On May 12, 2010, at 9:58 PM, Dr. Carl S. Milsted, Jr. wrote:
Condorcet is not simpler! Describing the results of a Condorcet
election requires an NxN matrix. Makes my eyes glaze and I have a
math degree. Most citizens do not. The potential for cycles is also
frightening. Bush/Gore was bad enough.
So a BS was the limit of what I could afford. Years after that I
finally got to see a real computer and my later life included
maintenance and language design for such. None of which makes my eyes
glaze more or less when I think of explaining how Condorcet uses such
matrices to a reasonably intelligent audience.
What some are doing with cycles is a bigger challenge, but
understanding their potential should be adequate and possible for most.
But even the ballot for Condorcet is complicated. Writing in the
numerical ordering for a ten candidate ballot is challenging. You
better allow pencils, since we'll need the ability to erase -- which
is not good if you like tamper proof ballots.
For either Range or Condorcet, and desiring to vote for more than 2 or
3 candidates, it makes sense to prepare before going near a ballot.
Any ballot design usable for Range would be suitable for the single
digit numbers adequate for Condorcet.
The deciding? Ordering per magnitude of liking does it for Condorcet;
Range requires distributing ratings to get desired gaps between them.
Netflix manages to have a user friendly form for ordering (time)
preference for movies, but it requires quite a bit of JavaScript and
animation to work. Would require rather sophisticated voting
machines to manage that interface.
Above I cover designing a ballot and reading what the voter did. Here
deciding what it means is unique to Condorcet, but not especially
complex programming
Range voting can be done using the same "fill in the dot" interface
we all remember from standardized tests. I guess we still have some
people too old to have taken such, but everyone below 50 at least
has mastered this interface. Range voting is much simpler than
Condorcet.
IRV usage of 3 ranks demonstrates such to be usable for Condorcet.
Range NEEDS more than 3 ratings. Condorcet would be happy with as
many ranks as get offered to Range ratings.
Moreover, I have experimented with both Range and Condorcet at the
club meeting level. Condorcet was a nightmare. Range was easier than
Robert's Rules.
Again, ballot design and reading what got voted is at least as easy
for Condorcet as for Range. Processing what got read is different,
but not especially difficult for Condorcet (unless given an especially
complex set of rules for resolving cycles).
Dave Ketchum
On Wed, 12 May 2010 19:00:14 -0400, Dave Ketchum wrote:
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
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