Proving the symmetry of self complementary graphs with n=4k+1

[MupptMath]

I am trying to prove the following:

Let G be a graph
Let |V(G)|=n=4k+1, for k an integer
Let G be isomorphic to G complement

Claim: Given degree sequence for G, d1>=d2>=...>=dn, prove d(i)+d(n-i+1)=n-1 for i=1,2,...,n

Now, we know for any vertex v in G, d(v)=(n-1)-D(v), where D(V) is the degree in G complement. So it seems a matter of showing that D(V) = d(n-i+1). Just not sure how.

There are some other basic things we can deduct, but I cannot create a positional relation ie relate the i to n-i+1 vertices - and show there is symmetry about a point in the degree sequence.

Any ideas??

Thanks in advance!

 

[Valkarie]

Let G be a graph with n=4k+1 vertices, with degree sequence d1>=d2>=...>=dn. Since G is isomorphic to its complement, we know that for any vertex v in G, d(v)=(n-1)-D(v), where D(V) is the degree in G complement. Now, let i be an arbitrary index in the degree sequence, such that i=1,2,...,n. We want to show that d(i)+d(n-i+1)=n-1.We can denote d(i) as di and d(n-i+1) as dn-i. Since G is isomorphic to its complement, we know D(v)=n-1-di and D(v)=n-1-dn-i. Since these are equal, we can add them together to get:di + dn-i = 2(n-1)Substituting in the value of n=4k+1, we get:di + dn-i = 8k + 2Rearranging, we get:di + dn-i = n-1Thus, we have shown that d(i)+d(n-i+1)=n-1.