discrete-mathematics-assignment-8

This graph is directed. Because the adjacency matrix is not symmetric

The two graphs are isomorphic.

• Base Case (n = 1):
  For a cube of dimension n = 1, we have 2 vertices connected by an edge. We can easily divide these two vertices into two separate sets, forming a bipartite graph.
  For a cube of dimension n = 2, we have 4 vertices connected by an edge. We can easily divide these two vertices into two separate sets like forming a bipartite graph.
• Inductive Hypothesis:
  Assume that a cube of dimension k is bipartite, where k ≥ 1.
• Inductive Step:
  Now we need to show that a cube is also bipartite. Consider a cube , which is obtained by adding one layer of vertices to the cube . Let's call this additional layer "L". Each vertex in layer L is connected to exactly one vertex in the cube . Now, we can divide the vertices of into two sets, S₁ and S₂.
  Next, we assign each vertex in layer L to the opposite set of its connected vertex in . In other words, if a vertex in L is connected to a vertex in S₁, we assign it to S₂, and vice versa. This assignment ensures that no two vertices within the same set (S₁ or S₂) are adjacent, including the newly added layer L.
  Therefore, we have successfully divided the vertices of into two sets, S₁ and S₂, such that no two vertices within the same set are adjacent. Hence, by induction, we can conclude that a cube of dimension n is bipartite for n ≥ 1.
• This completes the proof.

a)

It's clear that the graph is planar. it can be drawn in the plane without any edges crossing
b) It has Euler Circuit cause the thorem A connected multi-graph has an Euler circuit if and
only if each vertex has even degree. The vertex in the graph all has 4 edges. Therefore, it must have Euler Circuit.
c) It has Hamilton Circuits such that 1->2->3 ->4 ->5 ->6->1

Consider the following graph G = (V, E), where V = {1, 2, 3, 4} and E = {(1, 2), (2, 3), (3, 4), (4, 1), (1, 3)}.
This graph satisfies the condition that for any two non-adjacent vertices u and v in G,
For example, if we take u = 1 and v = 4, then
However, this graph does not have a Hamilton circuit.
Therefore, we have shown that the converse of Ore's theorem is false by providing a counterexample.