tag:blogger.com,1999:blog-2811876938195306723.post1666922619328024253..comments2023-07-07T01:27:13.382-07:00Comments on Absolutely Regular: Tighter Hamming ball volume?Aryehhttp://www.blogger.com/profile/14913393383227385317noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2811876938195306723.post-21276307759829057762009-05-04T08:37:00.000-07:002009-05-04T08:37:00.000-07:00It seems like there are a lot of expressions that ...It seems like there are a lot of expressions that look kind of like the bound you want in Alon and Spencer, <I>The Probabilistic Method</I>, 2nd edition (2000), especially Chapter 14 on "Codes, Games, and Entropy". That may be a good place to start looking.Michael Lugohttps://www.blogger.com/profile/01950197848369071260noreply@blogger.comtag:blogger.com,1999:blog-2811876938195306723.post-8739302344629143012009-05-04T08:25:00.000-07:002009-05-04T08:25:00.000-07:00Thanks Michael. I kind of wanted to avoid re-provi...Thanks Michael. I kind of wanted to avoid re-proving this, since it <B>must</B> be well-known!.. I'll keep hoping that some kind soul will throw a reference my way...Aryehhttps://www.blogger.com/profile/14913393383227385317noreply@blogger.comtag:blogger.com,1999:blog-2811876938195306723.post-70682769574823480342009-05-04T06:56:00.000-07:002009-05-04T06:56:00.000-07:00I don't have a reference off the top of my hea...I don't have a reference off the top of my head.<br /><br />But for large n, B(n,d) = C(n,d) (1 + O(1/n)). So it's enough to bound C(n,d). <br /><br />And C(n, d) = n^d/d! (1 + O(1/n)) for large n.<br /><br />Finally, d! ~ sqrt(2πd) (d/e)^d (Stirling), giving<br /><br />C(n,d) ~ (2πd)^(-1/2) (en/d)^d.<br /><br />I won't bother making this more precise because I'm not sure how n and d are going to infinity in this constant. But this gives an improvement of the well-known bound by a factor of (2πd)^(1/2).Michael Lugohttps://www.blogger.com/profile/01950197848369071260noreply@blogger.com