Figure 1: A cycle's covering. Figure : Two dierent coverings for the same graph. A lot of properties can be easily proved on coverings. Co

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1 Covering and spanning tree of graphs Anne Bottreau LaBRI-Universit Bordeaux I 351 cours de la Lib ration Talence cedex FRANCE tel: (+33) , fax:(+33) Abstract: In this note we recall some well-known results about covering and k-covering of graphs. Then, we introduce a new notion of spanning tree which we call the p-spanning tree. This notion allows us to reduce the upper bound for the minimum integer k for which a connected graph G has no nontrivial k-covering. Key-words: Graphs, covering of graph, k-covering of graph, permutations, cycles, spanning tree. This note deals with the notion of covering of graphs and more precisely with the notion of k-covering of graphs. In the rst part, we introduce the notion of covering and recall the Reidemeister theorem. In the second part, we focus on the notion of k-covering of graphs. We introduce then the p-spanning tree. 1 Covering of graph 1.1 Denition and properties The notion of graph covering is based on the notion of bijection between neighbourhoods. If we are able to nd a surjective homomorphism between two graphs G and H which maps the vertices of H onto the vertices of G with respect of adjacency, then H is a covering of G. Denition 1 Let G be a connected graph and H be a graph, H is a covering of G if there exists a surjective homomorphism from V (H) onto V (G) such that for any v in V (H), the restriction of to N H (v) is a bijection between N H (v) and N G ((v)). Our examples use dierent geometric shapes for the vertices in order to code the homomorphism : any vertex of the covering has the same shape as its image by the homomorphism. Example 1 The cycle with six vertices is a covering of the cycle with three vertices (cf. Figure 1). In Figure we give a graph with six vertices having at least two dierent coverings with twelve vertices. 1

2 Figure 1: A cycle's covering. Figure : Two dierent coverings for the same graph. A lot of properties can be easily proved on coverings. Consider G a connected graph and H a covering of G via the homomorphism. There exists an integer q such that V (H) = qv (G). This integer is called the number of tiles. This statement can be proved thanks to the following sequence of lemmas. Two vertices of H having the same image in G have disjoint neighbourhoods (Lemme 1). Any edge of G has a set of disjoint edges for inverse image (Lemme ). Any simple path in G has a set of disjoint simples paths in H for inverse image. Thus, we deduce that there exists an integer q such that V (H) = q V (G) (Lemme 4). Lemma 1 Let v 1 and v be two dierent vertices of H. If (v 1 ) = (v ), then N H (v 1 ) \ N H (v ) = ;. Lemma Let e be an edge of G. The inverse image of e via is a set of disjoint edges of H. Let us now dene a simple path p by a sequence of vertices and edges (v 0 ; e 1 ; v 1 ; ; e n ; v n ) with 8i; e i = fv i?1 ; v i g and 8i 6= i 0 ; v i 6= v 0 i. Lemma 3 Let p be a simple path of length n in G. The inverse image of p by is a set of disjoint simple paths isomorphic to p. Lemma 4 There exists an integer q such that 8u V (G) card(?1 (u)) = q.

3 Let H be a graph consisted of q disjoint connected components isomorphic to G. From the denition of a covering, it follows that H is a covering of G with q tiles. Moreover this covering is called trivial. Thanks to the previous lemma concerning the inverse image of an edge, it is rather easy to demonstrate that trees only have trivial coverings (the proof can be done by induction on the number of vertices of the tree). Lemma 5 Let A be a tree, every connected covering of A is a tree isomorphic to A. Thus, Corollary 1 Any covering of a tree is a trivial covering. Moreover, Corollary Let A be a spanning tree of G. The inverse image of A by is a set of disjoint trees isomorphic to A. 1. Reidemeister theorem K. Reidemeister introduced in [Rei3] a method for the construction of all the coverings of a given graph. He proved that any covering of a graph G can be obtained thanks to a set of copies of a spanning tree A of G and some permutations associated to the edges of E(G)nE(A) Covering with q tiles Let G be a connected graph, A a spanning tree of G and q an integer. We denote by F the set of edges of G which are not in A and we associate to each edge of F a permutation of f1; ; qg. Consider now the graph obtained by q copies of A connected to each other by the q copies of each edge of F obtained by the exchange of end-points (depending on the associated permutation). This graph is a covering of G with q tiles. Example We give here the construction of a covering with 3 tiles. Graph G A a spanning tree of G. F = ffa; bg; fc; dgg a b c d 3

4 Consider three copies of the spanning tree : q = 3. For the edge fa; bg, we choose fa;bg = (1; ; 3), we add three edges. a 1 b 1 a b a 3 b 3 c 1 c c 3 d 1 d d 3 For the edge fc; dg, we choose fc;dg = (1; 3)(). a1 b1 a b a3 b3 c c1 c3 d1 d d3 The choice for the permutations of S 3 is arbitrary. We obtain a covering with 3 tiles of G. Let S q be the group of permutations on f1; ; qg. For i f1; ; qg, the i th copy of A, denoted A i, is dened by : V (A i ) = f (x; i) = x V (A)g; E(A i ) = f f(x; i); (y; i)g = fx; yg E(A)g: To each edge fx; yg of F we associate a permutation of S q, denoted xy, and a set of q edges F xy : = f f(x; i); (y; xy (i))g = i f1; ; qgg: F xy Let us denote the set of these permutations : = f xy = fx; yg F g: We dene the graph R q; thanks to the q copies of A and the sets of edges given by and F : the q copies of A are connected by the q copies of each edge of F. V (R q; ) = S q i=1 V (A i); E(R q; ) = S fx;ygf F xy S ( S q i=1 E(A i)) : Lemma 6 The graph R q; obtained as it is explained previously with and q copies of A is a covering of G with q tiles. 4

5 Proof: The graph R q; is dened by the following sets of edges and vertices : V (R q; ) = f (x; i) = x V (A) and i f1; ; qgg; E(R q; ) = f f(x; i); (y; i)g = i f1; ; qg; fx; yg E(A)g S f f(x; i); (y; xy (i))g = i f1; ; qg; fx; yg F and xy g: Let us dene the application by : : R q;! G (x; i) 7! x Let f(x; i); (y; j)g be an edge of R q;. By the denition of E(R q; ), fx; yg is an edge of E(G). Moreover we have : ((x; i)) = x and ((y; j)) = y. Thus, is a surjective homomorphism. Consider a vertex (x; i) of R q;. We denote N ((x; i)) the neighbourhood of (x; i) in R q;. We have : N ((x; i)) = f (y; j) =f(x; i); (y; j)g E(R q; )g: N ((x; i)) = f (y; i) = fx; yg E(A)g [ f (y; xy (i)) = fx; yg F g: Therefore, jn ((x; i))j = jn G (x)j: Regarding the denition of, the restriction of to N ((x; i)) is a bijection between N ((x; i)) and N G (x). Thus the graph R q; is a covering of G with q tiles. 1.. Reidemeister Theorem Theorem 1 (Reidemeister, 193) Let G be a connected graph, A be a spanning tree of G. We denote F the set of edges of G which are not in A. For any graph H, H is a covering of G if and only if there are an integer q and a set of permutations of S q associated to each edge of F such that H is isomorphic to the graph R q;. Proof: that : Consider that H is a covering of G via. By the lemma 4: there is an integer q such 8 x V (G); q = card(?1 (x)): Let A be a spanning tree of G, by the corollary, the inverse image of A by is a set of q disjoint trees, A i, isomorphic to A:?1 (A) = A 1 [ A [ [ A q : Let now fu; vg be an edge of G which is not in A : fu; vg F: By the lemma, the inverse image by of fu; vg is a set of disjoint edges of H. The vertices u and v have q inverse image by, one in each tree A i :?1 (u) = fu 1 ; u ; ; u q g; where u i A i : 5

6 As H is a covering of G by, each of the inverse image of u is connected to a unique inverse image of v in H. Therefore, we can dene the set?1 (fu; vg) :?1 (fu; vg) = f fu 1 ; v l1 g; fu q ; v lq g g; with i; l i f1; ; qg: These edges are disjoint i.e. 8r 6= s; l r 6= l s. If we consider that 8i, l i = (i), then is a permutation of S q. Thus, for any edge of F, we dene a permutation of S q. Let us denote the set of these permutations. The graph H is isomorphic to R q;. Conversely, let q be an integer and a set of permutations of S q associated to each edge of F. Suppose that R q; is isomorphic to H. The lemma 6 states that this graph is a covering of G. Therefore, H is isomorphic to a covering of G, i.e. H is a covering of G. k-covering of graph If in the denition of graph covering we remove the notion of bijection between neighbourhoods by the notion of isomorphism between centered balls of radius k, then we obtain the notion of k-covering. This notion was introduced in [LMZ95]..1 Denition and rst properties We denote by B H (v; k) (the ball of center v and of radius k) the subgraph of H induced by the set of vertices having their distance to v at most k. Denition Let G be a connected graph and k be a positive integer. A graph H is a k- covering of G via the homomorphism if is an homomorphism from H onto G and if for any vertex v of H, the restriction of to the ball B H (v; k) is an isomorphism between B H (v; k) and B G ((v); k). From this denition it easily follows that given a graph G and an integer k : Lemma 7 Any k-covering of G is a covering of G. Lemma 8 Any k-covering of G is a h-covering of G for any h k. Moreover, a trivial covering of G is a k-covering of G, for any integer k. Thus, a k-covering of G is called trivial if it is isomorphic to a trivial covering of G. Example 3 Previously we give a covering with tiles for the cycle C 3 (cf. gure 1). This covering is not a 1-covering of C 3. There are bijections between neighbourhoods of vertices, but there aren't isomorphisms between balls of radius 1. N() N() 6

7 B(,1) B(,1) If we consider now the graph with six vertices of the gure, then we are able to nd a 1-covering with tiles (cf. 3). Figure 3: A 1-covering. Our interest is now to adapt Reidemeister theorem to the notion of k-covering. Reidemeister gave the description of coverings thanks to permutations. In order to have a description concerning k-coverings, we prove that the permutations must be the solutions of a set of equations related to the cycles of length at most k + 1. For the sake of clarity, we denote a cycle by the sequence of its vertices (without edges) : C = (x 1 ; x ; ; x m ) is equivalent to C = (x 1 ; e 1 ; ; x m ; e m+1 ) with e i = fx i ; x i+1 g; 8i < m and e m+1 = fx m ; x 1 g. It will be the same for simple paths issued from a cycle. We denote Id q the identity permutation of the integer set f1; ; qg. We will now drop the parentheses and the comma when denoting a vertex of R q;. Thus, (u; i) and u i will have the same meaning (we believe that this convention improves the clarity of exposition). Let G be a connected graph, k an integer and A a spanning tree of G. Let R q; be a covering of G with q tiles. Proposition 1 If R q; is a k-covering of G with q tiles, then for any cycle C = (x 1 ; x ; ; x m ) in G of length m k + 1, we have xmx 1 xm?1 x m x x 3 x1 x = Id q ; with xi x i+1 = Id q si fx i ; x i+1 g E(A) and xi x i+1 =?1 x i+1 x i. Proof: Consider a cycle C in G of length m, m < k + 1. This cycle has at least one edge fu 1 ; u g which is not an edge of A. Let C = (u 1 ; u ; ; u m ). There exists a vertex u l on the cycle C such that the paths P 1 = (u 1 ; u ; ; u l ) and P = (u 1 ; u m ; ; u l ) have a length at most k. 7

8 These paths have for inverse image in R q; a set of q disjoint paths obtained by the permutations of : 8j; P 1 (j) = (u j 1 ; uu 1 u (j) ; ; u u l?1 u l u 1 u (j) P (j) = (u j 1 ; uu 1 m um (j) l ) um (j) um (j) ; u umu m?1 u 1 m?1 ; ; u u l+1 u l u 1 l ): Both paths P 1 (j) and P (j) have a length at most k. Thus, the ball of center u j and 1 of radius k in R q; contains these two paths. As R q; is a k-covering of G, B Rq; (u j 1 ; k) is isomorphic to B G (u 1 ; k) i.e. it contains only one vertex having u l for image in G. Therefore, 8j; ul?1 u l u1 u (j) = ul+1 u l u1 u m (j). I.e. umu1 um?1 u m u u 3 u1 u = Id q. Let us now consider the case of a cycle of length exactly k + 1 (it contains at least one edge which is not in A) : C = (u 1 ; u ; ; u k ; ; u k+1 ). Both paths P 1 = (u 1 ; u ; ; u k ) and P = (u 1 ; u k+1 ; u k ; ; u k+1 ) have a length equal to k. The inverse image of these paths are the following disjoint paths : 8j; P 1 (j) = (u j 1 ; uu 1 u (j) ; ; u u k?1 u k u 1 u (j) k ) P (j) = (u j 1 ; uu 1 u k+1 (j) ; u u k+1 u k u 1 u k+1 (j) ; ; u u k+ u k+1 u 1 u k+1 (j) ): k+1 k+1 The balls B Rq; (u j 1 ; k) and B G(u 1 ; k) are isomorphic. between u u k?1 u k u 1 u (j) k Thus, k and u u k+ u k+1 u 1 u k+1 (j) k+1. Therefore, there exists an edge 8j; uk u k+1 uk?1 u k u1 u (j) = uk+ u k+1 u1 u k+1 (j) I:e: uk+1 u 1 uk u k+1 u1 u = Id q : Proposition If R q; is such that for any cycle C = (x 1 ; x ; ; x m ) of G having a length m k + 1, we have xmx 1 xm?1 x m x x 3 x1 x = Id q, with xi x i+1 = Id q if fx i ; x i+1 g E(A) and xi x i+1 =?1 x i+1 x i, then R q; is a k-covering of G with q tiles. Proof: Given R q; a covering of G having this property related to cycles, we now prove that for any vertex u of G, B G (u; k) is isomorphic to B Rq; (u j ; k) for any j f1; ; qg. Let u V (G). Let v V (B G (u; k)), there exists at least one vertex v h in B Rq; (u j ; k). There exists indeed in B G (u; k) a path from u to v: P = (u; u 1 ; u ; ; u s ; v). Consider that h = usv us?1 u s uu1 (j). We suppose that there exists h 0 such that v h0 is in the ball of radius k centered on u j. It means that there exists in G another path coming from u to v, denoted P 0 = (u; u 0 1 ; u0 ; ; u0 s ; v) with h0 = u 0 s v u 0 s?1 u 0 s uu 0 (j). 1 If P and P 0 form a cycle (i.e. the intersection of P and P 0 is restricted to their end-points), then this cycle is (u; u 1 ; ; u s ; v; u 0 s ; ; u0 1 ) of length at most k + 1. By using our property on this cycle, we obtain that h = h 0. If the intersection of both paths contains more than the vertices u and v, then there are cycles of length less than k + 1 in P [ P 0. In this way we can prove (using the same means) that h = h 0. Thus for any v B G (u; k), there is only one vertex v h B Rq; (u j ; k). 8

9 Consider now fv 1 ; v g an edge of B G (u; k). There is in G a path from u to v 1 which is denoted P 1 = (u; u 1 ; u ; ; u s ; v 1 ), and a path from u to v denoted P = (u; u 0 1 ; u0 ; ; u0 t ; v ). We suppose that fv 1 ; v g create a cycle i.e. P 1 and P have an intersection restricted to u (we might have the same demonstration in the case where both paths weren't disjoint). Let us consider h 1 = usv 1 u1 u uu1 (j) and h = u 0 t v u 0 1 u 0 uu 0 1 (j). As fv 1; v g is an edge of G, in R q; we have the edges fv h 1 1 ; vv 1 v (h 1) g and fv h ; vv v 1 (h ) 1 g. Therefore, v1 v (h 1 ) = v1 v usv 1 u1 u uu1 (j): Using the hypothesis on the cycles of length at most k + 1: v1 v (h 1 ) = u 0 t v u 0 1 u 0 uu 0 1 (j): It means : v1 v (h 1 ) = h and v v 1 (h ) = h 1. There is only one edge in B Rq; (u j ; k) having fv 1 ; v g for image. These two propositions lead us to an extension of the Reidemeister Theorem related to the k-coverings with q tiles of a connected graph G [Bot97]. Theorem Let G be a connected graph and H be a graph. The graph H is a k-covering with q tiles of G if and only if H is isomorphic to a graph R q; such that for any cycle C = (x 1 ; x ; ; x m ) of G having a length m, m k + 1, we have : xmx 1 xm?1 x m x x 3 x1 x = Id q, with xi x i+1 = Id q if fx i ; x i+1 g E(A) and xi x i+1 =?1 x i+1 x i, Consider now a graph which doesn't contain any cycle of length at most k + 1. Regarding Theorem and Lemma 7, this graph has a k-covering. Lemma 9 Let k be a positive integer. Let G be a graph which doesn't contain any cycle of length at most k + 1. Any covering of G is indeed a k-covering of G. It directly comes that : Corollary 3 Any covering of a bipartite graph is a 1-covering.. On the k-covering problem What we call the k-covering problem can be stated in the following way : For a given integer k, can we decide whether or not a connected graph G has a nontrivial k-covering? As this question is hard, we deal with the minimal integer k such that a graph G has no k-covering dierent from the trivial one : rev(g) = minfk=g has a no nontrivial k? coveringg We denote D(G) the diameter of G and g(g) the girth of G (the minimum length of a cycle in G). As we are interested in nontrivial k-coverings, we have trivial lower and upper bounds for rev(g). 9

10 Lemma 10 Let G be a connected graph with a girth g(g) and a diameter D(G). We have g(g)? 1 rev(g) D(G) Proof: From Lemma 9, we get the lower bound. From Theorem, we get the upper bound. We introduce a new notion of spanning trees of graphs which allows us to reduce the upper bound of rev(g)...1 Some useful tools We dene here a notion of binary relation useful in the denition of our special spanning trees. Denition 3 Let n be a positive integer, n > 1. We dene the binary relation?! by : two n connected graphs G and H are in relation by?! (which is denoted G?! H) i V (H) = n n V (G), E(H) = E(G) [ feg and e belongs to a cycle of length at most n + 1 in H. We denote by?! n the reexive and transitive closure of?! n. Denition 4 Let G be a connected graph and A be a spanning tree of G. The spanning tree A is a p-spanning tree of G if A?! G, for p > 1. p The circumference of a graph G, denoted c(g), is the maximal length of a cycle of G (cf. [Bol78]). Proposition 3 Let G be a connected graph. If the circumference of G is at most p + 1, then each spanning tree of G is a p-spanning tree. Proof: Consider A a spanning tree of G. We denote by C e the cycle obtained by adding e to A with e E(G) n E(A). As c(g) p + 1, for any edge e E(G) n E(A) the length of C e is at most p + 1. Thus, A?! A + e for any e E(G) n E(A). Therefore, A?! G. p p Let us now consider the number T (G) as the minimum integer p such that G has a p- spanning tree : T (G) = minfp = G has a p?spanning tree:g It obviously comes that : Lemma 11 For any connected graph G with at least one cycle, g(g)? 1 T (G) c(g)? 1: 10

11 .. k-covering and k-spanning tree Let us rst study the case of the connected graphs having no nontrivial k-covering. This property is strongly linked to the existence of cycles. In a connected graph G, a cut-edge is an edge such that if we delete it in the graph then connected components appear. Proposition 4 Let G be a connected graph. If G has no nontrivial k-covering, then every edge of G is in a cycle of length at most k + 1 or is a cut-edge. Proof: Suppose that G has no nontrivial k-covering and that there exists an edge e in G which is not a cut-edge and which is not an edge of a cycle of length at most k + 1. Then, it means that there is a cycle of length l > k + 1 which contains e. As e is in a cycle, there exists a spanning tree of G which doesn't contain e. By Theorem, we can give any permutation for the edge e. We could take e 6= I q. This leads to a contradiction with the fact that G has no nontrivial k-covering. Proposition 5 Let G be a connected graph. If G has no nontrivial k-covering then the number of cycles of length at most k + 1 in G is greater than je(g)j? jv (G)j + 1. Proof: We denote by #C the number of cycles of length at most k + 1 in G. In order to construct a k-covering of G, we need a spanning tree A of G, and we need permutations associated to each edge of E(G) n E(A). These permutations must be solutions of a set of equations. We have je(g)j? jv (G)j + 1 permutations for #C equations. The graph G has no nontrivial k-covering. It means that the set of equations has a unique solution (identity permutations). If #C < je(g)j? jv (G)j + 1, then we obtain an innite set of solutions. Therefore, #C je(g)j? jv (G)j + 1: The main result is the fact that the existence of a k-spanning tree forbids the existence of nontrivial k-covering as we prove it now : Proposition 6 Let G be a connected graph. If there exists a k-spanning tree for G, then G has no nontrivial k-covering. Proof: Consider a connected graph G and a k-spanning tree of G, denoted A. We denote fe 1 ; e ; ; e n g the set E(G) n E(A). Assume that there is a nontrivial k-covering of G, denoted H. It follows from Theorem that : 1. H is isomorphic to a graph R q; issued from A.. The equations of permutations given by the cycles of length at most k + 1 are satised by. All the permutations associated to the edges of A are equal to the identity of S q (by denition). As A is a k-spanning tree, we have : A?! A + e 1?! A + e 1 + e?!?! A + e 1 + e + + e n k k k k 11

12 It means that there is a cycle of length at most k + 1 containing e 1 and edges of A. Thus, the equation for this cycle leads us to the fact that the permutation of e 1 is the identity. And so on for all the edges of E(G) n E(A). Therefore, it means that R q; consists of q connected components isomorphic to G i.e. H is a trivial k-covering which is a contradiction. From the denition of rev(g) and T (G) and from the proposition 6, we get : Lemma 1 Let G be a connected graph with at least one cycle. We have : g(g)? 1..3 Some families of graphs rev(g) T (G) : For some families of graphs it is rather easy to compute T (G). Thus, we obtain exact results for rev(g) (thanks to the trivial lower bound). complete graphs K n : T (K n ) = thus rev(k n ) = 1. bipartite complete graphs K p;q : T (K p;q ) = 3 thus rev(k p;q ) =. hypercubs Q n : T (Q n ) = 3, thus rev(q n ) =. rings C n : T (C n ) = n? 1 thus rev(c n ) = n?1. maximal outerplanar graphs : T (G) = and rev(g) = 1. References [Bol78] B. Bollob s. Extremal graph theory. Academic Press, [Bot97] A. Bottreau. R critures de graphe et calculs distribu s. PhD thesis, Universit Bordeaux I, LaBRI, juin [LMZ95] I. Litovsky, Y. M tivier, and W. Zielonka. On the recognition of families of graphs with local computations. Information and computation, 115(1):1119, [Rei3] K. Reidemeister. Einf hrung in die Kombinatorische Topologie. Vieweg, Brunswick,