On subsets of the hypercube with prescribed Hamming distances

Transcription

1 O subses of he hypercube wh prescrbed Hammg dsaces Hao Huag Oleksy Klurma Cosm Pohoaa Absrac A celebraed heorem of Klema exremal combaorcs saes ha a colleco of bary vecors {0, 1} wh dameer d has cardaly a mos ha of a Hammg ball of radus d/2 I hs paper, we gve a algebrac proof of Klema s Theorem, by carefully choosg a pseudo-adjacecy marx for cera Hammg graphs, ad applyg he Cvekovć boud o depedece umbers Ths mehod also allows us o prove several exesos ad geeralzaos of Klema s Theorem o oher allowed dsace ses, parcular blocks of cosecuve egers ha do o ecessarly grow learly wh We also mprove o a heorem of Alo abou subses of F p whose dfferece se does o ersec {0, 1} orvally 1 Iroduco A rough verso of he sodamerc equaly (see [12] saes ha R, amog all bodes of a gve dameer, he -dmesoal ball has he larges volume Aalogues of he sodamerc equaly are cosdered he dscree segs Recall ha he Hammg dsace d( x, y bewee wo vecors x ad y s he umber of coordaes whch hey dffer Solvg a cojecure of Erdős, [19] Klema proved he followg mpora heorem exremal se heory I ca also be vewed as a sodamerc equaly for dscree hypercubes Theorem 11 Suppose F s a colleco of bary vecors {0, 1}, such ha he Hammg dsace bewee ay wo vecors s a mos d < The ( ( 0 + ( 1 + +, for d = 2; F 2 (( ( , for d = Boh equales are sharp Whe d = 2, he upper boud s aaed by a Hammg ball of radus d/2, for sace, a colleco of bary vecors havg a mos 1-coordaes For d = 2 + 1, a opmal example s gve by he Caresa produc of {0, 1} ad he ( 1- dmesoal Hammg ball of radus, or aleravely, a Hammg ball of radus d/2 ceered a (1/2, 0,, 0 Klema s Theorem ca be reduced o he celebraed Kaoa erseco heorem [18], usg he squashg operao (see [2] or [14] Geeralzaos of hs heorem have bee suded for he -dmesoal grd [m] wh Hammg dsace [3, 13]; as well as [m] ad he -dmesoal orus Z m wh Mahaa dsace [1, 5, 8] Deparme of Mah ad CS, Emory Uversy, Alaa, USA Research suppored par by he Collaborao Gras from he Smos Foudao Deparme of Mahemacs, KTH Royal Isue of Techology, Sockholm, Swede Deparme of Mahemacs, Calfora Isue of Techology, Pasadea, USA 1

2 I hs paper, we gve a ew proof of Klema s Theorem, based o he followg boud by Cvekovć: he depedece umber α(g s bouded from above by he umber of o-egave (resp o-posve egevalues of he adjacecy marx of G Ths resul, whe exeded o pseudo-adjacecy marces, s sll correc A careful choce of a proper pseudo-adjacecy marx would lead o a algebrac proof of Klema s sodamerc heorem Ths mehod also allows us o prove a smlar upper boud, whe he allowed Hammg dsaces bewee wo vecors s a se of cosecuve egers I parcular, we show he followg ew esmae Theorem 12 For gve egers > s 0, suppose F s a colleco of bary vecors {0, 1}, such ha for every x, y F, d( x, y L, wh L = {2s + 1,, 2}, he for suffcely large, F ( + s s Smlarly, f L = {2s + 1,, 2 + 1}, he F (2 + o(1 ( s 0 Subses of he hypercube wh Hammg dsaces a prescrbed se of cosecuve egers appear he codg heory leraure he regme whe s ad are lear, for sace he coex of ɛ-balaced codes (of legh These are subses F of {0, 1} wh parwse Hammg dsaces bewee 1 ɛ 2 R va ad 1+ɛ 2 By mappg hem o vecors o he u sphere (v 1,, v 1 { (( 1 v1, ( 1 v2,, ( 1 v 1, 1 }, oe ca easly oe ha hs case esmag F amous o esmag he legh of a cera sphercal code, for whch oher mehods are useful We refer o [4] for more deals For our geeral rage ( parcular whe s ad are small compared o, he problem of upper boudg F s of a dffere aure, ad resuls abou sphercal codes do o apply I Seco 3, we dscuss several such exesos of Klema s Theorem o oher dsace ses A umber of oher echques cludg he Croo-Lev-Pach lemma esablshes asympocally sharp bouds for hese problems I Seco 4, we cosder a somewha dffere exremal se heory problem A se H Z + s ersecve f wheever A s a subse of posve upper desy of Z, we have (A A H I he lae 1970s, Sárközy [24], ad depedely Furseberg [16, 17], proved ha he se of perfec squares s ersecve A quaave verso has also bee cosdered Deoe by D(H, N he maxmum sze of a subse A {1,, N} such ha (A A H = I s o hard o see ha a se s ersecve f ad oly f D(H, N = o(n For may ersecve H, he asympocs of D(H, N have bee suded, here we refer he readers o a survey of Lê [20] Oe parcular eresg exremal problem s he aalogue of hs oo o vecor spaces over fe felds Cosder a N-dmesoal lace F N p, wh F p beg he fe feld of p elemes for some prme p Le J = {0, 1} N ad D Fp (J, N be he maxmum cardaly of H F N p such ha (H H J = A use of Sperer s Theorem shows ha D Fp (J, N = o(p N ad Alo (see [20] proved he followg bouds (he secod equaly beg a sace of he polyomal mehod: (p 1 N p N D F p (J, N (p 1 N I hs paper we also use he specral mehod o slghly mprove hs upper boud 2

3 Theorem 13 D Fp (J, N ( 1 1 ( 1 1 p (p 1 N 2 p 1 2 A algebrac proof of Klema s Theorem To pu hgs o moder coex, we sar hs seco wh a algebrac argume ha comes very close o provg Theorem 11, bu oly eds up gvg a weaker boud The ma dea s o use he followg lemma of Croo, Lev, ad Pach [6] Lemma 21 Le P F 2 [x 1,, x ] be a mullear polyomal of degree a mos d, ad le M deoe he 2 2 marx wh eres M x, y = P ( x + y for x, y F 2 The rak F2 (M 2 d/2 Ths was he ma grede he rece cap-se problem breakhrough [9] ad he drvg force behd may rece developmes addve combaorcs We refer he reader o [6] for he orgal applcao for whch was developed ad o [22] for a beer accou of s rece hsory ad a comprehesve ls of refereces For hs cpe dscusso, we oly address he case d = 2 We eumerae he elemes of F 2 ad cosder he 2 2 marx M defed by d( x, y 1 M x, y = := 2 (d( x, y 1 (d( x, y 2 (2! for every x, y F 2 Le M deoe he 2 2 bary marx obaed from M by reducg each eleme modulo 2 Noe ha every wo dsc vecors x, y F has Hammg dsace {1,, 2}, ad z 1 2 equals 0 for z {1,, 2}, ad o-zero for z = 0 Therefore he marx M resrcs o F F o a full-rak sub-marx, ad hus rak M rak M F O he oher had, here s a polyomal p F 2 [ 1,, ] wh deg p 2 so ha d( x, y 1 p( x y = mod 2, for every x, y F 2 2 Ths polyomal s gve explcly by p( 1,, = Ideed, oe ha for every x, y F 2, S {1,,}, S k S d( x, y 1 2 d( x, y = ( 1 l 2 l S =l S l=0 Furhermore, F 2 we also have ha ( d( x, y (x y = l, 3

4 so d( x, y 1 = 2 S {1,,}, S 2( 1 S S as clamed Lemma 21 he mmedaely mples F 2 (x y = S {1,,}, S 2 S Whe d = 2 + 1, s o o hard o adap he above argume o show ha ( 1 1 F (x y, Oe ca however mprove o hs rak argume ad esablsh he precse verso of Theorem 11 I fac, we wll prove Theorem 31, bu o keep hgs smple for he res of hs seco we wll sck o he case whe s = 0 whch recovers Theorem 11 We sar wh a few lemmas volvg smple lear algebra Le M,k be a 2 2 marx, whose rows ad colums are dexed by vecors {0, 1} The ( x, y-h ery of M,k s equal o 1 f ad oly f x ad y dffer exacly k coordaes, ad 0 oherwse For example, M,1 s he adjacecy marx of he -dmesoal hypercube, ad M,k s he adjacecy marx of a Hammg-ype graph whch wo verces are adjace f hey are a dsace k The followg lemma deermes he specrum of all M,k for all 1 k Lemma 22 The specrum of M,k cosss of K k (; wh mulplcy (, for = 0,, Here K k (; s he Krawchouk polyomal wh parameer 2: K k (; = k ( 1 j j k j j=0 For example, whe k = 1, s easy o check ha he egevalues of he -dmesoal hypercube are K 1 (; = 2 wh mulplcy ( The lemma ca be foud [21] (Theorem 301 For compleeess, we clude s proof below usg he Fourer rasform o hypercubes as egevecors Throughou he proof we use he oao d(u, V for he Hammg dsace bewee he dcaor vecors of U ad V, for wo subses U, V [] Proof Le v S be a vecor R 2 defed as (wh s 2 coordaes vewed as subses of []: ( v S T = ( 1 S T I s o hard o show ha { v S } S [] form a orhogoal bass O he oher had, (M,k v S T = (M,k T,U ( v S U = U [] U:d(U,T =k ( 1 S U Noe ha he umber of ses U wh he propery ha U ad T dffer j coordaes S s equal o ( S S j k j For each of such U, ( 1 S U = ( 1 S T ( 1 j, 4

5 sce S U = S T + (T U S 2 S T U Therefore S S S (M,k v S T = ( 1 S T ( 1 j = K k ( S ; ( v S T j k j j=0 Ths mmedaely shows ha K k (; are egevalues of M,k wh mulplcy ( From he proof of Lemma 22, observe ha for fxed, he egespace decomposo of M,k s he same for every k Hece s sraghforward o esablsh he followg resul Lemma 23 Suppose f(1,, f( s a sequece of real umbers ad le M = k=1 f(km,k The he specrum of M cosss of wh mulplcy (, for = 0,, λ = f(kk k (; k=1 The followg well-kow heorem sudes he relao bewee he specrum of a symmerc marx ad ha of s prcpal mor Lemma 24 (Cauchy s Ierlacg Theorem Le A be a symmerc marx of sze, ad B s a prcpal mor of A of sze m Suppose he egevalues of A are λ 1 λ 2 λ, ad he egevalues of B are µ 1 µ m The for 1 m, we have λ + m µ λ The followg corollary of he Cauchy s Ierlacg Theorem was dscovered earler by Cvekovć [7] I provdes a useful echque o boud he depedece umber of a graph Corollary 25 Le G be a -verex graph, ad M be a symmerc marx such ha M j = 0 wheever j E(G (such M s ofe called a pseudo-adjacecy marx of G Le 0 (M (resp 0 (M be he umber of o-posve (resp o-egave egevalues of M The he depedece umber of G sasfes α(g m{ 0 (M, 0 (M} Proof Suppose I s a maxmum depede se of G wh I = α(g The I aurally correspods o a all-zero prcpal mor B of M Ad he egevalues of B are µ 1 = = µ α(g = 0 Le λ 1 λ be he egevalues of M By Cauchy s Ierlacg Theorem, 0 = µ α(g λ α(g So M has a leas α(g o-egave egevalues Smlarly, λ 1+ α(g µ 1 = 0, whch mples ha M has a leas α(g o-posve egevalues Now we are ready o prove Theorem 11 5

6 Proof of Theorem 11 For gve, d, we defe a graph G whose verex se V (G = {0, 1}, ad wo verces are adjace f her Hammg dsace s a leas d + 1 Klema s problem s ow equvale o deermg he depedece umber α(g We sar wh he eve case d = 2 By Corollary 25 appled o G, suffces o fd real umbers f(k for k = 2 + 1,, ad defe M = k=2+1 f(km,k, such ha eher he umber of o-posve or o-egave egevalues of M s a mos ( A hs po, s perhaps mpora o meo ha choosg f(k = k 1 recovers he 2 2 symmerc marx M defed by M x,y = d(x,y 1 2 from he Croo-Lev-Pach approach, however hs s o gog o be he choce we are gog o make for he sequece f(1,, f( We choose f(k = ( l f k = 2l + 1 or k = 2l + 2 Equvalely f(k = (k 1/2 By Lemma 23, he egevalue of M wh mulplcy ( s equal o λ = k=2+1 f(k k ( 1 j, j k j j=0 for = 0,, Alhough compug he exac value of λ s mgh o be easy, urs ou ha we ca deerme her sgs a raher sraghforward way We clam ha for every, we have ( 1 λ > 0 for = 0,, λ = λ +1 for = 0,, 1 λ +1 = λ +2 = = λ = ( 1 +1 To show he above clams, we use geerag fucos ad observe ha λ s equal o he cosa erm of he followg formal power seres: ( ( f(kx k ( 1 j x j x l j l k=2+1 j=0 Here he geerag fuco, we may exed he sum ad he doma of f o all he egers greaer or equal o 2 + 1, wh f(k = (k 1/2 as before Ths would o affec he cosa erm sce for ( j ad l o be o-zero, oe mus have j ad l So f(k for oly hose k up o may corbue o he cosa erm A quck calculao shows ha + 1 f(kx k = (x (2+1 + x (2+2 + (x (2+3 + x (2+4 + k=2+1 ( + 1 = x (2+1 (1 + x 1 + x 2 + = x (2+1 (1 + x 1 (1 x 2 (+1 = x + 1 (x , l=0 whch he power seres coverges whe x > 1 Noe ha ( 1 j x j = (1 x ad j j=0 x l = (1 + x l l=0 6

7 Therefore, λ s equal o he cosa erm of he followg power seres: x + 1 (x (1 x (1 + x = ( 1 +1 (1 + x (1 x 1 For + 1, boh ad 1 are oegave, so he cosa erm s equal o ( 1 +1 For 0, oe eeds o cosder he cosa erm of +1 (1 + x (1 + x ( 1 = ( 1 (1 x +1 (x 1 +1 = ( 1 x (+1 (1 + x (1 1 x (+1 Obvously he expaso of (1 + x (1 1 x (+1 for x > 1, all he coeffces are posve So ( 1 λ s posve sce + 1 For >, oe ha a power seres, subsug x by x does o chage he cosa erm Therefore leg = + 1 j, λ s equal o he cosa erm of +1 (1 + x (+1 (1 + x( j ( 1 = ( 1+1 (1 x ( (1 x, (+1 j whch s exacly λ j = λ +1 Therefore for eve = 2m, he oly o-egave egevalues are λ 0, λ 2,, λ 2m, λ 1, λ 3,, λ (2m 1, ad her mulplces add up o ( Smlarly whe = 2m+1, he oly o-posve egevalues are λ 1,, λ 2m+1, λ,, λ 2m, ad her oal mulplcy equals as well Ths fshes he proof for he eve case The proof for he odd case d = works a smlar fasho, excep ha we have o choose f(k for k = 2 + 2,, a slghly dffere way Here we defe f(k = 0 for odd k, ( for eve k By a smlar argume, he egevalue λ wh mulplcy ad f(k = ( k/2 1 s equal o he cosa erm of he followg formal sum: ( f(kx k ( 1 j x j j k=2+2 j=0 I s equal o ( + 1 x (2+2 + x 2 + ( + 2 = x (2+2 (1 x 2 (+1 (1 x (1 + x = ( 1 +1 (1 x 1 (1 + x 1 ( x l l l=0 x 4 + (1 x (1 + x Oce aga, for + 1 1, he cosa erm equals ( 1 +1 For 0, s equal o ( 1 +1 (1 x (+1 (1 + x 1 = ( 1 x (+1 (1 + x 1 (1 1 x (+1 Aga oe ha he expasos of boh (1+x 1 ad (1 1 x (+1 oly coss of posve coeffces Therefore ( 1 λ > 0 Smlar as before, oe ca show ha for, λ = λ Now we apply Corollary 25 oce aga Noe ha oe of λ s s zero We oly eed o show ha eher he umber of posve or egave egevalues s small For eve = 2m, he oly posve λ are λ 0, λ 2,, λ 2m ad λ, λ 2,, λ 2m, whose oal mulplcy s 7

8 equal o 2 m ( 2 For odd = 2m + 1, he oly egave egevalues are λ1, λ 3,, λ 2m+1 ad λ 1, λ 3,, λ (2m+1, whose mulplcy s 2 m ( 2+1 Fally, s easy o check boh sum equals he sum Klema s Theorem for he case d = 2 + 1, og ha for eve = 2m, m ( m ( = 2 + = 2, ad for odd = 2m + 1, m ( m 2 = (( = Exesos o arbrary dsace ses I hs seco, we dscuss a few geeralzaos of Klema s heorem o oher ses of allowed dsaces The ex heorem shows ha a boud smlar o Klema s holds for all whe he se of allowed dsaces cosss of cosecuve egers Theorem 31 For gve egers > s 0, suppose F s a colleco of bary vecors {0, 1}, such ha for every x, y F, d( x, y L, wh L = {2s + 1,, 2}, he for all, F s s Proof We follow he proof of Theorem 11 for a dffere pseudo-adjacecy marx M = k=1 f(km,k Here for every eger l 0, we ake f(2l + 1 = f(2l + 2 = l s s By exedg he defo of bomal coeffces o he whole se of egers, we have ha f(k 0 f k 2+1, or 1 k 2s Therefore M s a pseudo-adjacecy marx for our purpose of boudg depedece umber The remag ask s o calculae he egevalues of M Usg smlar argumes as Theorem 11, we have ha M has a egevalue λ of mulplcy (, ad λ s equal o he cosa erm he followg formal power seres, for x > 1, ( f(kx k (1 x (1 + x k=1 Le g s (x = ( k s k=0 s x k, we wll frs show by duco (o s ha for x < 1, coverges o ( j=s j (x 1 j h s (x = (1 x s+1 For s = 0, g s (x = x + ( + 1 x +1 + = x /(1 x +1, 8

9 whch equals h s (x Assume for s 0, g s (x = h s (x, he ( ( h s+1 (x = h s (x(1 x ( 1 s = g s (x(1 x ( 1 s s s k s k s = x k x k+1 ( 1 s s s s k=0 k=0 k s k s 1 = x k x k ( 1 s s s s k=0 k=1 s 1 k s 1 = + x k ( 1 s = g s+1 (x s s 1 s k=0 Ths complees he proof ha g s (x = h s (x Now we have ( f(kx k (1 x (1 + x = (1 x (1 + x (x 1 + x 2 g s (x 2 k=1 ( = (1 x (1 + x +1 x 2 j=s j (x 2 1 j (1 x 2 s+1 (1 = ( 1 s+1 (1 + x j+s (1 x j+s 1 x 2(j s j Recall ha he egevalue λ of mulplcy ( s equal o he cosa erm he power seres Noe ha he sum s over all egers j bewee s ad I hs rage, j+s ( s, j + s 1 1 ( s, ad 2(j s s srcly greaer ha 0 excep for j = s So for ( s + 1 ( s, he produc he sum s a polyomal dvsble by x ad hus s cosa erm s 0 Oly j = s would corbue o he cosa erm Therefore for hs rage, λ = ( 1 s+1 s I oher words, for ( s + 1 ( s, λ have he same sg Ths would mmedaely mply j=s s α(g m{ 0 (M, 0 (M} + = s s = ( s+1 Remark For suffcely large, by a slghly more careful aalyss, oe ca acually remove he facors of 2 he saeme of Theorem 31, ad show ha F + + +, s s whch gves he exac same boud as Theorem 31 To acheve hs goal, oe ca show ha whe s s odd, λ 2 > 0 wheever 0 2 s ad λ 2 1 > 0 wheever 2 1 > ( s; 9

10 ad whe s s eve, λ 2+1 < 0 wheever s, ad λ 2 < 0 wheever 2 > ( s The applyg Corollary 25 gves he desre upper boud The calculaos are a b edous, so we decded o om he deals, sce sll gves a upper boud ha s asympocally he same, (1 + o(1 ( s Moreover, usg smlar echques, oe ca show ha f he se of allowed dsaces are {2s + 1,, 2 + 1}, he F (2 + o(1 ( s, geeralzg Klema s Theorem for odd dameers Whe L = {2s + 1,, 2}, Theorem 31 gves a upper boud whch s O( s for fxed s, ad large The followg heorem shows ha hs upper boud s gh up o a cosa facor o(1 ( Theorem 32 For suffcely large, here exss a famly F of (1/ ( s s bary vecors {0, 1}, such ha for every wo vecors x, y F, d( x, y L = {2s + 1,, 2} Proof We wll defe a famly F cossg of some vecors wh 2 1-coordaes For wo such vecors x ad y, deoe by X ad Y he -ses hey aurally correspod o The d( x, y {2s+1,, 2} s equvale o 4 2 X Y {2s+1,, 2}, e X Y {,, 2 s 1} By he famous resul of Rödl [23] o he Erdős-Haa Cojecure [11], for suffcely large, here exss a packg of m = (1 o(1 ( s / s copes of complee ( s-uform hypergraphs K s K s Suppose he verex se of hese hyperclques are V 1,, V m The V = ad V V j {0,, s 1} Take F = V { + 1,, } ad F = {F 1,, F m } I s easy o check ha F = 2 ad F F j {,, 2 s 1} For a se L of egers, le f L ( be he maxmum umber of bary vecors {0, 1} wh parwse Hammg dsace L The heorems above show ha (1 o(1 / f {2s+1,,2} ( (1 + o(1 s s s For s = 0 he upper ad lower bouds agree, as show by Theorem 11 For geeral s ad, s plausble ha he lower boud s asympocally gh We are able o verfy hs cojecure for he specal case L = {2s + 1, 2s + 2} We sar wh he followg lemma o subses of resrced erseco szes Lemma 33 Gve egers > j 1, suppose F s a colleco of -subses of [], whose parwse erseco has sze exacly j, he F (1 + o(1/( j Proof Suppose F = {F 1,, F m }, ad whou loss of geeraly assume F 1 = {1,, } For every j-subse S of [], le F S = {F : F [] = S}, he by he assumpo F = {F 1 } ( S F S Le F S = {F \S : F F S} The each o-empy F S cosss of parwse dsjo ( j-subses of { + 1,, } Ths mmedaely gves F S ( /( j We clam ha for wo dsc j-ses S ad T, f boh F S ad F T are o-empy, he hey boh coa a mos j ses Ths s because S T < j ad hus F S ad F T are cross-ersecg, ad ha a se U F S of sze j ca oly ersec wh a mos j parwse dsjo subses Therefore for all bu a mos oe se S, F S j Hece ( F 1 + F S ( j + = (1 + o(1 j j j S:S [], S =j Ths complees he proof 10

11 Theorem 34 For egers s 0, f {2s+1,2s+2} ( = (1 + o(1 s + 1 Proof Le F be a famly of m vecors {0, 1} wh parwse Hammg dsace eher 2s + 1 or 2s + 2 Whou loss of geeraly assume oe of hese vecors s he all-zero vecor, he he remag m 1 vecors are he dcaor vecors of subses of [] of sze 2s + 1 or 2s + 2 Deoe by A he famly of hese (2s + 1-ses, ad B he famly of (2s + 2-ses For wo ses A 1, A 2 A, we have A 1 + A 2 2 A 1 A 2 = A 1 A 2 {2s + 1, 2s + 2} By cosderg he pary, hs gves A 1 A 2 = s Smlar argumes show ha for wo ses B 1, B 2 B, B 1 B 2 = s + 1 Ad for A A, B B, A B = s + 1 Now we cosruc a ew famly A of subses of [+1], by addg he eleme +1 o each se A I s sraghforward o check ha C = A B sasfes he propery ha every se coas 2s + 2 elemes, whle every wo subses ersec exacly s + 1 elemes Now applyg Lemma 33 for C, we have C (1 + o(1/(s + 1, ad he same upper boud o F ad f {2s+1,2s+2} ( follows O he oher had, Theorem 32 wh = s + 1 gves f {2s+1,2s+2} ( (1 o(1/(s + 1 ad hs complees he proof Noe ha {2s + 1,, 2} s a se cossg of 2( s egers I s empg o speculae ha he order of magude of f L ( solely depeds o he sze of he se L of allowed dsaces However hs s false For example, suppose L oly cosss of odd dsaces The f L ( 2 sce f he famly coas hree vecors, her correspodg subses A, B, C sasfy 2( A B C A B C = A B + A C + B C 1 (mod 2, resulg a coradco Ths observao mmedaely leads o he followg smple upper boud for geeral L Theorem 35 Le L be a se of dsc posve egers Suppose c of hem are eve umbers The whe eds o fy, f L ( = O( c Proof Suppose L = {l 1,, l s }, ad F s a famly of vecors {0, 1} wh parwse Hammg dsaces L Whou loss of geeraly assume 0 F, ad he res of he vecors correspod o subses a famly A The every subse A has sze L = {l 1,, l s } Defe A = {A : A = l, A A}, for = 1,, s The for wo dsc subses X, Y A, her correspodg vecors have Hammg dsace equal o 2l 2 X Y = X + Y 2 X Y = X Y L Sce here are c eve umbers L, X Y belogs o a se of a mos c possble erseco szes By he Frakl-Wlso Theorem [15], A ( c + + ( 0, ad herefore F = 1 + A = 1 + s A = O( c =1 11

12 Alhough he problem of deermg he order of magude for every fxed dsace se L ad suffcely large seems beyod our reach, we ca sll esablsh asympocally sharp bouds for some oher specal dsace ses I fac, we have already esablshed oe a he begg of Seco 2 Recall ha we sared by usg he Croo-Lev-Pach Lemma o show a weaker verso of Theorem 11 Smlarly, we ca also prove he followg asympocally sharp esmae for a dffere ype of arhmec cosra o L Theorem 36 For gve egers, k such ha 2 k, le L coss of all he egers bewee 1 ad ha are o dvsble by 2 k The, for suffcely large, we have f L ( = (2 + o(1 2 k 1 1 The reader should compare hs o Theorem 35 Ths was also recorded depedely by Elleberg [10] Proof We sar by provg he upper boud Take a 2 2 marx M, whose rows ad colums correspod o -dmesoal bary vecors, ad M x, y = g( x y Here g : F 2 F 2 s he he followg polyomal: g( z = k 1 j=0 ( 1 ( z 2 j Here s he Hammg orm, so d( x, y = x y Suppose F s a famly of vecors such ha her parwse Hammg dsace s o dvsble by 2 k Therefore for dsc x, y F, he bary represeao of x y, he las k dgs are o all 0 A hs po, we recall he classcal Lucas heorem Lemma 37 Gve wo posve egers A B ad a prme umber p, suppose her p-ary represeao are A = s a p ad B = s b p, he A B s ( a b (mod p By Lemma 37, for some j {0,, k 1}, ( z 2 1 (mod 2 Therefore M x, y = g( x y 0 j (mod 2 O he oher had, obvously M x, x 1 (mod 2 Therefore he famly F aurally duces a submarx of M, whch s a u marx F 2 ad has full rak As a cosequece, F s upper-bouded by he F 2 -rak of M Noe ha deg(g = k 1 j=0 2j = 2 k 1 Lemma 21 mmedaely mples F 2 2 k 1 1 The lower boud ca be obaed aga by he same exremal cosruco for Klema s heorem, whe he allowed dsace se s {1,, 2 k 1} Theorem 11 gves f L ( 2 2 k 1 1 j=0 ( 1 j 12

13 4 O ersecve ses F N p I hs seco, we prove Theorem 13 Here we brefly skech he dea We cosruc a graph G wh verex se F N p, wo verces x ad y are adjace f x y or y x s J = {0, 1} N The α(g = D Fp (J, N We wll choose a pseudo-adjacecy marx for G ad apply Corollary 25 The followg lemma compues he specrum of a famly of marces, ha are aural caddaes for he pseudo-adjacecy marx of G Lemma 41 Le ω = e 2π/p ad M be a p N p N marx whose rows ad colums are dexed by vecors F N p, ad M u, v = f( u v, where f s a fuco mappg F N p o R The he fuco χ v : F N p C wh χ v ( u = ω u, v, whe vewed as a vecor, s a egevecor of M, correspodg o he egevalue f( xω v, x Moreover all of hem form a bass of R (pn Proof We frs verfy χ v s a egevecor of M We have x (Mχ v z = y M z, y χ v ( y = y f( z y ω v, y = f( xω v, z x = ω v, z x x = χ v ( z f( xω v, x I s sraghforward o show ha χ v are learly depede Now we are ready o prove Theorem 13 x f( xω v, x Proof of Theorem 13 From he dscussos a he begg of hs seco, we oly eed o upper boud he depedece umber of G We defe M o be a p N p N marx wh rows ad colums dexed by vecors F N p We le M u, v = ( 1 c( u v, for vecors u v wh eher u v or v u {0, 1} N ; ad 0 oherwse Here he fuco c maps a vecor F N p o s umber of o-zero coordaes Clearly M s a pseudo-adjacecy marx of G By Lemma 41, for every v = (v 1,, v N F N p, χ v s a egevecor of M wh egevalue equal o ( 1 c( x ω v, x + ( 1 c( x ω v, x Noe ha x {0,1} N \ 0 x {0,1} N ( 1 c( x ω v, x = = x {0,1} N ( 1 (1 ω v =1 x {0,1} N \ 0 N =1 x ω v, x = x {0,1} N =1 ( 1 x ω vx 13

14 Smlarly oe ca show ha x {0,1} N ( 1 c( x ω v, x = Therefore χ v correspods o he egevalue =1 (1 ω v =1 (1 ω v + (1 ω v 2 Whe v j = 0 for some dex j, ω vj = ω vj = 1, so hs gves egevalue 2 Oherwse all he v j {1,, p 1} Ths already shows ha he umber of o-egave egevalues s a mos (p 1 N, ad Corollary 25 gves a upper boud machg Alo s boud Bu fac we ca esmae he umber of o-egave egevalues more carefully Noe ha Smlarly, =1 =1 (1 ω vj = (1 cos(2πv j /p + s(2πv j /p = 2 s(πv j /p e (π/2 πvj/p = 2 s(πv j /p e (πn/2 π N (1 ω vj = 2 s(πv j /p e (πn/2 π N =1 vj/p vj/p Therefore N (1 ω v + (1 ω v 2 = 2 2 s(πv j /p cos πn/2 π v j /p 2 If s o-egave, he sce s(πv j /p 0 for v j {1,, p 1}, mus hold ha N cos πn/2 π v j /p > 0 Noe ha hs equaly cao hold for boh (v 1,, v N ad (u 1,, u N = (p 1 v 1,, p 1 v p, p v p+1,, p v N Sce N N cos πn/2 π u j /p = cos πn/2 π v j /p Therefore here are a leas half of hose (v 1,, v N [p 2] p [p 1] N p correspod o egave egevalues Therefore α(g 0 (M (p 1 N 1 2 (p 2p (p 1 N p ( = 1 1 ( 1 1 p (p 1 N 2 p 1 14

15 Whe p he cosa facor eds o 1 1/(2e Remark We beleve ha for geeral p, a more carefully aalyss of( he sgs of hese egevalues should show ha for a mos half of (v 1,, v N [p 1] N, cos πn/2 π N v j/p > 0 Ths would mprove he cosa o 1/2 For some small values of p, we ca acually oba beer cosas For example, whe p = 3, he same mehod gves α(g (1/3 + o(12 N 5 Cocludg Remarks I he frs par of hs paper, we gve a lear algebrac proof for Klema s damerc heorem, ad sudy several exesos ad geeralzaos Below are some observaos ad relaed problems We use he Cvekovć specral boud o fd a gh upper boud for he sze of a famly of bary vecors wh resrced dameer For he -dmesoal lace [m] equpped wh Hammg dsace, Ahlswede ad Khachara [3] showed ha maxmum famly of vecors wh parwse dsace a mos 2d s aaed by oe of he followg famles, smlar o her celebraed Complee Ierseco Theorem For a vecor v [m], le S v = { : v = 0} The he famles are F = { v : v [m], S v [ 2] d }, for = 0,, d I parcular, for fxed m, d ad suffcely large, F 0, e he Hammg ball of radus d, has he maxmum sze amog F s I would be eresg o see wheher he specral echque we used o prove Klema s Theorem ca also be appled hs case Noe ha he proof of Theorem 11 works almos he same f we replace he geerag fuco k=2+1 f(kx k = (x + 1/(x by he raoal fuco ((m 1x + 1/((x 1((m 1x The clams regardg sgs of λ sll work, bu uforuaely oly leads o a upper boud F + (m 1 d 1 + m 2 +, sead of he desred upper boud F F 0 = + (m (m 1 d 0 1 d Bu s sll possble ha by properly choosg a geerag fuco f (or equvalely a pseudo-adjacecy marx, oe ca prove he gh upper boud I Seco 3, we show ha f {2s+1,,2} ( s of he order Θ( s, ye we oly deerme he exac cosa facor for s = 0 (Klema s Theorem, ad for = s + 1 (Theorem 34 We beleve ha he lower boud cosruco Theorem 32 s asympocally bes possble Noe ha by our mehod of cosderg he pseudo-adjacecy marx of he form k f(km,k, he Cvekovć boud always gves a sum of bomal coeffces he form of ( Maybe pckg a more complcaed pseudo-adjacecy marx so ha s (S, T -ery does o solely depeds o S T would be useful here I Seco 3, we also esablsh a upper boud o f L ( for geeral L, usg he umber of eve umbers L I s o hard o see ha he upper boud Theorem 35 s gh up 15

16 o a cosa facor, for all L = {2s + 2, 2s + 4,, 2} L, wh L oly cossg of odd umbers We are curous wheher here are oher ses L of allowed dsaces ha sasfy hs propery Also would be eresg o show ha he lm log f L ( lm log exss for all L Ad perhaps he lm mus be of eger value Klema s damerc heorem may also be geeralzed he followg way Gve a coeced smple graph G = (V, E, he dsace of wo dsc verces u, v s he legh of he shores pah coecg hem Le f(g, d be he maxmum sze of a subse of verces wh parwse dsaces a mos d Ad he subse of verces ha play he role of Hammg ball s eher N d/2 (v, all he verces a dsace a mos d/2 from v, for some v V (G whe d s eve; or N (d 1/2 (u N (d 1/2 (v for some edge uv E(G, whe d s odd Oe could ask a geeral queso: for fxed eger d, wha graph G sasfes he sodamerc equaly, e f(g, d s aaed by oe of he ses defed above? We call such graphs G d-sodamerc For example, Klema s Theorem says ha he hypercube Q s d-sodamerc for d 1 I s easy o see ha a graph s 1-sodamerc f ad oly f s ragle-free Is possble o characerze all d-sodamerc graphs, or a leas amog Cayley graphs or verex-rasve graphs? Ackowledgmes We would lke o hak Noga Alo, Fedor Perov ad Wll Saw for several useful dscussos Refereces [1] R Ahlswede, N Ca, ad Z Zhag, Damerc heorems sequece spaces, Combaorca 12(1 (1992, 1 17 [2] R Ahlswede, G O H Kaoa, Corbuos o he geomery of Hammg spaces, Dscree Mah 17 (1977, 1 22 [3] R Ahlswede, L H Khachara, The damerc heorem Hammg spaces opmal acodes, Advaces Appled Mahemacs 20 (1998, [4] N Alo, Perurbed dey marces have hgh rak: proof ad applcaos, Comb Probab Compu 18 (2009, 3 15 [5] B Bollobás, I Leader, Maxmal ses of gve dameer he grd ad he orus, Dscree Mahemacs 122 (1993, [6] E Croo, V Lev ad P Pach, Progresso-free ses Z 4 are expoeally small, A of Mah 185(1 (2017, [7] D M Cvekovć, Chromac umber ad he specrum of a graph, Publ Is Mah (Beograd 14 (28 (1972, [8] D Z Du, D J Klema, Dameer ad radus he Mahaa merc, Joural of Dscree ad Compuaoal Geomery, 5(4 (1990, [9] J S Elleberg, D Gjswj, O large subses of Z q wh o hree-erm arhmec progresso, A of Mah 185(1 (2017,

17 [10] J S Elleberg, Dfferece ses mssg a hammg sphere, blog pos a hps://quomodocumquewordpresscom/2017/02/11/dfferece-ses-mssg-a-hammgsphere/ [11] P Erdős, H Haa, O a lm heorem combaoral aalyss, Publ Mah Debrece, 10 (1963, [12] L C Evas, R F Garepy, Measure heory ad fe properes of fucos, CRC Press, 1991 [13] P Frakl, Z Fured, The Erdős-Ko-Rado heorem for eger sequeces, SIAM J Algebrac Dscree Mehods, 1(4 (1980, [14] P Frakl, N Tokushge, Exremal problems for fe ses, STML 86, Amerca Mahemacal Socey [15] P Frakl ad R M Wlso, Ierseco heorems wh geomerc cosequeces, Combaorca 1 (1981, [16] H Furseberg, Recurrece Ergodc Theory ad Combaoral Number Theory, Prceo Uv Press, 1981 [17] H Furseberg, Ergodc behavor of dagoal measures ad a heorem of Szemeréd o arhmec progressos, J D Aalyse Mah, 71 (1977, [18] G O H Kaoa, Ierseco heorems for sysems of fe ses, Aca Mah Hugar 15 (1964, [19] D Klema, O a combaoral cojecure of Erdős, J Comb Theory Ser A 43 (1986, [20] T H Lê, Problems ad resuls o ersecve ses, Combaoral ad addve umber heory, Sprger Proceedgs Mahemacs & Sascs 101 (2014, [21] J H va L ad R M Wlso A Course Combaorcs, Cambrdge Uversy Press, 2001 [22] L M Lovász ad L Sauerma, A lower boud for he k-mulcolored sum-free problem Z m, Proceedgs of he Lodo Mahemacal Socey, o appear [23] V Rödl, O a packg ad coverg problem, Europea J Comb 6 (1985, [24] A Sárközy, O dfferece ses of sequeces of egers, I, Aca Mah Acad Sc Hugar 31 (1978,