Definition 7.2.1• Hamiltonian graph: A graph with a spanning cycle

(also called a Hamiltonian cycle).

Hamiltonian graph

Hamiltonian cycle

Theorem 7.2.8• 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 K2 is not Hamiltonian but satisfies δ(K2) ≥ n(K2)/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 7.2.8

4. When uv in G, G has spanning path u=v1,…,vn=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, say vi+1, directly follows a neighbor of v, say vi, on the path.

Theorem 7.2.8

6. We show that there is a common index in the sets

S={i: uvi+1} and T={i: vvi}.

7. |ST| + |ST| = |S| + |T| = d(u) + d(v) ≥ n .

8. Neither S nor T contains the index n.

9. Thus |ST| < n, and hence |ST| ≥ 1.

Lemma 7.2.9 • 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 7.2.8.

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 7.2.12• The closure of G is well-defined.

Proof: 1. Let e1,…,er and f1,…,fs be sequences of edges added in forming C(G), the first yielding G1 and the second G2.

2. f1, being initially addable to G, must belong to G1.

3. If f1, …, fi–1 E(G1), then fi becomes addable to G1 and therefore belongs to G1. Hence, G1 G2.

4. Similarly, G2 G1.

Theorem 7.2.11• A simple n-vertex graph is Hamiltonian if an only if

its closure is Hamiltonian.

Theorem 7.2.13• Let G be a simple graph with vertex degree d1 ≤ … ≤

dn, where n ≥ 3. If i < n/2 implies that di > i or dn-i ≥ n-i (Chvatal’s condition), then G is Hamiltonian.

Proof: 1. G is Hamiltonian if and only if C(G) is Hamiltonian. (Theorem 7.2.11)

2. It suffices to show C(G) = G’ = Kn.3. G’ satisfies Chavata’s condition because adding edges to form the closure reduces no

entry in the degree sequence.

Theorem 7.2.134. We prove the contrapositive: if G’ is not a complete

graph, there exists i < n/2 such that

(1) di i (at least i vertices have degree at most i), and

(2) dn-i < n-i (at least n-i vertices have degree less than n–i).

5. With G’ ≠ Kn, let nonadjacent vertices u and v have maximum degree sum.

6. Because G’=C(G), uv implies that d(u) + d(v) < n.

7. Let d(u) ≤ d(v). Then d(u) < n/2.

Theorem 7.2.138. 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) di i (at least i vertices have degree at most i)

Theorem 7.2.1312. 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) dn-i < n-i (at least n-i vertices have degree less than n–i)

Hamiltonian Path

• Hamiltonian path: A spanning path.

Hamiltonian path

Remark 7.2.16• A graph G has a spanning path if and only if the

graph G K1 has a spanning cycle.

Theorem 7.2.17

• Let G be a simple graph with vertex degrees d1 ≤ … ≤dn. If i < (n+1)/2 implies (di ≥ i or dn+1–i ≥ n–i), then G has a spanning path.

Proof: Let G’ = G K1, let n’ = n+1, and let d1’,…,dn’’ 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) di’ > i, (2) dn’-i’ ≥ n’–i.

Theorem 7.2.17

3. Since the new vertex is adjacent to all of V(G), we have dn’’ = n and dj’ =dj +1 for j < n’.

4. For i < n’/2 = (n+1)/2, we have

(1) di’ – i = di+1 – i ≥ i+1 – i > 0 or

(2) dn’-i’ – (n’–i) = dn+1-i+1 – (n’–i) ≥ n–i+1 – (n’– i) = 0.