We prove that if two graphs of girth atleast 6 have isomorphic squares, then the graphs themselves are isomorphic. This is the best possible extension of the results of Ross and Harary on trees and the results of Farzad et al. on graphs of girth at least 7. We also make a remark on reconstruction of graphs from their higher powers

Adamaszek, Anna

Adamaszek, Michał

English

Anna Adamaszek

Micha L Adamaszek

CiteSeerX

Uniqueness of graph square roots of girth six

Warwick Research Archives Portal Repository

 University of Warwick institutional repository: http://go.warwick.ac.uk/wrap  This paper is made available online in accordance with publisher policies. Please scroll down to view the document itself. Please refer to the repository record for this item and our policy information available from the repository home page for further information.  To see the final version of this paper please visit the publisher’s website. Access to the published version may require a subscription. Author(s): Anna Adamaszek and Michal Adamaszek  Article Title: Uniqueness of graph square roots of girth six Year of publication: 2011 Link to published article:  http://www.combinatorics.org/ Publisher statement:  First published in The Electronic Journal of Combinatorics in Vol.18 No.1, 2011, published by the American Mathematical Society.  Uniqueness of graph square roots of girth sixAnna Adamaszek∗Department of Computer Scienceand Centre for Discrete Mathematics and its Applications (DIMAP)University of Warwick, Coventry, CV4 7AL, UKannan@mimuw.edu.plMicha l Adamaszek∗Warwick Mathematics Instituteand Centre for Discrete Mathematics and its Applications (DIMAP)University of Warwick, Coventry, CV4 7AL, UKaszek@mimuw.edu.plSubmitted: Dec 11, 2009; Accepted: Jun 22, 2011; Published: Jul 1, 2011Mathematics Subject Classifications: 05C12, 05C75AbstractWe prove that if two graphs of girth at least 6 have isomorphic squares, then thegraphs themselves are isomorphic. This is the best possible extension of the resultsof Ross and Harary on trees and the results of Farzad et al. on graphs of girthat least 7. We also make a remark on reconstruction of graphs from their higherpowers.1 IntroductionFor a simple, undirected, connected graph H its square G = H2 is the graph on the samevertex set in which two distinct vertices are adjacent if their distance in H is at most 2.In this case H is called the square root of G. Also, recall that the girth of a graph is thelength of its shortest cycle (or ∞ for a tree). The neighbourhood NH(u) of u will be theset consisting of u and its adjacent vertices in H . By distH(u, v) we denote the distancebetween two vertices in H .We investigate the uniqueness of square roots of graphs. Ross and Harary [5] provedthe following theorem:(1) If T1 and T2 are two trees such that T21and T 22are isomorphic, then T1 and T2 areisomorphic.∗Research of both authors supported by the EPSRC award EP/D063191/1.the electronic journal of combinatorics 18 (2011), #P139 1This was recently improved by the authors of [1] who proved:(2) If H1 and H2 are two graphs of girth at least 7 such that H21and H22are isomorphic,then H1 and H2 are isomorphic.In the next section we prove the best possible result, which is:(3) If H1 and H2 are two graphs of girth at least 6 such that H21and H22are isomorphic,then H1 and H2 are isomorphic.The key idea behind (1) and (2) is that each maximal clique of the square corresponds tothe neighbourhood of some vertex in the root. This fails in the case of roots of girth 6.For example, the vertices 1, 3, 5 of the cycle C6 form a maximal clique in C26even thoughthey do not induce a star in C6. This is where we will need a new idea to prove (3).2 Proof of the theoremLet H be a graph of girth at least 6 on the vertex set V and let G = H2. We have thefollowing easy observations:(*) If there is a path from u to v in H of length exactly 3, then u 6∈ NG(v) (otherwisethere would be a cycle in H of length at most 5).(**) If uv ∈ E(H) then NG(u) ∩ NG(v) = NH(u) ∪ NH(v). Indeed, the inclusion ⊇ isobvious. To prove ⊆ note that if some vertex w ∈ NG(u) ∩ NG(v) was adjacent toneither u nor v in H , then it would be in distance 2 from both of them, which wouldyield a 5-cycle in H .We start with a lemma which can also be deduced from [1]. The notation H1 = H2means that two graphs are equal (the same vertex set and the same edges), not justisomorphic.Lemma 2.1. Let H1 and H2 be graphs of girth at least 6 on the vertex set V . Supposethat G = H21= H22and that u, v, w ∈ V are three vertices such that uvw is a path in bothH1 and H2. Then H1 = H2.Proof. Let H be any graph of girth at least 6 such that G = H2. The following statementsfollow easily from (*) and (**):• If xyz is a path in H , then (see Fig. 2)NH(x) = (NG(x) ∩NG(y)) \NG(z) ∪ {x, y}.• If y is of degree 1 in H and xy ∈ E(H) thenNH(x) = NG(y).the electronic journal of combinatorics 18 (2011), #P139 2x y z(NG(x) ∩NG(y)) \NG(z)NG(x) ∩NG(y)Figure 1: The structure of the neighbourhoods of x and y in H.With the above formulas, given one path uvw of H one can recursively compute allthe edges of H using only the information from G, so the square root of G with thisdistinguished path is unique. This ends the proof.Clearly, it suﬃces to prove our main result with the assumption “H21and H22areisomorphic” replaced by “H21= H22”. This is what we now prove:Theorem 2.2. Suppose H1 and H2 are two graphs of girth at least 6 such that G = H21=H22. Then H1 and H2 are isomorphic.Proof. Let V be the common vertex set of H1, H2 and G. If uvw is a path in both H1and H2 for some u, v, w then H1 = H2 by the previous lemma. Therefore we may assumethat for every v the set Xv = {u : uv ∈ E(H1) ∩ E(H2)} has at most 1 element. Deﬁnethe following map f : V −→ V :• if |Xv| = 0 then f(v) = v,• if |Xv| = 1 then f(v) is the unique element of Xv.Clearly f is an involution.We shall ﬁrst prove two statements:• (A) if uv ∈ E(H1), |Xv| = 1 and u 6= f(v) then |Xu| = 0,• (B) if uv ∈ E(H1) and |Xv| = 0 then |Xu| = 1.Proof of (A). Let v be a vertex with |Xv| = 1 and let f(v) = w, meaning that vw is anedge in both H1 and H2. Let u be any neighbour of v in H1, other than w. We will showthat |Xu| = 0. Suppose, on the contrary, that z ∈ Xu (Fig.2.). Then distH1(w, z) = 3,so, by (*), z 6∈ NG(w). Since u and v are not neighbours in H2, but u ∈ NG(v) ∩NG(w),the property (**) implies that uw is an edge in H2. However uz is also an edge in H2, soz ∈ NG(w). This contradiction proves that |Xu| = 0 for all neighbours u of v in H1 otherthan w.Proof of (B). Let v be a vertex with |Xv| = 0. Let u be adjacent to v in H1. We willshow that |Xu| = 1. Suppose, on the contrary, that |Xu| = 0. In H2 the vertex v mustthe electronic journal of combinatorics 18 (2011), #P139 3zuvw zuvwH1 : H2 :Figure 2: Illustration for the proof of (A) in Theorem 2.2. The bold edges are present in bothH1 and H2.be in distance 2 from u, so there is an x such that uxv is a path in H2. In particular,x ∈ NG(v) ∩ NG(u), so by (**) we have x ∈ NH1(u) ∪ NH1(v). This is a contradictionsince x would be adjacent to either u or v in both H1 and H2, which is impossible byXv = Xu = ∅.Proof of the theorem. Now we prove that f (treated as a map of graphs H1 −→ H2)maps edges to edges. Let uv ∈ E(H1).If |Xu| = |Xv| = 1 then, by (A), f(u) = v, f(v) = u and uv is an edge in both graphs,so f takes uv to an edge vu in H2.If |Xu| = 0 and |Xv| = 1, then let w = f(v). Since uv 6∈ E(H2) and u ∈ NG(v)∩NG(w),we have from (**) that uw ∈ E(H2) and f takes uv ∈ E(H1) to f(u)f(v) = uw ∈ E(H2).The case |Xu| = |Xv| = 0 is not possible by (B).To prove that f−1 maps edges to edges one simply inverts the roles of H1 and H2in the above argument (the deﬁnition of f was symmetric with respect to H1 and H2).Therefore f is an isomorphism.3 Remarks and modificationsThis result is optimal in the sense that it cannot hold for girth at least 5 because K1,42 =C52 = K5.The r-th power Hr of a graph is deﬁned analogously, that is edges in Hr correspondto pairs of vertices in distance at most r in H . Observe that regardless of the girthrestriction, there can be no analogous general result for higher graph powers, becausethere exist non-isomorphic trees whose r-th power is a complete graph for all r ≥ 3. Thisand the work of [2, 3] suggest that one may beneﬁt from forbidding vertices of degree onein the root. Consider the following problem: what are the minimal values of g1(r) andg2(r) for which the following statements hold:(1) For any two graphs H1 and H2 of girth at least g1(r) with no vertices of degree one,if Hr1= Hr2then H1 and H2 are isomorphic.(2) For any two graphs H1 and H2 of girth at least g2(r) with no vertices of degree one,if Hr1= Hr2then H1 = H2.For example g2(2) = 7, as proved in [3]. Our work proves that g1(2) ≤ 6 and this is, infact, optimal: there exist two non-isomorphic graphs of girth 5 and no degree one verticesthe electronic journal of combinatorics 18 (2011), #P139 401234567891011121314150123456789101112131415Figure 3: Two non-isomorphic graphs of girth 5, minimal vertex degree 2 and the same square.having the same squares. The smallest such example is a pair of graphs on 16 verticesshown in Fig.3 (found with [4]). Therefore g1(2) = 6.It is known that 2r + 3 ≤ g2(r) ≤ 2r + 2⌈(r− 1)/4⌉+ 1 (see [2]) and conjectured thatg2(r) = 2r + 3 for all r. Any nontrivial result about g1(r) (possibly in relation to g2(r))would be very interesting.References[1] Babak Farzad, Lap Chi Lau, Van Bang Le, Nguyen Ngoc Tuy, Computing Graph RootsWithout Short Cycles, Proc. 26th STACS 2009 (2009) 397-408[2] V. I. Levenshtein, A conjecture on the reconstruction of graphs from metric balls oftheir vertices, Discrete Mathematics 308(5-6): 993-998 (2008)[3] V. I. Levenshtein, E.V. Konstantinova, E. Konstantinov, S. Molodtsov, Reconstructionof a graph from 2-vicinities of its vertices, Discrete Applied Mathematics 156(9): 1399-1406 (2008)[4] B.D. McKay, The Nauty graph automorphism package,http://cs.anu.edu.au/~bdm/nauty/[5] D. J. Ross, F. Harary, The square of a tree, Bell System Technical Journal 39 (1960),641-647the electronic journal of combinatorics 18 (2011), #P139 5

A conjecture on the reconstruction of graphs from metric balls of their vertices,

Lap Chi Lau, Van Bang Le, Nguyen Ngoc Tuy, Computing Graph Roots Without Short Cycles,

Reconstruction of a graph from 2-vicinities of its vertices,

The Nauty graph automorphism package,

The square of a tree,

Uniqueness of graph square roots of girth six

Abstract

Similar works

Full text

Available Versions

CiteSeerX

Warwick Research Archives Portal Repository