prizzyjade
  • 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.
Pre-Algebra
schrodinger
  • schrodinger
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zzr0ck3r
  • 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
  • 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
  • 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
  • zzr0ck3r
prizzyjade
  • prizzyjade
Im Sorry.for not replying I still have class sorry
zzr0ck3r
  • zzr0ck3r
no problem. Class is important :)
prizzyjade
  • prizzyjade
thanks for the help ....
prizzyjade
  • prizzyjade
I'm having a hard time analyzing the problems coz i cant focus having so many math subjects ..
zzr0ck3r
  • zzr0ck3r
what all are you taking?
prizzyjade
  • prizzyjade
Calculus with analytic geometry ,physics(i know its science but still it uses math ) and graph theory...
zzr0ck3r
  • zzr0ck3r
yep, that is a full boat

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