Is it possible to construct a graph with 12 vertices such that 2 of the vertices have degree 3 and the remaining vertices have degree 4?

Let the number of edges of Kn be M. The degree of each vertex is (n − 1). There are n vertices. Thus the sum of the degrees of the n vetices is n(n − 1), which is 2M.

Is it possible for a graph to have 12 edges and at the same time every vertex has degree 3?

Solution. (a) No. Euler’s theorem says that a graph has an Euler circuit if and only if every node has even degree, which is not the case here.

Is it possible to have a graph with exactly one node of odd degree and all the rest having even degree?

Solution: This is not possible by the handshaking theorem, because the sum of the degrees of the vertices 3 ⋅ 5 = 15 is odd. Theorem: An undirected graph has an even number of vertices of odd degree. This sum must be even because 2m is even and the sum of the degrees of the vertices of even degrees is also even.

How many edges are there in a graph with 12 vertices of degree six?

Solution: the sum of the degrees of the vertices is 6 ⋅ 10 = 60. The handshaking theorem says 2m = 60. So the number of edges is m = 30.

Is it possible to construct a graph with 12 vertices?

If it’s a finite graph, then yes. the only (finite) trees with no vertices of degree ≥ 3 are the paths, and they have at most two terminal points. So a graph meeting your conditions must have at least one vertex of degree ≥ 3.

Can a simple graph have five vertices and twelve edges?

{3 marks} Can a simple graph have 5 vertices and 12 edges? If so, draw it; if not, explain why it is not possible to have such a graph. ANSWER: In a simple graph, no pair of vertices can have more than one edge between them.

How many edges are there in a graph with 12 vertices each of degree four?

Start with a cycle of size 12, then for every vertex i in the cycle connect it to i+2 and i-2. This graph has 24 edges and each vertex has degree 4.

Can a graph exist with 15 vertices each of degree five?

Therefore by Handshaking Theorem a simple graph with 15 vertices each of degree five cannot exist.

Can it be possible for a graph to have an odd number of odd vertices?

It can be proven that it is impossible for a graph to have an odd number of odd vertices. The Handshaking Lemma says that: In any graph, the sum of all the vertex degrees is equal to twice the number of edges.

How many vertices are necessary to construct a graph with exactly 12 edges in which each vertex is of degree three?

Describe an unidrected graph that has 12 edges and at least 6 vertices. 6 of the vertices have to have degree exactly 3, all other vertices have to have degree less than 2.

Can a simple graph have 5 vertices and 12 edges if so draw it if not explain why it is not possible to have such a graph?

If so, draw it; if not, explain why it is not possible to have such a graph. ANSWER: The maximum number of edges in the complete graph containing 5 vertices is given by K5: which is C(5, 2) edges = “5 choose 2” edges = 10 edges. Since 12 > 10, it is not possible to have a simple graph with more than 10 edges.

What is the difference between nodes and arcs in a graph?

The nodes are represented by numbered circles and the arcs by arrows. The arcs are assumed to be directed so that, for instance, material can be sent from node 1 to node 2, but not from node 2 to node 1. Generic arcs will be denoted by i– j, so that 4–5 means the arc from node 4 to node 5.

How do you find the degree of an undirected graph?

Approach: For an undirected graph, the degree of a node is the number of edges incident to it, so the degree of each node can be calculated by counting its frequency in the list of edges.

How do you prove a graph is not a simple graph?

Basically, it goes like this (using the degree sequence [3 2 2 1] as an example): If any degree is greater than or equal to the number of nodes, it is not a simple graph. Handshaking lemma: if the number of vertices with odd degrees is odd, it is not a simple graph.

Is there such a graph with a vertex with degree 4?

The answer for c is that there cannot be such a graph – since there are 2 vertices with degree 4, they must be connected to all other vertices. Therefore, the vertex with degree one, is an impossibility. (a) 3,3,3,3,2 – YES! Graph Justifies claim (d)2,2,2,1,1 – YES! Graph Justifies Claim