A subset K of candidates covers the entire set of candidates kappa iff no member of kappa-K (the complement of K) defeats every member of K. Let W be the matrix whose j_th entry in row i is 1 or 0 depending on whether or not candidate i is ranked ahead of candidate j on more ballots than not. This W is called the pairwise win matrix. Let C be a set of columns of W. Then AND(C) is a vector in the column space of W formed by applying the logical AND operator component-wise to the set of columns. When dealing with zeros and ones, AND is the same as both muliplication and minimization: 1×1=1, 1×0=0, min(1,1)=1, min(1,0)=0. It is easy to see that AND(C) is the zero vector iff the set of candidates K contributing the set of columns C covers the entire set of candidates kappa. In particular, the j_th and k_th columns of W will have a dot product of zero iff candidates j and k together cover kappa. This fact is of practical importance because ... if kappa is the set of candidates in a public election, there will practically always be a pair of candidates that together cover kappa. A single candidate covers kappa iff its column in W is the zero vector ... which means it is unbeaten pairwise. So not every ballot set has a Condorcet Winner, but every public election has a covering pair ... with practical certainty ... no known public ballot set has ever required more than two candidates to cover it ... and it's not easy to devise a plausible theoretical one. These considerations are the practical basis for the following method: If there is no single covering candidate (i.e. Condorcet Candidate) ... ... then within 24 hours after the votes have been tallied, each candidate must submit a proposed covering pair. Any proposed pair that does not cover the candidates is thrown out. The majority winner of each confirmed covering pair is then determined. Finally, the winner with the most votes is elected. That's it ... except ... legally the theoretical possibility of non-existence of a covering pair must be "covered". In that vein I suggest including a clause that says ... if no candidate is able to submit a valid covering pair, an additional 24 hours is allowed for each candidate to propose a covering triple. Among the respective IRV winners of the confirmed covering triples, elect the one with the most transferred votes. -Forest

Can anybod find a simpler example of a candidate set that cannot be covered by some pair of its members? This is a cycle of cycles created by cloning each member of an ABCA cycle ... replacing A with an a1a2a3a1 cycle, B with a b1b2b3b1 cycle, and C with a c1c2c3c1 cycle. It's tedious but not impossible to construct a ballot set that creates this cycle of cycles. No pair covers it. But the triple {a1,a2,b1} does cover it, as does {a3,b2,c1}, for example. Two questions: Q1: Is there a simpler example of a candidate set not covered by some pair of candidates? Q2: What is the likelihood of such a pair appearing in a public election? -Forest On Thu, Mar 9, 2023, 4:13 PM Forest Simmons <forest.simmons21@gmail.com> wrote: > A subset K of candidates covers the entire set of candidates kappa iff no > member of kappa-K (the complement of K) defeats every member of K. > > Let W be the matrix whose j_th entry in row i is 1 or 0 depending on > whether or not candidate i is ranked ahead of candidate j on more ballots > than not. > > This W is called the pairwise win matrix. > > Let C be a set of columns of W. Then AND(C) is a vector in the column > space of W formed by applying the logical AND operator component-wise to > the set of columns. > > When dealing with zeros and ones, AND is the same as both muliplication > and minimization: 1×1=1, 1×0=0, min(1,1)=1, min(1,0)=0. > > It is easy to see that AND(C) is the zero vector iff the set of candidates > K contributing the set of columns C covers the entire set of candidates > kappa. > > In particular, the j_th and k_th columns of W will have a dot product of > zero iff candidates j and k together cover kappa. > > This fact is of practical importance because ... if kappa is the set of > candidates in a public election, there will practically always be a pair of > candidates that together cover kappa. > > A single candidate covers kappa iff its column in W is the zero vector ... > which means it is unbeaten pairwise. > > So not every ballot set has a Condorcet Winner, but every public election > has a covering pair ... with practical certainty ... no known public ballot > set has ever required more than two candidates to cover it ... and it's not > easy to devise a plausible theoretical one. > > These considerations are the practical basis for the following method: > > If there is no single covering candidate (i.e. Condorcet Candidate) ... > ... then within 24 hours after the votes have been tallied, each candidate > must submit a proposed covering pair. > > Any proposed pair that does not cover the candidates is thrown out. > > The majority winner of each confirmed covering pair is then determined. > > Finally, the winner with the most votes is elected. > > That's it ... except ... legally the theoretical possibility of > non-existence of a covering pair must be "covered". > > In that vein I suggest including a clause that says ... if no candidate is > able to submit a valid covering pair, an additional 24 hours is allowed for > each candidate to propose a covering triple. > > Among the respective IRV winners of the confirmed covering triples, elect > the one with the most transferred votes. > > -Forest > > > >