The isoperimetric problem on the hypercube

Transcription

1 The isoperimetric problem o the hypercube Prepared by: Steve Butler November 2, The isoperimetric problem We will cosider the -dimesioal hypercube Q Recall that the hypercube Q is a graph whose vertices ca be represeted by biary words of legth 1, ie, V = {0, 1} The edges of Q joi vertices whose words differ i eactly oe coordiate Aother iterpretatio is that the vertices are the possible subsets of a -set, ie, V = {s : s []}, ad edges joi two subsets s, t if the symmetric differece of the two subsets, (s t) ( s t), is a oe elemet set The origial isoperimetric problem deals with fidig a figure i the Euclidea plae that will eclose a give area with miimal perimeter (i other words, of all figures with the same area, fid the miimal perimeter) To geeralize this to graphs we eed to decide what to cosider as the area ad perimeter The area for the hypercube will ituitively be the umber of vertices (Note the hypercube is regular ad we are leaig o that for our ituitio, for oregular graphs we might try to capture more vertices of higher degree ad so the volume as defied i previous lectures for graphs would become a better choice) For the perimeter (or boudary) we have two choices, we ca use either edges or vertices The verte boudary of a set S is δ(s) = {v / S : v u, u S}, the vertices which are adjacet to a elemet i S but are ot i S This is the miimum umber of vertices that must be removed to separate S from the other vertices The edge boudary is similarly defied as all edges which coect a verte i S to a verte ot i S ad is the miimum umber of edges whose removal discoects S from the other vertices of the graph δ(s) S Q 1

2 It is importat to ote that the verte boudary ad edge boudary have very differet behaviors ad so we must aalyze them separately As a eample if we wat to cut Q i half the the optimal way to do it usig a edge boudary is to split the vertices ito two subcubes ad the remove all edges betwee the cubes O the other had if we wated to split the cube i half usig verte boudaries the the optimal thig is to use Hammig balls (which we defie below) 11 Hammig balls Sice we have a otio of distace o a graph (ie, the legth of the shortest path joiig the two vertices), the the otio of a ball geeralizes easily Sice the ball solved the isoperimetric problem i Euclidea space it is reasoable to epect that it might agai solve the problem i graphs So let B r (v) = {u : d(u, v) r} deote a ball of radius r cetered at v i our graph, i Q these are called Hammig balls For the hypercube it is easy to calculate the umber of vertices cotaied i such a ball Namely we have b r = B r (v) = r r 1 0 [Here we ca eploit the high symmetry of the hypercube ad ote that the behavior at oe poit is the same as ay other poit by a easy automorphism, i this case the volume of the ball is idepedet of v] The volume equatio follows by recallig that every verte ca be represeted by a biary word of legth ad it is easy to show that the distace betwee two vertices i Q is the umber of etries i their ide which disagree So i the above calculatio ( r) is the umber of vertices distace r away from v (ie, we choose r out of etries at which their words disagree), ad i geeral ( k) is the umber of vertices at distace k away from v 12 The isoperimetric problem defied We will aswer the followig questio: Give m > 0 fid f(m) = mi S V { δ(s) : S = m, S V (Q )} I other words, what is the smallest umber of vertices eeded to separate m vertives from the rest of the graph We wat to determie both f(m) ad the sets S which achieve this miimum value While i geeral this is difficult to calculate for a graph, for the hypercube the results have log bee kow Whe m = b r, a good guess would be that the set S is a Hammig ball ad that f(m) = ( r+1) (the et layer of vertices from the ceter verte) We will show that is the case ad that i geeral that the sets S are ear Hammig balls 2

3 2 Mai results We have two ways to state the mai results, the first will be a rough estimate but is useful The secod will be a precise estimate but is somewhat messy ( ) ( ) Theorem 1 Suppose m = b r + < b r+1 the f(m) b r+1 m + r 1 r 2 I the statemet of the theorem the i ( k) eed ot be a iteger I geeral we have for real ad k 0 that ( ) = k ( 1) ( k + 1) k! We must be careful i usig these biomial coefficiets as some properties o loger hold i this more geeral settig, for istace ( ) ( k k) i geeral Theorem 2 Suppose ( ) m = ( ) ( ) s, 0 s < 1 k + 1 Let a k > a k 1 > > a t t 1 be as large as possible ad satisfyig ak ak 1 at s = k k 1 t ak ak 1 at The f(m) = b k m k 1 k 2 t 1 ( ) k [Note: the epressio for f(m) is very suggestive about what to iclude i the boudary] Our first step will be to show that the right object to study are the ear Hammig balls (these are Hammig balls which are possibly missig some vertices i the last level) I particular, the above results will the follow Theorem 3 Give m, the f(m) = mi{ δ(s) : S = m, S a ear Hammig ball} 21 Pushig Hammig balls To show Theorem 3 the idea will be to push ay set ito a Hammig ball without icreasig the size of the verte boudary I particular we eed the followig result Theorem 4 Give A, B V (Q ), with d(a, B) = mi{d(a, b) : a A, b B} beig the miimal distace betwee the sets The there are two atipodal ear Hammig balls A, B with A = A, B = B ad d(a, B ) d(a, B) 3

4 Atipodal meas that the ceter of the two ear Hammig balls are at vertices whose idices are complemets of each other The vague idea behid this is that if we have two sets we push oe to the top ad oe to the bottom ad our distace ca oly icrease We ow show that Theorem 3 follows from Theorem 4 Suppose that we have a set A with A = m, δ(a) = f(m) The let B = V A δ(a), ote that d(a, B) = 2 (ie, the shortest path is go from a verte i A adjacet to the boudary to the boudary to a verte i B) By Theorem 4 there eists sets A, B which are ear-hammig balls where A = A = m ad B = B with d(a B ) d(a, B) = 2 It suffices to show that the set A has δ(a ) f(m) Note that poits i δ(a ) caot be i A ad caot be i B (sice A ad B are at least distace 2 apart) ad so ad the result follows δ(a ) 2 A B = 2 A B = δ(a) = f(m), Proof of Theorem 4 We first cosider the followig sets S 1 = {(a, a ) : a A, a / A, ad a < a }, S 2 = {(b, b ) : b B, b / B, ad b > b } By a we mea either the umber of 1 s i the biary word ideig the verte, or the size of the subset ideig the verte (depedig o which iterpretatio we use) Ituitively we are tryig to push A up to be cetered aroud the all 1s verte (respectively []) ad pushig B dow to be cetered aroud the atipodal poit of the all 0s verte (respectively ) If S 1 = the A is a ear Hammig ball cetered at the top ad if S 2 = the B is a ear Hammig ball cetered at the bottom If both S 1 ad S 2 are empty the we are doe If oe of them is oempty the we ca push, we ow assume that we are i the case where oe of the sets is oempty Without loss of geerality let us suppose that (a, a ) S ad (a, a ) has miimum symmetric differece, ie, (a a ) (a a) is miimized over all elemets i S 1 ad S 2 Let = a a ad y = a a where < y We defie two operators U for a A ad D for b B (these are the up ad dow operators respectively) For the followig we will thik of the subset covetio of deotig vertices { (a y) if a, y a =, (a y) / A; U(a) = a otherwise { (b ) y if y b, b =, (b ) y / B; D(b) = b otherwise What is happeig is that for a A we (if possible) cut out the small set ad replace it by the larger set y movig that elemet up (we kow that this is possible for oe such elemet ad so we will reduce the size of S 1 ) Similarly for b B we (if possible) cut out the large set y ad replace it by the smaller set movig that elemet dow 4

5 Because we do ot move a elemet if the ew elemet already eists i the set it is easy to see that U(A) = A ad D(B) = B Sice we also have reduced the size of S 1 we ca repeat this procedure oly fiitely may times util S 1 = ad S 2 = at which poit we will have atipodal Hammig balls The oly remaiig step is to check that for a A, b B with a = U(a) ad b = D(b) that d(a, b ) d(a, b ) for some a A, b B [Note: i geeral we eed ot have a = a ad b = b ] Eercise 1 Complete the proof of Theorem 4 (Hit: there are a few cases to cosider, but all of the key ideas are already give above) 5