the edge reconstruction number of a disconnected graph

The Edge Reconstruction Number of a Disconnected Graph

Robert Molina DEPARTMENT OF MATHEMATICS

COLORADO STATE UNIVERSI7Y FORT COLLINS. COLORADO

e-mail: molina@alma. edu

ABSTRACT

The edge reconstruction number of a graph G, RN(G), is the minimum number of edge deleted subgraphs required to determine G up to isomorphism. We prove the following results for a disconnected graph G with at least two nontrivial components. If G has a pair of nontrivial nonisomorphic components then RN(G) I 3. If G has a pair of nontrivial nonisomorphic components, is not a forest, and contains a nontrivial component other than K3 or Kl,3 then RN(G) I 2. Finally, if all nontrivial components of G are isomorphic to a graph with kedges, then RN(G) I k + 2. The edge reconstruction results in this paper are similar to the vertex reconstruction results stated by Myr- vold ("The Ally-Reconstruction Number of a Disconnected Graph," Ars Combhatoria, Vol. 28 119891 pp. 123-1271, but a significant difference is that the edge reconstruction number of a disconnected graph is often two, while the vertex reconstruction number of a graph is always three or more. 0 1995 John Wiley & Sons, Inc.

1. INTRODUCTION

Let G be a simple finite undirected graph. The edge deck of G, denoted E D ( G ) , is the set of edge deleted subgraphs of G, and we refer to the elements of E D ( G ) as cards. The edge reconstruction number of G, denoted R N ( G ) , is the size of a smallest subset of E D ( G ) that determines G up to isomorphism, if such a subset exists. More precisely, if R N ( G ) = k , then there are edges e l , e2,. . . , e k in G such that if H is a graph with edges el,e2, . . . ,e[, where G - ei = H - e( for i = 1,2,. . . , k , then G = H . Graphs for which R N ( G ) is defined, i.e., those graphs that are determined by their edge deck, are said to be edge reconstructible. The vertex reconstruction number of a graph is defined similarly with vertex deletions instead of edge

Journal of Graph Theory, Vol. 19, No. 3, 375-384 (1995)

I 1

0 1995 John Wiley & Sons, Inc. CCC 0364-9024/95/030375-10

376 JOURNAL OF GRAPH THEORY

deletions. The vertex reconstruction number of a graph was first defined by Harary and Plantholt in [ 2 ] , and is referred to as the ally reconstruction number by Myrvold in [4].

In this paper we compute R N ( G ) for disconnected graphs with at least three edges and at least two nontrivial components. Since such graphs are known to be edge reconstructible [I] , we know that R N ( G ) is defined. In [4] Myrvold proved that if G is disconnected with at least two nonisomorphic components, then the vertex reconstruction number of G is 3. Although the proof in that paper was flawed (see [ 3 ] ) , many of the arguments used there for the vertex case are adapted for use here in the edge case. A significant difference between the edge reconstruction results contained in this paper and the vertex reconstruction results contained in [4] is that the edge reconstruction number of a disconnected graph is often two, while the vertex reconstruction number of a graph is always three or more.

We now introduce some terminology and notation. If G - e is a card of E D ( G ) and if H is a graph such that G - e 2: H - e’ for some edge e’ in H , then a graph isomorphic to H can be obtained by adding an edge en to G - e . Thus H = (G - e ) + e”, and since a graph isomorphic to H is obtained by replacing e by e”, we refer to the edge elr as a replacing edge. In the arguments that follow, if G - e and H - e’ are isomorphic, then “the replacing edge e”” always refers to an edge e” with the property that H = (G - e ) + e”. If K is a subgraph of G - e and if the replacing edge e” is placed between nonadjacent vertices in K , then we say that d l joins K . We note that if e’ lies in some component C of H , then the replacing edge e” joins a subgraph of G - e that is isomorphic to C - e’ to form a component isomorphic to C.

Let G be a disconnected graph with at least two nontrivial components. We also assume that G has at least three edges since disconnected graphs with only two edges are not reconstructible. We number the nontrivial components CI , Cz, . . . , C,. Let e ( C , ) denote the number of edges in C, and assume that e ( C , ) 2 . . . 2 e(C,,). We now prove a number of propositions that show that a disconnected graph is usually determined by 2 or 3 of its edge deleted subgraphs. We use the following lemma repeatedly.

Lemma 1. Suppose that CI + Cn for some k . Let el and e2 be edges of C1 and Ck, respectively, and suppose that there exists a graph H with distinct edges el and e: such that G - e l = H - e: and G - e2 H - e:. Then e: is in some component of H - ei that is isomorphic to C, .

Proof. ( H - e : ) - ei = ( H - e:) - e l . and H - e: has more CI type components than ( H - e ; ) - el does.

2. G IS A FOREST

Proposition 1. and at least one component is not a star, then R N ( G ) I 3 .

Suppose e(CI) > e ( C , ) . If all components of G are trees

EDGE RECONSTRUCTION NUMBER 377

Proof. We pick el an endline of C1 and e2 any edge from C,. Also, if it is possible to pick e l so that the nontrivial component of CI - el is isomorphic to C,, then we do so. We pick e3 so that one of these three edges in not an endline. If H is a graph with G - e j = H - e: for i = 1,2,3, then since H - e: has one more isolate than one of the other cards, we know that e{ is an endline of H, and thus the components of H are trees. Since lemma 1 implies that e{ is in a component with at least e(Cl) edges, the replacing edge e:‘ joins an isolate and a nontrivial component in G - e l . Thus G and H have the same number of components, the same number of isolates, and all nontrivial components of H have at least e(C,) edges. We may assume then that the replacing edge e:’ joins the two components of C, - e2 to form some tree with e(C,) edges. So H has components Ci, . . . , CL, where Ci = Ci for i # n, and we may assume that the components of H are numbered so that ei is in C ; .

Now we compare G - e l with H - el . Since these cards are isomorphic, either C, = CL or C, is isomorphic to the nontrivial component of C{ - e{ . But if the latter is true then it was possible to pick el so that the nontrivial component of CI - el was isomorphic to C,, and thus we still have C, = CA. But C , = C; implies G = H.

Proposition 2. If e(C1) > e(C,) and each component Ci is a star, then RN(G) 5 3 .

Proof. In this case C1 = K l , , , for some m 2 2. We pick el and e2 from CI and e3 from C,. Let H be a graph with G - ei = H - e: for i = 1,2,3. We first consider the case C, = K2. In this case H - e{ and H - e; have different numbers of isolates and thus e: is an endline. Let k be the number of isolates in G. If the replacing edge eS’ joins an isolate and a nontrivial component on G - e3. then the resulting graph has at most one edge deleted subgraph with k + 1 isolated vertices. Since H has two such cards, we see that ey must form a K2 and thus G = H.

Suppose C, =+ K2. Then m 2 3 and since H - el has more degree m vertices than H - ei, the replacing edge e:’ must join a vertex of degree m - 1 on G - e l . Because the components of G - e l are all stars, the replacing edge must be a bridge. Since the cards H - el, i = 1,2,3, all have the same number of isolates, the associated edges are either all endlines or all nonendlines. But if the replacing edge e:’ joins two nontrivial components, then the resulting component has at most two edges that are not endlines. Therefore the replacing edge e:’ joins a KI, , , -I and an isolated vertex to form a K l , m and thus G = H.

3. G CONTAINS A CYCLE

Proposition 3. edges are trees, then R N ( G ) 5 2.

If G contains a cycle and all components of G with e(C1)


Proof. We pick el an endline of CI and ez an edge from any cycle in G. If H is any graph with edges e: where G - ei = H - e : , i = 1,2, then e{ is an endline and e: lies in a cycle. By Lemma 1, ell is in a subgraph of H isomorphic to CI . The replacing edge ey does not form a component with a cycle since then this component would have at most e(C1) edges and thus no subgraph isomorphic to C1 since CI is a tree. So the replacing edge ey forms a tree with at least e(Cl) edges. But H has no trees with e ( C , ) + 1 edges since H - ei does not, and thus the replacing edge e’: joins an isolate and a tree with e(Cl) - 1 edges to form a tree isomorphic to CI. Therefore the components of H have at most e(CI) edges, and those components of H that do have e(C1) edges are trees. It follows that the replacing edge e:’ joins a component with at most e(C1) - 2 edges, and thus the collection of trees with e(Cl) - 1 or more edges is the same in G and H . Therefore there is only one way up to isomorphism for the replacing edge el’ to join G - e l and thus G = H .

Proposition 4. If e (Cl ) > e (C , ) and CI is bridgeless, then R N ( G ) 5 2

Proof. We pick e l an edge from CI and e2 an edge from C,. Also if it is possible to pick e l so that CI - e l is isomorphic to C, , then we do so. If H is a graph with G - ei = H - e! for i = 1,2, then by Lemma 1 we know that el is in some component of H - e: that is isomorphic to C1, and thus el is not a bridge. We known then that G and H have the same number of components and the same number of isolates, and that all nontrivial components of H have at least e(C, ) edges. Thus we may assume that the replacing edge ey joins C, - ez to form some connected component C:l. But now we can show that C, = C:, by using the same argument we used at the end of the proof of Proposition 1 , and thus G = H .

Proposition 5. Suppose e(CI) > e(C, ) . If there are edges x and y in G where x is in C l , y is in C,,x is an endline or nonbridge, and the set { x , y } contains both a bridge and a nonbridge, then R N ( G ) 5 2.

Proof. We pick the edges x and y to be el and e2. Also if it is possible to pick e l and e2 so that in addition to satisfying the conditions set on x and y, CI - e l (or the nontrivial component of CI - el if e l is an endline) is isomorphic to C,, then we do so. If H is a graph with G - ei = H - el for i = 1,2, then since H - el and H - e; have different numbers of components, we know that el or e; is not a bridge, and that G and H have the same number of components and the same number of isolates. We also know that the replacing edge el‘ forms a component with at least e(Cl) edges and thus all nontrivial components of H have at least e(C, ) edges. Thus we may assume that the replacing edge ey joins the subgraph C, - e2 to form a component with e(C, ) edges. So H has components Ci, . . . , CA, where Ci = Ci for i # n, and we may assume that the components of H


are numbered so that el is in Ci. Since G - el H - e ; , we also know that e: is an endline if el is, and that ei is a nonbridge if el is, and thus if Ci - el has a component isomorphic to C,, then e l was selected so that CI - el has a component isomorphic to C,. By the same argument used in Propositions 1 and 4, we can show that C, = C; and thus G = H.

Proposition 6. Suppose e(C1) > e(C,). If CI has a bridge but no endlines and if C, is bridgeless, then RN(G) I 2.

Case 1. There is a bridge x in CI and an edge y in C, where CI - x has no component isomorphic to C, - y.

We pick the edges x and y to be el and e2. If it is also possible to pick e l and e2 so that in addition to satisfying the conditions set on x and y in the statement of Case 1, CI - e l has a component isomorphic to C,, then we do so. Let en denote the component C, - e2. If H is any graph with edges e: where G - ei = H - e: , i = 1,2, then since H - ei and H - e; have different numbers of nontrivial components, we know that e; is a bridge but not an endline and that e$ is not a bridge. Since H - el has no components isomorphic to e,, we see that the replacing edge e:’ must join t,, to form a component with e(C,) edges. Thus H has components C [ , C : ,... ,CA, where Cl = Ci for i < n, and where ei is in Ci. Let A and B denote the components of C1 - e l , and let A’ and B‘ denote the components of C{ - el. Since G - e l = H - e l , the sets {A, B , C,} and {A’, B’, CA} must be equal. But e l was selected so that CI - e l has at least as many components isomorphic to C, as C{ - el, and thus we have C, = CA and G = H.

Case 2. For every bridge x in C1 and edge y in C,, C, - y is isomorphic to one of the components of CI - x .

Let x be a bridge in CI and e2 an edge of C, . Let en denote the component C, - e2 and let A and B denote the components of C1 - x where A = en. Our choice of el depends on the component B.

Subcase a: B # C,.

In this case we pick e l to be an edge of B that is not a bridge. Let H be a graph where G - ei = H - el for some edges el, i = 1,2. Since H - e; and H - e: have the same numbers of components, we know that el is a bridge if and only if e: is. Suppose both edges are bridges. Let C{ denote the component of H - e: that e: is in, and let A’ and B’ denote the components of Ci - e ; . Also denote the component of H - e: that is isomorphic to en by e:. Since Ci = CI, we may assume that A’ = e, and that B’ # C,. Since H - e{ has no components isomorphic to en the edge e: is a bridge


between A’ and CL in H. But then the only way for H - el and G - el to be isomorphic is for B’ = C,, which is a contradiction. Thus the edges e{ and e; are not bridges and the replacing edge e: must join en to form some component CL. Since the replacing edge e:‘ joins a component with more than e(C,) edges to form a component isomorphic to CI, the components with e(C,) edges in C and H are the same. Thus CA = C, and G = H.

Subcase b: B = C,.

In this case we pick e l to be an edge of A. Since C, is bridgeless we see that x is the only bridge in CI. Let H be a graph where G - e; = H - ei for some edges el, i = l ,2 . If the edges el and e: are not bridges, then as before we can show that the replacing edge e; joins en to form some component isomorphic to C,, and then G = H. Otherwise e{ and e: are both bridges. If we define C { , A‘, B’, and I?; as above, then as before the edge ei is a bridge between A’ and t; in H. But now G - el = H - e{ only if CI - el is isomorphic to the component of H - el consisting of A’, c:, and the bridge e:. But this is a contradiction since these two components are nonisomorphic. Thus el and e: are not bridges and G = H.

4. ALL COMPONENTS HAVE SAME NUMBER OF EDGES

Lemma 2. in C j , then C; = C j .

If C; - x = C j - y for every edge x in C; and every edge y

Proof. One can verify that this result holds if e (Ci ) 5 3. We are using the fact that the components we are considering are connected since otherwise there are counterexamples with both two and three edges. Since regular graphs with at least four edges are reconstructible, we may assume that these components are not regular. The only other possibility is that Ci and C j are bipartite biregular graphs and each edge in either graph is adjacent to a vertex of degree 6 and a vertex of degree A, where S < A. If 6 = 1 then C; is a star and thus reconstructible. So we may assume that S 2 2. But now Ci can be obtained from any card C; - x by locating the unique vertex of degree S - 1 and the unique vertex of degree A - 1 that has no neighbors of degree A, and then placing an edge between them. Thus Ci is reconstructible in this case also.

Corollary. and y such that Ci - x + C j - y.

If Ci + C j for some integers i a n d j then there exist edges x

Proposition 7. Assume that all nontrivial components of G have k edges. If G has a pair of nontrivial nonisomorphic components and an edge that is a bridge but not an endline then R N ( G ) 5 2.


Proof. Since G has a pair of nontrivial nonisomorphic components, k must be at least three. We may assume that the components of G are labeled so that CI contains a bridge that is not an endline and that CI is not isomorphic to C2. Let el be a bridge that is not an endline in C1 and let e2 be a nonbridge or endline in C2. If H is a graph such that G - ei = H - e( for i = 1,2, then by Lemma 1 we know that ei is in a component of H - e: that is isomorphic to C1 and that e: is in a component H - e{ that is isomorphic to C2. Since the replacing edge e t forms a component with at least k edges, we know that the nontrivial components of H have at least k - 1 edges. It follows that the replacing edge ey must join C1 - e l to form a component isomorphic to C1, and thus G = H .

Lemma 3. Assume that all nontrivial components of G have k edges and that CI is not isomorphic to C2. Also assume that e l is a nonbridge in CI and that e2 is an edge of C2 such that G - e2 does not have a component isomorphic to C1 - e l . Finally, assume that if H is a graph for which G - e; = H - e( for i = 1,2, then el is not a bridge in H . If all of these conditions are satisfied, then R N ( G ) 5 2.

Proof. Suppose G satisfies the conditions listed in Lemma 3 and that H is a graph where G - ei = H - e: for i = 1,2. By Lemma 1 the edge ei lies in a component with a subgraph isomorphic to CI. Since the edge el is not a bridge, the replacing edge e:’ forms a component with either k or k + 1 edges. But if H has a component with k + 1 edges, then it also has one with k - 1 edges that is isomorphic to C1 - e l , and e{ and e: must both lie in the component with k + 1 edges. But then a component isomorphic to CI - el appears on both H - el and H - e:, which is a contradiction. Thus the component formed by the replacing edge ell’ has k edges and is isomorphic to CI. Therefore G = H .

Proposition 8. Assume that all nontrivial components of G have k edges where k > 3. If G has a pair of nontrivial nonisomorphic components, then R N ( G ) 5 2.

Proof. By Proposition 7 we may assume that all bridges in G are endlines. There are two cases to consider.

Case I . There is a bridgeless component in G.

We may assume that C , is bridgeless and that C2 is not isomorphic to CI. We want to select edges e l in C1 and e2 in C2 so that G - e2 does not have a component isomorphic to C1 - el. Note that if G had components with 3 edges, then such a choice of edges might not be possible since, for example, CI might be a triangle and C2 a K1.3. But since k > 3 we claim that such a choice exists. For if C2 is a star, then since CI is bridgeless and


has 4 or more edges, an edge deleted from CI does not give a star, and thus any choice of el in C1 and e2 in C2 will do. If C2 is bridgeless, then we can find edges el and e2 so that CI - e l .ji. C2 - e2 by the corollary to Lemma 2. Finally, if C2 has endlines but is not a star, then we can find edges e l and e2 so that G - e2 does not have a component isomorphic to C1 - el by either taking e2 to be a bridge or nonbridge as needed. Now if H is a graph where G - e j = H - e: for i = 1,2, then by Lemma 1 the edge e; lies in a subgraph of H that is isomorphic to CI and thus e: is not a bridge. Since all the conditions of Lemma 4 have been met, we see that R N ( G ) 5 2.

Case 2. All nontrivial components of G have endlines.

We will refer to an edge of a graph that is not an endline as an internal edge. We note that the internal edges of G are not bridges since we are assuming that all bridges are endlines. From the set of components of G that are not stars, we select an element with a minimal number of internal edges. We may assume that this component is C1 and that C2 is a component that is not isomorphic to CI. Let el be an internal edge of CI such that C1 - el is not a star, and let e2 be an endline of C2. Then G - e2 does not have a component isomorphic to C , - e l since CI - e l is not a star and has fewer internal edges than any of the nontrivial nonstar components of G - e2. If H is a graph where G - ei = H - e: for i = 1,2, then since H - e; has fewer components than H - ei, e; is not a bridge in H. By Lemma 3 we see that R N ( C ) 5 2.

Proposition 9. If G has exactly two types of nontrivial components, namely those isomorphic to K 3 and those isomorphic to K1.3. then R N ( G ) 5 3.

Proof. We select e l from a triangle and e2 and e3 to be endlines. If H is a graph where G - ei = H - el for i = 1,2,3, then el is not a bridge and the replacing edge ey must join a star on G - e l . But if ey forms a component with four edges, then G and H have only two cards in common. Thus the replacing edge ey forms a triangle and G = H.

Proposition 10. If C j = CI for i = 2,3 ,..., n, then R N ( G ) 5

e(C1) + 2.

Proof. Let e(CI) = k. One can verify that this result holds if k = 1. Suppose k 2 2. Let G - e be a card from ED(G) whose nontrivial components all have at least k - 1 edges, and H is a graph where G - e L-

H - e’. We consider the possibilities for the replacing edge e”, keeping in mind that each card in E D ( G ) has n - 1 components isomorphic to C1 plus one subgraph isomorphic to some member of ED(C1). If the replacing edge


el’ forms a component with more than k + 1 edges, then this component has at least 2k edges, and thus E D ( G ) and E D ( H ) can share at most 2 cards. If the replacing edge el’ forms a component with exactly k + 1 edges, then ED(G) and E D ( H ) can share at most k + 1 cards. If the replacing edge e” forms a component with k edges that is not isomorphic to C1, then ED(G) and E D ( H ) can share at most k cards. If the replacing edge el/ forms a component with a single edge, then ED(G) and E D ( H ) can share at most k cards since in this case if an edge is deleted from a component of H that has k edges, the resulting card would have three nontrivial components with fewer than k edges. The only other possibility is that e” forms a component that is isomorphic to C1 in which case G = H.

5. THE MAIN THEOREM

We now summarize the results of the previous sections. Let S denote the family of graphs that have exactly two types of nontrivial components, namely those isomorphic to K3 and those isomorphic to K1,3. If G has a pair of nontrivial nonisomorphic components, contains a cycle, and is not in S, then we have shown that R N ( G ) 5 2 . For suppose G is such a graph. If the components of G all have the same number of edges, then Propositions 7 and 8 together with the fact that G is not in S imply that RN(G) 5 2. Suppose e (C , ) > e(C,). By Propositions 3 and 4 we can assume that CI contains both a bridge and a nonbridge. If CI contains an endline or if C , contains a bridge, then we can find edges x and y as described in Proposition 5. Otherwise C, is bridgeless and CI has a bridge but no endlines and we may appeal to Proposition 6. We have the following theorem.

Theorem. Let G be a disconnected graph with at least three edges and at least two nontrivial components.

(a) If G has a pair of nontrivial nonisomorphic components, then R N ( G ) 5 3. If such a graph G contains a cycle and is not in S, then R N ( G ) 5 2 .

(b) If all nontrivial components of G are isomorphic to some graph C, then R N ( G ) 5 e(C) + 2.

We conclude by giving some examples that show that the upper bounds on R N ( G ) are tight. In the following example, R N ( G ) = 3. To see this, note that for every pair of cards in ED(G) one of the graphs Hi has the same pair in its edge deck.


Here is another example of a graph with R N ( G ) = 3, but with components that are not all stars.

We note that in this example there are 3 different kinds of cards in ED(G) and thus six graphs Hi are required to show that R N ( G ) = 3. The details are left to the reader. The reader may also wish to verify that any graph in S has R N ( G ) = 3, and thus the conclusion of Proposition 9 can not be improved.

Finally, if G consists of two copies of Kl,,, and if H has one KI,,+I component and one Kl ,m- l component, then R N ( G ) = m + 2 since every m + I-subset of ED(G) is also a subset of E D ( H ) . The case m = 4 is shown.

We note that in [4] Myrvold gives the example G = 2K,, for which the vertex reconstruction number is m + 2. Our example above corresponds exactly to Myrvold’s since the line graph of ~ K I , , is 2Km.

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the edge reconstruction number of a disconnected graph

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