Babai, Kantor and Lubotzky proved in 1989 the following theorem ([Sciencedirect link to article](https://www.sciencedirect.com/science/article/pii/S0195669889800678)).


THEOREM 1.1. There is a constant $C$ such that every nonabelian finite simple group $G$ has a set $S$ of at most 7 generators for which the diameter of $\mathrm{Cay}(G,S)$ is at most $C\log|G|$.


Then they remark that 


"A crude estimate for $C$ is $10^{10}$, but we will not include the bookkeeping required to
estimate $C$."


This is my question.


"Is there a finite simple group $G$ for which there exists a generating set $S$ which satisfies the conditions in the above theorem for some reasonably small $C$ (comparing to the order of $G$)?"



Diameter of Cayley graphs of finite simple groups
There are two examples, $\mathrm{Alt}\_n$ and $\mathrm{PSL}\_2(q)$, in [this](https://pages.uoregon.edu/kantor/PAPERS/STOCdiameter.pdf) paper (p.861). 


For $\mathrm{Alt}\_n$, the authors used 3 generators and achieved diameter at most $(1+o(1))4n\log n$.


For $\mathrm{PSL}\_2(q)$, an upper bound is $12\log\_4(q)$ (Every integer in $\{0 .. q\}$ can be represented by $(...(a\_m·4+a\_{m-1})4+...)4+a\_0$, where $m < \log\_4(q)+1$, and $a\_k\in\{-1,0,1,2\}$ for $k\in\{0..m\}$. Representing each $a\_k$ costs at most $2$ generators, and multiplying by $4$ costs $2$ generators. There are $3$ numbers need to be represented, as $\mathrm{PSL}\_2(q)$ has $3$ degrees of freedom).


The bound can be improved to $12\log\_5(q)$ if $q$ is a prime of which $5$ is a quadratic residue: just replace "multiplying by $4$" by "multiplying by $5$". 


I believe there are much better bounds if we exploit the full power of $7$ generators.


