Maximising x(G) Let $G$ be a graph on $n$ vertices. To each vertex, we assign zero or more colours such that


*

*Any two vertices sharing an edge must have a common colour.

*Any two vertices not sharing an edge do not share any common colours.


Let $x(G)$ be the least number of colours needed to fulfil the condition. Across all possible $G$, what's the maximum value of $x(G)$? 

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Idk it is true or not ...
We can construct a  complete bipartite graph which is balanced as early as possible .
Such construction is per Tarun's construction.so maximising x(G) can be done by finding maximal triangle free vertices.
Case1 if n(even) then we can construct biparite graph with lenth $n/2$.hence this creates$n²/4$ vertices.
Observation  two vertices are in same 
Side (as they are not adjecent) we can't color any 3 points with same color .so it need distinct color .
$X(G)\ge n²/4$
Case2
If n is odd then there are (n+1)(n-1)/4 vertices with partition size n+1/2
 So  $ x(G)>= n²-1/4$
So ,max{x(G)}= n²/4 for enen n
                = n²-1/4 ,for odd n
