Can there be a simple graph that has 7 vertices all of different degrees Why or why not?
Since there are n vertices, if they all have different degrees, they must be 0,1,2,…,(n-1). But then we have that the vertex of degree (n-1) must have an edge to all other vertices, and the vertex of degree 0 has no edges. This is a contradiction so no such graph can exist.
Does there exist a simple Eulerian graph on even number of vertices and odd number of edges?
You can probably run wild with this idea but it’s clear that the answer to the question is yes, a connected Eulerian graph can have an even number of vertices and an odd number of edges.
Does every graph with a Euler circuit have an even number of vertices?
If a graph G has an Euler circuit, then all of its vertices must be even vertices. If the number of odd vertices in G is anything other than 0, then G cannot have an Euler circuit.
Can a 3-regular graph have 6 vertices?
All the six vertices have constant degree equal to 3. The edges of the graph are indexed from 1 to nd 2 = 6×3 2 = 9.
How many 3 graphs does 6 vertices have?
Two 3-regular graphs with 6 vertices.
How many edges are there in a graph with 10 vertices each of degree 6?
30 edges
Example: How many edges are there in a graph with 10 vertices, each of degree 6? Solution: The sum of the degrees of the vertices is 610 = 60. According to the Handshaking Theorem, it follows that 2e = 60, so there are 30 edges.
What is the maximum number of edges in a simple graph with 7 vertices?
The maximum number of edges possible in a single graph with ‘n’ vertices is nC2 where nC2 = n(n – 1)/2. The number of simple graphs possible with ‘n’ vertices = 2nc2 = 2n(n-1)/2.
What is the maximum number of edges present in a simple graph with 7 vertices?
Discussion Forum
| Que. | What is the maximum number of edges present in a simple directed graph with 7 vertices if there exists no cycles in the graph? |
|---|---|
| b. | 7 |
| c. | 6 |
| d. | 49 |
| Answer:6 |
Which graph has a Euler circuit?
A graph has an Euler circuit if and only if the degree of every vertex is even. A graph has an Euler path if and only if there are at most two vertices with odd degree.
Is vertices 3 and 4 Eulerian graph?
Because as can be seen vertices, 3 and 4 have degree of 3. So, any idea what that one is actually Eulerian graph? If no, can someone tell me whether an Eulerian graph can be found for the odd vertices, even edges case?
Can a simple connected graph of 5 vertices with 7 edges?
The 6th edge connects two vertices, A and B. The 7th edge critter connects one of these, say A, to a third vertex, C, or connects two other vertices D and E. If A to C then B and C have degree 3. If D to E then A, B, C and D all have degree 3. Thus a simple connected graph of 5 vertices with 7 edges cannot be Eulerian.
What is the complement of a graph with 3 disjoint edges?
The complement of your graph should have 6 vertices, only 3 edges, and there cannot be an independent set of size 4 (i.e., four vertices with no edges between them). Clearly, each graph with three disjoint edges has the desired property – regardless of how you pick the four vertices, you have to pick both endpoints of (at least) one of the edges.
Do all vertices of an Euler cycle have degree 2?
In order to have an Euler cycle, all vertices have to have even degree. Since it’s connected, all vertices must have positive degree. So each vertex has to have degree at least 2. However, if all vertices have degree I though the answer was no. But I’m wrong. Consider a square and a triangle which share a vertex.