On 03/08/2023 08:01, Richard Lung wrote: > > My feeling is that "Method X" is doomed to failure, while based on > competing elimination methods. Eliminations being based on rules > outside the power of the voters, are bound to be arbitrary, and > arbitrary rules cannot provide a definitive method X. > > Tho I think it is good that you look for a definitive method. > > > Regards, > >  Richard Lung. > > > On 03/08/2023 01:54, Kevin Venzke wrote: >> I plan to implement Method X when I get a moment. I'm not completely >> skeptical about its monotonicity since it doesn't seem like there's >> an obvious argument why it shouldn't be monotone. >> >> For Landau, I use some code like this, which I base on the >> explanation on Wikipedia: >> >> let landau = []; >> for(x in candidates) { >> let i_am_covered = false; >> for(z in candidates) { >> if( x != z ) { // X can't cover themselves >> let found_y = false; >> for(y in candidates) { >> // Note that beats_or_ties should evaluate as true if the two >> operands are the same candidate >> if( x beats_or_ties y && y beats_or_ties z ) { >> // alternative to the above line which should be right: >> // if( not( y beats x || z beats y ) ) { >> found_y = true; // found a counterexample to Z covering X >> break; // cease checking for Ys for this Z >> } //if x... >> } //for y >> if( not found_y ) { >>   i_am_covered = true; // Z evidently covers X >> break; // cease checking for Zs for this X >> } // if not found_y >> } // if x!=z >> } // for z >> if( not i_am_covered ) landau.append( x ); >> } // for x >> return landau; >> >> Kevin >> (end) >> >> >> Le mercredi 2 août 2023 à 09:54:45 UTC−5, Forest Simmons >> <forest.simmons21@gmail.com> a écrit : >> On Wed, Aug 2, 2023, 4:15 AM Kristofer Munsterhjelm >> <km_elmet@t-online.de> wrote: >>> I tried to check Landau//X. It had plenty of Other strats, which is not >>> unexpected given your proof about independence of covered alternatives >>> plus first preference tiebreak implying nonmonotonicity... but I think >>> my implementation of Landau is suspect, so I'm not including it here. >> Let M be the matrix whose (j,k) element is one (else zero) if >> candidate j defeats candidate k. >> >> Then candidate j is uncovered iff the j_th row of (I+M)^2 is devoid >> of zeros  ... because the (j,k) element of this matrix is the number >> of distinct short beatpaths (i.e. of length 0,1, or 2) from candidate >> j to candidate k. >> ---- >> Election-Methods mailing list - see https://electorama.com/em for >> list info