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Dirac's And Ore's Theorem
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Dirac's Theorem
Ore's Theorem
Dirac's Theorem
Theorem 1 (Dirac's Theorem): If G=(V(G),E(G)) is connected graph on n-vertices so that
for every x,y∈V(G), where x≠y, and deg(x)+deg(y)≥n for all x,y∈V(G), then G is a
Hamiltonian graph.
Let's verify Dirac's theorem by testing to see if the following graph is Hamiltonian:
Clearly the graph is Hamiltonian. However, let's test all pairs of vertices:
deg(x)+deg(y)≥n True/False ?
deg(a)+deg(b)≥6, so 5+5≥6 True
deg(a)+deg(c)≥6, so 5+4≥6 True
deg(a)+deg(d)≥6, so 5+5≥6 True
deg(a)+deg(e)≥6, so 5+5≥6 True
deg(a)+deg(f)≥6, so 5+4≥6 True
deg(b)+deg(c)≥6, so 5+4≥6 True
deg(b)+deg(d)≥6, so 5+5≥6 True

2. deg(b)+deg(e)≥6, so 5+5≥6 True
deg(b)+deg(f)≥6, so 5+4≥6 True
deg(c)+deg(d)≥6, so 4+5≥6 True
deg(c)+deg(e)≥6, so 4+5≥6 True
deg(c)+deg(f)≥6, so 4+4≥6 True
deg(d)+deg(e)≥6, so 5+5≥6 True
deg(d)+deg(f)≥6, so 5+4≥6 True
So by Dirac's theorem, this graph must be Hamiltonian. We will now look at Ore's theorem.
Ore's Theorem
Ore's theorem is a vast improvement to Dirac's theorem:
Theorem 2 (Ore's Theorem): If G=(V(G),E(G)) is connected graph on n-vertices where
n≥3] so that for [[x,y∈V(G), where x≠y, and deg(x)+deg(y)≥n for each pair of non-
adjacent vertices x and y then G is a Hamiltonian graph.
Recall that two vertices are said to be adjacent if they are connected by an edge. Two vertices x
and y are said to be adjacent if {x,y}∈E(G). Thus, non-adjacent vertices x and y are vertices
such that {x,y}∉E(G).
Let's see if this theorem is accurate by testing the graph from earlier, let's call it G. This graph is
on 6 vertices and is clearly Hamiltonian. Let's verify this with Ore's theorem. The only pair of
non-adjacent vertices are c and f, since {c,f}∉E(G). For G to be Hamiltonian, it must follow
that:
(1)
deg(c)+deg(f)≥64+4≥68≥6
Thus according to Ore's theorem, the graph G is Hamiltonian.
Note that these theorems provide a condition for a graph to always be Hamiltonian. If a graph G
does not pass this test, then it doesn't imply that G is not Hamiltonian. However, if the graph G
does pass this test, then it is definitely Hamiltonian.
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