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Which of the following graphs has an Eulerian circuit?

  1. Any $k$-regular graph where $k$ is an even number.
  2. A complete graph on $90$ vertices.
  3. The complement of a cycle on $25$ vertices.
  4. None of the above
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5 Answers

Best answer
143 votes
143 votes

A connected Graph has Euler Circuit $\iff$ all of its vertices have even degree
A connected Graph has Euler Path $\iff$ exactly 2 of its vertices have odd degree

A. $k$-regular graph where $k$ is even number. 
     a $k$-regular graph need not be connected always. Example : The given below graph is a $2$ regular graph is not a Euler graph. This is so because there is no single walk  which covers all edges.

B. The complete graph of $90$ vertices
     In such a graph every vertex will have an odd degree = 89, Hence it cannot have a Euler path/Circuit.

C. To get degree of all vertices of the complement of cycle on $25$ vertices we need to subtract the degree of a complete graph of 25 vertices with degree of vertices in the original given graph i.e. cycle on $25$ vertices.

Degree of complement $= 24 - 2 = 22$. Since, every degree is Even, and it is connected also, therefore Graph Might be a Euler Cycle but we need to check graph is connected or not.

Now Question comes it is connected or not :

It is connected because, there is a theorem which says, "Let $G$ be a graph with $n$ vertices and if every vertex has a degree of at least $\frac{n-1}{2}$ then $G$ is connected." [check this] (Alternate Proof: We can also prove this theorem using contradiction. Suppose we have two disconnected components, then the least component (component with minimum vertices) has to have $ \leq \frac{n}{2}$ vertices. Now we can try to maximize the degree of any vertices in the least component, which can be at most $\frac{n}{2}-1 (=\frac{n-2}{2})$, but we already know that the minimum degree is $\frac{n-1}{2}$, hence contradiction. )

Here degree of each vertex is $22$, which is, of course, greater than $\frac{25-1}{2}=12 $.

Hence, Graph must be Euler graph.

Option C

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14 votes
14 votes

Euler Graph-For a graph to be Euler it must have Euler Line.

Euler Line-It is a closed walk(yes closed walk and not closed path because as per definition of a path you are not allowed to re-visit a vertex) which covers all the edges of graph G.

Now since a Euler graph always has a Euler line, we have a closed walk(walk in which initial and terminal vertices are same), and all the vertices in this closed walk are of degree 2. So, needless to say, every vertex in a Euler graph must have Even degree. For one edge enters and one edge leaves this walk.

So, all in all for a graph to be Euler it must certainly have 2 conditions

(1)It must be connected. (Otherwise, you cannot form Euler Line)

(2)All vertices must have even degree.

Okay now let's check options one by one

(A)Any k-regular graph where k is an even-number:

Okay, we have one condition satisfied for Graph to be Euler but here it is not told whether the graph is connected or not?. So with surety, we cannot comment about this graph being Euler.So, this option is rejected.

(B)A complete graph on 90 vertices: This graph would have the degree of every vertex = 90-1=89 which is odd and yes the graph is connected but we don't have even degree for all vertices. So this option rejected.

(C)The complement of a cycle on 25 vertices.

The Complement of a graph G is defined as a graph G` with |V(G)|=|v(G`)|

and |E(G`)| = $\frac{n(n-1)}{2}$ - |E(G)|.

So this graph G` here would have

$\frac{25(24)}{2}-25$ edges

which evaluates to 275 edges.

Now, the degree of all vertices of this graph would be 24-2=22.

We need to check whether this graph is connected or not.

For a complete graph on n vertices, we have $\frac{(n-1)!}{2}$ distinct cycles covering all vertices.

(Ref: https://math.stackexchange.com/questions/249817/how-many-hamiltonian-cycles-are-there-in-a-complete-graph-k-n-n-geq-3-why )

So for K25 we must be $\frac{24!}{2}$having  cycles from which we are removing only 1 so complement of a cycle on 25 vertices need to be connected.

Hence, the complement of a cycle on 25 vertices must be Eulerian.

Answer-(C)

12 votes
12 votes

Option is C

A) does not confirm that regular graph should be connected or not.

B)Degree of the each vertices will be 89 since it is complete graph.

C)The complement of the cycle graph with 25 graph where the degree of each vertex will be 24(even) so according to the theorem for the existence of Euler's circuit all the vertices should be of even degree  

6 votes
6 votes

In this answer, I want to present another proof for Option C being Euler Graph.

If a vertex $u$ has degree $d$ in graph $G$ then in $\overline{G},$ $degree(u) = (n-1) – d.$

Since, every vertex in $C_n$ has degree $2$, so, degree of every vertex in $\overline{C_n}$ in $n-3.$

So, degree of every vertex in $\overline{C_{25}}$ in $22.$

Theorem :

If $C_n$ is cycle graph on $n$ vertices then $\overline{C_n}$ is connected if $n \geq 5.$

Proof :

Take any two vertices $a,b$ in $\overline{C_n}.$ We show that they will be connected by a path.

Case 1 : $a,b$ are Not adjacent in $C_n :$

In this case, $a,b$ will be adjacent in $\overline{C_n}.$ So, they are connected by a path (of length one)

Case 2 : $a,b$ are adjacent in $C_n :$

Since $n \geq 5$ and every vertex in $C_n$ has degree 2, also $a,b$ are adjacent, so, there is at least one vertex, say  $x$ which is Not adjacent to $a,b$, So, in $\overline{C_n}$, $ax,bx$ will be edges, so, we have a path between $a,b.$

Hence proved. 


Since, we can manually check that $\overline{C_3}$, $\overline{C_4}$ are disconencted, so, we can say the following theorem :

Theorem :

If $C_n$ is cycle graph on $n$ vertices then $\overline{C_n}$ is connected if and only if $n \geq 5.$


So, $\overline{C_{25}}$ is connected and has even degree for every vertex, so, it is Euler graph.

We can generalize it as follows :

Complement of a cycle on $n$ vertices has Euler Circuit if and only if $n$ is odd and $n\geq 5.$

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