For a graph G, its rth power G^r has the same vertex set as G, and has an
edge between any two vertices within distance r of each other in G. We give a
lower bound for the number of edges in the rth power of G in terms of the order
of G and the minimal degree of G. As a corollary we determine how small the
ratio e(G^r)/e(G) can be for regular graphs of diameter at least r

Pokrovskiy, Alexey

English

arXiv

For a graph G, its rth power G^r has the same vertex set as G, and has an edge between any two vertices within distance r of each other in G. We give a lower bound for the number of edges in the rth power of G in terms of the order of G and the minimal degree of G. As a corollary we determine how small the ratio e(G^r)/e(G) can be for regular graphs of diameter at least r

Name not available

Edge growth in graph powersA. PokrovskiyLondon School of Economics and Political Sciencesa.pokrovskiy@lse.ac.ukDecember 10, 2018AbstractFor a graph G, its rth power Gr has the same vertex set as G, and has an edgebetween any two vertices within distance r of each other in G. We give a lower boundfor the number of edges in the rth power of G in terms of the order of G and theminimal degree of G. As a corollary we determine how small the ratio e(Gr)/e(G)can be for regular graphs of diameter at least r.1 IntroductionWe will consider both graphs that may have loops and graphs in which loops are explicitlyforbidden. Loopless graphs will be denoted by Roman italic letters, such as “G”, whilegraphs with loops allowed will be denoted by curly letters, such as “G”. For two vertices xand y (possibly x = y) we only allow one edge between x and y. The rth power of G,denoted Gr, is the graph with vertex set V (G), and xy an edge whenever x and y arewithin distance r of each other. The diameter of a connected graph is the smallest r forwhich Gr is complete. For all standard notation we refer to [5].For a connected graph of diameter at least r, one would expect Gr to have substantiallymore edges than G. In this note we examine how small the ratio e(Gr)/e(G) can be,focusing primarily on the case when G is a regular graph.The motivation for studying this comes from a corollary of the Cauchy-DavenportTheorem from additive number theory which we will now state. The Cayley graph of asubset A ⊆ Zp is constructed on the vertex set Zp. For two distinct vertices x, y ∈ Zp, wedefine xy to be an edge whenever x− y ∈ A or y − x ∈ A. The following is a consequenceof the Cauchy-Davenport Theorem (usually stated in the language of additive numbertheory).Theorem 1 (Cauchy, Davenport, [1, 2]). Let p be a prime, G the Cayley graph of a setA ⊆ Zp, and r an integer such that r < diam(G). Then we havee(Gr)e(G)≥ r. (1)1arXiv:1202.6085v1  [math.CO]  27 Feb 2012One could ask whether inequalities similar to (1) hold for more general families ofgraphs. Motivated by the fact that Cayley graphs are regular, Hegarty asked this questionfor regular graphs and proved the following theorem.Theorem 2 (Hegarty, [7]). Let G be a regular, connected graph, with diam(G) ≥ 3. Thenwe havee(G3)e(G)≥ 1 + . (2)Where  ≈ 0.087.The constant  has since been improved to 16by the author [8] and to 34by DeVos andThomasse´ [4]. The value  = 34is optimal in the sense that there exists a sequence ofregular graphs of diameter greater than 3, Gm, satisfyinge(G3m)e(Gm)→ 74as m → ∞ [4]. It isnatural to ask what happens for other powers of G.For G2, Hegarty showed that no inequality similar to (2) with  > 0 can hold forregular graphs in general, by exhibiting a sequence of regular, connected graphs of diametergreater than 2, Gm, satisfyinge(G2m)e(Gm)→ 1 as m → ∞ [7]. Goff [6] studied the 2nd powerof regular graphs further and showed that for any d-regular connected graph G such thatdiam(G) > 2, we have e(G2)e(G)≥ 1 + 32d− o (1d). For general d-regular connected graphs Gwith diam(G) > 2, the 32dterm in this result cannot be replaced with λdfor any λ > 32.However it is shown in [6] that with the exception of two families of exceptional graphs,we have e(G2)e(G)≥ 1 + 2d− o (1d)for all d-regular connected graphs with diam(G) > 2.In this note we consider all r ≥ 4 and determine how small e(Gr)e(G)can be for G a regular,connected graph of diameter at least r. We prove the following theorem.Theorem 3. Let G be a connected, regular graph, and r a positive integer such thatdiam(G) ≥ r.• If r ≡ 0 (mod 3), then we havee(Gr)e(G)≥ r + 33− 32(r + 3).• If r 6≡ 0 (mod 3), then we havee(Gr)e(G)≥⌈r3⌉.The case r = 3 of Theorem 3 is due to DeVos and Thomasse´ [4], and will not be provedhere. Theorem 3 gives a lower bound on the ratio e(Gr)e(G)for regular graphs. The bounds one(Gr)e(G)in Theorem 3 are optimal in the following sense. For each r, there exists a sequence ofregular, connected graphs of diameter at least r, Gm, such thate(Grm)e(Gm)tends to the bound2Figure 1: Graphs showing the cases “r = 8” and “r = 6” of Theorem 3 to be optimal.The grey ovals represent complete graphs of specified order. The black lines between thesets represent all the edges being present between them. The white cycle in the “r = 8”case represents the removal of a single cycle passing through all the vertices in the sets itintersects. The white matchings in the “r = 6” case represent a perfect matching beingremoved from the specified sets.given by Theorem 3 as m tends to infinity. We refer to Figure 1 for a diagram of thesequences that we construct.To see this for r 6≡ 0 (mod 3), we construct the following sequence of graphs Gm. Takedisjoint sets of verticesN0, ..., Nr, with |Ni| = m−1 if i ≡ 1 (mod 3) and |Ni| = 2 otherwise.Add all the edges between Ni and Ni+1 for i = 0, 1, . . . , r−1. Add all the edges within Nifor all i. Remove a cycle passing through all the vertices in N1∪ ...∪Nr−1. It is easy to seethat Gm is m-regular and has diameter r. If r ≡ 1 (mod 3) then |Gm| = 13(rm+ 2m+ 3r)will hold. Since Gm is m-regular, we have e(Gm) =16(rm + 2m + 3r)m. Since Grm iscomplete, we have e(Grm) =118(rm + 2m + 3r)(rm + 2m + 3r − 1). This implies thate(Grm)e(Gm)→ ⌈ r3⌉as m → ∞. A similar calculation can be used to show that the same limitholds when r ≡ 2 (mod 3).For r ≡ 0 (mod 3), we construct the following sequence of graphs Hm to show thatTheorem 3 is optimal. Take disjoint sets of vertices N0, ..., Nr+1. Let |N0| = |Nr+1| =2m + 1, |Ni| = 1 if i ≡ 2 (mod 3), and |Ni| = 2m otherwise. Add all the edges betweenNi and Ni+1 for i = 0, 1, . . . , r. Add all the edges within Ni for all i. Delete a perfectmatching from each of the sets N2 and Nr. This will ensure that Hm is 4m-regular andhas diameter r + 1. Note that |Hm| = 13(4rm + r + 12m + 6), and so we have e(Hm) =16(4rm+ r+ 12m+ 6)4m. The only edges missing from Hrm will be between N0 and Nr+1,so we have e(Hrm) =118(4rm+ r+ 12m+ 6)(4rm+ r+ 12m+ 5)− (2m+ 1)2. This impliesthat e(Hrm)e(Hm)→ r+33− 32(r+3)as m→∞. This construction is a generalization of one from [4].All the examples constructed above have their diameter close to r. If a graph G hasdiameter larger than r, it seems that the bounds of Theorem 3 can be improved. Someresults in this direction have been obtained DeVos, McDonald and Scheide [3].3The requirement of G being regular in the above theorems is quite restrictive. Follow-ing [4], we will instead assume that G has minimum degree δ(G), and give the followingbound on e(Gr) in terms of |G| and δ(G).Theorem 4. Let G be a connected graph, and r a positive integer such that diam(G) ≥ r.• If r ≡ 0 (mod 3), then we havee(Gr) ≥(r + 36− 34(r + 3))δ(G)|G|.• If r 6≡ 0 (mod 3), then we havee(Gr) ≥ 12⌈r3⌉δ(G)|G|.The case r = 3 of Theorem 4 is due to DeVos and Thomasse´ [4], and will not be provedhere. Theorem 4 easily implies Theorem 3.2 Proof of Theorem 4We will prove a version of Theorem 4 for graphs which may contain loops since in thatsetting the proof seems more natural.The neighbourhood of a vertex x, N(x), is defined as the set of vertices adjacent to x.(If there is a loop at x, then N(x) will contain x itself.) The degree of x is |N(x)|. Forgraphs with loops allowed, Gr is defined identically to how it was defined for looplessgraphs. Note that if G is a graph with loops allowed, then Gr always has a loop at eachvertex. For two sets of vertices X and Y , let d(X, Y ) denote the length of a shortest pathbetween a vertex in X and a vertex in Y . If X is a set of vertices, let N r(X) be the set ofvertices at distance at most r from X. We abbreviate N r({x}) as N r(x) and d({x}, {y})as d(x, y).We prove the following theorem, and then deduce Theorem 4 as a corollary. Severalideas in the proof of Theorem 5 are taken from [4]. In particular, Claims 11 and 12 areanalogues of claims proved in [4].Theorem 5. Let G be a connected graph, and r a positive integer such that r ≥ 6 anddiam(G) ≥ r.• If r ≡ 0 (mod 3), then we havee(Gr) ≥(r + 36− 34(r + 3))δ(G)|G|+ 12|G|.• If r 6≡ 0 (mod 3), then we havee(Gr) ≥ 12⌈r3⌉δ(G)|G|+ 12|G|.4Proof. For convenience, we will set δ = δ(G). If P is a path between two vertices x and y,we say that P is a geodesic if the length of P is d(x, y). The notion of a geodesic was usedin [4], and is useful because the neighbourhood of a geodesic must be quite large. This isquantified in the following claim.Claim 6. Let P be a length k geodesic. Then |N(P )| ≥ (⌊k3⌋+ 1)δ holds.Proof. If x0, x1, . . . , xk are the vertices of P (in the order in which they occur along thepath), then N(x0), N(x3), . . . , N(x3b k3c) are all disjoint, contained in N(P ), and of orderat least δ. This implies the result.We now prove the case “r 6≡ 0 (mod 3)” of the theorem.The diameter of G is at least r, so G contains a length r geodesic, P . Claim 6 impliesthat the following holds:|G| ≥ |N(P )| ≥(⌊r3⌋+ 1)δ ≥⌈r3⌉δ. (3)Note that Gr contains a loop at every vertex, so we have e(Gr) = ∑v∈V (G) (12 |N r(v)|+ 12).Thus to prove Theorem 5 it is sufficent to exhibit⌈r3⌉δ elements of N r(v) for each vertexv ∈ V (G).Let v be a vertex in G. Suppose that there exists a length r − 1 geodesic Pv startingfrom v. Then N(Pv) is contained in Nr(v), giving|N r(v)| ≥ |N(Pv)| ≥(⌊r − 13⌋+ 1)δ =⌈r3⌉δ.The second inequality is an application of Claim 6.Suppose that all the vertices in G are within distance r − 1 of v. In this case we haveN r(v) = V (G), which is of order at least ⌈ r3⌉δ by (3). This completes the proof of the case“r 6≡ 0 (mod 3)” of the theorem.For the rest of the proof fix r such that r ≡ 0 (mod 3) and r ≥ 6.If v is a vertex of G, we say that v is sufficient if |N r(v)| ≥ ( r3+ 1)δ. Otherwise wesay that v is insufficient.The following is a useful property of insufficient vertices.Claim 7. Let v be an insufficient vertex. Then there is some vertex at distance r + 1from v.Proof. Since diam(G) ≥ r, Claim 6 implies that |G| ≥ ( r3+ 1)δ. By assumption N r(v) <(r3+ 1)δ, so v cannot be within distance r from all the vertices in the graph.The following three claims will allow us to bound the number of insufficient verticesin G.Claim 8. If 2 < d(x, y) < r holds for x, y ∈ V (G), then either x or y is sufficient.5Proof. Suppose that x is insufficient. By Claim 7, we can find a length r geodesic startingfrom x with vertex sequence x, x1, x2, . . . , xr.Suppose that N(y)∩N(xi) 6= ∅ for some i with 3 ≤ i ≤ r−3. In this case N(x), N(x3),N(x6), . . . , N(xr) are all contained in Nr(y). There are r3+ 1 of these, they are all disjoint(since x, x1, x2, . . . , xr form a geodesic), and are of order at least δ. Hence y is sufficient.Otherwise N(y) ∩ N(xi) = ∅ for all 3 ≤ i ≤ r − 3. In this case N(x), N(y), N(x3),N(x6), . . . , N(xr−3) are all disjoint and contained in N r(x). This contradicts our initialassumption that x is insufficient.Claim 9. Let x and y be two vertices in G such that d(x, y) = r or d(x, y) = r+1. If thereexists a vertex z ∈ G such that d(z, x), d(z, y) ≥ r − 1, then either x or y is sufficient.Proof. Choose any z in N r−1({x, y}) \ N r−2({x, y}. This set is nonempty by the secondassumption of the claim. We will have d(z, x), d(z, y) ≥ r− 1 and either d(z, x) = r− 1 ord(z, y) = r− 1. Without loss of generality assume that d(z, x) = r− 1 and d(z, y) ≥ r− 1.We will show that x is sufficient. Let x, x1, . . . , xd(x,y)−1, y be a geodesic between xand y. For i = 1, . . . , d(x, y)− 1, the triangle inequality implies thatd(xi, z) ≥ d(x, z)− d(x, xi) = d(x, z)− i, (4)d(xi, z) ≥ d(y, z)− d(y, xi) = d(y, z)− d(x, y) + i. (5)Averaging (4) and (5), and use the inequalities d(z, x), d(z, y) ≥ r − 1 and d(x, y) ≤ r + 1givesd(xi, z) ≥ r − 32. (6)If r ≥ 9, then (6) implies that d(xi, z) ≥ 3 for all i. Hence N(x), N(z), N(x3),N(x6), . . . , N(xr−3) are all disjoint and contained in N r(x). Hence x is sufficient.If r = 6, then (4) and (5) imply that d(xi, z) ≥ 3 for all xi except possibly x3 or x4.In this case N(z), N(x2) and N(x5) are all disjoint and contained in N6(x). Hence x issufficient.Claim 10. If d(x, y) = r holds for x, y ∈ V (G), then either x or y is sufficient.Proof. Suppose that x and y are insufficient. By Claim 7 there exists z ∈ V (G) such thatd(x, z) = r + 1. Let x, x1, . . . , xr−1, y be a geodesic between x and y. Since x and y areinsufficient, Claim 9 implies that we have d(z, y) < r− 1. Note that d(x, z) = r+ 1 impliesthat N(z) ∩ N(xi) = ∅ for all i ≤ r − 2. Thus N(z), N(x1), N(x4), . . . , N(xr−2) are alldisjoint and contained in N r(y). This contradicts our assumption that y is insufficient.Let X be the set of insufficient vertices in G. We define an equivalence relation “∼”on X by letting x ∼ y if d(x, y) ≤ 2. For r ≥ 6, Claim 8 implies that this is an equivalencerelation. Let X1, . . . , Xl be the equivalence classes of ∼.The following claim gives a lower bound on the order of G.Claim 11. |G| ≥ ( r+36)δl6Proof. Claims 8 and 10 imply that d(Xi, Xj) ≥ r+ 1 for all i 6= j. If d(Xi, Xj) = r+ 1 forsome i and j, then Claim 9 implies that we have d(Xi, z) < r− 1 or d(Xj, z) < r− 1 for allz ∈ V (G). Then, Claim 8 implies that all the vertices outside of Xi and Xj are sufficient.This gives us two cases to consider:(i) d(Xi, Xj) ≥ r + 2 for all i 6= j.(ii) d(X1, X2) = r + 1.Suppose that (i) holds (this includes the case when l = 1). For each i, choose xi to be avertex in Xi. Note that Nb r2c(xi) contains a length⌊r2⌋geodesic, Pi. Using Claim 6 gives∣∣∣Nb r2c+1(Xi)∣∣∣ ≥ |N(Pi)| ≥ (⌊13⌊r2⌋⌋+ 1)δ ≥(r + 36)δ.For the last inequality we are using the fact that r ≡ 0 (mod 3). Note that (i) implies thatNb r2c+1(Xi) ∩Nb r2c+1(Xj) = ∅ for all i, j. This implies that the following holds:|V (G)| ≥l∑i=1∣∣∣Nb r2c+1(Xi)∣∣∣ ≥ (r + 36)δl.Suppose that (ii) holds. Using Claim 6 we obtain|V (G)| ≥(r3+ 1)δ =(r + 36)δl.When x is insufficient, the following claim gives a lower bound on the order of N r(x).Claim 12. Suppose that x is an insufficient vertex in the equivalence class Xi. Then,|N r(x)| ≥ |Xi|+ r3δ holds.Proof. By Claim 7, we can choose a length r geodesic from x. Let x, x1, . . . , xr be thevertices of this geodesic. Suppose that Xi ∩ N(xj) is nonempty for some xj. Choosey ∈ Xi ∩N(xj). Clearly j ≤ 1 must hold, since otherwise N(x), N(x3), N(x6), . . . , N(xr)would all be contained in N r(y), contradicting that y is insufficient (since y ∈ Xi).Hence Xi, N(x2), N(x5), . . . , N(xr−1) are all disjoint and contained in N r(x) provingthe claim.Claims 11 and 12 are all that is needed to prove Theorem 5, as follows72e(Gr)−(r + 33− 32(r + 3))δ|G| − |G| =∑x∈V (G)|N r(x)| −(r + 33− 32(r + 3))δ|G|≥ 32(r + 3)δ|G|+l∑i=1(|Xi|2 − |Xi|δ)≥ 14δ2l +l∑i=1(|Xi|2 − |Xi|δ)=l∑i=1(|Xi|2 − |Xi|δ + 14δ2)=l∑i=1(|Xi| − 12δ)2≥ 0.The first equality uses the fact that Gr contains a loop at every vertex, hence 2e(Gr) =∑x∈V (G) |N r(x)|+ |G|. The first inequality follows from the definition of “sufficent vertex”,Claim 12 and rearranging, while the second follows from Claim 11. This completes theproof.Proof of Theorem 4. Let G be a copy of G with a loop added at every vertex. Then Grwill be isomorphic to Gr with a loop added at every vertex. Note that we have e(Gr) =e(Gr) + |G|, and δ(G) = δ(G) + 1. Substitute these into Theorem 5 obtain the following.• If r ≡ 0 (mod 3), then we havee(Gr) ≥(r + 36− 34(r + 3))δ(G)|G|+(r + 36− 34(r + 3)− 12)|G|.• If r 6≡ 0 (mod 3), then we havee(Gr) ≥ 12⌈r3⌉δ(G)|G|+(12⌈r3⌉− 12)|G|.Note that for r ≥ 3, both r+36− 34(r+3)− 12and 12⌈r3⌉− 12are non-negative, so Theorem 4follows.AcknowledgmentThe author would like to thank his supervisors Jan van den Heuvel and Jozef Skokan foradvice and discussions.8References[1] A. L. Cauchy. Recherches sur les nombres. J. E´cole Polytech, 9:99–116, 1813.[2] H. Davenport. On the addition of residue classes. J. London Math. Soc., 7:30–32, 1935.[3] M. DeVos, J. McDonald, and D. Scheide. Average degree in graph powers.arXiv:1012.2950, 2010.[4] M. DeVos and S. Thomasse´. Edge growth in graph cubes. arXiv:1009.0343, 2010.[5] R. Diestel. Graph Theory. Springer-Verlag, 2000.[6] M. Goff. Edge growth in graph squares. arXiv:1112.5157, 2011.[7] P. Hegarty. A Cauchy-Davenport type result for arbitrary regular graphs. Integers, 11,2011.[8] A. Pokrovskiy. Growth of graph powers. Electron. J. Combin., 18, 2011.9

Edge growth in graph powers

For a graph G, its rth power G is defined as the graph with the same vertex set as G, and an edge between any two vertices whenever they are within distance r of each other in G. Motivated by a result from additive number theory, Hegarty raised the question of how many new edges G has when G is a regular, connected graph with diameter at least r. We address this question for r ≠ 3, 6. We give a lower bound for the number of edges in the rth power of G in terms of the order of G and the minimal degree of G. As a corollary, for r ≠3, 6, we determine how small the ratio e(G )/e(G) can be for regular, connected graphs of diameter at least r. r r 

Pokrovskiy, A

UCL Discovery

For a graph G, its rth power Gr has the same vertex set as G, and has an edge between any two vertices within distance r of each other in G. We give a lower bound for the number of edges in the rth power of G in terms of the order of G and the minimal degree of G. As a corollary we determine how small the ratio e(Gr)/e(G) can be for regular graphs of diameter at least r. 

Edge growth in graph powers

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