This is a very good idea. I'd guess that the shortest 0-vector sums possible will be among the ones occurring in the 2^n possibilities that Warren mentions. But it's pretty opaque to me how to take this further. --Dan Warren wrote: << I wrote:
Warren brings up the very interesting question: For any such "0-modification" of the set {-1,1}^n (i.e., so that some components of some vectors become 0, the rest remaining unchanged), what is the smallest number f(n) such that there is always a nonempty subset S of modified vectors summing to the 0 vector, with #(S) <= f(n) ???
-- . . . In the Amit proof the initial vector which Asimov had taken as all 1's, could instead be taken to be anything… . . . So there are really 2^n different such proofs and we can ask which one yields the smallest 0-sum set. . . .