On Apr 7, 2010, at 8:29 PM, robert bristow-johnson wrote:
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...
vs comparing per winning votes - each has backers.
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).
My thoughts are that supporting write-ins is worthy and doable, but am
not excited about how.
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.
So far we are just doing counting. After that, time to think of CW
and cycles, and for methods such as ranked pairs mattering.
Think of 10 candidates and a bunch of bullet voters. For each ballot
its candidate needs counting in each of its 9 pairs. I would have the
counters count such in its element in the N array, with all of the N
array added into the N*N matrix in one step later.
Try a voter ranking two of the 10 (should be common for those not
doing bullet). Stepping two entries in the N array will get 16
entries in the matrix properly stepped. One entry will get stepped as
if each ranked higher, so I have the counters adjusting for this as
part of counting that ballot.
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.
Explained above.
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.
Your preference over the matrix is interesting, provided you keep your
location of pairs understandable and get the same winner as the matrix
would achieve.
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.
Your many scans are something I would avoid.
For some of this computers vs humans matters - counting the bullet
votes is easier for humans to get right by doing a single entry in the
N array vs 9 in the matrix for each such ballot, while a computer only
needs simple programming.
In looking for the CW I eliminate a candidate via each carefully
selected comparison - 9 comparisons if 10 candidates. That does not
tell if there is a CW, so scan that candidate's pairs - if any losers
that is a cycle member and we now know all the cycle members it loses
to. Scan those members' pairs for the other members they lose to,
continuing until we know all the interesting pairs.
Most cycles only have three members. Anyway, we have the pairs that
compose the cycle and can use whatever method we choose to sort them
out.
(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.
For here copying from the array to the matrix can put noise in these
diagonals, so prettying seems nice, though of no real value.
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.
Most methods only care about cycle members and their relationships. I
find the members and only consider methods interested in such - and do
not argue here about which of such methods to actually use.
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.
Anyway, thanks for digging.
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 r...@audioimagination.com
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