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Definition 7.2.1

Definition 7.2.1. Hamiltonian graph: A graph with a spanning cycle (also called a Hamiltonian cycle). Hamiltonian cycle. Hamiltonian graph. 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.

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Definition 7.2.1

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  1. Definition 7.2.1 • Hamiltonian graph: A graph with a spanning cycle (also called a Hamiltonian cycle). Hamiltonian cycle Hamiltonian graph

  2. 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.

  3. Theorem 7.2.8 • 4. When uv 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.

  4. Theorem 7.2.8 6. We show that there is a common index in the sets S={i: uvi+1} and T={i: vvi}. 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.

  5. 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.

  6. 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.

  7. 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.

  8. Theorem 7.2.11 • A simple n-vertex graph is Hamiltonian if an only if its closure is Hamiltonian.

  9. 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.

  10. Theorem 7.2.13 4. 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), uv implies that d(u) + d(v) < n. 7. Let d(u) ≤ d(v). Then d(u) < n/2.

  11. Theorem 7.2.13 8. 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)

  12. Theorem 7.2.13 12. 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)

  13. Hamiltonian Path • Hamiltonian path: A spanning path. Hamiltonian path

  14. Remark 7.2.16 • A graph G has a spanning path if and only if the graph G  K1 has a spanning cycle.

  15. 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.

  16. 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.

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