Ken Kuhlman wrote: > PS: I had a conjecture that the pairwise matrix & independence matrix > combined contained enough information to re-construct the original > ballots (assuming fully ranked ballots). Would anyone be interested > in evaluating it? I could easily be embarrassingly wrong again, but > if not, it would be pretty exciting...
It's true when there are 3 candidates. The ballots a: A>B>C b: A>C>B c: B>A>C d: B>C>A e: C>A>B f: C>B>A produce the system of equations [ 1 1 -1 -1 1 -1] [a] [mAB] [ 1 1 1 -1 -1 -1] [b] [mAC] [ 1 -1 1 1 -1 -1] [c] = [mBC] [ 1 0 1 0 1 1] [d] [cAB] [ 0 1 1 1 1 0] [e] [cAC] [ 1 1 0 1 0 1] [f] [cBC] where mAB = pairwise margin of victory of A over B mAC = pairwise margin of victory of A over C mBC = pairwise margin of victory of B over C cAB = corr(A, B) wrt C cAC = corr(A, C) wrt B cBC = corr(B, C) wrt A The left-hand matrix is invertible, so therefore the original ballots can be reconstructed from the parwise array plus the correlation array. ---- election-methods mailing list - see http://electorama.com/em for list info
