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.