Ans: Eulerian Circuits in K5 Graph: Does it Exist?

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In summary, the complete graph with 5 vertices, denoted by K5, does not contain Eulerian circuits because not all vertices have even degree. The degree of each vertex in K5 is 4. K1, K2, K3, K4, and K5 all have degrees of 0, 1, 2, 3, and 4, respectively.
  • #1
Joystar77
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Consider the complete graph with 5 vertices, denoted by K5.

D.) Does K5 contain Eulerian circuits? (why?) If yes, draw them.

I know that Eulerian circuits are a circuit that uses every edge of a graph exactly once. These type of circuits starts and ends at the same vertex. If I find that the K5 has Eulerian circuits, then would I draw this on the graph in an oblong square shape? If I find that it doesn't wouldn't it have to do with the isomorphic representation because of the edges being counted twice?
 
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  • #2
As for whether $K_{5}$ has an Euler circuit, just look at whether it's connected, with all vertices of even degree. Then, if you want to draw the circuit, you can label the edges and then notate the path.
 
  • #3
Consider the complete graph with 5 vertices, denoted by K5.

Does K5 contain Eulerian circuits? (Why?) If yes, draw them.

Is it correct to say that K5 doesn't contain Eulerian circuits because all the vertices are not of even degree?
 
  • #4
Joystar1977 said:
Is it correct to say that K5 doesn't contain Eulerian circuits because all the vertices are not of even degree?

What is the degree of each vertex?
 
  • #5
I am not sure about what the degree of each vertex really is. I do know that Kn has N vertices. Each vertex has degree N-1. The sum of all degrees is N (N - 1). I do know that there is a complete graph with 5 vertices denoted by K5.

Ackbach said:
What is the degree of each vertex?
 
  • #6
Joystar1977 said:
I am not sure about what the degree of each vertex really is. I do know that Kn has N vertices. Each vertex has degree N-1. The sum of all degrees is N (N - 1). I do know that there is a complete graph with 5 vertices denoted by K5.

Perfect; I think you're sitting right on top of your answer. So $K_{5}$ has $5$ vertices. Based on what you just wrote, what is the degree of each vertex in $K_{5}$?
 
  • #7
Ackbach:

Am I suppose to add all the vertices in order to get the answer for what the degree of each vertex is in K5?

1 + 2 + 3 + 4 + 5 = 15

Is this correct?
 
  • #8
No, you don't want to add up the vertex numbers. That would have no meaning. What is your goal here? Stay focused on that goal.
 
  • #9
Ackbach: I don't quite understand and was trying to guess at what I am suppose to do. This textbook really sucks that I have for this course doesn't give any examples and no dictionary in the back of the book.

Consider the complete graph with 5 vertices, denoted by K5.

Does K5 contain Eulerian circuits? (Why?) If yes, draw them.

I am not sure about what the degree of each vertex really is. I do know that Kn has N vertices. Each vertex has degree N-1. The sum of all degrees is N (N - 1). I do know that there is a complete graph with 5 vertices denoted by K5.

I am completely lost and don't understand.
 
  • #10
Substitute the value for $n$ that you have. You know that $K_{n}$ has $n$ vertices, each of degree $n-1$. So $K_{5}$ has how many vertices? Each of them has what degree? (Just plug in what $n$ is for this case!) And what does the degree of each vertex tell you about whether $K_{5}$ has an Euler circuit?
 
  • #11
Please be patient while I try to understand this material. First, you say to substitute the value for n that you have. I have no idea what the value of n is and all I see is a numerical value for the vertices which is 5. I thought that in order to get the degree your suppose to add (v1, v2), v3, v4, v5) and then divide by 2. What do you mean by asking, "Each of them has what degree?" I do not understand you or what your asking me. How am I suppose to plug in something for n when there is no value for N? That doesn't make one bit of sense to me.
 
  • #12
I think there might have been some confusion because there was an uppercase $N$ and a lowercase $n$. Let's just stick to lowercase $n$.

Definition: $K_{n}$ is the complete graph with $n$ vertices. Complete means that every pair of vertices is connected by an edge. You can see the first seven complete graphs here.

Now $K_{5}$ is the complete graph with five vertices. So if we're comparing $K_{n}$ with $K_{5}$, we would say that $n=5$ for this particular complete graph.

Now the degree of a vertex in a graph is the number of edges connected to the vertex. If you have a pictorial representation of a graph, then for any vertex, all you have to do to find the degree of each vertex is count the number of edges going into it.

You have also said that the degree of every vertex in $K_{n}$ is $n-1$. So what would the degree of every vertex in $K_{1}$ be? How about $K_{2}$? How about $K_{3}$? $K_{4}$? $K_{5}$?

Finally, an indirected graph has an Euler circuit if and only if every vertex has even degree, and the graph is connected. Are both of these conditions satisfied for $K_{5}$? If so, could you construct an Euler circuit for $K_{5}$?
 
  • #13
I am not trying to confuse myself even more with the uppercase N or the lowercase n. I will stick with the lowercase n. K5 is a complete graph that has 5 vertices. Why are you comparing Kn with K5? What do you mean by count the number of edges going into it? Are you suppose to count the number of edges going into K5 or Kn? What do you mean by asking me the degree of K1, K2, K3, K4, and K5?

I am not sure about what the degree of each vertex really is. I do know that Kn has N vertices. Each vertex has degree n-1. The sum of all degrees is n(n - 1). I do know that there is a complete graph with 5 vertices denoted by K5.

I am totally and completely lost. Just so you know I do have a learning disability which is Epilepsy Seizures. This causes me to have a hard time comprehending and understanding things. Please explain this in a different way because I am not understanding you. I don't know if the conditions are satisfied for K5. I am not sure whether or not I could construct a Eulerian circuit for K5.
 
  • #14
Joystar1977 said:
I am not trying to confuse myself even more with the uppercase N or the lowercase n. I will stick with the lowercase n. K5 is a complete graph that has 5 vertices. Why are you comparing Kn with K5? What do you mean by count the number of edges going into it? Are you suppose to count the number of edges going into K5 or Kn? What do you mean by asking me the degree of K1, K2, K3, K4, and K5?
I am not sure about what the degree of each vertex really is. I do know that Kn has N vertices. Each vertex has degree n-1. The sum of all degrees is n(n - 1). I do know that there is a complete graph with 5 vertices denoted by K5.

Please study this webpage..

In the graph \(\displaystyle K_n\) there are \(\displaystyle n\) vertices; there are \(\displaystyle \frac{n(n-1}{2}\) edges; and each vertex has degree \(\displaystyle n-1\).

Therefore, if \(\displaystyle n\) is odd then \(\displaystyle K_n\) has an Euler Circuit.

Thus \(\displaystyle K_5\) does and \(\displaystyle K_8\) does not.
 

FAQ: Ans: Eulerian Circuits in K5 Graph: Does it Exist?

What is an Eulerian circuit?

An Eulerian circuit is a path in a graph that visits every vertex exactly once and ends at the starting vertex. It is named after the Swiss mathematician Leonhard Euler.

What is a K5 graph?

A K5 graph, also known as a complete graph with five vertices, is a graph in which every vertex is connected to every other vertex by an edge. It is the simplest type of complete graph.

Can an Eulerian circuit exist in a K5 graph?

No, an Eulerian circuit cannot exist in a K5 graph because it is not possible to visit every vertex exactly once and return to the starting vertex while only traveling along edges. This is because there are 10 edges in a K5 graph, and each vertex must have an even degree in order for an Eulerian circuit to exist.

Can a K5 graph have an Eulerian path?

Yes, a K5 graph can have an Eulerian path. An Eulerian path is a path that visits every edge exactly once, but does not necessarily start and end at the same vertex. In a K5 graph, a vertex can have an odd degree and still allow for an Eulerian path to exist.

What are some real-world applications of Eulerian circuits in K5 graphs?

Eulerian circuits in K5 graphs can be used in various fields such as transportation planning, network routing, and computer science. For example, in transportation planning, an Eulerian circuit can help determine the most efficient route for a delivery truck to visit all locations and return to the starting point without repeating any routes. In computer science, Eulerian circuits can be used in network routing algorithms to ensure all nodes are visited and connected without any dead ends.

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