Let's consider the case of one candidate ordering (i.e. ranking) per candidate. One way this can happen is by each voter supporting the ranking published by his or her favorite candidate. Another way that this can be achieved is by averaging together all of the ballots that rate a given candidate in top place. There are many other ways as well, but let's not worry about them. For now, we assume that we have one faction per candidate. Let's start with an example of three candidates A, B, and C, with three factions as follows: 45 A>B>C 40 B>C>A 15 C>A>B Choose a random permutation of the candidates so that we can refer to them as first candidate, second candidate, and third candidate. In this example, for simplicity, suppose that the permutation is just ABC, so the order is alphabetical. List all of the three digit binary numbers in numerical order (in the general case list all of the N digit, base (N-1) numbers in numerical order.): 000 001 ... 110 111 Each of these numbers is a code in whichthe k_th digit tells where to place the approval cutoff in the k_th candidate's ranking. For example, 000 means place the approval cutoff in the lower position in each faction, and 101 means place the cutoff in the upper position in the first and last factions, but in the lower position for the second faction. Now replace each number with the name of the candidate that would win, given the approval cutoffs encoded in that number: 000->B, 001->B, 010->B, 011->B, 100->A, 101->C, 110->A, 111->A In the resulting list group the names by twos BB|BB|AC|AA Use the last candidate's ranking to pick one candidate from each group: B B C A Now group these by twos BB|CA Use the second from last candidate's ranking to pick one from each of these groups: B C Use the first candidate's ranking to choose from among the remaining candidates: B wins. [This corresponds to the fact that if the A faction had to choose a cutoff before the B faction and C faction, then A would use the lower cutoff. Otherwiise, the B faction would use the lower cutoff, and C would use the upper, and win.] In general, if there are N candidates, we need (N-1)^N code numbers to encode the possibilities. For eleven candidates that's 10^11, or one hundred billion code numbers. For practical purposes, the candidate pool should be pared down to a dozen or so before by some other means before applying this procedure. That's all for now. Forest
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