[EM] UncAAO, DMC, ASM, and TACC
chrisjbenham at optusnet.com.au
Sat Mar 10 23:04:33 PST 2007
Forest W Simmons wrote (March 11, 2007):
>I'm sure that you are aware that in your example,
>37: c|a ,
>the UncAAO winner is c, so that DMC is not uniformly better than UnCAAO
>in the three candidate case, assuming that c "should" win.
>But I'm not so sure that c should win. In fact, I believe that more
>likely than not, a ballot set like this reflects a burial of b by the
>37 c faction, anticipating a victory for c from a method like ASM or
>Smith Approval. So I believe that a ballot set like this would occur
>much more rarely under DMC than under ASM.
I gave the example purely as something normative, not as something
likely to occur in practice or
to invite speculation about about which, if any, of these votes are
insincere. And I can't see any justification
for your speculation that the 37c faction are burying, or for any other
speculation that these votes aren't
In this fairly simple near symmetrical 3-candidate situation with no
truncation or equal-ranking, in comparing
C to B, C has more first preferences (37-32) and more second preferences
(32-31) and so of course fewer
bottom preferences (31-37) and on top of that more explicit approval
So the only way a case for B versus C can be made is to somehow make one
or more of the algorithms for
DMC or AWP a standard, which to me doesn't fly. Smith//Approval as a
standard has more appeal.
Without the approval cutoffs, I find the example is a good test for
mono-raise: "if c doesn't win, the method
fails"; because if the candidate with the most support doesn't win in a
deterministic method then there is
doubtless some way its supporters can make it win by lowering its voted
With the approval cutoffs, DMC (and AWP) come close to failing mono-raise.
A>B>C>A Approvals: A35, B32, C33.
A eliminates (doubly defeats) B, and C wins. (AWP measures
defeat-strengths by the number of ballots on
the winning side that approve the winner and not the loser, and so says
C's defeat is the weakest and so also
Now change the 4 A>>C ballots to C>>A
37: C>>A (4 were A>>C)
A>B>C>A Approvals: C37, B32, A31
Now C doubly defeats A, and B wins. (AWP also elects B)
In the case of ASM/Approval Margins, the winner changed from A to C
reflecting C's rise in support
versus A. TACC elects B both times.
>Another nice method, Jobst's TACC (Total Approval Chain Climbing), also
>gives b as winner.
>Recall that Jobst initializes a "chain" with the lowest approval
>candidate, and (moving up the approval list) adds only those candidates
>to the chain that pairwise defeat each of the other candidates
>currently in the chain.
From the horse's mouth (March 4, 2005):
> TACC (Total Approval Chain Climbing):
> 1. Sort the candidates by increasing total approval.
> 2. Starting with an empty "chain of candidates", consider each
> candidate in the above order. When the
> candidate defeats all candidates already in the chain, add her at the
> top of the chain. The last added
> candidate wins.
> Let us study shortly why these two methods are monotonic: Let us
> assume that the actual winner X
> is raised on a some individual ballots by moving either the approval
> cutoff or another candidate from
> directly above X to directly below X.
> Then what can happen is twofold: First, the order in step 1 either
> does not change or does only change
> in that X gets moved up one position. Second, the defeat do not change
> or do only change in that X now
> defeats some candidate Y which she was defeated by before.
> In either case, X still must win: It was the last candidate added to
> the chain. The new chain developes
> exactly as before: As the order did not change left of X, the chain
> evolves just as before until the original
> position of X. If X did not change position, it still defeats all
> candidates in the chain and so gets added.
> If X did change position with Y, then Y was beaten by X or some other
> candidate already in the chain
> since it was not added to the chain originally. If it is added now, it
> must hence be beaten by X, so that
> X still gets added after Y was added. In either case, when X was
> considered, the resulting chain is the
> old one except that perhaps Y is added. As no later candidate beat was
> added originally and X beats
> everything it beat before, also now no further candidate can be added,
> so that X is again the winner. QED.
> What do we learn from that? The crucial point in this proof is that by
> raising X, the only change to the order
> in step 1 can be that X is raised there, too, but the relative
> position of the other candidates in the order cannot
TACC seems to have the curious property that if there are three
candidates in a top cycle, the most approved
can't win unless it pairwise loses to the second-most approved
candidate. This doesn't seem to cause any
monotonicity problem, but it does seem to be much more vulnerable to
Burial than the other candidate methods:
44: B>>C (sincere is B or B>>A)
A>B>C>A. Approvals: A46, B44, C10.
TACC elects the buriers' candidate B (chain: B>C) while all our other
candidate methods (UncAOO, ASM,
DMC,AWP) elect A.
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