A few days ago, I wrote a program that was intended to determine the minimum number of values requires to tell disjoint winner sets apart, and I decided to test it with the Condorcet winner sets - i.e. determine how many values would be required for a summary that tells apart elections where X is the CW and Y is the CW, for every pair X and Y. And I couldn't get it to work because it was returning results suggesting that the number of dimensions required is much lower than the n^2 I was expecting. But then I thought a bit about it and I realized there does exist an O(n) summary that can be used to create a method that passes the Condorcet criterion... *as long as you give up neutrality*. It's so obvious in retrospect, but it was very surprising to me before the fact. Can you see how it would be done? -km

On 2024-01-26 14:19, Kristofer Munsterhjelm wrote: > But then I thought a bit about it and I realized there does exist an > O(n) summary that can be used to create a method that passes the > Condorcet criterion... *as long as you give up neutrality*. > > It's so obvious in retrospect, but it was very surprising to me before > the fact. Can you see how it would be done? Whoops, I have to retract this. The results didn't show what I thought they'd show, and thus I can't prove O(n) Condorcet -- only O(n) Condorcet loser (which is easy). (I then proceeded to trick myself into thinking it was obvious :-) -km