On 7/3/2012 11:38 AM, Warren Smith wrote:
Hmm, ok, seems I did not make as craftmanlike a problem as I thought, sorry.
DISCRETE VERSION: (2k)^n vectors, each n entries from +-{1,2,3,...,k}. Child replaces subset of vector entries by integers of same sign (or 0) and smaller absolute value. To prove: modified vectors still must have a nonempty subset which sums to 0-vector. SOLUTION: similar to the Alon Amit proof (which had been only for the case k=1).
CONTINUUM VERSION: Real n dimensional vectors, each entry from real interval [-1,1]. I guess we had better exclude 0 though, otherwise the answer "use the 0-vector, which is immune to modification" kinds of screws things up.
But if you exclude the 0 vector then the malicious child can just set all the vectors to zero except one; and then there is no remaining subset that sums to zero. I think you need to prohibit 0 as a substitution value. Brent Meeker
Child replaces subset of vector entries by reals of same sign (or 0) and smaller absolute value. To prove or disprove: modified vectors still must have a nonempty subset which sums to 0-vector. (NON)SOLUTION: Well... I should give you more time before reveal. But I believe I have a solution in which "sum" means what sum normally means, for any subset of cardinality which is either finite or countably infinite. I suspect one can also use same proof idea to work for where "sum" means "Lebesque integration" provided the child's mapping function obeys some kind of Lebesque-measurability postulates, but I'd need to bone up on that stuff. If the child's map function obeys other more restrictive postulates, then some other interesting stuff happens...