The procedure called individualization breaks symmetry arbitrarily. It chooses some nodes in the graph, arbitrarily, to give their own unique names.
Suppose there exists a set $S \subset V$, such that after individualising this set and running plane WL (Weisfeiler-Lehman) algorithm each vertex gets a unique label (i.e. discretised ). There may be other subsets of $V$ of same size as $S$, whose individualisation (after running WL procedure) does not assign unique label to each vertex of input graph.
I am not individualising $k$ vertices simultaneously, but individualising one and then running the WL procedure.
Is there any way to minimise the individualising set size? if it is given that after individualising $l$ many vertices are sufficient.
Question : After individualising one vertice of input graph and then running the WL, I will have say some equitable partitions of vertex set of input graph $G$. How to pick second vertex for individualisation process? To me it appears I should choose a vertex who has say almost same number of neighbours and non-neighbours. I have tried some other parameters like degree etc to pick second vertex, but don't seems working. The best choice seems to me the vertex, which belong to same orbit as first vertex.
Please note that I don't want to try all possible choices of $S$ and looking from theoretical point of view completely.
EDIT: If input graph is regular then choice of first vertex is not important. I mean to say there is always going to be an individualising set (minimum size) which contain this first vertex. As input graph is regular, the number of neighbours and non-neighbours set size will be same for every choice of first vertex.