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Nov 2, 2013 at 3:49 history edited Norouzi CC BY-SA 3.0
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Nov 21, 2011 at 1:03 comment added Norouzi Great! Let me know if you have some insight.
Nov 20, 2011 at 23:48 history edited Norouzi CC BY-SA 3.0
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Nov 20, 2011 at 18:21 comment added Gerhard Paseman I see my silly mistake now. Indeed the minimum will occur for s > 2/epsilon and less than Clog n. I hope to give details on C soon. Gerhard "Ask Me About System Design" Paseman, 2011.11.20
Nov 20, 2011 at 13:18 answer added Brendan McKay timeline score: 4
Nov 20, 2011 at 5:59 comment added Gerhard Paseman Pretend the sum is bounded by l*2^s and u*2^s. I get that the 1/s term dominates and that no minimum is achieved as s goes to infinity. In particular, passing from s to s+1, you will often get a decrease. If epsilon is small, a good lower bound is (s+1) choose epsilon*s. Gerhard "Ask Me About System Design" Paseman, 2011.11.19
Nov 20, 2011 at 3:21 history edited Norouzi CC BY-SA 3.0
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Nov 20, 2011 at 3:16 history asked Norouzi CC BY-SA 3.0