For a graph G, its rth power is constructed by placing an edge between two
vertices if they are within distance r of each other. In this note we study the
amount of edges added to a graph by taking its rth power. In particular we
obtain that either the rth power is complete or "many" new edges are added.
This is an extension of a result obtained by P. Hegarty for cubes of graphs.Comment: 6 pages, 1 figur

Pokrovskiy, Alexey

English

arXiv

For a graph G, its rth power is constructed by placing an edge between two vertices if they are within distance r of each other. In this note we study the amount of edges added to a graph by taking its rth power. In particular we obtain that either the rth power is complete or "many" new edges are added. This is an extension of a result obtained by P. Hegarty for cubes of graphs

Name not available

Growth of graph powersA. Pokrovskiy (LSE)November 9, 2018AbstractFor a graph G, its rth power is constructed by placing an edge between two vertices ifthey are within distance r of each other. In this note we study the amount of edges addedto a graph by taking its rth power. In particular we obtain that either the rth power iscomplete or “many” new edges are added. This is an extension of a result obtained byP. Hegarty for cubes of graphs.1 IntroductionThis note addresses some questions raised by P. Hegarty in [2]. In that paper he studiedresults about graphs inspired by the Cauchy-Davenport Theorem.All graphs in this paper are simple and loopless. For two vertices u, v ∈ V (G), denote thelength of the shortest path between them by d(u, v). For v ∈ V (G), define its ith neighborhoodas Ni(v) = {u ∈ V (G) : d(u, v) = i}. The rth power of a graph G, denoted Gr, is constructedfrom G by adding an edge between two vertices x and y when they are within distance r in G.Define the diameter of G, diam(G), as the minimal r such that Gr is complete (alternatively,the maximal distance between two vertices). Denote the number of edges of G by e(G). Forv ∈ V (G) and a set of vertices S, define er(v, S) = |{u ∈ S : d(v, u) ≤ r}|.The Cayley graph of a subset A ⊆ Zp is constructed on the vertex set Zp. For two distinctvertices x, y ∈ Zp, we define xy to be an edge whenever x − y ∈ A or y − x ∈ A. Thefollowing is a consequence of the Cauchy-Davenport Theorem (usually stated in the languageof additive number theory [1]).Theorem 1. Let p be a prime, A a subset of Zp, and G the Cayley graph of A. Then forany integer r < diam(G):e(Gr) ≥ r e(G).If we take A to be the arithmetic progression {a, 2a, . . . , ka}, then equality holds in thistheorem for all r < diam(G). We might look for analogues of Theorem 1 for more generalgraphs G. In particular since these Cayley graphs are always regular and (when p is prime)connected, we might focus on regular, connected G. In [2] Hegarty proved the followingtheorem:Theorem 2. Suppose G is a regular, connected graph with diam(G) ≥ 3. Then we havee(G3) ≥ (1 + ) e(G),with  ≈ 0.0871arXiv:1005.2542v1  [math.CO]  14 May 2010In other words, the cube ofG retains the original edges ofG and gains a positive proportionof new ones. In Section 3 we prove this theorem with an improved constant of  = 16 . Therequirement of regularity cannot be easily dropped, as shown in [2].Theorem 2 leads to the question of how the growth behaves for other powers of the G.Note that Theorem 2 cannot be used recursively to obtain such a result – since the cube ofa regular graph is not necessarily regular. In [2] it was shown that no equivalent of Theorem2 exists with G3 replaced by G2, and it was asked what happens for higher powers. In thisnote we address that question.2 Main ResultWe prove the following theorem:Theorem 3. Suppose G is a regular, connected graph, and r ≤ diam(G). Then we have:e(Gr) ≥(⌈r3⌉− 1)e(G).Proof. Let the degree of each vertex be d. Fix some v with Ndiam(G)(v) nonempty.Consider any vertex u ∈ V (G). Then for any j satisfying d(u, v)−r < j ≤ d(u, v), there isa wj ∈ Nj(v) such that d(u,wj) < r. For such a wj , all vertices x ∈ N1(wj) have d(u, x) ≤ r.All such x are contained in Nj−1(v) ∪Nj(v) ∪Nj+1(v), henceer(u,Nj−1(v) ∪Nj(v) ∪Nj+1(v)) ≥ d. (1)Note that each j ∈ {d(u, v)− 3, d(u, v)− 6, . . . , d(u, v)− 3 (⌈13 min{d(u, v), r}⌉− 1)} satisfiesd(u, v) − r < j ≤ d(u, v). Summing the bound (1) over all these j, noting that any edge iscounted at most once, we obtainer(u,N0(v) ∪ · · · ∪Nd(u,v)−2(v)) ≥⌈13min{d(u, v), r}⌉d− d.Now we sum this over all u ∈ G. Note that since the edges counted above go from someNi(v) to Nj(v) with j < i, each edge is counted at most once. Also we haven’t yet countedany of the original edges of G, so we might as well add them. Hencee(Gr) ≥∑u∈Ger(u,N0(v) ∪ · · · ∪Nd(u,v)−2(v)) + e(G)≥∑u∈G⌈13min{d(u, v), r}⌉d− |V (G)|d+ e(G)=∑u∈G⌈13min{d(u, v), r}⌉− e(G). (2)Obviously there was nothing particularly special about v. We can get a similar expresssionusing v′ ∈ Ndiam(G)(v), namelye(Gr) ≥∑u∈G⌈13min{d(u, v′), r}⌉− e(G). (3)2Averaging (2) and (3) we gete(Gr) ≥ 12∑u∈G(⌈13min{d(u, v), r}⌉+⌈13min{d(u, v′), r}⌉)d− e(G). (4)Note that for any u ∈ V (G) we have⌈13min{d(u, v), r}⌉+⌈13min{d(u, v′), r}⌉≥⌈r3⌉. (5)This is because d(u, v) + d(u, v′) ≥ d(v, v′) = diam(G) ≥ r. Putting the bound (5) into thesum (4) we obtaine(Gr) ≥ |V (G)|d2⌈r3⌉− e(G) =⌈r3⌉e(G)− e(G).Thus the theorem is proven.3 CubesNote that for r ≤ 6 the bounds in Theorem 3 are trivial. In particular it says nothing aboutthe increase in the number of edges of the cube of a regular, connected graph. Such anincrease was already demonstrated by Hegarty in Theorem 2. Here we give an alternativeproof of that theorem, yielding a slightly better constant.Theorem 4. Suppose G is a regular, connected graph with diam(G) ≥ 3. Then we havee(G3) ≥(1 +16)e(G).Proof. Let the degree of each vertex be d. Note that as G is regular, and not complete, everyv ∈ V (G) will have a non-neighbour in G. Together with connectedness this implies that eachv ∈ V (G) has at least one new neighbour in G2. This implies the theorem for d ≤ 6. For theremainder of the proof, we assume that d > 6. The proof rests on the following colouring ofthe edges of G: For an edge uv in G, colouruv red if |N1(u) ∩N1(v)| > 23d,uv blue if |N1(u) ∩N1(v)| ≤ 23d.Notice that if uv is a blue edge, then there are at least 43d−1 neighbours of u in G2. This isbecause u will be connected to everything in N1(u)∪N1(v) except itself, and |N1(u)∪N1(v)| ≥43d for uv blue. If, in addition, we have some x connected to u by an edge (of any colour),then x will be at distance at most 3 from everything in N1(u) ∪ N1(v) \ {x}. Hence x willhave at least 43d− 1 neighbours in G3.Partition the vertices of G as follows:B = {v ∈ V (G) : v has a blue edge coming out of it},R = {v ∈ V (G) : v /∈ B and there is a u ∈ B such that uv is an edge},S = V (G) \ (B ∪R).By the above argument, if v is in B ∪R, then e3(v, V (G)) ≥ 43d− 1. Recall that each u ∈ S3will have at least one new neighbour in G2, giving e3(u, V (G)) ≥ d+ 1. Summing these twobounds over all vertices in G, noting that any edge is counted twice, gives2e(G3) ≥(13d− 1)|B ∪R|+ (d+ 1)|S|=(43d− 1)|B ∪R|+ (d+ 1) (|V (G)| − |B ∪R|)=76d|V (G)|+ 13(|B ∪R| − 12|V (G)|)(d− 6)=73e(G) +13(|B ∪R| − 12|V (G)|)(d− 6) .Recall that we are considering the case when d > 6. Thus to prove that e(G3) ≥ 76e(G),it suffices to show that |B ∪R| ≥ 12 |V (G)|. To this end we shall demonstrate that |S| ≤ |R|.First however we need a proposition helping us to find blue edges in G.Proposition 5. For any v ∈ V (G) there is some b ∈ B such that d(v, b) ≤ 2.Proof. Suppose d(v, u) = 3. Then there are vertices x and y such that {v, x, y, u} forms apath between u and v. We will show that one of the edges vx, xy or yu is blue. This willprove the proposition assuming that there are any blue edges to begin with. However, it alsoshows the existence of blue edges because diam(G) ≥ 3.So, suppose that the edges vx and uy are red. Then we have |N1(v) ∩N1(x)| > 23d, and|N1(u) ∩N1(y)| > 23d. Using this and |N1(u) ∩N1(v)| = ∅ gives|N1(x) ∪N1(y)| ≥ |(N1(x) ∪N1(y)) ∩N1(v)|+ |(N1(x) ∪N1(y)) ∩N1(u)|≥ |N1(x) ∩N1(v)|+ |N1(y) ∩N1(u)|>43d.Therefore |N1(x) ∩ N1(y)| = 2d − |N1(x) ∪ N1(y)| ≤ 23d. Hence xy is blue, proving theproposition.Now we will show that |S| ≤ |R|. Suppose r ∈ R. By the definition of R, there is a b ∈ Bsuch that rb is an edge. This edge is neccesarily red as r /∈ B. Using N1(b) ⊆ B ∪R,we have|N1(r) ∩ (B ∪R)| ≥ |N1(r) ∩N1(b)| > 23d. Hence|N1(r) ∩ S| ≤ 13d. (6)Suppose s ∈ S. Proposition 5 implies that there is some r ∈ R such that sr is an edge.Since sr is red, we have |N1(s) ∩N1(r)| > 23d. Using this, the fact that N1(s) ⊆ R ∪ S, and(6), gives|N1(s) ∩R| ≥ |N1(s) ∩N1(r) ∩R|= |N1(s) ∩N1(r)| − |N1(s) ∩N1(r) ∩ S|≥ |N1(s) ∩N1(r)| − |N1(r) ∩ S|>13d. (7)4Double-counting the edges between S and R using the bounds (6) and (7) gives a contra-diction unless |S| ≤ |R|. Therefore |B ∪R| ≥ 12 |V (G)| as required.4 DiscussionTheorem 3 answers the question of giving a lower bound on the number of edges that aregained by taking higher powers of a graph. We obtain growth that is linear with r – just asin Theorem 1.• The constant ⌈13r⌉ in Theorem 3 cannot be improved to something of the form λr withλ > 13 . To see and consider the following sequence of graphs Hr(d) as d tends to infinity:Take disjoint sets of vertices N0, ..., Nr, with |Ni| = d− 1 if i ≡ 0 (mod 3) and |Ni| = 2otherwise. Add all the edges within each set and also between neighboring ones. So ifu ∈ Ni, v ∈ Nj , then uv is an edge whenever |i− j| ≤ 1 (see Figure 1).Figure 1: The graph H6(9).The number of edges in Hr(d) is at least the number of edges in the larger classes whichis⌈13(r + 1)⌉ (d−12).The rth power Hr(d)r has less than(|V (G)|2)edges which is less than(d 13(r+1)e(d+3)2).Therefore,lim supd→∞e(Hr(d)r)e(Hr(d))≤ limd→∞(d 13(r+1)e(d+3)2)⌈13(r + 1)⌉ (d−12) = ⌈13(r + 1)⌉.The graphs Hr(d) are not regular, but if r 6≡ 2 (mod 3), it is possible to remove a small(less than |V (G)|) number of edges from the graphs and make them d-regular withoutlosing connectedness (any cycle passing through all the vertices in N1∪ ...∪Nr−1 wouldwork). Call these new graphs Hˆr(d). By the same argument as before we havelim supd→∞e(Hˆr(d)r)e(Hˆr(d))≤⌈13(r + 1)⌉.If r ≡ 2 (mod 3), a similar trick can be performed, but we’d need to start with |Ni| =d− 1 if i ≡ 1 (mod 3) and |Ni| = 2 otherwise.So the factor of 13 cannot be improved for regular graphs. All these examples areinspired by one given in [2] to show that for any  there are regular graphs G withe(G2) < (1 + )e(G).5• Despite the above example, there is certainly room for further improvement in Theorems3 and 4. In particular, Theorem 4 doesn’t seem tight in any way. The graphs Hˆr(d)seem to give essentially the slowest possible growth for all powers of regular graphs.Considering the graphs H3(d) leads to the conjecture ofe(G3) ≥ 2e(G),for G regular, connected, and diam(G) ≥ 3.A shortcoming of Theorem 3 is that it only gives a good bound if the diameter of G isclose to r. When this is not the case, the number of edges in Gr seems to grow faster.It would be interesting to obtain a good lower bound on e(Gr) involving both r anddiam(G).• All the questions from this paper and [2] could be asked for directed graphs. In particularone can define directed Cayley graphs for a set A ⊆ Zp by letting xy be a directed edgewhenever x− y ∈ A. Then the Cauchy-Davenport Theorem implies an identical versionof Theorem 1 for directed Cayley graphs. In this setting it is easy to show that there isgrowth even for the square of an out-regular oriented graph D (a directed graph wherefor a pair of vertices u and v, uv and vu are not both edges). In particular, we havee(D2) ≥ 32e(D). (8)This occurs because every vertex v has |Nout2 (v)| ≥ 12 |Nout1 (v)| in an out-regular orientedgraph. It’s easy to see that this is best possible for such graphs. One can constructout-regular oriented graphs with an arbitrarily large proportion of vertices v satisfying|Nout2 (v)| = 12 |Nout1 (v)|.However if we insist on both in and out-degrees to be constant, (8) no longer seems tight.Such graphs are always Eulerian. In [3] there is a conjecture attributed to Jackson andSeymour that if an oriented graph D is Eulerian, then e(D2) ≥ 2 e(D) holds. If thisconjecture were proved, it would be an actual generalization of the directed version ofTheorem 1, as opposed to the mere analogues proved above.5 AcknowledgementI would like to thank my supervisors Jan van den Heuvel, and Jozef Skokan for much helpfuladvice and remarks.References[1] H. Davenport, On the addition of residue classes, J. London Math. Soc., Issue 10, 30-32,(1935).[2] P. Hegarty, A Cauchy-Davenport type result for arbitrary regular graphs.arXiv:0910.2250v2 [math.CO] , (2009).[3] B. D. Sullivan, A summary of results and problems related to the Caccetta-Ha¨ggkvist con-jecture. arXiv:math/0605646v1 [math.CO] , (2006).6

Growth of graph powers

For a graph G, its rth power is constructed by placing an edge between two vertices if they are within distance r of each other. In this note we study the amount of edges added to a graph by taking its rth power. In particular we obtain that, for r ≥ 3, either the rth power is complete or "many" new edges are added. In this direction, Hegarty showed that there is a constant ε > 0 such e(G3) ≥ (1 + ε)e(G). We extend this result in two directions. We give an alternative proof of Hegarty's result with an improved constant of ε = 1/6. We also show that for general

Pokrovskiy, A

UCL Discovery

Growth of graph powersA. Pokrovskiy ∗Departement of MathematicsLondon School of Economics and Political Sciences,Houghton Street, WC2A 2AELondon, United Kingdom.a.pokrovskiy@lse.ac.ukSubmitted: Nov 11, 2010; Accepted: Mar 23, 2011; Published: Apr 14, 2011Mathematics Subject Classification: 05C12AbstractFor a graph G, its rth power is constructed by placing an edge between twovertices if they are within distance r of each other. In this note we study the amountof edges added to a graph by taking its rth power. In particular we obtain that,for r ≥ 3, either the rth power is complete or “many” new edges are added. In thisdirection, Hegarty showed that there is a constant ǫ > 0 such e(G3) ≥ (1 + ǫ)e(G).We extend this result in two directions. We give an alternative proof of Hegarty’sresult with an improved constant of ǫ = 16 . We also show that for general r,e(Gr) ≥(⌈r3⌉− 1)e(G).1 IntroductionThis note addresses some questions raised by P. Hegarty in [4]. In that paper he studiedresults about graphs inspired by the Cauchy-Davenport Theorem.All graphs in this paper are simple and loopless. For two vertices u, v ∈ V (G), denotethe length of the shortest path between them by d(u, v). For v ∈ V (G), define its ithneighborhood as Ni(v) = {u ∈ V (G) : d(u, v) = i}. The rth power of a graph G,denoted Gr, is constructed from G by adding an edge between two vertices x and y whenthey are within distance r in G. Define the diameter of G, diam(G), as the minimalr such that Gr is complete (alternatively, the maximal distance between two vertices).Denote the number of edges of G by e(G). For v ∈ V (G) and a set of vertices S, defineer(v, S) = |{u ∈ S : d(v, u) ≤ r}|.The Cayley graph of a subset A ⊆ Zp is constructed on the vertex set Zp. For twodistinct vertices x, y ∈ Zp, we define xy to be an edge whenever x − y ∈ A or y − x ∈ A.∗Research supported by the LSE postgraduate research studentship scheme.the electronic journal of combinatorics 18 (2011), #P88 1The following is a consequence of the Cauchy-Davenport Theorem (usually stated in thelanguage of additive number theory [1, 2]).Theorem 1. Let p be a prime, A a subset of Zp, and G the Cayley graph of A. Then forany integer r < diam(G):e(Gr) ≥ r e(G).If we take A to be the arithmetic progression {a, 2a, . . . , ka}, then equality holds inthis theorem for all r < diam(G). We might look for analogues of Theorem 1 for moregeneral graphs G. In particular since these Cayley graphs are always regular and (whenp is prime) connected, we might focus on regular, connected G. In [4] Hegarty proved thefollowing theorem:Theorem 2. Suppose G is a regular, connected graph with diam(G) ≥ 3. Then we havee(G3) ≥ (1 + ǫ) e(G),with ǫ ≈ 0.087In other words, the cube of G retains the original edges of G and gains a positiveproportion of new ones. In Section 3 we prove this theorem with an improved constantof ǫ = 16. Since we announced this note, DeVos and Thomassé [3] further improved theconstant in Theorem 2 to ǫ = 34. They also show that the constant cannot be improvedfurther by exhibiting a sequence of regular graphs Gn, such thate(Grn)e(Gn)→ 74as n → ∞.Theorem 2 leads to the question of how the growth behaves for other powers of theG. Note that Theorem 2 cannot be used recursively to obtain such a result – since thecube of a regular graph is not necessarily regular. In [4] it was shown that no equivalentof Theorem 2 exists with G3 replaced by G2, and it was asked what happens for higherpowers. In this note we address that question.2 Main ResultWe prove the following theorem:Theorem 3. Suppose G is a regular, connected graph, and r ≤ diam(G). Then we have:e(Gr) ≥(⌈r3⌉− 1)e(G).Proof. Let the degree of each vertex be d. Fix some v with Ndiam(G)(v) nonempty.Consider any vertex u ∈ V (G). Then for any j satisfying d(u, v) − r < j ≤ d(u, v),there is a wj ∈ Nj(v) such that d(u, wj) < r. For such a wj, all vertices x ∈ N1(wj) haved(u, x) ≤ r. All such x are contained in Nj−1(v) ∪ Nj(v) ∪ Nj+1(v), henceer(u, Nj−1(v) ∪ Nj(v) ∪ Nj+1(v)) ≥ d. (1)the electronic journal of combinatorics 18 (2011), #P88 2Note that each j ∈ {d(u, v) − 3, d(u, v) − 6, . . . , d(u, v) − 3(⌈13min{d(u, v), r}⌉− 1)}satisfies d(u, v) − r < j ≤ d(u, v). Summing the bound (1) over all these j, noting thatany edge is counted at most once, we obtainer(u, N0(v) ∪ · · · ∪ Nd(u,v)−2(v)) ≥⌈13min{d(u, v), r}⌉d − d.Now we sum this over all u ∈ G. Note that since the edges counted above go fromsome Ni(v) to Nj(v) with j < i, each edge is counted at most once. Also we haven’t yetcounted any of the original edges of G, so we might as well add them. Hencee(Gr) ≥∑u∈Ger(u, N0(v) ∪ · · · ∪ Nd(u,v)−2(v)) + e(G)≥∑u∈G⌈13min{d(u, v), r}⌉d − |V (G)|d + e(G)=∑u∈G⌈13min{d(u, v), r}⌉d − e(G). (2)Obviously there was nothing particularly special about v. We can get a similar ex-presssion using v′ ∈ Ndiam(G)(v), namelye(Gr) ≥∑u∈G⌈13min{d(u, v′), r}⌉d − e(G). (3)Averaging (2) and (3) we gete(Gr) ≥12∑u∈G(⌈13min{d(u, v), r}⌉+⌈13min{d(u, v′), r}⌉)d − e(G). (4)Note that for any u ∈ V (G) we have⌈13min{d(u, v), r}⌉+⌈13min{d(u, v′), r}⌉≥⌈r3⌉. (5)This is because d(u, v) + d(u, v′) ≥ d(v, v′) = diam(G) ≥ r. Putting the bound (5) intothe sum (4) we obtaine(Gr) ≥|V (G)|d2⌈r3⌉− e(G) =⌈r3⌉e(G) − e(G).Thus the theorem is proven.3 CubesNote that for r ≤ 6 the bounds in Theorem 3 are trivial. In particular it says nothingabout the increase in the number of edges of the cube of a regular, connected graph.Such an increase was already demonstrated by Hegarty in Theorem 2. Here we give analternative proof of that theorem, yielding a slightly better constant.the electronic journal of combinatorics 18 (2011), #P88 3Theorem 4. Suppose G is a regular, connected graph with diam(G) ≥ 3. Then we havee(G3) ≥(1 +16)e(G).Proof. Let the degree of each vertex be d. Note that as G is regular, and not complete,every v ∈ V (G) will have a non-neighbour in G. Together with connectedness this impliesthat each v ∈ V (G) has at least one new neighbour in G2. This implies the theorem ford ≤ 6. For the remainder of the proof, we assume that d > 6. The proof rests on thefollowing colouring of the edges of G: For an edge uv in G, colouruv red if |N1(u) ∩ N1(v)| >23d,uv blue if |N1(u) ∩ N1(v)| ≤23d.Notice that if uv is a blue edge, then there are at least 43d − 1 neighbours of u inG2. This is because u will be connected to everything in N1(u) ∪ N1(v) except itself,and |N1(u) ∪ N1(v)| ≥43d for uv blue. If, in addition, we have some x connected tou by an edge (of any colour), then x will be at distance at most 3 from everything inN1(u) ∪ N1(v) \ {x}. Hence x will have at least43d − 1 neighbours in G3.Partition the vertices of G as follows:B = {v ∈ V (G) : v has a blue edge coming out of it},R = {v ∈ V (G) : v /∈ B and there is a u ∈ B such that uv is an edge},S = V (G) \ (B ∪ R).By the above argument, if v is in B ∪ R, then e3(v, V (G)) ≥ 43d − 1. Recall that eachu ∈ S will have at least one new neighbour in G2, giving e3(u, V (G)) ≥ d + 1. Summingthese two bounds over all vertices in G, noting that any edge is counted twice, gives2e(G3) ≥(43d − 1)|B ∪ R| + (d + 1)|S|=(43d − 1)|B ∪ R| + (d + 1) (|V (G)| − |B ∪ R|)=76d|V (G)| +13(|B ∪ R| −12|V (G)|)(d − 6)=73e(G) +13(|B ∪ R| −12|V (G)|)(d − 6) .Recall that we are considering the case when d > 6. Thus to prove that e(G3) ≥ 76e(G),it suffices to show that |B∪R| ≥ 12|V (G)|. To this end we shall demonstrate that |S| ≤ |R|.First however we need a proposition helping us to find blue edges in G.Proposition 5. For any v ∈ V (G) there is some b ∈ B such that d(v, b) ≤ 2.the electronic journal of combinatorics 18 (2011), #P88 4Proof. Suppose d(v, u) = 3. Then there are vertices x and y such that {v, x, y, u} forms apath between u and v. We will show that one of the edges vx, xy or yu is blue. This willprove the proposition assuming that there are any blue edges to begin with. However, italso shows the existence of blue edges because diam(G) ≥ 3.So, suppose that the edges vx and uy are red. Then we have |N1(v) ∩ N1(x)| >23d,and |N1(u) ∩ N1(y)| >23d. Using this and N1(u) ∩ N1(v) = ∅ gives|N1(x) ∪ N1(y)| ≥ |(N1(x) ∪ N1(y)) ∩ N1(v)| + |(N1(x) ∪ N1(y)) ∩ N1(u)|≥ |N1(x) ∩ N1(v)| + |N1(y) ∩ N1(u)|>43d.Therefore |N1(x) ∩ N1(y)| = 2d − |N1(x) ∪ N1(y)| ≤23d. Hence xy is blue, proving theproposition.Now we will show that |S| ≤ |R|. Suppose r ∈ R. By the definition of R, thereis a b ∈ B such that rb is an edge. This edge is neccesarily red as r /∈ B. UsingN1(b) ⊆ B ∪ R,we have |N1(r) ∩ (B ∪ R)| ≥ |N1(r) ∩ N1(b)| >23d. Hence|N1(r) ∩ S| ≤13d. (6)Suppose s ∈ S. Proposition 5 implies that there is some r ∈ R such that sr is an edge.Since sr is red, we have |N1(s) ∩ N1(r)| >23d. Using this, the fact that N1(s) ⊆ R ∪ S,and (6), gives|N1(s) ∩ R| ≥ |N1(s) ∩ N1(r) ∩ R|= |N1(s) ∩ N1(r)| − |N1(s) ∩ N1(r) ∩ S|≥ |N1(s) ∩ N1(r)| − |N1(r) ∩ S|>13d. (7)Double-counting the edges between S and R using the bounds (6) and (7) gives acontradiction unless |S| ≤ |R|. Therefore |B ∪ R| ≥ 12|V (G)| as required.4 DiscussionTheorem 3 answers the question of giving a lower bound on the number of edges that aregained by taking higher powers of a graph. We obtain growth that is linear with r – justas in Theorem 1.• The constant⌈13r⌉in Theorem 3 cannot be improved to something of the form λrwith λ > 13. To see this, consider the following sequence of graphs Hr(d) as d tendsto infinity:the electronic journal of combinatorics 18 (2011), #P88 5N0N1N2N 3N4N5N6Figure 1: The graph H6(8).Take disjoint sets of vertices N0, ..., Nr, with |Ni| = d − 1 if i ≡ 0 (mod 3) and|Ni| = 2 otherwise. Add all the edges within each set and also between neighboringones. So if u ∈ Ni, v ∈ Nj, then uv is an edge whenever |i − j| ≤ 1 (see Figure 1).The number of edges in Hr(d) is at least the number of edges in the larger classeswhich is⌈13(r + 1)⌉ (d−12).The rth power Hr(d)r has less than(|V (G)|2)edges which is less than(⌈ 13(r+1)⌉(d+3)2).Therefore,lim supd→∞e(Hr(d)r)e(Hr(d))≤ limd→∞(⌈ 13(r+1)⌉(d+3)2)⌈13(r + 1)⌉ (d−12) =⌈13(r + 1)⌉.The graphs Hr(d) are not regular, but if r 6≡ 2 (mod 3), it is possible to removea small (less than |V (G)|) number of edges from the graphs and make them d-regular without losing connectedness (any cycle passing through all the vertices inN1 ∪ ... ∪ Nr−1 would work). Call these new graphs Ĥr(d). By the same argumentas before we havelim supd→∞e(Ĥr(d)r)e(Ĥr(d))≤⌈13(r + 1)⌉.If r ≡ 2 (mod 3), a similar trick can be performed, but we’d need to start with|Ni| = d − 1 if i ≡ 1 (mod 3) and |Ni| = 2 otherwise.So the factor of 13cannot be improved for regular graphs. All these examples areinspired by one given in [4] to show that for any ǫ there are regular graphs G withe(G2) < (1 + ǫ)e(G).• All the questions from this paper and [4] could be asked for directed graphs. Inparticular one can define directed Cayley graphs for a set A ⊆ Zp by letting xy bea directed edge whenever x− y ∈ A. Then the Cauchy-Davenport Theorem impliesan identical version of Theorem 1 for directed Cayley graphs. In this setting it iseasy to show that there is growth even for the square of an out-regular orientedgraph D (a directed graph where for a pair of vertices u and v, uv and vu are notthe electronic journal of combinatorics 18 (2011), #P88 6both edges). In particular, we havee(D2) ≥32e(D). (8)This occurs because every vertex v has |Nout2 (v)| ≥12|Nout1 (v)| in an out-regularoriented graph. It’s easy to see that this is best possible for such graphs. One canconstruct out-regular oriented graphs with an arbitrarily large proportion of verticesv satisfying |Nout2 (v)| =12|Nout1 (v)|.However if we insist on both in and out-degrees to be constant, (8) no longer seemstight. Such graphs are always Eulerian. In [5] there is a conjecture attributed toJackson and Seymour that if an oriented graph D is Eulerian, then e(D2) ≥ 2 e(D)holds. If this conjecture were proved, it would be an actual generalization of thedirected version of Theorem 1, as opposed to the mere analogues proved above.AcknowledgementThe author would like to thank his supervisors Jan van den Heuvel and Jozef Skokan forhelpful advice and discussions.References[1] A. L. Cauchy. Recherches sur les nombres. J. Ecole Polytech, 9:99–116, 1813.[2] H. Davenport. On the addition of residue classes. J. London Math. Soc., 10:30–32,1935.[3] M. DeVos and S. Thomassé. Edge growth in graph cubes. arXiv:1009.0343v1[math.CO], 2010.[4] P. Hegarty. A Cauchy-Davenport type result for arbitrary regular graphs. Integers,11, Paper A19, 2011.[5] B. D. Sullivan. A summary of results and problems related to the Caccetta-Häggkvistconjecture. AIM Preprint, 2006.the electronic journal of combinatorics 18 (2011), #P88 7

Growth of Graph Powers

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Growth of graph powers

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