Definition Hamiltonian graph: A graph with a spanning cycle (also called a Hamiltonian cycle). Hamiltonian graph Hamiltonian cycle

Theorem If G is a simple graph with at least three vertices and δ(G) ≥ n(G)/2, then G is Hamiltonian. Proof: 1. The condition n(G) ≥ 3 must be included, since K 2 is not Hamiltonian but satisfies δ(K 2 ) ≥ n(K 2 )/2. 2. The proof uses contradiction and extremality. 3. Let G be the maximal no-Hamiltonian graphs with minimum degree at least n/2. 4. Adding any edge joining nonadjacent vertices in G creates a spanning cycle.

Theorem When u  v in G, G has spanning path u=v 1,…,v n =v, because G+uv has a spanning cycle which contains the new edge uv. 5. It suffices to show there is a neighbor of u (v i+1 ) directly follows a neighbor of v (v i ) on the path.

Theorem We show that there is a common index in the sets S={i: u  v i+1 } and T= {i: v  v i }. 7. |S  T| + |S  T| = |S| + |T| = d(u) + d(v) ≥ n. 8. Neither S nor T contains the index n. 9. Thus |S  T| < n, and hence |S  T| ≥ 1.

Lemma Let G be s simple graph. If u and v are distinct nonadjacent vertices of G with d(u) + d(v) ≥ n(G), then G is Hamiltonian if and only if G + uv is Hamiltonian. Proof: 1. (  ) Trivial. 2. (  ) The proof is the same as for Theorem

Hamiltonian Closure Hamiltonian closure of a graph G, denoted C(G): The graph with vertex set V(G) obtained from G by iteratively adding edges joining pairs of nonadjacent vertices whose degree sum is at least n, until no such pair remains.

Theorem The closure of G is well-defined. Proof: 1. Let e 1,…,e r and f 1,…,f s be sequences of edges added in forming C(G), the first yielding G 1 and the second G f 1, being initially addable to G, must belong to G If f 1, …, f i–1  E(G 1 ), then f i becomes addable to G 1 and therefore belongs to G 1. Hence, G 1  G Similarly, G 2  G 1.

Theorem A simple n-vertex graph is Hamiltonian if an only if its closure is Hamiltonian.

Theorem Let G be a simple graph with vertex degree d 1 ≤ … ≤ d n, where n ≥ 3. If i i or d n-i ≥ n-i (Chvatal’s condition), then G is Hamiltonian. Proof: 1. G is Hamiltonian if and only if C(G) is Hamiltonian. (Theorem ) 2. It suffices to show C(G) = G’ = K n. 3. G’ satisfies Chavata’s condition because adding edges to form the closure reduces no entry in the degree sequence.

Theorem We prove the contrapositive: if G’ is not a complete graph, there exists i < n/2 such that (1) d i  i (at least i vertices have degree at most i), and (2) d n-i < n-i (at least n-i vertices have degree less than n–i). 5. With G’ ≠ K n, let nonadjacent vertices u and v have maximum degree sum. 6. Because G’=C(G), u  v implies that d(u) + d(v) < n. 7. Let d(u) ≤ d(v). Then d(u) < n/2.

Theorem Let i = d(u). 9. Every vertex of V – {v} not adjacent to v has degree at most d(u)= i. 10. There are n -1 – d(v) such vertices. 11. d(u) + d(v) ≤ n -1 yields n -1 – d(v) ≥ d(u) = i. (1) d i  i (at least i vertices have degree at most i)

Theorem Every vertex of V –{u} not adjacent to u has degree at most d(v) (d(v) < n –d(u) = n – i). 13. There are n –1 – d(u) such vertices. 14. Since d(u) ≤ d(v), we can also add u itself to the set of vertices with degree at most d(v). 15. We thus obtain n- i vertices with degree less than n – i. (2) d n-i < n-i (at least n-i vertices have degree less than n–i)

Hamiltonian Path Hamiltonian path: A spanning path. Hamiltonian path

Remark A graph G has a spanning path if and only if the graph G  K 1 has a spanning cycle.

Theorem Let G be a simple graph with vertex degrees d 1 ≤ … ≤d n. If i < (n+1)/2 implies (d i ≥ i or d n+1–i ≥ n–i), then G has a spanning path. Proof: Let G’ = G  K 1, let n’ = n+1, and let d 1 ’,…,d n’ ’ be the degree sequence of G’. 2. Since a spanning cycle in G’ becomes a spanning path in G when the extra vertex is deleted, it suffices to show that G’ satisfies Chvatal’s condition. 3. We have to show if i < n’/2, at least one of the following conditions hold: (1) d i ’ > i, (2) d n’-i ’ ≥ n’–i.

Theorem Since the new vertex is adjacent to all of V(G), we have d n’ ’ = n and d j ’ =d j +1 for j < n’. 4. For i < n’/2 = (n+1)/2, we have (1) d i ’ – i = d i +1 – i ≥ i+1 – i > 0 or (2) d n’-i ’ – (n’–i) = d n+1-i +1 – (n’–i) ≥ n–i+1 – (n’– i) = 0.