Basic definitions. Remarks

Transcription

1 Basic definitions In a graph G(V, E) a Hamiltonian tour sometimes also called a Hamiltonian cycle is a closed trail that includes each of the graph s vertices exactly once. A graph that contains such a cycle is called a Hamiltonian graph. Remarks It s much, much harder to decide whether a graph is Hamiltonian than to decide whether a multigraph is Eulerian. Roughly speaking, we can be sure that a graph is Hamiltonian if it has lots of edges. The complete graph on n 3 vertices, K n, is clearly Hamiltonian as it includes the cycle {(v 1, v 2), (v 2, v 3),..., (v n 1, v n), (v n, v 1)} where all the edges are certain to exist as K n contains all possible edges.

2 Two ways to be sure a graph is Hamiltonian Theorem (Dirac (1952)) Let G be a graph with n 3 vertices. If each vertex of G has deg(v) n/2, then G is Hamiltonian. Theorem (Ore (1960)) Let G be a graph with n 3 vertices. If deg(u) + deg(v) n for every pair of non-adjacent vertices u and v, then G is Hamiltonian. We ll obtain both of these as corollaries of a more general result, the Bondy-Chvátal Theorem.

3 Main Result About Hamiltonian Cycles Suppose G is a graph on n vertices. Then the closure [G] is constructed by adding a new edge (u, v) to connect each pair of non-adjacent vertices u and v for which deg(u) + deg(v) n. ( ) One continues recursively, adding new edges according to ( ) until all non-adjacent pairs u, v satisfy deg(u) + deg(v) < n. Theorem (Bondy and Chvátal (1976)) A graph G is Hamiltonian if and only if its closure [G] is Hamiltonian.

4 Computing [G], the closure of G We ll compute the closure [G] of a graph G(V, E) by producing a sequence of graphs G j(v, E j) that all have the same vertex set as the original graph, but have different edge sets: G(V, E) = G 1(V, E 1),..., G K(V, E K) = [G]. The first graph in the sequence, G 1 is the graph whose closure we want to compute. Successive edge sets will differ from each other by a single edge, so that E j+1 = E j {(u, v)} where the extra edge (u, v) involves two vertices u and v that are: not adjacent in G j, so (u, v) / E j, satisfy deg Gj (u) + deg Gj (v) n, where the subscript is to indicate that we compute the degrees in the graph G j. We continue adding edges, one at a time, until we can t find any more edges to add. The final graph in the sequence is then [G].

5 An example: Building G 2 from G 1 The graph G = G 1 G Working through all possible new edges The edge i j (v i, v j)... deg G1 (v i) deg G1 (v j)

6 An example: Building G 2 from G 1 The graph G = G 1 G Working through all possible new edges The edge i j (v i, v j)... deg G1 (v i) deg G1 (v j) 1 2 Doesn t exist

7 An example: Building G 2 from G 1 The graph G = G 1 G Working through all possible new edges The edge i j (v i, v j)... deg G1 (v i) deg G1 (v j) 1 2 Doesn t exist

8 Finishing the construction Adding further edges G

9 Finishing the construction Adding further edges G deg G2 (v 1) + deg G2 (v 3) = nd G 2 G 3: add edge (v 1, v 3) as

10 Finishing the construction Adding further edges G deg G2 (v 1) + deg G2 (v 3) = nd G 2 G 3: add edge (v 1, v 3) as

11 Finishing the construction Adding further edges G deg G2 (v 1) + deg G2 (v 3) = nd G 2 G 3: add edge (v 1, v 3) as 2 5 3rd G 3 G 4: add edge (v 2, v 3) as deg G3 (v 2) + deg G3 (v 3) =

12 Finishing the construction Adding further edges G nd G 2 G 3: add edge (v 1, v 3) as deg G2 (v 1) + deg G2 (v 3) = rd G 3 G 4: add edge (v 2, v 3) as 3 6 deg G3 (v 2) + deg G3 (v 3) = th [G] = G 4: all pairs of non-adjacent vertices v i, v j satisfy deg G4 (v i) + deg G4 (v j) < 7

13 Proof of the Bondy-Chvátal Theorem The theorem says two things: If G is Hamiltonian then so is [G]. If [G] is Hamiltonian then so is G. The first of these is obvious: [G] contains all the edges that appear in G (and perhaps some others as well). Thus any Hamiltonian tour that appears G is automatically present in [G] too. The second statement is more subtle and we ll prove it by contradiction. Assume that somewhere in the sequence G 1(V, E 1),..., G j(v, E j), G j+1(v, E j+1),..., G K(V, E K) = [G] the graphs switch from being non-hamiltonian to being Hamiltonian. That is, there s some j for which G j doesn t contain a Hamiltonian tour, but G j+1 does, as do all subsequent graphs in the sequence, including [G].

14 Proof of the Bondy-Chvátal Theorem, continued The step from G j to G j+1 v n v 1 v 2 v 3 v n e v 1 v 2 v 3 v n-1 v n-1 v n-2 v n-2 G j G j+1 v k+1 v k-1 v k+1 v k-1 v k v k We suppose that G j does not have a Hamiltonian cycle, but G j+1 does, even though G j+1 differs from G j only by the single edge e = (v 1, v n). In the diagrams above the vertices are numbered according to their position in G j+1 s Hamiltonian cycle. The dashed lines are meant to indicate that G j may have some other, extra edges in addition to those that will form the cycle.

15 Proof of the Bondy-Chvátal Theorem, continued Define two subsets of E j, the edge set of the graph G j(v, E j) X = {v i (v i 1, v n) E j and 2 < i < n} and Y = {v i (v 1, v i) E j and 2 < i < n} v n v 1 v 2 v 3 v n - 2 v i v n - 1 In the diagram at left the blue vertex, v i, is in X because it is connected indirectly to v n, through its predecessor v i 1, while the orange vertex is in Y because it is connected directly to v 1. Note that, by construction, neither X nor Y contains any of the vertices v 1, v 2 or v n. v i-1

16 Proof of the Bondy-Chvátal Theorem, continued v n - 2 v n - 1 v n deg(v n ) -1 edges v 1 deg(v 1 ) -1 edges v 2 v 3 The set X has deg Gj (v n) 1 members, while the set Y has deg Gj (v 1) 1. So then X + Y = deg Gj (v n) + deg Gj (v 1) 2 n 2 where the inequality follows because we know deg Gj (v n) + deg Gj (v 1) n v k G j as the closure construction is going to add the edge e = (v 1, v n) when passing from G j to G j+1. v k-1 vi On the other hand, both X and Y were drawn from the set of vertices which has only n 3 members. {v i 2 < i < n}

17 Proof of the Bondy-Chvátal Theorem, continued v n v 1 v 2 v 3 We ve just seen that v n - 1 X + Y n 2 v n - 2 G j but that both X and Y are drawn from a set with only n 3 members: then by the pigeonhole principle, they must have at least one member in common. v k + 1 v k v k - 1 v k - 2 Say that v k is a shared member. Then it is possible to find a Hamiltonian tour in G j, as illustrated in red at left. The tour in G j: runs from v 1 to v k 1, in the same order as the tour found in G j+1, then jumps from v k 1 to v n: this is possible becase v k X. The tour then continues, passing from v n to v n 1 and on to v k, in the opposite order from the tour in G j+1 and concludes with a jump from v k to v 1: this is possible because v k Y.

18 Proof of the Bondy-Chvátal Theorem, concluded The Hamiltonian tour in G j that we found on the previous slide contradicts our initial assumption that G j is not Hamiltonian, but G j+1 is. This means no such G j can exist: the sequence of graphs in the closure construction can never switch from non-hamiltonian to Hamiltonian and so if [G] is Hamiltonian, then G must be too. Afterword Dirac and Ore s Theorems are both easy corollaries of the Bondy-Chvátal Theorem in that the hypotheses of both the older results ensure that [G] is a complete graph, which is clearly Hamiltonian. And then Bondy-Chvátal establishes that as [G] is Hamiltonian, so is G.