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Thm 4.6:  Let $G$ be a graph with $n$ vertices.  Suppose also that $\exists$ non-adjacent  vertices $u$ and $w$  such that $\delta(u) + \delta(w) \geq n$.

Then $G$ is Hamiltonian $\Longleftrightarrow$ $G + <u, w>$ is Hamiltonian.

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Proof  ($\Rightarrow$)  A Hamiltonian cycle in $G$ is also a Hamiltonian cycle in $G  + <u, w>$.

($\Leftarrow$)  Suppose $G  + <u, w>$ is Hamiltonian.

Proof by contradiction:  Suppose $G$ is not Hamiltonian.
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Thus the Hamiltonian  cycle in $G  + <u, w>$ must contain the edge $<u, w>$ (else this cycle would be a Hamiltonian cycle in $G$).
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Thus we can write the Hamiltonian cycle as

\centerline{$u, w, v_1, v_2, \dots, v_{n-2}, u$}

Suppose $<u, v_{i-1}>$ and $<w, v_i>$ are both edges in $G$.  Then
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\centerline{~~~~~$w, v_i, \dots, v_{n-2}, u, v_{i-1}, v_{i-2}, \dots, v_1, w$} 
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is a Hamiltonian cycle in $G$, a contradiction.

\eject

Thus $<u, v_{i-1}> \in E(G)$ implies $<w, v_i> \not\in E(G)$.

Suppose in $G$, $\delta(u) = k$, then $\delta(w) \leq $
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Thus in $G$, $\delta(u) + \delta(w) \leq $
\vfill
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Defn 4.3:  If $V(G) = n$, the {\bf closure} of $G$ is the graph obtained from $G$ by iteratively adding edges to $G$ joining non-adjacent  vertices $u$ and $w$ where $\delta(u) + \delta(w) \geq n$.

To create the closure of $G$, create the sequence

$G_0 = G, ~G_1 = G_0 + <u_0, w_0>$, ..., 

\rightline{$ G_k = G_{k-1} + <u_{k-1}, w_{k-1}>$}

where $<u_i, w_i> \not\in G_i$ and  
$\delta(u_i) + \delta(w_i) \geq n$ in $G_i$.

Moreover if $\delta(u) + \delta(w) \geq n $ in $G_k$,

\rightline{ then $<u, w> \in  G_k$.}

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Thm 4.7: A simple graph is Hamiltonian if and only if its closure is Hamiltonian.
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\eject 
Thm (Ore 1960). \hfil \break
If $G$ is a simple graph with $|V(G)| = n\geq  3$
 and \hfil \break
 if  $\forall$ non-adjacent  vertices $u$ and $w$,  $\delta(u) + \delta(w) \geq n$.

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\centerline{Then $G$ is Hamiltonian. }


Proof: The closure of $G$ is    ~~~~~~  and ~~~~~~~ is Hamiltonian.

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Corollary 4.5:  If $G$ is a simple graph with \break $|V(G)| = n\geq  3$ and each vertex $v$ has degree \break
$\delta(v) \geq {n \over 2}$, then $G$ is Hamiltonian. 











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