Cycle Double Cover Conjecture

Transcription

1 Cycle Double Cover Conjecture Paul Clarke St. Paul's College Raheny January 5 th 2014 Abstract In this paper, a proof of the cycle double cover conjecture is presented. The cycle double cover conjecture purports that if a graph is bridgeless, then there exists a list of cycles in the graph such that every edge in the graph appears in the list exactly twice. By applying induction on the number of edges in a bridgeless graph, I show that when an edge is added to a bridgeless graph, we can reform the cycle double cover to include that edge. By mathematical induction, this concludes the general CDC. Keywords: cycle double cover conjecture, bridgeless graph, graph theory, topological embeddings, induction 1. Introduction The cycle double cover conjecture was posed by George Szekeres in [1] Another mathematician, Paul Seymour, independently discovered the conjecture in [2] The cycle double cover conjecture purports that if a graph is bridgeless, then there exists a list of cycles in the graph such that every edge in the graph appears in the list exactly twice. Work on the cycle double cover conjecture has mainly focused on strengthening the conjecture in the areas of topological embeddings of bridgeless graphs. The following conjecture is already

2 known to imply the CDC, but the converse is not necessarily true. Conjecture 1 (Strong embedding conjecture): Every 2-connected graph has a strong embedding in some surface. See [3] for the definition of a strong embedding of a graph. Attempting to prove this conjecture for specific special cases has become a prime method for attempting to conclude the CDC. If we restrict the number of possible cycles that are allowed in a cycle double cover, another avenue for partial results reveals itself. Conjecture 2: Every bridgeless graph has a 5-cycle double cover. A. Huck showed in [4] that a smallest counterexample to Conjecture 2 must have girth at least ten. If the CDC is false, it has been shown in [5] that a minimal counterexample must be a snark. L. Goddyn showed in [6] that a smallest counterexample to the general CDC must have girth at least ten. A. Huck then improved this result to girth twelve computationally. [4] For more information on the history of the cycle double cover conjecture, please consult the following references: [7] [8] [9] In this paper, I attempt to conclude the CDC by an inductive procedure. In section two of this paper, relevant graph theoretical terminology is discussed and defined. In section three, some preliminary results on cycle double covers in graphs are presented. Finally, in section four, a final proof of the general CDC is concluded. 2. Terminology In this section, we define graph theoretical terminology relevant to the material in this paper. For graph theoretical terminology not defined here, consult the books of Trudeau [10] and Harary [11] We define a graph G = (V, E), where V is a finite non-empty set called the vertex set. The elements of V(G) are called vertices. The elements of E(G) are two element subsets of V(G) and are called edges. In general, n = V(G). Two vertices v, u V(G) are adjacent in G if and only if {v, u} E(G). The number of vertices adjacent to a vertex v, denoted deg(v), is the degree of v. A graph G' = (V, E) is a subgraph of G (denoted G' G) if V(G') V(G) and E(G') E(G). A walk of length k in a graph G from vertex v to vertex u is a subgraph W G whereby V(W) is a sequence of vertices {v = v 0, v 1, v 2,..., v k = u} in G such that {v i, v i+1 } E(G) for all 0 i < k. E(W) is defined as {{v i, v i+1 } 0 i < k, v i V(W)}. The two vertices v 1, v 2 V(W) whereby deg(v 1 ) = 1

3 and deg(v 2 ) = 1 are called the endpoints of W. A path P is defined as a walk whereby for every vertex v V(P), deg(v) 2. A cycle C in a graph is a path with an extra edge connecting the two endpoints of the path. A k-cycle is a cycle C whereby V(C) = k. If there exists a cycle C in G such that V(C) = V(G), G is Hamiltonian. The spanning cycle is called a Hamiltonian cycle in G. A cycle double cover of G is a set X of cycles in G such that for every edge e E(G), there exist exactly two cycles C i X and C j X such that e E(C i ) and e E(C j ). A graph G is connected if for every pair of vertices v and u in V(G), there exists a path in G with v and u as endpoints. G is called k-connected if the removal of fewer than k vertices from V(G) cannot disconnect G. The connectivity of a graph, denoted by k(g), is defined as the minimum number of vertices the removal of which from V(G) will disconnect G. A graph G is k-edgeconnected if the removal of fewer than k edges from E(G) cannot disconnect G. The edge connectivity of a graph G, denoted by λ(g), is defined as the minimum number of edges the removal of which from E(G) will disconnect G. G is bridgeless if λ(g) 2. The following definition of a snark can be found in [12]. A snark G is defined as a bridgeless, cubic graph. In addition, the edges of G cannot be colored with three colors such that no two adjacent edges share the same color. In this paper, we consider only finite, simple, undirected graphs. 3. Preliminary results In this section, two preliminary results on cycle double covers are presented. We begin our investigation of the CDC with a preliminary result on sequences and adjacency's of cycles in a bridgeless graph. Theorem 1: Let G be a bridgeless graph with a cycle double cover X. Then for every pair of cycles C 1 and C k in X there exists a sequence of cycles S = (C 1, C 2, C 3, C k ) in X such that V(C i ) V(C i + 1) {} for all 1 i < k and V(C i ) V(C j ) = {} for all 1 i k, 1 j k, j i, i + 1, i 1. Proof: Suppose that we have two arbitrary cycles C 1 and C k in X. We will build a sequence S = (C 1, C 2, C 3, C k ) of cycles in G such that V(C i ) V(C i + 1 ) {} for all 1 i < k and V(C i ) V(C j ) = {} for all 1 i k, 1 j k, j i, i + 1, i 1. Let S = (). Append C 1 to S. Let v 1 be some arbitrary vertex on C 1 and let v k be some arbitrary vertex on C k. As G is bridgeless, there exists some path with v 1 and v k as endpoints in G. Let P be the shortest

4 such path. Let v 2 be the first vertex on P such that v 2 V(C 1 ). If there is no such vertex, then the proposition trivially holds, as either C 1 = C k or v k V(C 1 ) V(C k ). Let v' V(C 1 ) be a vertex adjacent to v 2. As X is a cycle double cover, there exists some cycle C 2 X such that {v 2, v'} E(C 2 ). Append C 2 to S. Note that v' V(C 1 ) V(C 2 ), so V(C 1 ) V(C 2 ) {}. Note that C 1 C 2. Now let v 3 be the first vertex on P / C 1 such that v 3 V(C 2 ), and repeat the above process with v 3 in the place of v 2. Repeat the above process recursively until we form S = (C 1, C 2, C 3, C k ). Clearly, by the above procedure, V(C i ) V(C i + 1 ) {} for all 1 i < k and all C i are unique. Suppose now that V(C i ) V(C j ) {} for some i, j i, i + 1, i 1 in S. Assume without loss of generality that i < j. Consider the set R = {C m i < m < j}, where C m is a cycle in S. Remove R from S while preserving the order of S. Restart the indices of S by letting the index of C j be equal to the index of R[0]. Clearly, S still satisfies the condition that V(C i ) V(C i + 1 ) {} for all 1 i < k. By repeating this process for every incidence of the assumption, we will form S such that V(C i ) V(C j ) = {} for all 1 i k, 1 j k, j i, i + 1, i 1. Therefore, the proposition holds. As our choice of C 1 and C k was arbitrary, the desired result follows. QED Any minimal counterexample to the CDC must be a snark. [5] Therefore, if G is a minimal counterexample to the CDC, for every vertex v V(G), deg(v) = 3. This observation allows us to conclude the CDC for a special case of graphs, which is pivotal in my inductive argument for settling the CDC in general. Theorem 2: If G is a bridgeless graph whereby for every edge e E(G), G / e is not bridgeless, G has a cycle double cover. Proof: Suppose that Theorem 2 is not true. Let G be a minimal counterexample. As G is also a minimal counterexample to the general CDC, for every vertex v V(G), deg(v) = 3. We will build a set S of cycles in G. Let C 1 be an arbitrary cycle in G and append C 1 to S. Let v be a vertex adjacent to a vertex v' V(C 1 ) such that v V(C 1 ). As G is bridgeless, v and v' lie on a common cycle C 2. This result can be found in [11]. Clearly, V(C 1 ) V(C 2 ) {} and C 1 C 2. Append C 2 to S. Now repeat the above process with C 1 C 2 in the place of C 1. By repeating the above process recursively, we will find a set S of cycles such that the union of the vertex sets of the cycles in S is equal to V(G). Let S = k. Let G' be a graph formed by the union of all cycles in S. Clearly, G' is bridgeless, as G' is connected and for every edge e E(G'), e is on some cycle in G'. Consider C k S. As C k was the last cycle added to S, and as C k is a cycle, there must exist some vertex v V(C k ) whereby deg(v) =

5 2. Therefore, deg(v) = 2 in G'. As deg(v) = 3 in G, there must exist some edge e' E(G) such that v e' and e' E(G'). Therefore, G / e' is bridgeless, as G' G / e'. However, this is a contradiction, as by the proposition, for every edge e E(G), G / e is not bridgeless. Therefore, there can exist no minimal counterexample to the proposition, and thus the desired result follows. QED 4. Final proof In this section, we apply the results presented in section 3 to conclude the general cycle double cover conjecture through an inductive procedure. Theorem 3 (cycle double cover conjecture): If G is bridgeless, then G has a cycle double cover. Proof: We will prove the above statement by induction on m = E(G). Suppose that m = 3. Clearly, the only possible bridgeless graph whereby E(G) = 3 is K 3. By appending K 3 to an empty set X twice, X becomes a cycle double cover for G = K 3. Therefore, we have proven the proposition for m = 3. Assume that the proposition holds for E(G) = m. Suppose now that we have some bridgeless graph G whereby E(G) = m + 1. Suppose that G is a bridgeless graph whereby for every edge e E(G), G / e is not bridgeless. Then, by Theorem 2, G has a cycle double cover X, and we have proven the m + 1 case. Suppose that there exists some edge e E(G) such that G / e is bridgeless. Remove e = {v 1, v k } from E(G). As G is bridgeless and as E(G) = m, by the induction assumption, G has a cycle double cover X. As X is a cycle double cover, there exists some cycle C 1 X such that v 1 V(C 1 ) and some cycle C k X such that v k V(C k ). Suppose that C 1 = C k. Then v 1 V(C 1 ) and v k V(C 1 ). Suppose now that we replace e back into E(G). The addition of e creates two new cycles C 1 ' and C 1 * from C 1. (Fig. 1) Suppose that we replace C 1 with C 1 ' and C 1 * in X. Note that for every edge r E(C 1 ), r E(C 1 ') or r E(C 1 *). Also, e E(C 1 ') and e E(C 1 *). Therefore, as the rest of the graph is unchanged, for every edge r E(G), there still exist exactly two cycles C i X and C j X such that r E(C i ) and r E(C j ). Therefore, G has a cycle double cover, and the proposition holds for m + 1. Suppose that C 1 C k. As G is bridgeless, and as G has a cycle double cover X, by Theorem 1, there exists a sequence of cycles S = (C 1, C 2, C 3, C k ) in X such that V(C i ) V(C i + 1 ) {} for all 1 i < k and V(C i ) V(C j ) = {} for all 1 i k, 1 j k, j i, i + 1, i 1. (Fig. 2) Replace e back

6 into E(G). We will build two cycles C' and C* such that C' C* covers the same set of edges as S, and also covers e twice, thus reforming the cycle double cover X of G. Fig. 1: C 1 split by e into C 1 ' and C 1 * Choose v 1 such that v 1 V(C 1 ) V(C 2 ) and choose v k such that v k V(C k ) V(C k-1 ). This is permissible as if v 1 V(C 1 ) V(C 2 ), we could let C 2 = C 1 in S and re-index the sequence. Similarly for v k. Consider v 1. Let C' = {{}, {}} and let C* = {{}, {}}, where C' and C* are graphs. Append v 1 to V(C') and V(C*). As C 1 is a cycle, there exist two paths P' C 1 and P* C 1 between v 1 and V(C 1 ) V(C 2 ). (Fig. 3) Fig. 2: A sample sequence S of cycles between C 1 and C k

7 Let v' be the vertex on the endpoint of P' and let v* be the vertex on the endpoint of P*. Let C' = C' P' and let C* = C* P*. Suppose that E(C 1 ) E(C 2 ) E(C 1 ). Therefore, there must exist exactly two paths P 1 and P 2 between v' and v* such that V(P 1 ) V(C 1 ) and V(P 2 ) V(C 2 ). Let C' = C' P 1 and let C* = C* P 2. Note that C' is a path with endpoint v* and C* is a path with endpoint v'. As C 2 is a cycle, there exists a path P 3 between v* and V(C 2 ) V(C 3 ). Let v 3 be the vertex on the endpoint of this path. Similarly, there exists a path P 4 between v' and V(C 2 ) V(C 3 ). Let v 4 be the vertex on the endpoint of this path. Let C' = C' P 3 and let C* = C* P 4. Note that C' C* now covers the same set of edges as C 1 ( C 2 / C 2 C 3 ). Suppose that E(C 1 ) E(C 2 ) E(C 1 ). Therefore, there exists exactly one path P between v* and v' along V(C 1 ) V(C 2 ). Let C' = C' P and let C* = C* P. Similar to the above, we can now let C' = C' P 3 and let C* = C* P 4. Note that C' C* now covers the same set of edges as C 1 ( C 2 / C 2 C 3 ). We can now repeat the above process with v 3 and v 4 in the place of v' and v*. Repeat the process recursively until C' and C* cover C k-1. Clearly, C' and C* are walks. Let v r be the vertex on the endpoint of C' and let v s be the vertex on the endpoint of C*. When C' and C* cover C k-1, there exists a path P r between v r and v k and a path P s between v s and v k such that E(P r ) E(P s ) = E(C k ) / E(C k-1 C k ). Let C' = C' P r and C* = C* P s without loss of generality. By the above procedure, C' C* covers the same set of edges as the cycles of S. Suppose that C' is not a path. Therefore, as C' is a walk, there must exist some cycle C'' in G such that C'' C'. Let v'' be the first vertex whereby v'' V(C') and v'' V(C''). Append C'' to a new set Y and continue C' from v'', ignoring C''. Do this for all such cycles C'' C'. As C' is now a walk with no cycles, C' is a path. Similarly for C*. Fig. 3: Two paths P' and P* between v 1 and V(C 1 ) V(C 2 )

8 Note now that {C', C*} Y covers the same set of edges as the cycles of S. Add {v 1, v k } to E(C') and E(C*). Clearly, C' and C* are now cycles. Remove the cycles of S from X and add C', C* and the cycles of Y to X. As stated before, {C', C*} Y covers the same set of edges as the cycles of S. In addition, e is also covered twice. Therefore, we have reformed the cycle double cover X in G, and the proposition holds for m + 1. Therefore, in all possible cases, we have proven the proposition for the m + 1 case. By mathematical induction, the proposition holds for all m 3. As this clearly encapsulates all possible bridgeless graphs, the desired result follows. QED 5. References [1]. Szekeres, G. (1973) Polyhedral decomposition of cubic graphs. Austral. Math. Soc., 8, [2]. Seymour, P.D. (1979) Sums of circuits. in: Graph Theory and related topics, J.A. Bondy and U.S.R Murty, eds., New York: Academic Press, p [3]. Chan, M. (2009) A survey of the cycle double cover conjecture. Department of Mathematics, Princeton University. [4]. Huck, A. (2000) Reducible configurations for the cycle double cover conjecture. Discrete Applied Mathematics, 99, [5]. Jaeger, F. (1985) A survey of the cycle double cover conjecture. Annals of Discrete Mathematics 27 Cycles in Graphs, North-Holland Mathematics Studies 27, [6]. Goddyn, L. (1985) A girth requirement for the double cycle cover conjecture. Ann. Discrete Math. 27, [7]. Zhang, C.Q. (1997) Integer Flows and Cycle Covers of Graphs. New York: Marcel Dekker Inc.

9 [8]. Alspach, B., Goddyn, L., Zhang, C. (1994) Graphs with the Circuit Cover Property. Transactions of the American Mathematical Society, 344, 1, [9]. DeVos, M., Johnson, T., Seymour, P. Cut Coloring and Circuit Covering. J. Combin. Theory Ser. B. [10]. Trudeau, R.J. (1993) Introduction to Graph Theory. New York: Dover Publications. [11]. Harary, F. (1972) Graph Theory. Addison-Wesley Publishing Company, Inc. [12]. Gardner, M. (1976) Mathematical Games. Scientific American 4 (234):