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- prizzyjade

A simple graph that is isomorphic to its complement is self complementary.
a. Prove that if G is self complementary, then G has 4k or 4k+1 vertices where k is an integer.

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- prizzyjade

- schrodinger

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- zzr0ck3r

The statement should include \(4k\) not just\(4k+1\) (look at part b)
The complete graph with \(n\) vertices has \(\frac{1}{2}n(n-1)\) edges.
Why? Because for each of the \(n\) vertices, there is \(n-1\) edges, but we count them all twice.
If a graph is self complimentary there must be \(\frac{1}{2}(\frac{1}{2}n(n-1))\) edges because the compliment must have equal amount of edges.
Now if \(\frac{1}{4}n(n-1)\in \mathbb{Z}\) then it must be that \(4\) divides \(n(n-1)\).
In other words \(n=4k\) or \(n=4k+1\)

- zzr0ck3r

b) there is one self complimentary graph on 4 verts. Hint: it is super easy and I must draw the LINE at how much i will give.

- zzr0ck3r

For 5 verts. I am sure that if you CYCLE through all the graphs you have in your wheel HOUSE I bet you will find it.

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- zzr0ck3r

- prizzyjade

Im Sorry.for not replying I still have class sorry

- zzr0ck3r

no problem. Class is important :)

- prizzyjade

thanks for the help ....

- prizzyjade

I'm having a hard time analyzing the problems coz i cant focus having so many math subjects ..

- zzr0ck3r

what all are you taking?

- prizzyjade

Calculus with analytic geometry ,physics(i know its science but still it uses math ) and graph theory...

- zzr0ck3r

yep, that is a full boat

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