Dan, I still don't have an answer, but am wondering if the right approach is to come up with a formula for the number of subsets of the resulting vectors which sum to 0, and then show that it's odd (or maybe non congruent to 0 modulo some other convenient integer). I could imagine using some clever trick involving interchanging summations. I also wonder how this generalizes to other finite groups: given a finite abelian group G, you can write down the matrix of all the group characters (which is what you've done with the group Z_2^n) -- the rows being the characters, and the columns the group elements. As long as G is not the trivial group, the rows sum to 0. So you can do the same thing -- change some arbitrary subset of the values of this matrix to 0, and ask if there is a non empty subset of the rows of the resulting matrix that sum to 0. Victor On Fri, Jun 22, 2012 at 5:03 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I recently heard this cute puzzle:
<< For some integer n > 0 you have all 2^n vectors of dimension n whose entries are +1 or -1. Of course the sum of all 2^n of these vectors is the 0 vector.
Then your three-year-old child changes some of the entries of some of these vectors to 0.
Show that there is still a nonempty subset of the new set of 2^n vectors that sums to the 0 vector.
--Dan
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