The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+33 votes
4.1k views

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
asked in Graph Theory by Veteran (59.7k points)
edited by | 4.1k views
+3
Directly one can solve it.
If a vertex has degree k, its complement has n-k-1 degree. (n vertices) (result is given in Korth book exercise)
In the cyclic graph, all vertex has degree 2. C20 vertex has complement has degree 25-2-1= 22.
All degree even hence we must have euler circuit.

5 Answers

+73 votes
Best answer

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, "$G$ 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: U can also prove this theorem using contradiction. Suppose u have two disconnected components, then least component (component with minimum vertices) has to have $ \leq \frac{n}{2}$ vertices. Now 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 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

answered by Loyal (6.1k points)
edited by
0
got it..:) Thank you ....!
+1

This might help ...

+2

This one also ....

0
Why always complicated answers are chosen as the best answer??
0
Nice explanation!
+1
@vijay for n=5 it the formula gives 6 edges but graph with 4 edges is also connected to 5 vertex. It is a one way statement
0
nice explanation @ mithlesh
0
Can we use Dirac's theorem to prove that G' is connected?

According to Dirac's thm, "If G is simple graph, with n vertices such that degree of each vertex is atleast >=n/2, than graph is hamiltonian"

.So if G ' is hamiltonian,then definitely connected.
0
Such a beautiful answer!
0

https://math.stackexchange.com/questions/130425/hamiltonian-path-detection good read to check if graph is connected or not.

+8 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  

answered by Boss (11.5k points)
+1
can u provide mathematical explnation for option C .

[E] + [E (IN COMPLEMENT)]= N(N-1)/2  ,i know further not getting.
+2 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)

answered by Boss (18.8k points)
0

why are we considering cycle on n vertices with n edges only, i mean we can take more than n edges also right,

for examle if i have 25 vertices in graph, then cycle on 25 vertices can contain 26 edges also right, since it will remain a cycle!!

consider this cycle on 5 vertices

this is also a cycle , now when i take its complement,  degree of 3 vertices will be even right, but degree of 2 vertex will be odd (since max degree in a graph with 5 vertices can be 4), which violates the property of euler graph that all vertices should have the even degree(which is a necessary and sufficient condition);

therefore c also becomes false... isn't it?

0

@Gate fever-Complement of "A cycle on 25 vertices

I think here they clearly mean a simple cycle on 25 vertices and not more than that.

Take $K_4$(Complete graph on 4 vertices)-Can you say it "A cycle on 4 vertices"? NO.It has cycles of length 4 and cycles of length 3 also, but it a graph which contains such cycles and not a cycle on 4 vertices.

Here  a CYCLE on 25 Vertices implies they have given you the definition of what type the graph is and not what the graph contains.

Hope you got it.

0
yes , thanks
0 votes
C option
answered by (131 points)
–7 votes
Option A) Any k-regular graph where k is even number
answered by Junior (581 points)
+6
https://en.wikipedia.org/wiki/Regular_graph

Here is example of 2 Regular Graph which do not have euler circuit.
Answer:

Related questions

+2 votes
1 answer
2
asked Dec 28, 2016 in Graph Theory by thor Loyal (8k points) | 410 views
+6 votes
2 answers
3
asked Oct 10, 2016 in Graph Theory by Rahul Jain25 Boss (11.6k points) | 324 views
+1 vote
1 answer
7
asked Oct 10, 2016 in Graph Theory by Rahul Jain25 Boss (11.6k points) | 79 views


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

44,253 questions
49,749 answers
164,072 comments
65,845 users