TITLE: Can you cover the Boolean cube {0,1}^n with O(1) Hamming-balls each of radius n/2-c*sqrt(n)?
QUESTION [16 upvotes]: (where c>0 and the balls need not be disjoint?)
This is an embarrassingly simple question, yet somehow I couldn't find an answer (not even, "this is a well-known open problem") after spending some time googling the literature on covering codes.  
A simple probabilistic argument shows that you can cover the Boolean cube with O(n) Hamming balls of radius $n/2-c\sqrt n$ each, for any $c>0$.  My guess would be that you can't do it with (much) fewer---O(1) Hamming balls seems aggressively optimistic---but I don't know if it's known how to prove that.
(In the language of coding theory, I want to know whether $K_2(n,n/2-c\sqrt n)$, the minimum size of a binary covering code with radius $n/2-c\sqrt n$, can be upper-bounded by a constant depending only on c, not on n, at least for some c>0.)

REPLY [8 votes]: Spencer's famous "Six Standard Deviations Suffice" is essentially equivalent to what you're asking (for a statement and proof see, e.g., the Alon-Spencer book or the  new proof by Lovett and Meka). It shows, e.g., that with balls of radius $n/2-6\sqrt{n}$ you need n balls to cover the cube. It also gives $\Omega(n)$ for any radius $n/2-c\sqrt{n}$.