aa r X i v : . [ m a t h . C O ] S e p Problems and results in Extremal Combinatorics - IV
Noga Alon ∗ Abstract
Extremal Combinatorics is among the most active topics in Discrete Mathematics, deal-ing with problems that are often motivated by questions in other areas, including TheoreticalComputer Science and Information Theory. This paper contains a collection of problems andresults in the area, including solutions or partial solutions to open problems suggested by variousresearchers. The topics considered here include questions in Extremal Graph Theory, CodingTheory and Social Choice. This is by no means a comprehensive survey of the area, and ismerely a collection of problems, results and proofs, which are hopefully interesting. As the titleof the paper suggests, this is a sequel of three previous paper [5], [7], [8] of the same flavour.Each section of this paper is essentially self contained, and can be read separately.
We say that two n -vertex graphs G and G pack if there exists an edge-disjoint placement of themon the same set of n vertices. There is an extensive literature dealing with sufficient conditionsensuring that two graphs G and G on n vertices pack. A well known open conjecture on thesubject is the one of Bollob´as and Eldridge [16] asserting that if the maximum degrees in G and G are d and d , respectively, and if ( d + 1)( d + 1) ≤ n + 1 then G and G pack. Sauer andSpencer ([37], see also Catlin [17]), proved that this is the case if 2 d d < n . For a survey of packingresults including extensions, variants and relevant references, see [30].A natural extension of the packing problem is that of requiring a packing in which the girth ofthe combined graph whose edges are those of the two packed graphs is large, assuming this is thecase for each of the individual graphs. Indeed, in the basic problem the girth of each of the packedgraphs exceeds 2, and the packing condition is simply the requirement that in the combined graphthe girth exceeds 2. Here we prove such an extension, observe that it implies the old result of Erd˝osand Sachs [20] about the existence of high- girth regular graphs, and describe an application forobtaining an explicit construction of high-girth directed expanders. Theorem 1.1.
Let G = ( V , E ) and G = ( V , E ) be two n -vertex graphs, let d be the maximumdegree of G and let d be the maximum degree of G . Suppose the girth of each of the graphs G i ∗ Department of Mathematics, Princeton University, Princeton, NJ 08544, USA and Schools of Mathematics andComputer Science, Tel Aviv University, Tel Aviv 69978, Israel. Email: [email protected] . Research supported inpart by NSF grant DMS-1855464, ISF grant 281/17, BSF grant 2018267 and the Simons Foundation. s at least g > and let k be the largest integer satisfying d + d ) + ( d + d )( d + d −
1) + . . . + ( d + d )( d + d − k − < n (1) Then there is a packing of the two graphs so that the combined graph has girth at least min { g, k } . Note that for fixed d + d ≥ n , the number k above is (1 + o (1)) log n log( d + d − . Clearly we may assume that both G and G have edges, thus d and d are positive. If 2 d d ≥ n then d + d ≥ √ n implying that 1 + ( d + d ) + ( d + d )( d + d − ≥ √ n + √ n ( √ n −
1) =2 n + 1 > n , that is, the largest k satisfying (1) is at most 1. In this case the conclusion is trivialsince min { g, k } ≤ G , G will do. We thus may and will assume that2 d d < n . Since any placement will do even if k = 2 we assume that k ≥
3. By the result of [37], G , G pack. Among all possible packings choose one in which the girth m of the combined graphis maximum, and the number of cycles of length m in this combined graph is minimum (if the girthis infinite there is nothing to prove). Suppose this packing is given by two bijections f : V V and f : V V where V is the fixed set of n vertices of the combined graph, which we denote by H = ( V, E ). As G and G pack, m ≥
3. refe111). Let v = f − ( v ) be the preimage of v in V .Let f ′ : V V be the bijection obtained from f by swapping the images of u and v . Formally, f ′ ( u ) = v, f ′ ( v ) = u , and f ′ ( w ) = f ( w ) for all w ∈ V − { u , v } .We claim that in the embedding of G , G given by f ′ , f the girth of the combined graph, callit H ′ , is at least m and the number of cycles of length m in H ′ is smaller than the correspondingnumber in H , contradicting the minimality in the choice of f , f . To prove this claim put u = f − ( u ), v = f − ( v ). Let X denote the set of images under f of all the neighbors of u in G ,and let X denote the set of images under f of all neighbors of u in G . Similarly, let Y be theset of images under f of all the neighbors of v in G , and let Y be the set of images under f of all neighbors of v in G . Note that since m ≥ k + 1 ≥ X , X , Y , Y arepairwise disjoint. The cycles of length m in H and H ′ that do not contain any of the two vertices u, v are exactly the same cycles. On the other hand, the cycle C is of length m and it exists in H but not in H ′ , since all edges of H between u and X do not belong to H ′ , and C contains such anedge (as well as an edge from u to X ). Any cycle C ′ of H ′ that is not a cycle of H must containat least one edge either between u and Y or between v and X (or both). Consider the followingpossible cases. Case 1a: C ′ contains u but not v and contains two edges from u to Y . In this case the cycleof H obtained from C ′ by replacing u by v is of the same length as C ′ . This is a one-to-onecorrespondence between cycles as above of length m in H ′ and in H (if there are any such cycles). Case 1b: C ′ contains u but not v and contains an edge uy from u to Y and an edge ux from u to X . In this case the part of the cycle between y and x which does not contain u is a path in H between y and x . The length of this path is at least k −
1, since the distance in H between u and v is at least k + 1. Therefore, the length of C ′ is at least ( k −
1) + 2 = k + 1 > m .2 ase 1c: C ′ contains v but not u : this is symmetric to either Case 1a or Case 1b. Case 2a: C ′ contains both u and v and contains two edges uy , uy ′ from u to Y . If bothneighbors of v in C ′ belong to Y then each of the parts of C ′ connecting any of them to y or to y ′ is of length at least m −
2, since the girth of H is m , hence the total length of C ′ is at least2( m −
2) + 4 = 2 m > m . If both neighbors of v in C ′ are in X then since the distance in H between X and Y is at least k −
1, in this case the length of C ′ is at least 2( k −
1) + 4 = 2 k + 2 > m . Ifthe two neighbors of v in C ′ are y ∈ Y and x ∈ X then the cycle C ′ contains a path from y toeither y or y ′ , whose length is at least m −
2, and a path from x to either y or y ′ , of length atleast k −
1. Thus the total length of C ′ is at least ( m −
2) + ( k −
1) + 4 > m . Case 2b: C ′ contains both u and v and the two neighbors of u in C ′ are y ∈ Y and x ∈ X .In this case the path in C ′ from v to y is of length at least m − X ,and at least k if it passes through X , and the path from v to x is of length at least k if it doesnot pass through X and of length at least m − X . In all these cases thelength of C ′ is at least 2 + 2 min { m − , k } = 2 m > m (where here we used the assumption that m < k ). Case 2c: C ′ contains both u and v and at least one edge from v to X . This is symmetric toeither Case 2a or Case 2b.It thus follows that the number of cycles of length m in H ′ is smaller than that number in H ,contradicting the minimality in the choice of H and implying that m ≥ min { g, k } . This completesthe proof of the theorem. (cid:3) Remark:
The above proof is constructive, that is, it provides a polynomial algorithm to find apacking of given graphs G , G as above, with the asserted bound on the girth of the combinedgraph. Indeed, as long as the girth is too small we can find a shortest cycle C , take in it a vertex u as in the proof, find a vertex v far from it and swap their roles in the image of G . By the argumentabove this decreases the number of short cycles by at least 1. As the total number of such cycles isless than n by the choice of the parameters and by (1), this process terminates in polynomial time. By applying Theorem 1.1 repeatedly, starting with a cycle of length n , it follows that for every d and all large n there is a 2 d -regular graph on n vertices with girth at least (1 + o (1)) log n log(2 d − whichcan be decomposed into d Hamilton cycles. This is a (modest) strengthening of the result of Erd˝osand Sachs about the existence of regular graphs of high girth. A more interesting application ofTheorem 1.1 is a strengthening of a result proved in [12] about the existence of high-girth directedexpanders.
Theorem 1.2.
For every prime p congruent to modulo and any n > n ( d ) there is an explicitconstruction of a d -regular graph on n vertices with (undirected) girth at least ( − o (1)) log n log( d − nd an orientation of this graph so that for every two sets of vertices X, Y satisfying | X | n · | Y | n ≥ d (2) there is a directed edge from X to Y and a directed edge from Y to X . This improves the estimate on the girth in the result proved in [12] by a factor of 3, andalso works for all large n . The proof combines Theorem 1.1 with an argument from [12] and arecent result proved in [9]. An explicit construction here means that there is a polynomial timedeterministic algorithm for constructing the desired graphs. Proof.
An ( n, d, λ )-graph is a d regular graph on n vertices in which the absolute value of anynontrivial eigenvalue is at most λ . The graph is Ramanujan if λ = 2 √ d −
1. Lubotzky, Phillipsand Sarnak [33], and independently Margulis [34] gave, for every prime p congruent to 1 modulo4, an explicit construction of infinite families of d = p + 1-regular Ramanujan graphs. The girth ofthese graphs is at least (1 + o (1)) log d − n ′ , where n ′ is the number of vertices. In [9] it is shownhow one can modify these graphs by deleting a set of appropriately chosen n ′ − n vertices and byadding edges among their neighbors to get an ( n, d, √ d − o (1))-graph with exactly n verticeskeeping the girth essentially the same. Fix such a graph H . By Theorem 1.1 we can pack twocopies of it H , H keeping the girth of the combined graph at leastmin { (1 + o (1)) 23 log d − n, (1 + o (1)) log d − n } = (1 + o (1)) 23 log d − n, where here we used the fact that for all admissible d ,23 log( d − ≤ d − . Let G be the combined graph. Number its vertices 1 , , . . . , n and orient every edge ij with i < j from i to j if it belongs to the copy of H and from j to i if it belongs to the copy of H .It is well known (c.f. [13], Corollary 9.2.5) that if A, B are two subsets of an ( n, d, λ )-graph and | A || B | n > λ d then there is an edge connecting A and B . Let X and Y be two sets of vertices satisfying (2). Let x be the median of X (according to the numbering of the vertices), y the median of Y . Without lossof generality assume that x ≤ y . Let A be the set of all vertices of X which are smaller or equal to x , B the set of all vertices of Y that are larger or equal to y . Then | A | ≥ | X | / | B | ≥ | Y | / | A || B | n ≥ | X || Y | n ≥ d > (2 √ d − o (1)) d . Therefore there is an edge of H connecting A and B which, by construction, is oriented from A to B . Similarly there is an edge of H oriented from B to A . This completes the proof.4 Nearly fair representation
The approach described here was initiated in discussions with Eli Berger and Paul Seymour [15].Let G = ( V, E ) be a graph and let P be an arbitrary partition of its set of edges into m pairwisedisjoint subsets E , E , . . . , E m . The sets E i will be called the color classes of the partition. For anysubgraph H ′ = ( V ′ , E ′ ) of G , let x ( H ′ , P ) denote the vector ( x , x , . . . , x m ), where x i = | E i ∩ E ′ | is the number of edges of H ′ that lie in E i . Thus, in particular, x ( G, P ) = ( | E | , . . . , | E m | ) . In acompletely fair representation of the sets E i in H ′ , each entry x i of the vector x ( H ′ , P ) should beequal to | E i | · | E ′ || E | . Of course such equality can hold only if all these numbers are integers. Buteven when this is not the case the equality may hold up to a small additive error.In this section we are interested in results (and conjectures) asserting that when G is either thecomplete graph K n or the complete bipartite graph K n,n , then for certain graphs H and for anypartition P of E ( G ) into color classes E , . . . , E m , there is a subgraph H ′ of G which is isomorphic to H so that the vector x ( H ′ , P ) is close (or equal) to the vector x ( G, P ) | E ( H ′ ) || E ( G ) | . Stein [38] conjecturedthat if G = K n,n and P is any partition of the edges of G into n sets, each of size n , then thereis always a perfect matching M in G satisfying x ( M, P ) = n x ( G, P ), that is, a perfect matchingcontaining exactly one edge from each color class of P . This turned out to be false, a clever counter-example has been given by Pokrovskiy and Sudakov. In [35] they describe a partition of the edgesof K n,n into n sets, each of size n , so that every perfect matching misses at least Ω(log n ) colorclasses.In [1] it is conjectured that when G = K n,n , P is arbitrary, and H is a matching of size n , thenthere is always a copy H ′ of H (that is, a perfect matching H ′ in G ), so that k x ( H ′ , P ) − n x ( G, P ) k ∞ < . This is proved in [1] (in a slightly stronger form) for partitions P with 2 or 3 color classes. Herewe first prove the following, showing that when allowing a somewhat larger additive error (whichgrows with the number of colors m but is independent of n ) a similar result holds for partitionswith any fixed number of classes. Theorem 2.1.
For any partition P of the edges of the complete bipartite graph K n,n into m colorclasses, there is a perfect matching M so that k x ( M, P ) − n x ( K n,n , P ) k ∞ ≤ k x ( M, P ) − n x ( K n,n , P ) k < ( m − (3 m − / . It is worth noting that a random perfect matching M typically satisfies k x ( M, P ) − n x ( K n,n , P ) k ∞ ≤ O ( √ n ) . The main challenge addressed in the theorem is to get an upper bound independent of n .Theorem 2.1 is a special case of a general result which we describe next, starting with thefollowing definition. 5 efinition 2.1. Let G be a graph and let H be a subgraph of it. Call a family of graphs H (whichmay have repeated members) a uniform cover of width s of the pair ( G, H ) if every member H ′ of H is a subgraph of G which is isomorphic to H , the number of edges of each such H ′ which are notedges of H is at most s , every edge of H belongs to the same number of members of H , and everyedge in E ( G ) − E ( H ) belongs to the same positive number of members of H . An example of a uniform cover of width s = 2 for G = K n,n and H a perfect matching in it isthe following. Let the n edges of H be a i b i where { a , a , . . . , a n } and { b , b , . . . , b n } are the vertexclasses of G . Let H be the family of all perfect matchings of G obtained from H by omitting apair of edges a i b i and a j b j and by adding the edges a i b j and a j b i . The width is 2, every edge of H belongs to exactly (cid:0) n (cid:1) − ( n −
1) members of H , and every edge in E ( G ) − E ( H ) belongs to exactly1 member of H . Theorem 2.2.
Let G be a graph with g edges, let F be a subgraph of it with f edges, and supposethere is a uniform cover of width s of the pair ( G, F ) . Then for any partition P of the edges of G into m -subsets, there is a copy H of F in G so that k x ( H, P ) − fg x ( G, P ) k ∞ ≤ k x ( H, P ) − fg x ( G, P ) k ≤ ( m − ( m − / s m . Theorem 2.1 is a simple consequence of Theorem 2.2. A similar simple consequence is thefollowing.
Proposition 2.3.
For any partition P of the edges of the complete graph K n into m color classes,there is a Hamilton cycle C so that k x ( C, P ) − n − x ( K n , P ) k ∞ ≤ k x ( C, P ) − n − x ( K n , P ) k < ( m − (3 m − / . Similar statements follow, by the same reasoning, for a Hamilton cycle in a complete bipartitegraph, or for a perfect matching in a complete graph on an even number of vertices. We proceedto describe a more general application.For a fixed graph T whose number of vertices t divides n , a T -factor in K n is the graph consistingof n/t pairwise vertex disjoint copies of T . In particular, when T = K this is a perfect matching. Theorem 2.4.
For any fixed graph T with t vertices and q edges and any m there is a constant c = c ( t, q, m ) ≤ ( m − ( m − / ( qt ) m so that for any n divisible by t and for any partition P of theedges of the complete graph K n into m subsets, there is a T -factor H so that k x ( H, P ) − q ( n − t x ( K n , P ) k ∞ ≤ k x ( H, P ) − q ( n − t x ( K n , P ) k ≤ c. We start with the proof of Theorem 2.2. 6 roof.
Let P be a partition of the edges of G into m color classes E i . Put y = ( y , y , . . . , y m ) = fg x ( G, P ) . Let H be a copy of F in G for which the quantity k y − x k = P mj =1 ( y i − x i ) is minimum where x = ( x , x , . . . , x m ) = x ( H, P ). Let H be a uniform cover of width s of the pair ( G, H ). Supposeeach edge of H belongs to a members of H and each edge in E ( G ) − E ( H ) belongs to b > H ′ of H , let v H ′ denote the vector of length m defined as follows. Foreach 1 ≤ i ≤ m , coordinate number i of v H ′ is the number of edges in E ( H ′ ) − E ( H ) colored i minus the number of edges in E ( H ) − E ( H ′ ) colored i . Note that the ℓ -norm of this vector is atmost 2 s and its sum of coordinates is 0. Therefore, its ℓ -norm is at most √ s . Note also that x ( H ′ , P ) = x ( H, P ) + v H ′ .We claim that the sum S of all |H| -vectors v H ′ for H ′ ∈ H is a positive multiple of the vector( y − x ). Indeed, each edge in E ( G ) − E ( H ) is covered by b members of H , and each edge of E ( H ) iscovered by a members of H . In the sum S above this contributes to the coordinate correspondingto color number i , b times the number of edges of color i in E ( G ) − E ( H ) minus ( |H| − a ) times thenumber of edges of color i in H . Equivalently, this is b times the number of all edges of G colored i minus ( |H| + b − a ) times the number of edges of H colored i . Since the sum of coordinates of eachof the vectors v H ′ is zero, so is the sum of coordinates of S , implying that bg = ( |H| + b − a ) f , thatis, |H| + b − a = gf b . Since gf y = x ( G, P ) this implies that S = bgf ( y − x ), proving the claim.Since the vector y − x is a linear combination with positive coefficients of the vectors v H ′ itfollows, by Carath´eodory’s Theorem for cones, that there exists a set L of linearly independentvectors v H ′ so that y − x is a linear combination with positive coefficients of them. Indeed, startingwith the original expression of y − x mentioned above, as long as there is a linear dependence amongthe vectors v H ′ participating in the combination with nonzero (hence positive) coefficients, we cansubtract an appropriate multiple of this dependence and ensure that at least one of the nonzerocoefficients vanishes and all others stay non-negative (positive, after omitting all the ones withcoefficient 0). As each vector v H ′ has m coordinates and their sum is 0, it follows that | L | ≤ m − y − x = P z H ′ v H ′ with the variables z H ′ for v H ′ ∈ L . Note that it is enough to consider any | L | ≤ m − y − x and solve thesystem corresponding to these coordinates. By Cramer’s rule applied to this system each z H ′ is aratio of two determinants. The denominator is a determinant of a nonsingular matrix with integercoefficients, and its absolute value is thus at least 1. The numerator is also a determinant, and byHadamard’s Inequality its absolute value is at most the product of the ℓ -norms of the columnsof the corresponding matrix. The norm of one column is at most k y − x k (this can be slightlyimproved by selecting the | L | -coordinates with the smallest ℓ -norm, but we do not include thisslight improvement here). Each other column has norm at most (2 s ) / . Therefore each coefficient z H ′ satisfies 0 ≤ z H ′ ≤ k y − x k (2 s ) ( m − / . By taking the inner product with y − x we get k y − x k = X v H ′ ∈ L z H ′ h y − x, v H ′ i X v H ′ ∈ L, h y − x,v H ′ i > z H ′ h y − x, v H ′ i≤ ( m − k y − x k (2 s ) ( m − / max h y − x, v H ′ i . Therefore, there is a v H ′ so that k y − x k ( m − s ) ( m − / = k y − x k ( m − s ) ( m − / k y − x k ≤ h y − x, v H ′ i , that is, k y − x k ≤ ( m − s ) ( m − / h y − x, v H ′ i = ( m − ( m − / s m − h y − x, v H ′ i . (3)By the minimality of k y − x k k x + v H ′ − y k = k x − y k − h y − x, v H ′ i + k v H ′ k ≥ k x − y k , implying that 2 s ≥ k v H ′ k ≥ h y − x, v H ′ i . Plugging in (3) we get k y − x k ≤ ( m − ( m − / s m , and the desired results follows since k y − x k ∞ ≤ k y − x k .The assertions of Theorem 2.1 and Proposition 2.3 follow easily from Theorem 2.2. Indeed, asdescribed above there is a simple uniform cover of width s = 2 for the pair ( K n,n , M ) where M isa perfect matching. There is also a similar uniform cover H of width s = 2 for the pair ( K n , C )where C is a Hamilton cycle. The n ( n − / H are all Hamilton cycles obtained from C by omitting two nonadjacent edges of it and by adding the two edges that connect the resultingpair of paths to a cycle.To prove Theorem 2.4 we need the following simple lemma. Lemma 2.5.
Let T be a fixed graph with t vertices and q edges, suppose t divides n and let H bea T -factor in K n . Then there is a uniform cover of width at most qt of the pair ( K n , H ) .Proof. Let H be a fixed T -factor in K n , it consists of p = n/t (not necessarily connected) vertexdisjoint copies of T which we denote by T , T , . . . , T p . Let H be the set of all copies H ′ of the T -factor obtained from H by replacing one the copies T i by another copy of T on the same set ofvertices, in all possible t ! ways. Note that if T has a nontrivial automorphism group some membersof H are identical, and H is a multiset. By symmetry it is clear that each edge of H belongs tothe same number of members of H . Similarly, each edge connecting two vertices of the same T i which does not belong to H lies in the same positive number of members of H . Beside these twotypes of edges, no other edge of K n is covered by any member of H . Let H be the (multi)-setof all copies of the T -factor obtained from H by choosing, in all possible ways, t of the copies of T , say, T i , T i , . . . , T i t , removing them, and replacing them by all possible placements of t vertex8isjoint copies of T where each of the newly placed copies contains exactly one vertex of each T i j .Again by symmetry it is clear that each edge of H belongs to the same number of members of H .In addition, each edge of K n connecting vertices from distinct copies of T in H belongs to the same(positive) number of members of H . No other edges of K n are covered by any H ′ ∈ H . It is nowsimple to see that there are two integers a , b , so that the multiset H consisting of a copies of eachmember of H and b copies of each member of H is a uniform cover of the pair ( K n , H ). Thewidth of this cover is clearly qt , as every member of H contains qt edges not in E ( H ), and everymember of H contains at most 2 q edges not in E ( H ). This completes the proof.The assertion of Theorem 2.4 clearly follows from the last Lemma together with Theorem 2.2. • The statement of Theorem 2.4 holds for any graph H consisting of n/t (not necessarilyconnected) vertex disjoint components, each having t vertices and q edges. The proof applieswith no need to assume that all these components are isomorphic. • The proof of Theorem 2.2 is algorithmic in the sense that if the cover H is given then one canfind, in time polynomial in n and |H| , a copy H of F satisfying the conclusion. Indeed, theproof implies that as long as we have a copy H for which the conclusion does not hold, thereis a member H ′ ∈ H for which k x ( H ′ , P ) − fg x ( G, P ) k is strictly smaller than k x ( H, P ) − fg x ( G, P ) k . By checking all members of H we can find an H ′ for which this holds. Asboth these quantities are non-negative rational numbers smaller than n with denominator g < n , this process terminates in a polynomial number of steps. We make no attempt tooptimize the number of steps here. • The results can be extended to r -uniform hypergraphs by a straightforward modification ofthe proofs. • There are graphs H for which no result like those proved above holds when G is either acomplete or a complete bipartite graph even if the number of colors is small. A simpleexample is when G = K n , H = K , n − and m = 3. The edges of K n can be partitionedinto two vertex disjoint copies of K n and a complete bipartite graph K n,n . For this partition,every copy of the star H misses completely one of the color classes, although it’s fair share init is roughly a quarter of its edges. More generally, let H be any graph with a vertex coverof size smaller than m − H contains a set of less than m − K n into m − ⌊ n/ ( m − ⌋ vertices, and an additional classcontaining all the remaining edges. Then any copy of H in this graph cannot contain edgesof all those m − m − G = K n,n as well. • The discussion here suggests the following conjecture.9 onjecture 2.6.
For every d there exists a c ( d ) so that for any graph H with at most n vertices and maximum degree at most d and for any partition P of the edges of K n into m color classes, there is a copy H ′ of H in K n so that k x ( H ′ , P ) − | E ( H ) | E ( K n ) | x ( K n , P ) k ∞ ≤ c ( d ) . The analogous conjecture for bipartite bounded-degree graphs H with at most n verticesin each color class and for partitions of the edges of K n,n is also plausible. Note that theconjecture asserts that the same error term c ( d ) should hold for any number of colors m .Note also that c ( d ) must be at least Ω( d ) as shown by the example of a star H = K ,d andthe edge-coloring of K n with m = 3 colors described above. The choice number of a graph G is the smallest integer s so that for any assignment of a list of s colors to each vertex of G there is a proper coloring of G assigning to each vertex a color from its list.This notion was introduced in [45], [21]. Let K m ∗ k denote the complete k -partite graph with k colorclasses, each of size m . Several researchers investigated the choice number ch ( K m ∗ k ) of this graph.Trivially ch ( K ∗ k ) = 1 as K ∗ k is a k -clique. In [21] it is proved that ch ( K ∗ k ) = k . Kierstead [28]proved that ch ( K ∗ k ) = ⌈ (4 k − / ⌉ and in [29] it is proved that ch ( K ∗ k ) = ⌈ (3 k − / ⌉ . In [21] it is shown that as m tends to infinity ch ( K m ∗ ) = (1 + o (1)) log m . In [2] the authorshows that there are absolute constants c , c > c k ln m ≤ ch ( K m ∗ k ) ≤ c k ln m for all m and k . In [22] it is proved that for fixed k , as m tends to infinity, ch ( K m ∗ k ) = (1 + o (1)) ln m ln( k/ ( k − and in [36] it is proved that if both m and k tend to infinity and ln k = o (ln m ) then ch ( K m ∗ k ) =(1 + o (1)) k ln m . Our first result here is that the assumption that ln k = o (ln m ) can be omitted,obtaining the asymptotics of ch ( K m ∗ k ) when m and k tend to infinity (with no assumption on therelation between them). Theorem 3.1. If m and k tend to infinity then ch ( K m ∗ k ) = (1 + o (1)) k ln m. The proof is probabilistic, similar to the one in [2], where the main additional argument is inthe proof of the upper bound for values of k which are much bigger than m .Our second result is the following. Theorem 3.2.
For any fixed integer m ≥ the limit lim k →∞ ch ( K m ∗ k ) k exists (and is Θ(ln m ) ). m ≥
1, let c ( m ) denote the above limit. By the known results stated above c (1) = c (2) = 1, c (3) = 4 / c (4) = 3 / c ( m ) = (1 + o (1)) ln m . The problem of determining c ( m ) precisely forevery m seems very difficult.We prove Theorem 3.1 without trying to optimize the error terms. To simplify the presentation,we omit all floor and ceiling signs whenever these are not crucial. Proposition 3.3.
For every m, k ≥ ch ( K m ∗ k ) ≤ k (ln m + ln ln m + 20) . Proof:
Since ln m + ln ln m + 20 ≥
20 for all m ≥ m >
20. Weconsider two possible cases.
Case 1: k ≤
10 ln m .In this case we show that lists of size s = k (ln m + ln ln m + 3) suffice. Let G = K m ∗ k = ( V, E ),and suppose we assign a list S v of colors to each vertex v ∈ V , where | S v | = s for all v . Let S = ∪ v ∈ V S v be the union of all lists. Let S = T ∪ T . . . ∪ T k be a random partition of all colorsin S into k pairwise disjoint subsets, where each color x ∈ S is assigned, randomly, uniformly andindependently, to one of the subsets T j . We obtain a proper coloring of G by coloring each vertex v that lies in color class number j by a color from S v ∩ T j . Clearly, if there is indeed such a colorfor each vertex, then the resulting coloring is proper. The probability that for a fixed vertex v theabove fails is exactly (1 − k ) | S v | ≤ e − (ln m +ln ln m +3) < e m ln m < km . As there are mk vertices, the probability that there is a vertex for which the above fails is smallerthan 1, completing the proof in this case. Case 2: k >
10 ln m .Note that since by assumption m >
20 this implies that k ≥
30. In this case we show thatlists of size s = k (ln m + 20) suffice. Let G = K m ∗ k = ( V, E ), and suppose we assign a list S v of s colors to each vertex v ∈ V . As before, let S = ∪ v ∈ V S v be the union of all lists. Our strategynow is to first define a set of reserve colors R , these colors will be used to assign colors to thevertices that will not be colored by the procedure applied in Case 1. Let R be a random subset of S obtained by picking each color in S to lie in R with probability p = m +20 , where all choicesare independent. For a fixed vertex v , the random variable | S v ∩ R | is a Binomial random variablewith expectation sp = 10 k . By the standard estimates for Binomial distributions (see, e.g., [13],appendix A, Theorems A.1.11 and A.1.13), the probability that this random variable is smallerthan k is less than e − k/ and the probability it is larger than 20 k is less than e − k/ . Thus theprobability it is not between k and 20 k is less than2 e − k/ < e − k/ e − k/ < k m < mk , k ≥
10 ln m to conclude that e − k/ < m and the fact that k ≥ e − k/ < k . It follows that with positive probability k ≤ | S v ∩ R | ≤ k forevery vertex v ∈ V . Fix a set of colors R for which this holds. Now proceed as in Case 1. Let S − R = T ∪ T . . . ∪ T k be a random partition of all colors in S − R into k pairwise disjoint subsets,where each color in S − R is assigned, randomly, uniformly and independently to one of the subsets T j . If a vertex v of G lies in color class number j , and S v ∩ T j = ∅ , then color it by an arbitrarycolor in this intersection S v ∩ T j . The probability that v fails to have such a color is(1 − k ) | S v − R | ≤ (1 − k ) k ln m ≤ m , where here we used the fact that | S v ∩ R | ≤ k for all v . By linearity of expectation, the expectednumber of uncolored vertices at this stage is at most k , hence we can fix a splitting T , · · · , T k asabove so that there are at most k uncolored vertices. But now we can color these vertices one byone using the reserve colors. Since for each such vertex u , | S u ∩ R | ≥ k , each of these vertices hasat least k colors of R in its list and thus we will be able to assign to it a color that differs from allcolors of R assigned to previous vertices. This completes the proof of the upper bound. (cid:3) The proof of the lower bound is essentially the one in [2], with a more careful computation andchoice of parameters. For completeness, we sketch the details.
Proposition 3.4.
There exists an m so that for all m > m and every k ch ( K k ∗ m ) > t where t = ( k − − k ln m )(ln m − m )(1 − ln mm ) ( = (1 + o (1)) k ln m ) , where the o (1) -term tends to zero as m and k tend to infinity. Proof:
We consider two possible cases.
Case 1: k ≤ m .In this case we prove that ch ( K k ∗ m ) > s , where s = ( k − − k ln m )(ln m − m ) ( = (1 + o (1)) k ln m ) . Let S be a set of k (ln m ) colors, and let S , S , . . . , S m be m random subsets of S , each chosenindependently and uniformly among all subsets of cardinality s of S , where s is as above. We claimthat with positive probability there is no subset of S of cardinality at most | S | /k = (ln m ) thatintersects all subsets S i . This claim suffices to prove the assertion of the proposition in this case.Indeed, we simply assign the m vertices in each color class of G the m lists S i . If there wouldhave been a proper coloring of G assigning to each vertex a color from its list, then the set of allcolors assigned to vertices in one of the color classes of G must be of size at most | S | /k and it mustintersect all lists S i , contradiction. It thus suffices to prove the claim.12ix a set T of (ln m ) colors. The probability that a random subset of size s of S does notintersect T is (cid:0) | S |−| T | s (cid:1)(cid:0) | S | s (cid:1) = (cid:0) ( k −
1) ln ms (cid:1)(cid:0) k ln ms (cid:1) . This quantity is at least( ( k −
1) ln m − k ln mk ln m − k ln m ) s = (1 − k (1 − / ln m ) ) [ k (1 − m ) − m − m ) ≥ ( 1 e ) ln m − m = ln mm , where here we used the fact that for every q >
1, (1 − /q ) q − ≥ e . Therefore, the probability thatnone of the m random sets S i misses T is at most(1 − ln mm ) m < e − ln m . As the number of choices for T is only (cid:18) k ln m ln m (cid:19) ≤ ( ek ) ln m ≤ e (1+ o (1)) ln m , where here we used the assumption that k ≤ m , the desired claim follows, completing the proof ofCase 1. Case 2: k ≥ m .In this case, take first the previous construction with m and k ′ = ln m . Replace k by thelargest integer k ′′ which is at most k and is divisible by k ′ , that is: k ′′ = k ′ ⌊ k/k ′ ⌋ . Note thatas k ≥ m and k ′ = ln m , k ′′ ≥ k (1 − ln mm ). Now replace in the construction for k ′ = ln m every color by a group of k ′′ /k ′ colors, where all groups are pairwise disjoint, to get m lists, eachof size (1 + o (1)) k ′′ ln m = (1 + o (1)) k ln m , in a set of size k ′′ ln m , so that no subset of sizeln m , that is, a fraction of 1 /k ′′ of the colors, intersects all of them. This shows, as before. that ch ( K m ∗ k ′′ ) > (1 + o (1)) k ′′ ln m = (1 + o (1)) k ln m , and as ch ( K m ∗ k ) can be only larger (since itcontains K m ∗ k ′′ ) as a subgraph), this completes the proof. (cid:3) In this subsection we prove Theorem 3.2. A natural way to try and prove it is to show that forevery fixed m , the function f ( k ) = ch ( K m ∗ k ) is either sub-additive or super-additive. In thesescases the existence of the limit would follow from Fekete’s Lemma. Unfortunately this function isnot always super-additive, as shown by the case m = 3, since ch ( K ∗ ) = 3 and ch ( K ∗ k ) = ⌈ (8 k − / ⌉ < k. Similarly, the function is not always sub-additive, as shown by the case of large m , where ch ( K m ∗ ) = (1 + o (1)) log m and for large k , ch ( K m ∗ k ) = (1 + o (1))2 k ln m > (1 + o (1)) k log m. Lemma 3.5.
There is a positive integer s , so that for every integer s > s the following holds.For every real c satisfying / ≤ c ≤ / and for every integer t ≥ : c [( s / + 3) t / − − c [( s / + 3) t / − ≥ [( s / + 3)( ct ) / − . Proof:
Put X = c / [( s / + 3) t / − , Y = ( s / + 3)( ct ) / − . Then the above inequality is equivalent to the statement X − c / X ≥ Y , that is, to ( X − Y )( X + XY + Y ) ≥ c / X . Since 1 / ≤ c ≤ /
3, we have 0 . < c / < .
88. Thus X − Y = 3 − c / > .
36. For sufficientlylarge s , X > Y > . X >
XY > . X and Y > . X . Therefore( X − Y )( X + XY + Y ) > . · . X = 0 . X > . X > c / X . This completes the proof. (cid:3)
We also need the following simple corollary of Chernoff’s Inequality (see, e.g., [13], AppendixA.)
Lemma 3.6.
There exists an s > so that for every s > s , every integer t ≥ and everyreal c satisfying / ≤ c ≤ / , the probability that the Binomial random variable with parameters [( s / + 3) t / − and c is at most c [( s / + 3) t / − − c [( s / + 3) t / − is smaller than st ) . Proof: , By Chernoff this probability is smaller than e − Ω(( st ) / ) . (cid:3) Using the above, we prove the following.
Proposition 3.7.
For every fixed m there exists k = k ( m ) so that for all k > k the followingholds. If ch ( K m ∗ k ) = s then for every integer t ≥ ch ( K m ∗ kt ) ≤ [( s / + 3) t / − . roof: Since trivially ch ( K m ∗ k ) ≥ ch ( K ∗ k ) = k , we can choose k > m so that for k > k , s = ch ( K m ∗ k ) is sufficiently large to ensure that the assertions of Lemma 3.5 and Lemma 3.6 hold.Note also that for this k , s > m . With this k we prove the above by induction on t . For t = 1there is nothing to prove. Assuming the result holds for all integers t ′ < t we prove it for t . Let G = K m ∗ kt = ( V, E ) have the kt color classes U , U , · · · U kt , and suppose we have a list L v of[( s / + 3) t / − colors assigned to each vertex v ∈ V . Put t = ⌊ t/ ⌋ , t = ⌈ t/ ⌉ and split V into two disjoint sets V , V , where V consists of all vertices in the first t k color classes U j and V consist of all vertices in the last t color classes U j . Let G be the induced subgraph of G on V and G the induced subgraph of G on V . Thus G is a copy of K m ∗ kt and G is a copy of K m ∗ t .Let S ∪ v ∈ V L v be the set of all colors, and let S = S ∪ S be a random partition of it intotwo disjoint sets, where each color in S is chosen, randomly and independently, to lie in S withprobability t /t and to lie in S with probability t /t .Our objective is to use only the colors of S for the vertices in G and only those of S forthe vertices in G . Note that 1 / ≤ t /t ≤ t /t ≤ /
3. For each vertex v ∈ V the set L v ∩ S of colors in S that belong to the list of v is of size which is a binomial random variable withparameters [( s / + 3) t / − and t /t . Therefore, by Lemma 3.6 the probability that thissize is smaller than ( t /t )[( s / + 3) t / − − ( t /t )[( s / + 3) t / − is less than st ) . Bythe same reasoning the probability that for a vertex u ∈ V the size of L u ∩ V is smaller than( t /t )[( s / +3) t / − − ( t /t )[( s / +3) t / − is less than st ) . As s > m, s ≥ k the total numberof vertices is smaller than kst < ( st ) and hence with positive probability this does not happen forany vertex. By Lemma 3.5 in this case each vertex of G still has at least [( s / + 3)( t ) / − colors in its list (restricted to the colors in S ), and a similar statement holds for the vertices of G . We can now fix a partition S = S ∪ S for which this holds and apply induction to color G by the colors from S and G by the colors from S , completing the proof. (cid:3) Proof of Theorem 3.2:
Fix an integer m ≥
1. By the result of [2] stated in Section 1,lim inf k →∞ ch ( K m ∗ k ) k = q exists (and is Θ(ln m )). Fix a small ǫ > k > k be a large integer, where k is as inProposition 3.7, so that ch ( K m ∗ k ) k ≤ q + ǫ. Put s = ch ( K m ∗ k ). Then s ≤ k ( q + ǫ ). By Proposition 3.7 for every integer t ≥ ch ( K m ∗ kt ) ≤ [( s / + 3) t / − < [ s / e /s / t / ] = ste /s / . Suppose, further, that k is chosen to be sufficiently large to ensure that e /k / < (1 + ǫ ) . As s = ch ( K m ∗ k ) ≥ k in this case we have also e /s / < (1 + ǫ ) . t ≥ ch ( K m ∗ kt ) ≤ ste /s / < k ( q + ǫ ) t (1 + ǫ ) . It follows that for every large integer p , ch ( K m ∗ p ) ≤ k ( q + ǫ ) ⌈ p/k ⌉ (1 + ǫ ) ≤ k ( q + ǫ )( p + k ) /k (1 + ǫ ) . Thus ch ( K m ∗ p ) p ≤ k ( q + ǫ )( p + k ) / ( pk )(1 + ǫ )which for sufficiently large p is at most, say,( q + ǫ )(1 + ǫ ) . Since, by the result in [2], q = Θ(ln m ) and ǫ > p →∞ ch ( K m ∗ p ) p ≤ q = lim inf p →∞ ch ( K m ∗ p ) p , completing the proof. (cid:3) Let p be a prime, let w = e πi/p be the p th primitive root of unity, and define w j = w j for0 ≤ j ≤ p −
1. Let n be an integer divisible by p , and let B be the set of all p n vectors of length n in which each coordinate is in the set { , w , . . . , w p − } . Let K ( n, p ) denote the minimum k sothat there exists a set { v , v , . . . , v k } of members of B such that for every u ∈ B there is some1 ≤ j ≤ k so that the scalar inner product v i · u = 0.Heged˝us [23] proved that for every prime p and n divisible by p , K ( n, p ) ≥ ( p − n , extendinga result of [10] where the statement is proved for p = 2. He also conjectured that equality alwaysholds, as is the case for p = 2, by a simple construction of Knuth (c.f. [10]). Our first observationhere is that this conjecture is (very) false for every prime p ≥ n . Proposition 4.1.
For every prime p and every n divisible by pK ( n, p ) ≥ p n [( n/p )!] p n ! . (4) Therefore, for every fixed p and large nK ( n, p ) ≥ (1 + o (1)) (2 π ) ( p − / p p/ · n ( p − / . (5)The proof of Heged˝us is based on Gr¨obner basis methods. In particular, he established thefollowing result. 16 heorem 4.2 ([23]) . Let p be a prime and let P ( x ) = P ( x , x , . . . , x p ) be a polynomial over Z p which vanishes over all { , } vectors of Hamming weight p and suppose that there is a { , } -vector z of Hamming weight p so that P ( z ) = 0 (in Z p ). Then the degree of P is at least p . An elementary proof of this lemma, due to S. Srinivasan, is given in [11]. Here we describea variant of this proof providing a very short derivation of this lemma from the CombinatorialNullstellensatz proved in [4], which is the following.
Theorem 4.3.
Let F be an arbitrary field, and let f = f ( x , . . . , x n ) be a polynomial in F [ x , . . . , x n ] .Suppose the degree deg ( f ) of f is P ni =1 t i , where each t i is a nonnegative integer, and suppose thecoefficient of Q ni =1 x t i i in f is nonzero. If S , . . . , S n are subsets of F with | S i | > t i , then there are s ∈ S , s ∈ S , . . . , s n ∈ S n so that f ( s , . . . , s n ) = 0 . Proof of Proposition 4.1:
Let M be the collection of all vectors in B in which each w i appearsin exactly n/p coordinates and let m = | M | = n ![( n/p )!] p be its cardinality. We claim that M is the set of all vectors in B that are orthogonal to the vector j = (1 , , . . . , ∈ B . Indeed, it is a well known consequence of Eisenstein’s criterion that theminimal polynomial of w over the rationals is the polynomial 1 + x + x + · · · + x p − . Therefore, if P p − i =0 α i w i = 0 for some integers α i , then the polynomial 1 + x + x + · · · + x p − divides P p − i =0 α i x i ,implying that all the coefficients α i are equal. This implies the assertion of the claim.By the claim, the number of vectors in B orthogonal to j is exactly m , and this is clearly alsothe number of vectors in B orthogonal to any other fixed member of B . It follows that if eachvector in B is orthogonal to at least one vector in a subset of cardinality k = K ( n, p ) of B , then k ≥ p n /m , implying (4). The estimate in (5) follows from (4) by Stirling’s Formula. (cid:3) Proof of Theorem 4.2:
Without loss of generality assume that z is the vector starting with 3 p
1s followed by p P is at most p − f ( x , x . . . , x p ) = f − f where f = P ( x )[1 − ( p X i =1 x i ) p − ] x x · · · x p +1 (1 − x p +1 )(1 − x p +2 ) · · · (1 − x p )and f = P ( z ) x x · · · x p (1 − x p +1 )(1 − x p +2 ) · · · (1 − x p ) . The degree of the polynomial f is at most 4 p −
1, that of f is exactly 4 p , hence the degree of f is 4 p and the coefficient of Q pi =1 x i in it is P ( z ) = 0.By the Combinatorial Nullstellensatz (Theorem 4.3) with F = Z p , n = 4 p , t i = 1 for all i and S i = { , } for all i there is a vector y = ( y , y , . . . , y p ) ∈ { , } p so that f ( y , y , . . . y p ) = 0.17owever, the only vector with { , } coordinates in which f is nonzero is z , and as f ( z ) = f ( z ) = P ( z ), f ( z ) = 0. Thus y = z and f ( y ) = f ( y ). If the Hamming weight of y is not divisible by p then the term [1 − ( P pi =1 y i ) p − ] vanishes. If the Hamming weight of y is 2 p then the term P ( y )vanishes. If it is 0 or p , then the term y y · · · y p +1 = 0 and if it is 4 p or 3 p (and y = z ) thenthe term (1 − y p +1 )(1 − y p +2 ) . . . (1 − y p ) = 0. Therefore f ( y ) = f ( y ) = 0, contradiction. Thiscompletes the proof. (cid:3) In May 2019 Shay Moran showed me a question posted by a woman named Ruthi Shaham in aFacebook Group focusing on Mathematics. She wrote that her son has finished elementary schooland was about to move to high school. When doing so, each child lists three friends, and theassignment of children into classes ensures that each child will have at least one of these threefriends in his class. Ruthi further wrote that her son heard from five of his schoolmates that theyfound that they can make their selections in a way that will ensure that all five will be scheduledto the same class. She tried to check with a paper and pencil and couldn’t decide whether or notthis is possible, but she suspected it is impossible. She thus asked if this is indeed the case, and ifso, whether a larger group of children can form such a coalition ensuring they will all necessarilybe assigned to the same class.In this brief section we show that Ruthi has indeed been right, no coalition of five children canensure they will share the same class. Moreover, no coalition of any size can ensure to share thesame class. This is related to known problems and results in Graph Theory, as are several variantsof the problem mentioned below.Here is a more formal formulation of the problem, with general parameters. Let N = { , , . . . , n } be a finite set of size n , let k and r be integers, and suppose n ≥ k + 1. For any collection of subsets S i of N , (1 ≤ i ≤ n ), with i S i , and | S i | = k for all i , let P ( S , S , . . . , S n ) be a partition of N ,so that :For any part N i of the partition and for any j ∈ N, if j ∈ N i then S j ∩ N i = ∅ . (6)Here N denotes the group of children, S i is the list of friends listed by child number i , andthe partition of N into parts N i is the partition of the set of children into classes. The function P represents the way the children are partitioned into classes N i given their choices S i , and thecondition (6) is the one ensuring that each child will have at least one other child from his list inhis class.We say that a subset R ⊂ N is a successful coalition, if there are choices S i , i ∈ R of sets S i satisfying | S i | = k and i S i so that for any sets S j ⊂ N with | S j | = k for all j ∈ N − R , and forany function P satisfying the conditions above, all elements of R belong to the same part of thepartition f ( S , S , . . . , S n ). Note that by symmetry if a successful coalition of size r is possible thenany set of size r can form such a coalition, and hence we may always assume that R = { , , . . . , r } .18he question of Ruthi is whether or not for k = 3 there can be a successful coalition R of size | R | = 5. Theorem 5.1.
1. For k ≤ and every integer r > , every set R of size r can form a successful coalition.2. For any k ≥ and every r > no set of size r can form a successful coalition. Before presenting the general proof, here is a short argument showing that for k = 3 no successfulcoalition of size 5 is possible. This proof is a simple application of the probabilistic method. Claim:
Suppose n ≥ N = { , , . . . , n } , R = { , , . . . , } , and let S , . . . , S be subsets of N ,each of size 3, so that i S i for all 1 ≤ i ≤
5. Then there are subsets S j ⊂ N , for 5 ≤ j ≤ n andthere is a partition P ( S , . . . , S n ) of N into two disjoint parts N , N satisfying (6) such that R intersects both N and N . Proof:
Color the elements of N randomly red and blue, where each i ∈ N randomly and indepen-dently is red with probability 1 / /
2. The probability that all membersof R have the same color is 1 /
16. For each fixed i ≤
5, the probability that the color of i is differentthan that of all elements in S i is 1 /
8. Therefore, with probability at least 1 − / − / > R contains both red and blue elements,and every i ∈ R has at least one member of S i with the same color as i . Fix such a coloring.Without loss of generality 1 is colored red and 2 is colored blue. Let N be the set of all elementscolored red and let N be the set of all elements colored blue. For each j ∈ N − R let S j contain 1and for each j ∈ N − R let S j contain 2. It is easy to see that the partition N = N ∪ N satisfies(6) but R intersects both N and N , completing the proof. (cid:3) Note that the above proof does not work for r ≥
8, thus the proof of Theorem 5.1 requires adifferent method, which we show next.
Proof of Theorem 5.1:
The case k ≤ k = 1 simply define S i = { ( i +1)(mod r ) } to see that the coalition R = { , , . . . , r } is successful. For k = 2 and r = 2, S = { , } , S = { , } show that { , } is successful. For any larger r add to the above S i = { , } for all 3 ≤ i ≤ r .The more interesting part is the proof that for k ≥ r > r < k here is simple. One possible proof is to repeat the probabilistic argument describedabove for the case k = 3 , r = 5. Since for 1 < r < k , k ≥ r − + r k ≤
12 + k − k ≤
12 + 28 < k ≥ r ≥ k consider the digraph whose set of vertices is N , where for each vertex i and each j ∈ S i , ij is a directed edge. Thus every outdegree in this digraph is exactly k . Given19he sets S , . . . , S r of outneighbors of the vertices in R = { , , . . . , r } (representing the childrenattempting to form a successful coalition), define the sets S j for j > r in such a way that theinduced subgraph on N − R is acyclic. (For example, we can define S j = { , , . . . , k } for each j > r , or S j = { j − , j − , . . . , j − k } for each j > r . Note that here we used the fact that r ≥ k ).The crucial result we use here is a theorem of Thomassen ([42], see also [3] for an extension).This Theorem asserts that any digraph with minimum outdegree at least 3 contains two vertexdisjoint cycles. Let A and B be the sets of vertices of these two cycles. Note that both A and B must contain a vertex of R (as N − R contains no directed cycles). Let A ′ , B ′ be two sets of verticessatisfying A ⊂ A ′ , B ⊂ B ′ with | A ′ | + | B ′ | maximum subject to the constraint that every outdegreein A ′ is at least 1 and every outdegree in B ′ is at least 1. We claim that A ′ ∪ B ′ is the set N of allvertices. Indeed, otherwise, every v in C = N − ( A ′ ∪ B ′ ) has no outneighbors in A ′ ∪ B ′ (otherwisewe could have added it to either A ′ or B ′ contradicting maximality), so has at least k ≥ > C and then we can replace A ′ by A ′ ∪ C contradicting maximality. This proves theclaim. The assignment to two groups is now N = A ′ and N = B ′ . Since both A ⊂ A ′ and B ⊂ B ′ contain elements of R , this shows that R is not a successful coalition, completing the proof. (cid:3)
1. What if every child is ensured to have at least two of his choices with him in his class ? Inthis case, even if k is arbitrarily large (but r is much larger) we do not know to prove thata coalition of r cannot ensure they are all in the same group. This is identical to one of theopen questions in [6], which is the following. Question:
Is there a finite positive integer k such that every digraph in which all oudegreesare (at least) k contains two vertex disjoint subgraphs, each having minimum outdegree atleast 2 ?On the other hand it is easy to see that this is impossible if r − + r (1+ k )2 k <
1. Indeed, ifso we can split the group of children randomly into two sets, red and blue. With positiveprobability the specific set of r children trying to form a coalition is not monochromatic, andalso for any child in the coalition there are at least two of his choices in his group. We cannow fix the choices of all others outside the coalition to ensure they will also be happy withthis partition. It follows that if in this version of the problem a successful coalition of size r is possible, then r has to be at least exponential in k .2. Suppose we change the rules, and each child lists k other children that he does not like, andwishes not to have many of them in his class. It can then be shown that for any k there isan example of choices of the children in which each one lists k others he prefers to avoid, sothat in any partition of the group of children into 2 classes, there will always be at least onepoor child sharing the same class with all the k he listed ! This is based on another resultof Thomassen [43]: for every k there is a digraph with minimum outdegree k which containsno even directed cycle. If D = ( N, E ) is such a digraph, and N = V ∪ V is a partition20f its vertex set into two disjoint parts, then, as observed in [6], there is a vertex in one ofthe classes having all its out-neighbors in the same class. Indeed, otherwise, starting at anarbitrary vertex v we can define an infinite sequence v , v , v , . . . , where each pair ( v i , v i +1 )is a directed edge with one end in V and one in V . As the graph is finite, there is a smallest j such that there is i < j with v i = v j , and the cycle v i , v i +1 , . . . , v j = v i is even, contradiction.On the other hand, by splitting the group of children into s ≥ k/s of the k children he wants toavoid. This follows from a result of Keith Ball described in [6]. ℓ -balls and projections of linear codes A remarkable known property of the Binomial distribution
Bin ( n, p ) is that its median is alwayseither the floor or the ceiling of its expectation np . In particular, if the expectation is an integerthen this is also the median. The following more general result is proved by Jogdeo and Samuelsin [27]. Theorem 6.1 ([27], Theorem 3.2 and Corollary 3.1) . Let X = X + X + . . . + X n be a sum ofindependent indicator random variables where for each i , P r ( X i = 1) = p i and P r ( X i = 0) = 1 − p i .Then the median of X is always the floor or the ceiling of its expectation P ni =1 p i . This theorem can be used to derive several interesting results. Here we describe one quickapplication and another more complicated one in which it is convenient (though not absolutelynecessary) to use it, combined with several additional ingredients. ℓ -balls and Hamming balls in the discrete cube If n is even, d = n/ x = (1 / , / , . . . , /
2) is the center of the n -dimensional real unit cube[0 , n , then the ℓ -ball of radius d centered at x contains all the 2 n points of the discrete cube { , } n . On the other hand, any Hamming ball of radius d centered at a vertex y of this discretecube contains only P di =0 (cid:0) ni (cid:1) = ( + o (1))2 n points of the cube, where the o (1)-term tends to 0 as n tends to infinity. Madhu Sudan [39] asked me whether a similar bound holds for any ℓ -ball ofintegral radius. The precise statement of the question is as follows:Is it true that for any positive integer d and for any ℓ -ball B (centered at any real point in R n )there is a Hamming ball of the same radius d centered at a point in { , } n that contains at leasthalf the points in B ∩ { , } n ?The following stronger result shows that this is indeed the case. Theorem 6.2.
For any real x = ( x , x , . . . , x n ) in R n and for any subset A of points of B ( x, d ) ∩{ , } n , where B ( x, d ) is the ℓ -ball of radius d centered at x , and d is an integer, there is y ∈ { , } n so that | A ∩ B ( y, d ) | ≥ | A | / . Proof.
Note, first, that we may assume that x i ∈ [0 ,
1] for all i . Indeed, otherwise, replace x i by 1if x i > x i <
0. This modification only decreases the ℓ -distance between x and any21oint in { , } n . Therefore A is a subset of the ball B ( x, d ) for the modified vector x too. We thusmay and will assume that x ∈ [0 , n . Let y = ( y , y , . . . , y n ) be a random binary vector obtainedby choosing, for each i, randomly and independently, y i to be 1 with probability x i and 0 withprobability (1 − x i ). For each point a ∈ A , the ℓ -distance between y and a is a random variablewhich is a sum of independent Bernoulli random variables and its expectation is exactly the ℓ distance between a and x , which is at most d . By Theorem 6.1 of Jogedo and Samuels stated abovethe probability that this random variable is at most d is at least a half. It follows by linearity ofexpectation that the expected number of points of A within distance at most d from y is at least | A | /
2, and thus there is a y as needed. Let F be a finite or infinite field, and let V be a linear code of length n , dimension k and minimumrelative distance at least δ over F . Thus V is a subspace of dimension k of F n , and the numberof nonzero coordinates of any nonzero codeword v ∈ V is at least δn . Let m be an integer. Aprojection of V on m random coordinates is obtained by selecting a random (multi)set I of m coordinates of [ n ], chosen with repetitions. With this random choice of I let V m ⊂ F m be thevector space over F consisting of all vectors { ( v i ) i ∈ I : v = ( v , v , . . . , v n ) ∈ V } . One may expectthat if m is large, then typically the vector space V m , considered as a linear code of length m over F , will have dimension k and minimum distance not much smaller than δm . This is easy to proveby a standard application of Chernoff’s Inequality and the union bound, provided m is sufficientlylarge as a function of | F | , k and δ . It is, however, not clear at all that this is the case for m of sizeindependent of the size of the field F (which may even be infinite). Such a statement is proved bySaraf and Yekhanin in [40]. Theorem 6.3 ([40], Theorem 3) . Let V be a linear code of dimension k , length n and minimumrelative distance δ over an arbitrary field F . If m is at least c ( δ ) k and V m is a projection of V on m random coordinates then with probability at least − e − Ω( δm ) the dimension of V m is k and itsminimum distance is at least δm/ . One can check that the estimate the proof in [40] provides for c ( δ ) is b log(1 /δ ) δ for a sufficientlylarge absolute constant b . Note, however, that the minimum relative distance obtained is only δ/ ε -nets and ε -approximationsin range spaces with finite Vapnik-Chervonenkis dimension to get an improved version of the abovetheorem in which the relative minimum distance obtained can be arbitrarily close to δ . Theorem 6.4.
There exists an absolute positive constant c so that the following holds. Let B > be an integer, and let V be a linear code of dimension k , length n and minimum relative distance δ over an arbitrary field F . If m is at least c B kδ log( B/δ ) and V m is a projection of V on m randomcoordinates then with probability at least − e − Ω( δm/B ) the dimension of V m is k and its minimumdistance is at least ( B − B +1 ) δm . B to be a large fixed constant we get that typically the minimum relative distance of V m is close to δ , and the estimates for m and for the failure probability are essentially as in Theorem6.3.We start with a quick reminder of the relevant facts about VC-dimension. The Vapnik-Chervonenkis dimension
V C ( C ) of a (finite) family of binary vectors C is the maximum cardinalityof a set of coordinates I such that for every binary vector ( b i ) i ∈ I there is a C ∈ C so that C i = b i for all i ∈ I . (In this case we say that the set I is shattered by C ). Suppose the vectors in thefamily are of length n . An ε -net for the family is a subset I ⊂ [ n ] such that for every C ∈ C ofHamming weight at least εn there is an i ∈ I so that C i = 1. An ε -approximation for the family isa sub(multi)set I ∈ [ n ] so that for every C ∈ C| | P ni =1 C i | n − | P ni ∈ I C i || I | | < ε. A basic result proved by Vapnik and Chervonenkis [44] (with a logarithmic improvement by Ta-lagrand [41]), is that if
V C ( C ) ≤ d then a random set of Θ( dε ) coordinates is typically an ε -approximation. A similar result, proved by Haussler and Welzl [25], is that for such a C a randomset of Θ( dε log(1 /ε )) coordinates is typically an ε -net. Another basic combinatorial result is theSauer-Perles-Shelah Lemma: if V C ( C ) ≤ d then the number of distinct projections of the set ofvectors in C on any set of t coordinates is at most g ( d, t ) = P di =0 (cid:0) ti (cid:1) . The relevance of the VC-dimension to projections of linear codes is the following simple obser-vation.
Claim 6.5.
Let F be an arbitrary field, and let V ⊂ F n be a linear subspace of dimension k over F . For each vector v ∈ C let C = C ( v ) denote the indicator vector of the support of v , that is, C i = 1 if v i = 0 and C i = 0 is v i = 0 . Put C = { C ( v ) : v ∈ V } . Then V C ( C ) = k .Proof. Since the dimension of V is k it contains a set of k vectors v ( i ) such that there is a set I = { i , i , . . . , i k } of k coordinates so that v ( i ) i j is 1 for i = j and 0 otherwise. The supports of theset of all linear combinations with { , } -coefficients of these vectors shatter the set I , implyingthat V C ( C ) ≥ k . Conversely, if there is a set of coordinates J shattered by the vectors in C , thenfor each j ∈ J there is a vector in V with v j = 0 and v i = 0 for all i ∈ J − j . These | J | vectors areclearly linearly independent, implying that | J | ≤ k and completing the proof.The above claim and the known result stated above about ε -approximation for families of vectorswith finite V C -dimension suffice to prove a version of Theorem 6.4 with m = Θ( B kδ ). Indeed, wesimply consider a 2 δ/ ( B + 1)-approximation for the set C corresponding to V . Similarly, the resultabout ε -nets shows that typically the dimension of V m is m .In order to prove the improved estimate for m stated in the theorem we show that in the settinghere the bound can be improved to be closer to that in the theorem about δ -nets. This is proved inthe following result, which applies to general collections of vectors with a bounded VC-dimension. Proposition 6.6.
There exists an absolute positive constant c > such that the following holds. Let C be a family of binary vectors of length n , and assume that V C ( C ) ≤ d . Let X be a random multiset f m coordinates, with m = c B dε log( B/ε ) , where B > is an integer. Then with probability atleast − e − Ω( εm/B ) , for every C ∈ C satisfying P ni =1 C i ≥ εn we have P i ∈ X C i ≥ B − B +1 εm. In order to prove the above statement, we need some standard estimates for large deviations ofthe hypergeometric distribution. The estimate we use here was first proved by Hoeffding [24], seealso [26], Theorem 2.10 and Theorem 2.1.
Lemma 6.7 (Hoeffding [24], see also [26]) . Let H be the hypergeometric distribution given by thecardinality | R ∩ S | where S is a random subset of cardinality m in a set of size N containing asubset R of cardinality pN . Then the probability that H is smaller than pm − t is at most e − t / pm . Proof of Proposition 6.6:
Let m be as in the statement of the proposition and let X =( x , . . . , x m ) be a random multiset obtained by m independent random choices, with repetitions,of elements of [ n ]. For C ∈ C we let | C | denote P ni =1 C i and let | C ∩ X | denote |{ i : C x i = 1 }| . Let E be the following event: E = {∃ C ∈ C : | C | ≥ εn, | C ∩ X | < B − B + 1 εm } To complete the proof we have to show that the probability of E is as small as stated in theproposition. To do so, we make an additional random choice and define another event as follows.Independently of the previous choice, let T = ( y , . . . , y Bm ) be obtained by Bm independent randomchoices of elements of [ n ]. Let E be the event defined by E = (cid:26) ∃ C ∈ C : | C | ≥ εn, | C ∩ X | < B − B + 1 εm, | C ∩ T | ≥ ⌊ Bεm ⌋ (cid:27) Claim 6.8.
P r ( E ) ≥ P rE .Proof. It suffices to prove that the conditional probability
P r ( E | E ) is at least 1 /
2. Suppose thatthe event E occurs. Then there is a C ∈ C such that | C | ≥ εn and | C ∩ X | < B − B +1 εm . Theconditional probability above is clearly at least the probability that for this specific C , | C ∩ T | ≥⌊ Bεm ⌋ . However | C ∩ T | is a binomial random variable with expectation at least Bεm , andtherefore, by Theorem 6.1 its median is at least the floor of that, implying the desired result.
Claim 6.9.
P r ( E ) ≤ g ( d, ( B + 1) m )2 − ǫm/ B +1) Proof.
The random choice of X and T can be described in the following way, which is equivalent tothe previous one. First choose X ∪ T = ( z , . . . , z ( B +1) m ) by making ( B + 1) m random independentchoices of elements of [ n ] (with repetitions), and then choose randomly precisely m of the elements z i to be the set X , where the remaining elements z j form the set T . For each member C ∈ C satisfying | C | ≥ εn , let E C be the event that | C ∩ T | ≥ ⌊ Bεm ⌋ and | C ∩ X | < B − B + 1 εm.
24 crucial fact is that if
C, C ′ ∈ C are two ranges, | C | ≥ εn and | C ′ | ≥ εn and if C ∩ ( X ∪ T ) = C ′ ∩ ( X ∪ T ), then the two events E C and E C ′ , when both are conditioned on the choice of X ∪ T ,are identical. This is because the occurrence of E C depends only on the intersection C ∩ ( X ∪ T ).Therefore, for any fixed choice of X ∪ T , the number of distinct events E C does not exceed thenumber of different sets in the projection of C on the coordinates X ∪ T . Since the VC-dimensionis at most d , this number does not exceed g ( d, ( B + 1) m ), by the Sauer-Perles-Shelah Lemma.Let us now estimate the probability of a fixed event of the form E C , given the choice of X ∪ T .This probability is at most the probability that a hypergeometric random variable counting thesize of the intersection of a random set of m elements with a subset R of size at least ⌊ Bεm ⌋ ina set of size N = ( B + 1) m is smaller than B − B +1 ǫm . By Lemma 6.7, and using the fact that thechoice of m implies that ⌊ Bεm ⌋ > ( B − / εm this probability is smaller than e − εm/ B +1) .By Claims 6.8 and 6.9, P r ( E ) ≤ g ( d, ( B + 1) m )2 − εm/ B +1) . The assertion of the theoremfollows using the fact that g ( d, ( B + 1) m ) < ( 2 e ( B + 1) md ) d . (cid:3) Proof of Theorem 6.4:
Let V be a linear code of length n , dimension k and minimum relativedistance δ . Let C be the set of all indicator vectors of supports of vectors in V . By Claim 6.5 theVC-dimension of C is at most k , and by definition the Hamming weight of each member C of C isat least δn . The desired result thus follows from Proposition 6.6. (cid:3) The first result in this section was obtained in joint discussions with Michael Krivelevich [31].Let G = ( V, E ) be a connected graph. Let γ ( G ) denote the minimum size of a dominatingset in it, that is, the minimum cardinality of a set of vertices X ⊂ V so that each v ∈ V − X has at least one neighbor in X . Let γ c ( G ) denote the minimum size of a connected dominatingset of G , that is, the minimum cardinality of a dominating set of vertices X so that the inducedsubgraph of G on X is connected. One of the reasons this parameter has been studied extensivelyis the fact that | V | − γ c ( G ) is exactly the maximum possible number of leaves in a spanning treeof G . It is well known that if the minimum degree in G is k and its number of vertices is n ,then γ ( G ) ≤ n (ln( k +1)+1) k +1 . See [32] or [13], Theorem 1.2.2 for a proof. As mentioned in [13] this isasymptotically tight for large k , see, e.g., [14] for a proof that for any ε > k > k ( ε ) a random k -regular graph on n vertices is unlikely to contain a dominating set of size at most (1 − ε ) n ln kk .Caro, West and Yuster [19] proved that for every connected graph G with n vertices andminimum degree k , γ c ( G ) is also not much larger than n ln( k +1) k +1 . The precise statement of theirresult is as follows. Theorem 7.1 ([19]) . Let G be a connected graph with n vertices and minimum degree at least k . hen γ c ( G ) ≤ n (ln( k + 1) + 0 . p ln( k + 1) + 145) k + 1Here we first prove a similar result with a slightly better estimate. Theorem 7.2.
Let G be a connected graph with n vertices and minimum degree at least k . Then γ c ( G ) ≤ n (ln( k + 1) + ln ⌈ ln( k + 1) ⌉ + 4) k + 1 . The main merit here is not the improved estimate, but the proof, which is much simpler thanthe one in [19]. Like the proof in [19], it provides a simple efficient algorithm for finding a connecteddominating set of the required size for a given input graph. As a byproduct of the proof we get anupper bound for the difference between γ ( G ) and γ c ( G ), as stated in the following theorem.Define a function f = f n,k mapping [1 , ∞ ) to [0 , ∞ ) as follows. For any real x ≥
1, let x = ( y + z ) nk +1 with y ≥ z ∈ [0 ,
1] a real:1. If y = 0 then f ( x ) = nk +1 z − .
2. If y = 1 then f ( x ) = nk +1 ( zy + 2) − .
3. If y ≥ f ( x ) = nk +1 ( zy + y − + · · · + + 2) − . The function f is piecewise linear and monotone increasing. Its derivative, which exists in all pointsof (1 , ∞ ) besides the integral multiples of nk +1 , is (weakly) decreasing, thus f is concave. In additionit satisfies the following. For every x = ( w + z ) nk +1 > nk +1 with w ≥ z ∈ [0 ,
1] areal, and for every w ′ satisfying w ≤ w ′ ≤ x − f ( x ) ≥ f ( x − w ′ ) + 1 (7)Indeed, the derivative of f ( z ) is at least w for every z in ( x − w ′ , x ] (besides the integral multiplesof nk +1 ), and thus f ( x ) − f ( x − w ′ ), which is the integral of this derivative from x − w ′ to x , is atleast w ′ · w ≥ Theorem 7.3.
Let G be a connected graph with n vertices, minimum degree at least k and domi-nation number γ = γ ( G ) . Then γ c ( G ) ≤ γ + f n,k ( γ ) . Therefore γ c ( G ) < γ + nk + 1 (ln ⌈ ln( k + 1) ⌉ + 3) . We also describe an improved argument that provides a better estimate than the ones in The-orems 7.1, 7.2.
Theorem 7.4.
Let G be a connected graph with n vertices and minimum degree at least k . Then γ c ( G ) ≤ nk + 1 (ln( k + 1) + 4) − . The proof here too provides an efficient randomized algorithm for finding a connected dominat-ing set with expected size as in the theorem. This algorithm can be derandomized and convertedinto an efficient deterministic algorithm. 26 .1 Proofs
In the proofs we use the following simple lemma.
Lemma 7.5.
Let G = ( V, E ) be a connected graph with n vertices and minimum degree at least k . Let S ⊂ V be a dominating set of G , let H be the induced subgraph of G on S , and supposethe number of its connected components is x = ( y + z ) nk +1 where y is a nonnegative integer and ≤ z ≤ is a real. Then γ c ( G ) ≤ | S | + f ( x ) , where f = f n,k is the function defined in the previoussubsection. Proof:
Starting with the dominating set S we prove, by induction on x , that it is always possibleto add to it at most f ( x ) additional vertices to get a connected dominating set. For x = 1 the givenset is already connected, and as f (1) = 0 the result in this case is trivial. If 1 < x ≤ nk +1 we notethat as long as there are at least two components, each one C can be merged to another one byadding at most two vertices. Indeed, every vertex in the second neighborhood of C is dominated,hence adding the two vertices of a path from C to any such vertex merges C to another component.This means that by adding at most 2( x −
1) = f ( x ) vertices to S we get a connected dominatingset, as needed.If x > nk +1 pick arbitrarily one vertex v = v ( C ) in each of the x connected components of H andlet N ( v ) denote its closed neighborhood consisting of v and all its neighbors in G . This set is of sizeat least k + 1. Therefore there is a vertex u of G that belongs to at least ⌈ ( k + 1) x/n ⌉ of these closedneighborhoods. (This can in fact be slightly improved as none of the vertices of the dominatingset belongs to more than one such closed neighborhood, but we do not use this improvement here).Define S ′ = S ∪ { u } and note that adding u merges at least ⌈ ( k + 1) x/n ⌉ components. Therefore, if x > w nk +1 for an integer w ≥
1, then the number of connected components of the induced subgraphof G on the dominating set S ′ is x − w ′ for some w ′ ≥ w . By induction one can add to S ′ at most f ( x − w ′ ) additional vertices to get a connected dominating set, and the desired result follows from(7). (cid:3) The proof clearly supplies an efficient deterministic algorithm for finding a connected dominatingset of the required size, given the initial dominating set S . Proof of Theorem 7.3:
This is an immediate consequence of Lemma 7.5 together with theobvious fact that if γ ( G ) = γ then G contains a dominating set S of size γ with at most | S | = γ connected components. The known fact that γ ≤ nk +1 (ln( k + 1) + 1) implies that γ ≤ nk +1 ( y + z )with y = ⌈ ln( k + 1) ⌉ and z = 1. The definition of the function f = f n,k thus implies that f n,k ( γ ) ≤ nk + 1 ( 1 y + 1 y − . . . + 11 + 2) − < nk + 1 (ln y + 3) , completing the proof. (cid:3) Proof of Theorem 7.2:
This follows from Theorem 7.3 together with the fact that γ ( G ) ≤ nk +1 (ln( k + 1) + 1). (cid:3) In order to prove Theorem 7.4 we need two simple lemmas. The first one is a known fact, c.f.,e.g., [18], Formula (3.2). for completeness we include a short proof.27 emma 7.6.
For a positive integer k and a real p ∈ (0 , , let B ( k, p ) denote the Binomial randomvariable with parameters k and p . Then the expectation of B ( k,p )+1 satisfies E [ 1 B ( k, p ) + 1 ] = 1( k + 1) p − (1 − p ) k +1 ( k + 1) p . Proof:
By definition E [ 1 B ( k, p ) + 1 ] = k X i =0 i + 1 (cid:18) ki (cid:19) p i (1 − p ) k − i = (1 − p ) k k X i =0 i + 1 (cid:18) ki (cid:19) ( p − p ) i . By the Binomial formula (1 + x ) k = P ki =0 (cid:0) ki (cid:1) x i . Integrating we get(1 + x ) k +1 − k + 1 = k X i =0 i + 1 (cid:18) ki (cid:19) x i +1 . Dividing by x and plugging x = p − p the desired result follows. (cid:3) Lemma 7.7.
Let H = ( V, E ) be a graph. For every v ∈ V let d H ( v ) denote the degree of v in H .Then the number of connected components of H is at most D ( H ) = P v ∈ V d H ( v )+1 . Proof:
The contribution to D ( H ) from the vertices in any connected component C of H with m vertices is X v ∈ C d ( v ) + 1 ≥ X v ∈ C m = 1 . (cid:3) Proof of Theorem 7.4:
Recall that the function f = f n,k defined in the previous subsectionis concave. Therefore, by Jensen’s Inequality, for every positive random variable X , E [ f ( X )] ≤ f ( E [ X ]).Let G = ( V, E ) be a connected graph with n vertices and minimum degree at least k . ByLemma 7.5 if there is a dominating set S of G and the induced subgraph of G on S has x connectedcomponents, then γ c ( G ) ≤ | S | + f ( x ) . (8)For a dominating set S , let H = H ( S ) be the induced subgraph of G on S , and put D ( H ) = P v ∈ S d H ( v )+1 where d H ( v ) is the degree of v in H . By Lemma 7.7 the number of connectedcomponents of H is at most D ( H ), and since the function f = f n,k defined above is monotoneincreasing this implies, by (8), that γ c ( G ) ≤ | S | + f ( D ( H )) = | S | + f ( X v ∈ S d H ( v ) + 1 ) . (9)We next describe a random procedure for generating a dominating set S and complete the proofby upper bounding the expectation of the right-hand-side of (9). The procedure is the standardone described in [13], Theorem 1.1.2 for generating a dominating set. Define p = ln( k +1) k +1 and let T
28e a random set of vertices of G obtained by picking, randomly and independently, each vertex of G to be a member of T with probability p . Let Y = Y T be the set of all vertices of G that are notdominated by T , that is, all vertices in V − T that have no neighbors in T . The set S defined by S = T ∪ Y T is clearly dominating. The expected size of T is np . The expected size of Y T is at most n (1 − p ) k +1 , since for any vertex v the probability it lies in Y T is exactly (1 − p ) d G ( v )+1 ≤ (1 − p ) k +1 ,and the bound for the expectation of | Y T | follows by linearity of expectation. We proceed to boundthe expectation of f ( P v ∈ S d H ( v )+1 ) . By Jensen’s Inequality and the convexity of f mentionedabove this is at most f ( E [ P v ∈ S d H ( v )+1 ]) . Since f is monotone increasing it suffices to bound theexpectation E [ P v ∈ S d H ( v )+1 ].Fix a vertex v . The probability it belongs to Y T (and hence has degree 0 in H ) is (1 − p ) d +1 ,where d is its degree in G . The probability it belongs to T and has degree i in H is p (cid:0) di (cid:1) p i (1 − p ) d − i .Therefore, the expectation of d H ( v )+1 is, by Lemma 7.6,(1 − p ) d +1 + p ( 1( d + 1) p − (1 − p ) d +1 ( d + 1) p ) < (1 − p ) k +1 + 1 k + 1 . Since (1 − p ) k +1 ≤ e − p ( k +1) = k +1 this implies, by linearity of expectation, that E [ X v ∈ S d H ( v ) + 1 ] ≤ nk + 1 . Using, again, linearity of expectation and the fact that f n,k ( nk +1 ) = 3 nk +1 − np + n (1 − p ) k +1 + 3 nk + 1 − ≤ nk + 1 (ln( k + 1) + 4) − . Therefore there is a dominating set S for which this expression is at most the above quantity,completing the proof. (cid:3) The proof of Theorem 7.4 clearly supplies a randomized algorithms generating a connected domi-nating set of expected size at most as in the theorem in any given connected input graph G = ( V, E )with n vertices and minimum degree at least k . This algorithm can be derandomized using themethod of conditional expectations, yielding a polynomial time deterministic algorithm for find-ing such a connected dominating set. Here is the argument. Let v , v , . . . , v n be an arbitrarynumbering of the vertices of G . The algorithm generates a dominating set S satisfying | S | + f ( D ( H )) = | T | + | Y T | + f ( X v ∈ S d H ( v ) + 1 ) ≤ nk + 1 (ln( k + 1) + 4) − , where f = f n,k is the function defined in the proof of Theorem 7.4, H is the induced subgraph of G on S = T ∪ Y T and D ( H ) = P v ∈ S d H ( v )+1 . Once such an S is found it is clear that the proof ofthe theorem provides an efficient way to construct a connected dominating set of the required sizeusing it. 29he algorithm produces S as above by going over the vertices v i in order, where in step i thealgorithm decides whether or not to add v i to S . Let S i denote S ∩ { v , v , . . . , v i } . Thus S = ∅ .For each i , 0 ≤ i ≤ n , define a potential function ψ i in terms of the conditional expectationsof | S | = | T | + | Y T | given S i , which is denoted by E [ | S || S i ] and the conditional expectation of P v ∈ S d H ( v )+1 given S i , denoted by E [ P v ∈ S d H ( v )+1 | S i ]. In this notation ψ i = E [ | S || S i ] + f ( E [ D ( H ) | S i ] = E [ | T || S i ] + E [ | Y T || S i ] + f ( E [ X v ∈ S d H ( v ) + 1 | S i ]) . Given the graph G and the set S i , it is not difficult to compute ψ i in polynomial time. Indeed, bylinearity of expectation, the conditional expectation E [ | T || S i ] is computed by adding the contribu-tion of each vertex v = v j to it. For j ≤ i this contribution is 1 if v j ∈ T and 0 if v j T . For j > i the contribution is p . The contribution of v j to E [ Y T | S i ] is 0 if v j is already dominated by a vertexin S i , and if it is not, then it is (1 − p ) s , where s is the number of neighbors of v j (including v j itself if j > i ) in the set V − { v , v , . . . , v i } .The conditional expectation E [ P v ∈ S d H ( v )+1 | S i ] is also computed using linearity of expectation,where the contribution of each vertex v j is E [ d H ( v j +1) | S i ]. This is also simple to compute in allcases. We describe here only one representative example. If j > i , q of the neighbors of v j appearin S i , and the number of its neighbors in G which lie in V − { v , v , . . . , v i } is s , then E [ 1 d H ( v j + 1) | S i ] = p · s X a =0 (cid:18) sa (cid:19) p a (1 − p ) s − a q + 1 + a . A similar expression exists in every other possible case.Put ψ i = ψ ( T ) i + ψ ( Y ) i + ψ ( f ) i , where ψ i ( T ) = E [ | T || S i ], ψ i ( Y ) = E [ | Y T || S i ], and ψ ( f ) i = f [ E ( D ( H ) | S i ]. By the definition of conditional expectation ψ ( T ) i = pE [ | T | | S i +1 = S i ∪ v i +1 ] + (1 − p ) E [ | T | | S i +1 = S i ] (10)and ψ ( Y ) i = pE [ | Y T | | S i +1 = S i ∪ v i +1 ] + (1 − p ) E [ | Y T | | S i +1 = S i ] (11)Similarly, using the fact that the function f is concave ψ ( f ) i = f ( pE [ X v ∈ H d H ( v j ) + 1) | S i +1 = S i ∪ v i +1 ] + (1 − p ) E [ X v ∈ H d H ( v j ) + 1) | S i +1 = S i ]) ≥ pf ( E [ X v ∈ H d H ( v j ) + 1) | S i +1 = S i ∪ v i +1 ]) + (1 − p ) f ( E [ X v ∈ H d H ( v j ) + 1 ) | S i +1 = S i ]) ≥ min { f ( E [ X v ∈ H d H ( v j ) + 1) | S i +1 = S i ∪ v i +1 ]) , f ( E [ X v ∈ H d H ( v j ) + 1) | S i +1 = S i ]) . Let ψ + i +1 denote the value of ψ i +1 with S i +1 = S i ∪ v i +1 and ψ − i +1 denote the value of ψ i +1 with S i +1 = S i . 30y adding the last inequality and (10),(11) we conclude that ψ i ≥ min { ψ + i +1 , ψ − i +1 } . Therefore, if the algorithm decides in each step i + 1 whether or not to add v i +1 to S i in order toget S i +1 by choosing the option that minimizes the value of ψ i +1 , then the potential function ψ i is a monotone decreasing function of i . Since ψ is at most nk +1 (ln( k + 1) + 4) − ψ n . However, ψ n is exactly | S | + f ( D ( H )) for the dominating set S constructedby the algorithm. This completes the description of the algorithm and its correctness. We conclude with the following problem.
Problem:
Determine or estimate the maximum possible value of the difference γ c ( G ) − γ ( G ),where the maximum is taken over all connected graphs G with n vertices and minimum degree atleast k .By Theorem 7.3 this maximum is at most nk +1 (ln ⌈ ln( k + 1) ⌉ + 3). It is not difficult to show thatit is at least ⌊ nk +1 ⌋ −
1. To see this assume, for simplicity, that k + 1 divides n and put m = nk +1 .For each 0 ≤ i < m let K i be the graph obtained from a clique on k + 1 vertices by deletinga single edge x i y i . Let G be the k -regular graph obtained from the vertex disjoint union of the m graphs K i by adding the edges y i x i +1 for all 0 ≤ i < m , where x m = x . For this cycleof cliques G , γ ( G ) = m = nk +1 as shown by a dominating set consisting of one vertex in each K i − { x i , y i } - this is a minimum dominating set as G is k -regular. On the other hand the inducedsubgraph on any connected dominating set must contain at least m − y i x i +1 andtheir endpoints, and it is not difficult to check that it must contain at least one additional vertex.Thus γ c ( G ) = 2 m − nk +1 −
1. It will be interesting to close the ln ln( k + 1) gap between theupper and lower bounds and decide whether or not the above maximum is Θ( nk +1 ). Acknowledgment
I thank Eli Berger, Michael Krivelevich, Shay Moran, Shubhangi Saraf, MadhuSudan and Tibor Szab´o for helpful discussions.
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