24 May
2008
24 May
'08
10:14 p.m.
On Sat, May 24, 2008 at 9:01 PM, <rcs@xmission.com> wrote:
We could solve NP problems if we had an efficient way to calculate either
a) F(N,D) = sum(1<=n<=N, 1<=d<=D) floor[n/d]
or
b) G(N,D) = sum(1<=n<=N) #{ divisors d of N such that 1<=d<=D }
Did you mean "divisors d of (lowercase) n"? I've never seen divisibility associated with NP completeness before; where did these expressions come from? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com