2have density at leastα.
ThenA+A+Acontains an affine subspace of codimensionO(log(1/α)). One can slightly weaken the Polynomial Bogolyubov Conjecture by re- placingA+A+AwithkAfor an integerk>3. It is known that any such weakening (for fixed finitek) is enough to imply the PFR Conjec- ture. • Sanders [ San10b] has the best result in the direction of these conjec- tures, showing that ifA? ?n 2has density at leastαthenA+Acontains
99% of the points in a subspace of codimensionO(log4(1/α)), and hence
4Acontains all of this subspace. This suffices to give the Freiman-
Ruzsa Conjecture with 2
O(log4C)in place of poly(C).
• Green and Tao [ GT09] have proved the Polynomial Freiman-Ruzsa
Conjecture in the case thatAis monotone.
Mansour"s Conjecture
Statement:Letf:{-1,1}n→{-1,1}be computable by a DNF of sizes>1 and let??(0,1/2]. Thenf"s Fourier spectrum is?-concentrated on a collectionF Source:[
Man94]
Remarks:
• Weaker version: replacingsO(log(1/?))bysO?(1). • The weak version with boundsO(1/?)is known to follow from the Fourier Entropy-Influence Conjecture.
• Proved for almost all" polynomial-size DNF formulas (appropriately defined) by Klivans, Lee, and Wan [ KLW10].
• Mansour [ Man95] obtained the upper-bound (s/?)O(loglog(s/?)log(1/?)). Bernoulli Conjecture
Statement:LetTbe a finite collection of vectors in ?n. Defineb(T)= E x≂{-1,1}n[maxt?T?t,x?], and defineg(T) to be the same quantityexcept withx≂ ?nGaussian. Then there exists a finite collection of vectorsT?such that Source:[
Tal94]
5 Remarks:
• The quantityg(T) is well-understood in terms of the geometry ofT, thanks to Talagrand"s majorizing measures theorem. • Talagrand offers a $5000 prize for proving this, and a $1000prize for disproving it. Fourier Entropy-Influence Conjecture
Statement:There is a universal constantCsuch that for anyf:{-1,1}n→ S?f(S)2log21
?f(S)2is the spectral entropy andI[f] is the total influence. Source:[
FK96] Remarks:
• Proved for almost all" polynomial-size DNF formulas (appropriately defined) by Klivans, Lee, and Wan [ KLW10].
• Proved for symmetric functions and functions computable by read-once decision trees by O"Donnell, Wright, and Zhou [ OWZ11].
• An explicit example showing thatC≥60/13 is necessary is known. (O"Donnell, unpublished.) • Weaker version: the Min-Entropy-InfluenceConjecture",which states that there existsSsuch that?f(S)2≥2-C·I[f]. This conjecture is strictly stronger than the KKL Theorem, and is implied by the KKL Theorem in the case of monotone functions. Majority Is Least Stable Conjecture
Statement:Letf:{-1,1}n→{-1,1}be a linear threshold function,nodd. Then for allρ?[0,1],Stabρ[f]≥Stabρ[Majn]. Source:[
BKS99]
Remarks:
2 π?δ+o(?δ).
• The best result towards the weaker version is Peres"s Theorem [ Per04],
2 π?δ+O(δ3/2).
• By takingρ→0, the conjecture has the following consequence, which is also open: Letf:{-1,1}n→{-1,1}be a linear threshold function withE[f]=0. Then?n i=1?f(i)2≥2 π. The best known lower bound here
is 1 2, which follows from the Khinchine-Kahane inequality; see [GL94].
6 Optimality of Majorities for Non-Interactive Correlation Distillation
Statement:Fixr?
?,nodd, and 0<1/2. Forf:{-1,1}n→{-1,1}, define P(f)=Pr[f(y(1))=f(y(2))=···f(y(r))], wherex≂{-1,1}nis chosen uniformly and then eachy(i)is (independently) an?-noisy copy ofx. Is it true thatP(f) is maximized among odd functionsfby the Majority function Majkonsome odd number of inputsk? Source:[
MO05] (originally from 2002)
Remarks:
• It is possible (e.g., forr=10,n=5,?=.26) for neither the Dictator (Maj 1) nor full Majority (Majn) to be maximizing.
Noise Sensitivity of Intersections of Halfspaces
Statement:Letf:{-1,1}n→{-1,1}be the intersection (AND) ofklinear logk)·?δ. Source:[
KOS02]
Remarks:
• The boundO(k)·? δfollows easily from Peres"s Theorem and is the best known. • The Gaussianspecial case" follows easily from the work ofNazarov[ Naz03].
• An upper bound of the form polylog(k)·δΩ(1)holds if the halfspaces are sufficiently regular" [ HKM10].
Non-Interactive Correlation Distillation with Erasures Statement:Letf:{-1,1}n→{-1,1}be an unbiased function. Letz≂ {-1,0,1}nbe a random restriction" in which each coordinateziis (indepen- dently)±1 with probabilityp/2 each, and 0 with probability 1-p. Assuming p<1/2 andnodd, is it true thatEz[|f(z)|] is maximized whenfis the ma- jority function? (Here we identifyfwith its multilinear expansion.) Source:[
Yan04]
Remarks:
• Forp≥1/2, Yang conjectured thatEz[|f(z)|] is maximized whenfis a dictator function; this was proved by O"Donnell and Wright [ OW12].
• Mossel [ Mos10] shows that iff"s influences are assumed at mostτthen E Triangle Removal in
?n 2Statement:LetA?
?n 2. Suppose that?2nelements must be removed
fromAin order to make it triangle-free" (meaning there does not exist 7 x,y,x+y?A). Is it true thatPrx,y[x,y,x+y?A]≥poly(?), wherexandy are independent and uniform on ?n 2? Source:[
Gre05a]
Remarks:
• Green [ Gre05a] showed the lower bound 1/(2↑↑?-Θ(1)). • Bhattacharyya and Xie [ BX10] constructed anAfor which the proba-
bility is at most roughly?3.409. Subspaces in Sumsets
Statement:Fix a constantα>0. LetA?
?n 2have density at leastα. Is it
true thatA+Acontains a subspace of codimensionO(? n)? Source:[
Gre05a]
Remarks:
• The analogousproblem for the groupZNdates back to Bourgain [ Bou90].
n)}, it is easy to show that codimensionO(? n) cannot be improved. This example is essentially due to Ruzsa [ Ruz93], see [Gre05a].
• The best bounds are due to Sanders [ San10a], who shows thatA+A
must contain a subspace of codimension?n/(1+log2(1-α 1-2α))?. Think-
ing ofαas small, this means a subspace ofdimensionroughlyα ln2· n. Thinking ofα=1/2-?for?small, this is codimension roughly n/log2(1/?). In the same work Sanders also shows that ifα≥1/2- .001/? nthenA+Acontains a subspace of codimension 1. • As noted in the remarks on the Polynomial Freiman-Ruzsa/Bogolyubov Conjectures, it is also interesting to consider the relaxedproblem where we only require thatA+Acontains 99% of the points in a large sub- space. Here it might be conjectured that the subspace can have codi- mensionO(log(1/α)). Aaronson-Ambainis Conjecture
Statement:Letf:{-1,1}n→[-1,1] have degree at mostk. Then there existsi?[n] withInfi[f]≥(Var[f]/k)O(1). Source:[
Aar08,AA11]
Remarks:
• True forf:{-1,1}n→{-1,1}; this follows from a result of O"Donnell, Schramm, Saks, and Servedio [
OSSS05].
• The weaker lower bound (Var[f]/2k)O(1)follows from a result of Dinur, Kindler, Friedgut, and O"Donnell [
DFKO07].
8 Bhattacharyya-Grigorescu-Shapira Conjecture
Statement:LetM?
?m×k 2andσ?{0,1}k. Say thatf:
?n 2→{0,1}is (M,σ)-
freeif there does not existX=(x(1),...,x(k)) (where eachx(j)? ?n 2is a row
vector) such thatMX=0 andf(x(j))=σjfor allj?[k]. Now fix a (possi- bly infinite) collection{(M1,σ1),(M2,σ2),···}and consider the propertyPn of functionsf: ?n 2→{0,1}thatfis (Mi,σi)-free for alli. Then there is a
one-sided error, constant-query property-testing algorithm forPn. Source:[
BGS10]
Remarks:
• The conjecture is motivated by a work of Kaufman and Sudan [ KS08] which proposes as an open research problem the characterization of testability for linear-invariant properties of functionsf: ?n 2→{0,1}.
The properties defined in the conjecture are linear-invariant. • Every property family (Pn) defined by{(M1,σ1),(M2,σ2),···}-freeness issubspace-hereditary; i.e., closed under restriction to subspaces. The converse also essentially" holds. [ BGS10].
• ForMof rank one, Green [ Gre05a] showed that (M,1k)-freeness is
testable. He conjectured this result extends to arbitraryM; this was confirmed by Král", Serra, and Vena [ KSV08] and also Shapira [Sha09].
Austin [
Sha09] subsequentlyconjectured that (M,σ)-freeness is testablequotesdbs_dbs1.pdfusesText_1
Source:[
Tal89]
Remarks:
• Talagrand in fact suggests the boundO(1 t?logt). • Talagrand offers a $1000 prize for proving this. • Even the special case" whenf"s domain is ?nwith Gaussian measure is open. In this Gaussian setting, Ball, Barthe, Bednorz, Oleszkiewicz, 3 and Wolff [ BBB+10] have shown the upper boundO(1t?logt) forn=1 and the boundO(loglogt t?logt) for any fixed constant dimension.Sensitivity versus Block Sensitivity
where sens[f] is the (maximum) sensitivity, maxx|{i?[n]:f(x)?=f(x?i)}|.Source:[
CFGS88,Sze89,GL92,NS94]
Remarks:
where bs[f] is the block sensitivity". However the version with degree is equally old, and in any case the problems are equivalent since it is known that bs[f] and deg(f) are polynomially related. • The best known gap is quadratic ([CFGS88,GL92]) and it is suggested
GL92]) that this may be the worst possible.
Gotsman-Linial Conjecture
Statement:Among degree-kpolynomial threshold functionsf:{-1,1}n→ {-1,1}, the one with maximal total influence is the symmetric onef(x)= sgn(p(x1+···+xn)), wherepis a degree-kunivariate polynomial which al- ternates sign on thek+1 values ofx1+···+xnclosest to 0.Source:[
GL94]Remarks:
• The casek=1 is easy. • Slightly weaker version: degree-kPTFs have total influenceO(k)·? n. • Even weaker version: degree-kPTFs have total influenceOk(1)·? n. • The weaker versions are open even in the casek=2. Thek=2 case may be related to the following old conjecture of Holzman: Ifg:{-1,1}n→ ?has degree 2 (forneven), thenghas at most?n n/2?local strict minima. • It is known that bounding total influence byc(k)·? nis equivalent to a boundingδ-noise sensitivity byO(c(k))·? • The Gaussian special case" was solved by Kane [Kan09].
• The best upper boundsknown are 2n1-1/2kand 2O(k)·n1-1/O(k)[DHK+10].
Polynomial Freiman-Ruzsa Conjecture (in the
?n2setting)
Statement:Suppose??=A?
?n ered by the union of poly(C) affine subspaces, each of cardinality at most|A|.Source:Attributed to Marton in [
Ruz93]; for the?n
2version, see e.g. [Gre05b]
Remarks:
4 • The following conjecture is known to be equivalent: Supposef: ?n2→
?n2satisfiesPrx,y[f(x)+f(y)=f(x+y)]≥?, wherexandyare in-
dependent and uniform on ?n2. Then there exists a linear function
f: ?n2→
?n2such thatPr[f(x)=?(x)]≥poly(?).
• The PFR Conjecture is known to follow from thePolynomial Bo- golyubov Conjecture[GT09]: LetA??n
2have density at leastα.
ThenA+A+Acontains an affine subspace of codimensionO(log(1/α)). One can slightly weaken the Polynomial Bogolyubov Conjecture by re- placingA+A+AwithkAfor an integerk>3. It is known that any such weakening (for fixed finitek) is enough to imply the PFR Conjec- ture. • Sanders [ San10b] has the best result in the direction of these conjec- tures, showing that ifA? ?n2has density at leastαthenA+Acontains
99% of the points in a subspace of codimensionO(log4(1/α)), and hence
4Acontains all of this subspace. This suffices to give the Freiman-
Ruzsa Conjecture with 2
O(log4C)in place of poly(C).
• Green and Tao [GT09] have proved the Polynomial Freiman-Ruzsa
Conjecture in the case thatAis monotone.
Mansour"s Conjecture
Statement:Letf:{-1,1}n→{-1,1}be computable by a DNF of sizes>1 and let??(0,1/2]. Thenf"s Fourier spectrum is?-concentrated on a collectionFSource:[
Man94]
Remarks:
• Weaker version: replacingsO(log(1/?))bysO?(1). • The weak version with boundsO(1/?)is known to follow from the FourierEntropy-Influence Conjecture.
• Proved for almost all" polynomial-size DNF formulas (appropriately defined) by Klivans, Lee, and Wan [KLW10].
• Mansour [ Man95] obtained the upper-bound (s/?)O(loglog(s/?)log(1/?)).Bernoulli Conjecture
Statement:LetTbe a finite collection of vectors in ?n. Defineb(T)= E x≂{-1,1}n[maxt?T?t,x?], and defineg(T) to be the same quantityexcept withx≂ ?nGaussian. Then there exists a finite collection of vectorsT?such thatSource:[
Tal94]
5Remarks:
• The quantityg(T) is well-understood in terms of the geometry ofT, thanks to Talagrand"s majorizing measures theorem. • Talagrand offers a $5000 prize for proving this, and a $1000prize for disproving it.Fourier Entropy-Influence Conjecture
Statement:There is a universal constantCsuch that for anyf:{-1,1}n→S?f(S)2log21
?f(S)2is the spectral entropy andI[f] is the total influence.Source:[
FK96]Remarks:
• Proved for almost all" polynomial-size DNF formulas (appropriately defined) by Klivans, Lee, and Wan [KLW10].
• Proved for symmetric functions and functions computable by read-once decision trees by O"Donnell, Wright, and Zhou [OWZ11].
• An explicit example showing thatC≥60/13 is necessary is known. (O"Donnell, unpublished.) • Weaker version: the Min-Entropy-InfluenceConjecture",which states that there existsSsuch that?f(S)2≥2-C·I[f]. This conjecture is strictly stronger than the KKL Theorem, and is implied by the KKL Theorem in the case of monotone functions.Majority Is Least Stable Conjecture
Statement:Letf:{-1,1}n→{-1,1}be a linear threshold function,nodd. Then for allρ?[0,1],Stabρ[f]≥Stabρ[Majn].Source:[
BKS99]
Remarks:
2π?δ+o(?δ).
• The best result towards the weaker version is Peres"s Theorem [Per04],
2π?δ+O(δ3/2).
• By takingρ→0, the conjecture has the following consequence, which is also open: Letf:{-1,1}n→{-1,1}be a linear threshold function withE[f]=0. Then?n i=1?f(i)2≥2π. The best known lower bound here
is 12, which follows from the Khinchine-Kahane inequality; see [GL94].
6 Optimality of Majorities for Non-Interactive CorrelationDistillation
Statement:Fixr?
?,nodd, and 0<1/2. Forf:{-1,1}n→{-1,1}, define P(f)=Pr[f(y(1))=f(y(2))=···f(y(r))], wherex≂{-1,1}nis chosen uniformly and then eachy(i)is (independently) an?-noisy copy ofx. Is it true thatP(f) is maximized among odd functionsfby the Majority function Majkonsome odd number of inputsk?Source:[
MO05] (originally from 2002)
Remarks:
• It is possible (e.g., forr=10,n=5,?=.26) for neither the Dictator (Maj1) nor full Majority (Majn) to be maximizing.
Noise Sensitivity of Intersections of Halfspaces
Statement:Letf:{-1,1}n→{-1,1}be the intersection (AND) ofklinear logk)·?δ.Source:[
KOS02]
Remarks:
• The boundO(k)·? δfollows easily from Peres"s Theorem and is the best known. • The Gaussianspecial case" follows easily from the work ofNazarov[Naz03].
• An upper bound of the form polylog(k)·δΩ(1)holds if the halfspaces are sufficiently regular" [HKM10].
Non-Interactive Correlation Distillation with Erasures Statement:Letf:{-1,1}n→{-1,1}be an unbiased function. Letz≂ {-1,0,1}nbe a random restriction" in which each coordinateziis (indepen- dently)±1 with probabilityp/2 each, and 0 with probability 1-p. Assuming p<1/2 andnodd, is it true thatEz[|f(z)|] is maximized whenfis the ma- jority function? (Here we identifyfwith its multilinear expansion.)Source:[
Yan04]
Remarks:
• Forp≥1/2, Yang conjectured thatEz[|f(z)|] is maximized whenfis a dictator function; this was proved by O"Donnell and Wright [OW12].
• Mossel [ Mos10] shows that iff"s influences are assumed at mostτthen ETriangle Removal in
?n2Statement:LetA?
?n2. Suppose that?2nelements must be removed
fromAin order to make it triangle-free" (meaning there does not exist 7 x,y,x+y?A). Is it true thatPrx,y[x,y,x+y?A]≥poly(?), wherexandy are independent and uniform on ?n 2?Source:[
Gre05a]
Remarks:
• Green [ Gre05a] showed the lower bound 1/(2↑↑?-Θ(1)). • Bhattacharyya and Xie [BX10] constructed anAfor which the proba-
bility is at most roughly?3.409.Subspaces in Sumsets
Statement:Fix a constantα>0. LetA?
?n2have density at leastα. Is it
true thatA+Acontains a subspace of codimensionO(? n)?Source:[
Gre05a]
Remarks:
• The analogousproblem for the groupZNdates back to Bourgain [Bou90].
n)}, it is easy to show that codimensionO(? n) cannot be improved. This example is essentially due to Ruzsa [Ruz93], see [Gre05a].
• The best bounds are due to Sanders [San10a], who shows thatA+A
must contain a subspace of codimension?n/(1+log2(1-α1-2α))?. Think-
ing ofαas small, this means a subspace ofdimensionroughlyα ln2· n. Thinking ofα=1/2-?for?small, this is codimension roughly n/log2(1/?). In the same work Sanders also shows that ifα≥1/2- .001/? nthenA+Acontains a subspace of codimension 1. • As noted in the remarks on the Polynomial Freiman-Ruzsa/Bogolyubov Conjectures, it is also interesting to consider the relaxedproblem where we only require thatA+Acontains 99% of the points in a large sub- space. Here it might be conjectured that the subspace can have codi- mensionO(log(1/α)).Aaronson-Ambainis Conjecture
Statement:Letf:{-1,1}n→[-1,1] have degree at mostk. Then there existsi?[n] withInfi[f]≥(Var[f]/k)O(1).Source:[
Aar08,AA11]
Remarks:
• True forf:{-1,1}n→{-1,1}; this follows from a result of O"Donnell,Schramm, Saks, and Servedio [
OSSS05].
• The weaker lower bound (Var[f]/2k)O(1)follows from a result of Dinur,Kindler, Friedgut, and O"Donnell [
DFKO07].
8Bhattacharyya-Grigorescu-Shapira Conjecture
Statement:LetM?
?m×k2andσ?{0,1}k. Say thatf:
?n2→{0,1}is (M,σ)-
freeif there does not existX=(x(1),...,x(k)) (where eachx(j)? ?n2is a row
vector) such thatMX=0 andf(x(j))=σjfor allj?[k]. Now fix a (possi- bly infinite) collection{(M1,σ1),(M2,σ2),···}and consider the propertyPn of functionsf: ?n2→{0,1}thatfis (Mi,σi)-free for alli. Then there is a
one-sided error, constant-query property-testing algorithm forPn.Source:[
BGS10]
Remarks:
• The conjecture is motivated by a work of Kaufman and Sudan [ KS08] which proposes as an open research problem the characterization of testability for linear-invariant properties of functionsf: ?n2→{0,1}.
The properties defined in the conjecture are linear-invariant. • Every property family (Pn) defined by{(M1,σ1),(M2,σ2),···}-freeness issubspace-hereditary; i.e., closed under restriction to subspaces. The converse also essentially" holds. [BGS10].
• ForMof rank one, Green [Gre05a] showed that (M,1k)-freeness is
testable. He conjectured this result extends to arbitraryM; this was confirmed by Král", Serra, and Vena [KSV08] and also Shapira [Sha09].
Austin [
Sha09] subsequentlyconjectured that (M,σ)-freeness is testablequotesdbs_dbs1.pdfusesText_1[PDF] indeed
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