Dirac theorem: For a simple network with n 3 vertices, if each vertex has degree at least n/2 (i.e., ki n/2, i), then the network has a Hamiltonian cycle,. Proof: Let G be the connected graph Let L = (i1,iL) = (i1, i2, . . . , iL) be the longest path. The neighbors of vertex i1 must lie on (i1, iL); the same is true for vertex iL, since if say i is not a neighbour, i i1 i2 ...iiL would be the longest path contradicting our selection of the longest path. The same for iL also. So both i1 and iL are adjacent to vertices in the longest path L only, Since degree of i1>k/2 and i1 is not adjacent to itself, k> n/2 +1 Let us assume a claim There is some value of j (1jk) such that: Incase claim is not true, Since all vertices adjacent to L1 or Lk lie on L, there must be at least deg(L1) vertices on P not adjacent to Lk. Since all the vertices adjacent to Lk and pk itself also lie on P, the path must have at least deg(L1)+deg(Lk)+1n+1 vertices. But L has only n vertices: a contradiction. This gives a cycle C=Lj+1Lj+2…LkLjLj1…L2L1Lj+1. Suppose LC is nonempty. Then since L is connected, there must be a vertex vGC adjacent to some Li. So the path from v to Li and then around C to the vertex adjacent to Li is longer than L, contradicting the definition of L. Hence our assumption/claim is true. Therefore all vertices in L are contained in C, making C a Hamilton cycle. Solution Dirac theorem: For a simple network with n 3 vertices, if each vertex has degree at least n/2 (i.e., ki n/2, i), then the network has a Hamiltonian cycle,. Proof: Let G be the connected graph Let L = (i1,iL) = (i1, i2, . . . , iL) be the longest path. The neighbors of vertex i1 must lie on (i1, iL); the same is true for vertex iL, since if say i is not a neighbour, i i1 i2 ...iiL would be the longest path contradicting our selection of the longest path. The same for iL also. So both i1 and iL are adjacent to vertices in the longest path L only, Since degree of i1>k/2 and i1 is not adjacent to itself, k> n/2 +1 Let us assume a claim There is some value of j (1jk) such that: Incase claim is not true, Since all vertices adjacent to L1 or Lk lie on L, there must be at least deg(L1) vertices on P not adjacent to Lk. Since all the vertices adjacent to Lk and pk itself also lie on P, the path must have at least deg(L1)+deg(Lk)+1n+1 vertices. But L has only n vertices: a contradiction. This gives a cycle C=Lj+1Lj+2…LkLjLj1…L2L1Lj+1. Suppose LC is nonempty. Then since L is connected, there must be a vertex vGC adjacent to some Li. So the path from v to Li and then around C to the vertex adjacent to Li is longer than L, contradicting the definition of L. Hence our assumption/claim is true. Therefore all vertices in L are contained in C, making C a Hamilton cycle..

1. Dirac theorem:
For a simple network with n 3 vertices, if each vertex has degree
at least n/2 (i.e., ki n/2, i), then the network has a Hamiltonian cycle,.
Proof: Let G be the connected graph
Let L = (i1,iL) = (i1, i2, . . . , iL) be the longest path.
The neighbors of vertex i1 must lie on
(i1, iL); the same is true for vertex iL, since if say i is not a neighbour, i i1 i2 ...iiL would be the
longest path contradicting our selection of the longest path.
The same for iL also.
So both i1 and iL are adjacent to vertices in the longest path L only,
Since degree of i1>k/2 and i1 is not adjacent to itself, k> n/2 +1
Let us assume a claim
There is some value of j (1jk) such that:
Incase claim is not true,
Since all vertices adjacent to L1 or Lk lie on L, there must be at least deg(L1) vertices on P not
adjacent to Lk.
Since all the vertices adjacent to Lk and pk itself also lie on P, the path must have at least
deg(L1)+deg(Lk)+1n+1 vertices.
But L has only n vertices: a contradiction.
This gives a cycle C=Lj+1Lj+2…LkLjLj1…L2L1Lj+1.
Suppose LC is nonempty.
Then since L is connected, there must be a vertex vGC adjacent to some Li.
So the path from v to Li and then around C to the vertex adjacent to Li is longer than L,
contradicting the definition of L.
Hence our assumption/claim is true.
Therefore all vertices in L are contained in C, making C a Hamilton cycle.
Solution
Dirac theorem:
For a simple network with n 3 vertices, if each vertex has degree
at least n/2 (i.e., ki n/2, i), then the network has a Hamiltonian cycle,.
Proof: Let G be the connected graph
Let L = (i1,iL) = (i1, i2, . . . , iL) be the longest path.

2. The neighbors of vertex i1 must lie on
(i1, iL); the same is true for vertex iL, since if say i is not a neighbour, i i1 i2 ...iiL would be the
longest path contradicting our selection of the longest path.
The same for iL also.
So both i1 and iL are adjacent to vertices in the longest path L only,
Since degree of i1>k/2 and i1 is not adjacent to itself, k> n/2 +1
Let us assume a claim
There is some value of j (1jk) such that:
Incase claim is not true,
Since all vertices adjacent to L1 or Lk lie on L, there must be at least deg(L1) vertices on P not
adjacent to Lk.
Since all the vertices adjacent to Lk and pk itself also lie on P, the path must have at least
deg(L1)+deg(Lk)+1n+1 vertices.
But L has only n vertices: a contradiction.
This gives a cycle C=Lj+1Lj+2…LkLjLj1…L2L1Lj+1.
Suppose LC is nonempty.
Then since L is connected, there must be a vertex vGC adjacent to some Li.
So the path from v to Li and then around C to the vertex adjacent to Li is longer than L,
contradicting the definition of L.
Hence our assumption/claim is true.
Therefore all vertices in L are contained in C, making C a Hamilton cycle.