[EM] Why Not Condorcet?

[Dave Ketchum]

There is considerable agreement that awarding the CW as winner is  
desirable - yet also claims that some method deserves use in spite of  
its inability to find the CW.

I  back Condorcet:
      Conceding that ability to resolve cycles  is part of deciding  
which method  to use.
      Agreeing that having several winners in a single race, such as  
of legislature members,  requires a different method.
      That Condorcet ranking lets the voter rank several candidates,  
assigning equal liking and/or showing liking some more than others.

Some comparisons:
      IRV deserves no mention, except as being less desirable than  
methods being backed - unlike IRV, Condorcet uses all that the voters  
say.
      Score/Range, with its ratings, is competitive with Condorcet,  
though I claim Condorcet ranking is simpler.
      Bucklin also competes, with its own complexity.
      Plurality and Approval - simpler.  Any voter finding either of  
these acceptable can vote such with Condorcet - these simply being  
part of the ability of Condorcet.

Races vary as to the thinking they inspire in voters.  Note that  
Condorcet requires no extra effort for Plurality (bullet) voting or  
for approving, but supports full use of its ability for any voter  
desiring this, permitting all in the same races.

Ballot must support ranking, voters need understanding, and many have  
some of this via IRV or Bucklin.

Now some thought about keeping it simple, yet doable.

I will lean toward Ranked Pairs with margins, but amending toward  
other Condorcet methods should be doable.

Voting:  Voter can rank one or more candidates.
      Bullet voting, ala Plurality - simply rank one.  For the debate  
I claim this is a suitable vote for many voters in many races.
      Approval - just give them the same rank.
      Condorcet - Equal ranking permitted.  Counters care only which  
of any pair of candidates ranks higher, not how voter decides on  
ranking.
      Rank below unranked candidates?  We sometimes wish such for  
those we most hate - not difficult if we figure out how voter should  
ask for such.
      Rank, but number not clear - rules could have counters treat  
such as a rank below lowest real rank.
      Write-ins permitted (if few write-ins expected, counters may  
lump all such as if a single candidate - if assumption correct the  
count verifies it; if incorrect, must recount - if many expected for  
one person, that name could be added in for counting).

Counting:  An N array makes this simpler - simply count each ranked  
candidate into the array.  When this is later copied into the N*N  
matrix it will supply exactly what is needed for pairs with no  
ranking, for pairs with one ranked, and for winner if both ranked.
      For pairs with a winner and loser, give loser a negative count  
now to adjust; for ties you can leave both winning; or mark both  
losing via negative counts.
      (For example, a ballot with 3 ranks gets 3 counts in N, and  
adjustments for 3 pairs in N*N - even if there are a dozen pairs on  
the ballot)

Completing matrix N*N:
      I had thought of doing adjustments if N was different in  
different matrices.  Having trouble with picking a need for this, but  
it is doable - add an empty element to N and an empty row and column  
to N*N.
      Shortest path is to sum all the matrices and all the arrays.   
Then add each array element into its matrix row as wins by its  
candidate in each of its pairs.
      Can want a matrix for a district - same idea as above.
      Either way the diagonals (A,A thru N,N) should be zero - make  
them thus.

Finding the winner.  What I suggest here is less labor than gets  
described for many methods, especially if there are many candidates -  
so, for candidates A-N:
      A single loss disqualifies a candidate from being CW, so start  
with A vs B.  If A loses, B continues; if B loses A continues; for a  
tie try a different pair among not-yet-losers (if any; else punt).
      Check final row when all but one have lost.  If no losers found  
we have the CW; else we have a cycle member.
      By checking each cycle member found for such losers, we make a  
list of all such (for the simplest cycles each loses to one other).
      There are many methods for resolving cycles.  For RP I see  
deleting the smallest margins from the list until what remains is not  
a cycle, but does identify a winner.

Dave Ketchum