If G is self complementary on n vertices then 4|n(n-1)
by adityaguharoy, Sep 11, 2017, 1:33 PM
A graph
is called self complementary if and only if it is isomorphic to it's own complement.
Prove that if a graph
with
vertices, is self complementary if and only if
is a multiple of
.
Proof

Prove that if a graph




Proof
First note that :
If a graph
with
vertices is self complementary then it must have
vertices. And thus,
is proved.
Now, for the other side : we use induction
For
it is obvious.
For
and
see the attached drawing.
Now let
be a graph with
vertices that is self complementary.
Then consider a path
of length
.
Then join
and
to every vertex of
.
(Such a construction is called
path addition.)
Then observe that the new graph
must be self complementary.
.................
Hence our construction is done !!
Thus, we conclude that :
If for a positive integer
there is a self complementary graph
with
vertices then there must be a graph
with
vertices that is also self complementary.
This completes the induction and proves our result.
The drawing shows self complementary graphs of
and
vertices.
(these are not the unique self complementary graphs of
and
vertices -- for example a path
of length
is also self complementary, the graphs whose vertices when joined look like
and
are also self complementary etc..)
If a graph




Now, for the other side : we use induction
For

For


Now let


Then consider a path



Then join



(Such a construction is called

Then observe that the new graph

.................
How does the complement of
look like ?
Take small values of
, draw and see for yourself.
~Often a diagram/drawing does the work of a thousand words.

Take small values of

~Often a diagram/drawing does the work of a thousand words.
Hence our construction is done !!
Thus, we conclude that :
If for a positive integer





This completes the induction and proves our result.
The drawing shows self complementary graphs of


(these are not the unique self complementary graphs of






This post has been edited 2 times. Last edited by adityaguharoy, Sep 11, 2017, 1:35 PM