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40 votes
40 votes
A cycle on $n$ vertices is isomorphic to its complement. The value of $n$ is _____.
in Graph Theory retagged by
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4 Comments

Why not 4.

Take l_l and its complement.

I can not type  its complement here, but try to form it.

Both are isomorphic.
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A cycle with n vertices has n edges.
For isomorphism, both graphs should have an equal number of edges.
If G is a simple graph with n vertices than #edges in G + #edges in G' = #edges in complete Graph.
i.e n + n = n(n-1)/2.

If we put 4 edges in this equation it will not satisfy the condition hence it is false, whereas 5 edges satisfy the condition.
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Bro it's already mentioned in question graph is cycle but the example with u are trying to counter is not a cycle.
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6 Answers

115 votes
115 votes
Best answer
A cycle with $n$ vertices has $n$ edges.
Number of edges in cycle $=n$
Number of edges in its complement $=\frac{n(n-1)}{2} - n.$

Fo isomorphism, both graphs should have equal number of edges.
$\begin{align}\text{This gives: }  \\ &\frac{n(n-1)}{2} - n = n. \\ &\Rightarrow n = 5.\end{align}$
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4 Comments

Sorry I took $\binom{n}{2}$ as degree.

It can be solved like degree of vertex of complement: (degree of complete graph with n vertices) -2(cycle) = n-1-2= n-3

Now as it is isomorphic and contains cycle n-3=2 hence n=5. 🙂
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So, points to be noted.

1. Cycles are defined only for graphs having $n\geq 3$, where $n$ is the no. of vertices.  Why :(

2. Also, the question doesn't explicitly mention if the graph is s1mple and undirected.  So, it's a valid assumption unless otherwise mentioned ?!!

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good explanation
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32 votes
32 votes

It is $n=5$ only.

Only $C_{5}$ is isomorphic to its complement.

Here $C_4$ and its complement are not isomorphic because both are having different structures.

Here, $C_5$ is isomorphic to its complement. Just imagine the reorganisation of vertices in the complement of $C_5$, it will become the same as $C_5 $.

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1 comment

yes,C5 is the only self complementary graph.

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20 votes
20 votes

No of edges in graph + no of edges in complement of graph = n(n-1)/2  // No Of edges in complete graph of N vertices

5 + 5 = n(n-1)/2 // Here in C5 edges are there and in isomorphic graph the no of edges should be equal to no of edges in graph.

so solving it we get n = 5 .

2 Comments

From where you got the above formula
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@Abhishek

This is General approach. If there is a Graph G which have some edges and if we want to find out the complement of the graph then we will take those edges in complimented graph which are not in Graph G.

Number of Edges in Graph G + number of edges in Graph G'= Maximum Edges possible with given vertices.

Cycle graph is a graph which has n vertices and n edges.

In this Question, they are saying," A cycle on n vertices is isomorphic to its complement"

so the number of edges in Graph G and its complimented Graph is same. Let us suppose, n edges are there.

So according to our formula:

n + n =n(n-1)/2

on solving we get

n=5
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9 votes
9 votes
Another Approach:

(degree of compliment) = (degree of complete graph with given 'n' vertices) - (degree of given graph with 'n' vertices)                                                  

 =>         2 =  (n-1) - 2           ( Given graph = a cycle so degree will be 2 and this will also be the degree of its compliment)
 =>         n = 2 +2 + 1 = 5

Hence, n=5.
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