[EM] Condorcet How?

[robert bristow-johnson]

On Apr 7, 2010, at 6:25 PM, Dave Ketchum wrote:

> This is some thought about keeping it simple, yet doable.
>
> I will lean toward Ranked Pairs with margins,

not sure what "with margins" does.  i'll read below...

> but amending toward other types of Condorcet should be doable.
>
> Voting:  Voter can rank one or more candidates.  Equal ranking  
> permitted.  Counters care only which of any pair of candidates ranks  
> higher, not how voter decides on ranking.  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).

no sweatsky.  and if some amazing political event is happening and  
some write-in may be winning, hand recount, examining the write-in  
entry and sorting is necessary in any case presently.  this is so  
improbable, but if it were to happen, it would likely be common  
knowledge locally who the write-in insurgent candidate is.   
nonetheless, rules could be drawn such as: in the case that the  
aggregate "write-in" wins with the machine count, each precinct digs  
out the paper ballots (i think everyone should use optical scan which  
has "paper backup" inherent) and begins to examine and count who  
"write-in" is.  as soon as, say, 10 write-ins that are for the same  
named candidate (say "Joe Schmoe"), then all ballots with write-in  
marked are separated as those for that named candidate and those that  
are not.  so then you get a pile for "write-in A" who has a name, a  
pile for "write-in B" who need not be the same name, and a pile for  
all other ballots (with no write-in marked).  Combine the first and  
last piles so now you have two piles: the first is just like a machine  
countable ballot but we know that "write-in" is "Joe Schmoe".  the  
second pile has "write-in" ranked, but the name written in is not "Joe  
Schmoe".

then the two piles can be run through the optical scan machine  
separately with the counters reset (zeroed) between each run.  it  
would be like two sub-precinct subtotals, where tallies for every  
candidate pair (not involving write-in) can be summed, but not those  
involving "write-in A" and "write-in B" which are now treated as two  
separate candidates.  but like all race pairs, the precinct subtotals  
for candidate pairs involving "write-in A" can summed, because (as  
Kathy would like) Condorcet is precinct summable.  what would be  
*really* unlikely is that "write-in B" wins, and then you would have  
to likewise examine the pile of ballots with "write-in B" marked and  
repeat the write-in recount mess.

the above is an idea for language dealing with the very unlikely case  
that "write-in" becomes the CW.  otherwise, we don't care (but Freedom  
of Information Act should apply and the media should be able to  
examine the ballots to find out who that loser write-in candidate was).

> Counting:  Besides the N*N matrix, I would add an N array to  
> optimize this.  Count each ranked candidate in the array.  Later the  
> array will be added into the matrix rows as if the ranked candidates  
> won in every one of their pairs.  This is correct for pairs with no  
> ranking, and for pairs with one ranked.

already this is complicated and someone in the "One person, one vote"  
crowd (the anti-IRVers) in Burlington would say that you're trying to  
pull one over on them.

Ranked pairs need not be that complicated.

>  For pairs w/winner and loser,

every pair has a winner and loser, unless they tie

> give loser a negative count to adjust;

no, just leave him alone.

> for ties you can leave both winning; or mark both losing via  
> negative counts.

you don't need to be adjusting any other counts in the N*(N-1)/2  
pairs, which is my preference of expressing the "N*N matrix".  i still  
find the "N*N matrix" to be a useless visual tool.  i want to see  
N*(N-1)/2 pairs of numbers.  that's how you visualize in a glance how  
Condorcet decides an election.  this "N*N matrix", such as it is, is  
just useless.

the N*(N-1)/2 pairs is all of the numerical information you need, and  
all you need to do is compute, for each pair, the abs value of the  
difference of vote counts. so you have 3 numbers for each of the  
pairs.  then there are N*(N-1)/2 - 1 scans, each identifying and  
removing the remaining pair with the highest abs(diff).  then, with  
each of these removed pairs in order of their removal, construct the  
"who beats who" paths out of nodes starting with the highest  
abs(diff), and as Tideman sez, ignoring pairs that are contradictions,  
resulting in a cycle after "committing" (i would use that word instead  
of "locking") to the race pairs previously accepted.

that is simpler.  both as language of law, and to program into a  
computer.

>     (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:
>      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.

whew!  already you lost me, here or earlier, Dave.

> Find the winner.  What I read below sounds like an excessive amount  
> of labor, especially if many candidates - so, for candidates A-N (I  
> have not specified response for ties):
>      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.   
> Check final row when all but one have lost.  If no losers found we  
> have the CW; else we have a cycle member.
>      Cycle members lose to other cycle members.  Based on this, make  
> a list of all such.
>      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.

it sure seems to me that the existing Tideman RP is simpler and at  
least as "meaningful" in reflecting voter preference.

sorry Dave, if i'm not resonating with this idea of yours.  i  
certainly am for Condorcet.  i think that it's likely that in the  
worst case of a goddamn cycle, that probably Markus's method would  
better reflect the will of the voters than Tideman, but the two don't  
disagree with a cycle of 3 in the Smith set, and i think that if  
either were adopted, it would be a few millennia before there would be  
a Condorcet-decided election that would be decided differently between  
Tideman and Schulze.  so Markus, despite your concise English language  
for how this could be legislated, i still think Tideman is more  
transparent.  but i think, if we could get the yahoo's here in the  
hinterland to understand and trust the Schulze method (presently,  
these yahoos don't trust anything other than "simple majority" which  
is really "simple plurality", because they don't seem to understand  
anything else), i think that would be better.

these "candidate elimination" schemes i earlier brought up as a way to  
resolve a Condorcet cycle, was only to come up with something the  
"yahoo's" could understand and that meaningfully dealt with the  
cycle.  and i'm still thinking (using the Burlington 2009 data) that  
cycles would be so rare (the 2009 case had an unambiguous Condorcet  
ordering of all 5 candidates, all the way from the CW to the Condorcet  
Loser), that any decent and sensible method for resolving the cycle  
would be fine with me.  but, if i were king of the world, it would be  
Schulze (but, then, if i were king of the world, we wouldn't be doing  
elections, at least not for king).

but Markus, i still think Tideman would be an easier sell.  and right  
now, after the bruising IRV battle we had in Burlington, all this is  
just a pipe dream.  looks like, for the time being, we just need to  
accept that God herself ordained the "traditional ballot" where we  
mark just one candidate.  it was a pretty close election, too.  the  
bastards.

oh well.

bestest,

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."