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.

Question

zzr0ck3r · Answer

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 · Answer

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 · Answer

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.

zzr0ck3r · Answer

@prizzyjade ?

prizzyjade · Answer

Im Sorry.for not replying I still have class sorry

zzr0ck3r · Answer

no problem. Class is important :)

prizzyjade · Answer

thanks for the help ....

prizzyjade · Answer

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

zzr0ck3r · Answer

what all are you taking?

prizzyjade · Answer

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

zzr0ck3r · Answer

yep, that is a full boat