[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[fsimmons]

A few years ago Jobst invented Total Approval Chain Climbing or TACC for short.

At the time I was too young (not yet sixty) to really appreciate how good it 
was.  It is a monotonic. clone 
free, Condorcet Efficinet method which always elects from the "Banks Set," a 
nice subset of the Smith 
Set (if not the entire Smith Set).

It is easy to describe:

(1) Initialize the variable S as the empty set  S = { }.

(2)  While some alternative beats every member of S pairwise, augment the list 
S with the lowest 
approval alternative that does so.

(3) Elect the last alternative added to S, i.e. the member of S with the 
greatest approval.

That's it.

Obviously the method will elect the CW when there is one.

If the Smith set consists of a cycle of three alternatives, say A beats B beats 
C beats A,, then this 
method (TACC) will will elect either the member of this cycle with the greatest 
approval or the one with 
the second greatest approval, depending on whether or not the cyclic order goes 
up or down the approval 
list.

What I didn't appreciate in my younger days was how beautifully resistant the 
method is to strategic 
manipulation.

Scenario 1:

49 C
27 A>B
24 B  (sincere is B>A)

The sincere CW is A, but the B faction creats an ABCA cycle by rruncation.

Assuming that "approval" is the same as "ranked" in each of the factions, the 
approva order is (from 
greatest to least)  B, C, A .  Since this is in the same cyclic order as the 
cycle,  C wins.  If the B voters 
are rational, they will not truncate A!

Now look at the burial temptation scenario:

49  A>B (sincere is A>C)
27  B>C
24  C>A

The sincere CW is C.

Now suppose that the A faction buries C as indicated above:

TACC will elect B. whether or not the A faction approves B. 

Forest 


Election-Methods mailing list - see http://electorama.com/em for list info

[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[C.Benham]

31: A>B
32: B>C
37: C>A

Approvals:  B63,  A68,  C69.   A>B>C>A.

TACC elects A, but  C is positionally the dominant candidate and 
pairwise beats A.


For a Condorcet method with pretension to mathematical elegance, I don't 
see how that

can be justified.

Chris Benham

PS:  Could someone please refresh our memories: What is the "Banks Set"?


From Jobst Heitzig (March 2005):


ROACC (Random Order Acrobatic Chain Climbing):
--
1. Sort the candidates into a random order.
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.

The good thing about ROACC is that it is both
- monotonic and
- the winner is in the Banks Set,
in particular, the winner is uncovered and thus the method is Smith-, 
Pareto-, and Condorcet-efficient.


Until yesterday ROACC was the only way I knew of to choose an 
uncovered candidate in a monotonic way. But Forest's idea of needles 
tells us that it can be done also in another way.
The only difference is that in step 1 we use approval scores instead 
of a random process:


TACC (Total Approval Chain Climbing):

1. Sort the candidates by increasing total approval.
2. Exactly as above.

The main differences in properties are: TACC is deterministic where 
ROACC was randomized, and TACC respects approval information where 
ROACC only uses the defeat information.
And, most important: TACC is clone-proof where ROACC was not! That was 
something Forest and I tried to fix without violating monotonicity but 
failed. More precisely, ROACC was
only weakly clone-proof in the sense that cloning cannot change the 
set of possible winners but can change the actual probabilites of 
winning. With TACC, this makes no difference since it
is deterministic and so the set of possible winners consists of only 
one candidate anyway.





Election-Methods mailing list - see http://electorama.com/em for list info

[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[C.Benham]

Chris wrote ...


/ 31: A>B

/>/ 32: B>C
/>/ 37: C>A
/>/
/>/ Approvals: B63, A68, C69. A>B>C>A.
/>/
/>/ TACC elects A, but C is positionally the dominant candidate and
/>/ pairwise beats A.
/>/
/>/ For a Condorcet method with pretension to mathematical elegance,
/>/ I don't/ /see how that/ /can be justified.
/>/
/>/ Chris Benham
/>/
/>/ PS: Could someone please refresh our memories: What is the
/>/ "Banks Set"?
/
Forest Replies:



As you know C is the DMC winner, and would be a slightly better winner, given
that the ballots are sincere.  But DMC is not as burial resistant and truncation
resistant as TACC.  


It is interesting that DMC and TACC have opposite rules for which of the top two
approval members of the top cycle (of three) wins.  DMC awards the win to the
one (of these two) that beats the other.  TACC awards the win to the one that is
beaten by the other.



Chris: 
I have long since abandoned the  "Definite Majority Choice" (DMC) method in
favour of Smith//Approval (as my preferred Condorcet method), which also elects C 
here.


I still like the Definite Majority criterion, which says that no candidate that 
is
pairwise beaten by a more approved candidate is allowed to win.

I think that (in isolation) meeting the Condorcet criterion is desirable, but 
not so
holy that on discovering there is no voted CW the method should proceed on the 
assumption
that there is really a "sincere CW" that has been victimised by strategists the 
method
should try to frustrate or punish.

Condorcet methods are vulnerable to Burial, period. Futile attempts to address 
this
should not be at the expense of producing winners that can have no philosophical
justification on the assumption that all the votes are sincere (or are all 
equally
likely to be sincere).

The TACC winner A simply has no shred of justification versus the 
Smith//Approval
winner C.

Forest:


I've come around to the belief that most Condorcet cycles in ordinary elections
are artificial, so chances are that this cycle was created from the burial of B
by the C faction.  Giving C the win only rewards this manipulation.



Chris:
I can't see any remotely rational justification for assuming that this is the 
case
rather than, say, the cycle was created by the A voters burying C.

BTW, I also like the version of Smith//Approval that allows voters to indicate 
an
approval threshold so they can rank among approved candidates.

I think on balance I prefer IBIFA to any of the Condorcet methods.

Thanks for explaining the Banks Set. I'll look more into it.


Chris Benham
















Election-Methods mailing list - see http://electorama.com/em for list info

[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[fsimmons]

> Chris wrote ...
> 
> >/ 31: A>B
> />/ 32: B>C
> />/ 37: C>A
> />/
> />/ Approvals: B63, A68, C69. A>B>C>A.
> />/
> />/ TACC elects A, but C is positionally the dominant candidate and
> />/ pairwise beats A.
> />/
> />/ For a Condorcet method with pretension to mathematical elegance,
> />/ I don't/ /see how that/ /can be justified.
> />/
> />/ Chris Benham
> />/
> />/ PS: Could someone please refresh our memories: What is the
> />/ "Banks Set"?
> /
> Forest Replies:
> 
> 
> >As you know C is the DMC winner, and would be a slightly better 
> winner, given
> >that the ballots are sincere. But DMC is not as burial 
> resistant and truncation
> >resistant as TACC. 
> >
> >It is interesting that DMC and TACC have opposite rules for 
> which of the top two
> >approval members of the top cycle (of three) wins. DMC awards 
> the win to the
> >one (of these two) that beats the other. TACC awards the win 
> to the one that is
> >beaten by the other.
> 
> 
> Chris: 
> I have long since abandoned the "Definite Majority Choice" 
> (DMC) method in
> favour of Smith//Approval (as my preferred Condorcet method), 
> which also elects C 
> here.
> 
> I still like the Definite Majority criterion, which says that no 
> candidate that is
> pairwise beaten by a more approved candidate is allowed to win.
> 
> I think that (in isolation) meeting the Condorcet criterion is 
> desirable, but not so
> holy that on discovering there is no voted CW the method should 
> proceed on the assumption
> that there is really a "sincere CW" that has been victimised by 
> strategists the method
> should try to frustrate or punish.
> 
> Condorcet methods are vulnerable to Burial, period. Futile 
> attempts to address this
> should not be at the expense of producing winners that can have 
> no philosophical
> justification on the assumption that all the votes are sincere 
> (or are all equally
> likely to be sincere).
> 
> The TACC winner A simply has no shred of justification versus 
> the Smith//Approval
> winner C.
> 
> Forest:
> 
> >I've come around to the belief that most Condorcet cycles in 
> ordinary elections
> >are artificial, so chances are that this cycle was created from 
> the burial of B
> >by the C faction. Giving C the win only rewards this manipulation.
> 
> 
> Chris:
> I can't see any remotely rational justification for assuming 
> that this is the case
> rather than, say, the cycle was created by the A voters burying C.
> 

Well, usually the largest faction is the one with the best chance of getting 
away with it, and would get 
away with it under Beatpath, Ranked pairs, etc. unless the Condorcet supporters 
took defensive action.  
With TACC no defensive action was necessary.

Now let's consider the possibility that you suggest, namel;y that the true 
preferences were

31 A>C
32 B>C
37 C>A

If there is enough information for the A faction to think it is safe to bury C, 
then there is enough 
information for the C faction to take the precaution of truncating A 
defensively.

The we have

31 A>B
32 B>C
37 C

That's why I think this scenario is less likely than the one i suggested.

To continue this scenario, we see that A now has the leasrt approval.

Under TACC, as long as the cyclic order of the cycle is A beats B beats C beats 
A,, and A has the least 
approval of these three alternatives, C will win.  In other words the defensive 
action saves the day for the 
sincere CW.

In this case the burial doesn't backfire on the defecting faction, but it 
doesn't do any good either.  

Forest

Election-Methods mailing list - see http://electorama.com/em for list info

[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[C.Benham]

Regarding my example

31: A>B
32: B>C
37: C>A

Forest:
>
> >I've come around to the belief that most Condorcet cycles in
> ordinary elections
> >are artificial, so chances are that this cycle was created from
> the burial of B
> >by the C faction. Giving C the win only rewards this manipulation.
>
>
> Chris:
> I can't see any remotely rational justification for assuming
> that this is the case
> rather than, say, the cycle was created by the A voters burying C.

Forest:

Well, usually the largest faction is the one with the best chance of 
getting away with it, and would get
away with it under Beatpath, Ranked pairs, etc. unless the Condorcet 
supporters took defensive action. 
With TACC no defensive action was necessary.


Now let's consider the possibility that you suggest, namely that the 
true preferences were


31 A>C
32 B>C
37 C>A

If there is enough information for the A faction to think it is safe 
to bury C, then there is enough
information for the C faction to take the precaution of truncating A 
defensively.


The we have


31 A>B
32 B>C
37 C


That's why I think this scenario is less likely than the one I suggested.


Chris:
Yes, but only somewhat. This all assumes that the A faction's pairwise 
preferences are all about equally
strong, that accurate pre-poll data is available to all the factions, 
and that the C faction voters are strategically

minded.

It is nothing like a sufficient counter to my original central point, 
that on the ballots as voted C is the strongest

candidate and solidly dominates the TACC winner A.

Interpreting the ballots as 3-slot ratings, C has the highest Approval 
score and the the highest Top-Ratings

score. And C pairwise beats A

Let me modify my example to further strengthen C and weaken A:

25: A>B  ("sincere" is A)
06: A>C
32: B>C
27: C>A
10: C

Approvals: B57,   A58,   C75.  ("Top-Ratings" as before). 
A>B 58-32,   B>C 57-43,  C>A 69-31.


TACC still elects A.

C is the "sincere CW". C's voted pairwise defeat is the weakest (as 
measured by both Winning Votes and

Margins) while A's is (by those same measures) is the strongest.

Chris Benham








Election-Methods mailing list - see http://electorama.com/em for list info

[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[C.Benham]

Forest wrote (13 Nov. 2010):


I'm not a die hard Condorcet supporter. In fact my truly favorite 
methods are neither Condorcet efficient
nor deterministic; hence the title of this thread is intended to 
connote a deliberate restriction of attention
to lesser evil methods that might be acceptable to Condorcet 
enthusiasts.  So far most Condorcet
supporters seem to think that we have to have cycles, and 
therefore.the important thing is how to deal

with them rather than how to prevent them.



Nor am I a die-hard Condoret supporter, but I'm intolerant of methods 
that aren't deterministic.


I have sympathy for the philosophical view that the winner must come 
from the Smith or Schwartz set.,
but not for the view that there aren't other desirable "representative" 
criteria regarding which member

of that set we elect.

25: A>B
06: A>C
32: B>C
27: C>A
10: C

TACC's election of  A here is unacceptably silly because C is so 
dominant over A.


I consider not electing C here somewhat embarrassing, but I have 
defended a couple

of methods that elect B: IRV and  Smith,IRV.

But IRV is completely invulnerable to Burial strategy, and Smith,IRV  is 
a Condorcet method
that keeps some of that IRV quality: Mutual Dominant Third candidates 
are invulnerable to

Burial.

In the example above we can see that C could be a sincere DMT candidate 
that has been

successfully buried by the 25 A>B voters (sincere may be A or A>C)  in TACC.

I think that  if for the sake of  defensive strategy  and/or higher 
Social Utility we encourage voters to
truncate, then it is better to dump the Condorcet criterion in favour 
of  the Favourite Betrayal criterion

(while "making do" with other representative criteria compliances.)

So I certainly prefer IBIFA (my favourite FBC method) to TACC and 
Winning Votes and Margins.


Chris Benham

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Jameson Quinn]

2010/11/8 

> A few years ago Jobst invented Total Approval Chain Climbing or TACC for
> short.
>
> At the time I was too young (not yet sixty) to really appreciate how good
> it was.  It is a monotonic. clone
> free, Condorcet Efficinet method which always elects from the "Banks Set,"
> a nice subset of the Smith
> Set (if not the entire Smith Set).
>
> It is easy to describe:
>
> (1) Initialize the variable S as the empty set  S = { }.
>
> (2)  While some alternative beats every member of S pairwise, augment the
> list S with the lowest
> approval alternative that does so.
>
> (3) Elect the last alternative added to S, i.e. the member of S with the
> greatest approval.
>
> That's it.
>
> Obviously the method will elect the CW when there is one.
>
> If the Smith set consists of a cycle of three alternatives, say A beats B
> beats C beats A,, then this
> method (TACC) will will elect either the member of this cycle with the
> greatest approval or the one with
> the second greatest approval, depending on whether or not the cyclic order
> goes up or down the approval
> list.
>
> What I didn't appreciate in my younger days was how beautifully resistant
> the method is to strategic
> manipulation.
>
> Scenario 1:
>
> 49 C
> 27 A>B
> 24 B  (sincere is B>A)
>
> The sincere CW is A, but the B faction creats an ABCA cycle by rruncation.
>
> Assuming that "approval" is the same as "ranked" in each of the factions,
> the approva order is (from
> greatest to least)  B, C, A .  Since this is in the same cyclic order as
> the cycle,  C wins.  If the B voters
> are rational, they will not truncate A!
>
> Now look at the burial temptation scenario:
>
> 49  A>B (sincere is A>C)
> 27  B>C
> 24  C>A
>
> The sincere CW is C.
>
> Now suppose that the A faction buries C as indicated above:
>
> TACC will elect B. whether or not the A faction approves B.
>

But if the A faction votes A>B>C (ie, if they approve C), then C wins. So I
think that this method would work best with only 3 rating levels (only 2
approval levels) available.

>
> Forest
>
> 
> Election-Methods mailing list - see http://electorama.com/em for list info
>

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Kristofer Munsterhjelm]

Jameson Quinn wrote:



2010/11/8 mailto:fsimm...@pcc.edu>>

A few years ago Jobst invented Total Approval Chain Climbing or TACC
for short.

At the time I was too young (not yet sixty) to really appreciate how
good it was.  It is a monotonic. clone
free, Condorcet Efficinet method which always elects from the "Banks
Set," a nice subset of the Smith
Set (if not the entire Smith Set).

It is easy to describe:

(1) Initialize the variable S as the empty set  S = { }.

(2)  While some alternative beats every member of S pairwise,
augment the list S with the lowest
approval alternative that does so.

(3) Elect the last alternative added to S, i.e. the member of S with
the greatest approval.

That's it.

Obviously the method will elect the CW when there is one.

(...)


49  A>B (sincere is A>C)
27  B>C
24  C>A

The sincere CW is C.

Now suppose that the A faction buries C as indicated above:

TACC will elect B. whether or not the A faction approves B.


But if the A faction votes A>B>C (ie, if they approve C), then C wins. 
So I think that this method would work best with only 3 rating levels 
(only 2 approval levels) available.


Would this method be any good if "Approval" was changed into some other 
method that's burial resistant and monotone, like Plurality or Bucklin?


For Plurality, the "score" would simply be the number of first place 
votes, breaking ties by the number of second place votes, breaking ties 
by the number of third place votes, etc.


For Bucklin, the score for a candidate would be the number of candidates 
minus the round at which it attains a majority, breaking ties by how far 
above the majority it got. QLTD might produce even fewer ties.


Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Kevin Venzke]

Hi Kristofer,

--- En date de : Mar 9.11.10, Kristofer Munsterhjelm  a 
écrit :
> Would this method be any good if "Approval" was changed
> into some other method that's burial resistant and monotone,
> like Plurality or Bucklin?

Bucklin maybe, but the trick seems to be that it's hard to bury without
altering approval order. In a cycle TACC elects the candidate who defeats
the approval loser. I doubt there are any scenarios where burial succeeds
whether or not the final approval order is changed.

I'm not fond of the minimal defense failure... It seems to me the
"train wreck" disincentive to defect is more likely to lead to favorite
betrayal (and nomination disincentive) than cooperation.

That is:

49 Bush
24 Gore
27 Nader>Gore

If this is a Bush win, and people think the votes might actually look like
this, it takes much less convincing to make the Nader voters create a
Gore win via compromising, than to make Gore voters add tons of Nader 
second preferences. It's not even clear here that Gore voters are
interested in Nader, in which case a Bush win is purely a favorite 
betrayal incentive.

One might argue that Nader voters won't compromise when Nader seems
stronger. But if Gore voters are reluctant to vote for Nader (no matter
what the reason is), Nader isn't stronger in any meaningful way. He can't
win the election.

Kevin



  

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[fsimmons]

Chris wrote ...

> 31: A>B
> 32: B>C
> 37: C>A
>
> Approvals: B63, A68, C69. A>B>C>A.
>
> TACC elects A, but C is positionally the dominant candidate and
> pairwise beats A.
>
> For a Condorcet method with pretension to mathematical elegance,
> I don't
> see how that
> can be justified.
>
> Chris Benham
>
> PS: Could someone please refresh our memories: What is the
> "Banks Set"?

Forest Replies:

As you know C is the DMC winner, and would be a slightly better winner, given
that the ballots are sincere.  But DMC is not as burial resistant and truncation
resistant as TACC.  

It is interesting that DMC and TACC have opposite rules for which of the top two
approval members of the top cycle (of three) wins.  DMC awards the win to the
one (of these two) that beats the other.  TACC awards the win to the one that is
beaten by the other.

DMC is slightly better for elections where no voter would think of burial or
truncation.  But in the real world, I think the small sacrifice in goodness of
the result for sincere voting is worth the greatly increased
truncation and burial resistance.

I've come around to the belief that most Condorcet cycles in ordinary elections
are artificial, so chances are that this cycle was created from the burial of B
by the C faction.  Giving C the win only rewards this manipulation.  I think
that within the constraints of deterministic methods, TACC's resistance to
burial and truncation is more valuable than DMC's slightly better results for
sincere ballots.

With a stochastic analog of TACC that we could call RBOCC for Random Ballot
Order Chain Climbing, we get a better result for this scenario than either TACC
or DMC:

Prob(A wins)=37%
Prob(B wins)=31%
Prob(C wins)=32%

Since A has a higher winning probability than C, the C faction will regret
burying B provided that (to them) B was higher than about 46% (fraction
32/(32+37)) of the way between A and C in utility.  In other words if they
valued B more than a lottery in which there was only a 46 percent probability of
getting C, but a 54 percent probability of getting A, then they cannot
rationally bury B.

Banks:

An alternative B is in the Banks set iff there exists some subset S of
alternatives such that (1) S is linearly ordered by the pairwise "beats"
relation, and (2) no alternative outside of S pairwise beats all of the members
of S, and (3) B pairwise beats every member of S except itself.

In other words, B is the max member of a maximally extended chain.

It follows that every Banks alternative is uncovered, but there are cases where
some uncovered alternatives are not Banks alternatives.  So the category Banks
is more exclusive than the category Uncovered.  In particualr, if an alternative
A is merely uncovered, but not a member of Banks, then it would be irrational
for a player in the following Rivest Game
to bet on A:

Two players choose one alternative each.  Then a random ballot is drawn.  The
player whose choice is preferred over the other player's choice on the random
ballot must give a dollar to the other player.

Forest

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[fsimmons]

> I'm not fond of the minimal defense failure... It seems to me the
> "train wreck" disincentive to defect is more likely to lead to
> favoritebetrayal (and nomination disincentive) than cooperation.
>
> That is:
>
> 49 Bush
> 24 Gore
> 27 Nader>Gore
>
> If this is a Bush win, and people think the votes might actually
> look like
> this, it takes much less convincing to make the Nader voters
> create a
> Gore win via compromising, than to make Gore voters add tons of
> Nader
> second preferences. It's not even clear here that Gore voters are
> interested in Nader, in which case a Bush win is purely a
> favorite
> betrayal incentive.
>
> One might argue that Nader voters won't compromise when Nader seems
> stronger. But if Gore voters are reluctant to vote for Nader (no
> matterwhat the reason is), Nader isn't stronger in any
> meaningful way. He can't
> win the election.

Forest Replies:

As we know, in the above scenario the result that is best assuming sincere votes
is not the same as the result that is best assuming that the Gore faction
truncated in order to create the cycle.

Assuming sincere votes giving Bush the win encourges favorite betrayal. 
Assuming insincere truncation of Nader in the second faction and giving Gore the
win rewards that truncation.  Nader cannot be given the win because of the
Plurality criterion.

So there is no completely satisfactory deterministic solution.  

However, on the whole, if we want determinism and Condorcet efficiency,  it
seems more important to prevent insincere cycle producing truncation than
worrying about lack of FBC compliance, since (1) the FBC and Condorcet
efficiency are incompatible, any way, and (2) sincere Condorcet cycles are rare
(in my opinion).  

Now consider the following propositions (relative to the above scenario):

(1) Woodall's Plurality Criterion says that Nader should have less probability
than Bush.  

(2) It seems that Gore and Nader together should have more probability than 
Bush.

(3) Gore should not have more probability than Bush, because that would
encourage truncation in another case where the sincere ballots of the middle
faction turn out to be Gore>Nader .

No deterministic method can satisfy these three conditions.

But the following probabilities are consistent with all three:

Prob(Bush)=38.5%
Prob(Gore)=35.5%
Prob(Nader)=26%

These probabilities came from the stochastic analog of TACC that I call RBOCC
for Random Ballot Order Chain Climbing.

Forest

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Kathy Dopp]

> Date: Mon, 08 Nov 2010 23:58:43 + (GMT)
> From: fsimm...@pcc.edu
>
> A few years ago Jobst invented Total Approval Chain Climbing or TACC for 
> short.
>
> At the time I was too young (not yet sixty) to really appreciate how good it 
> was.  It is a monotonic. clone
> free, Condorcet Efficinet method which always elects from the "Banks Set," a 
> nice subset of the Smith
> Set (if not the entire Smith Set).
>
> It is easy to describe:
>
> (1) Initialize the variable S as the empty set  S = { }.
>
> (2)  While some alternative beats every member of S pairwise, augment the 
> list S with the lowest
> approval alternative that does so.
>
> (3) Elect the last alternative added to S, i.e. the member of S with the 
> greatest approval.
>
> That's it.
>
> Obviously the method will elect the CW when there is one.
>
> If the Smith set consists of a cycle of three alternatives, say A beats B 
> beats C beats A,, then this
> method (TACC) will will elect either the member of this cycle with the 
> greatest approval or the one with
> the second greatest approval, depending on whether or not the cyclic order 
> goes up or down the approval
> list.
>
> What I didn't appreciate in my younger days was how beautifully resistant the 
> method is to strategic
> manipulation.
>
> Scenario 1:
>
> 49 C
> 27 A>B
> 24 B  (sincere is B>A)
>
> The sincere CW is A, but the B faction creats an ABCA cycle by rruncation.
>
> Assuming that "approval" is the same as "ranked" in each of the factions, the 
> approva order is (from
> greatest to least)  B, C, A .  Since this is in the same cyclic order as the 
> cycle,  C wins.  If the B voters
> are rational, they will not truncate A!
>
> Now look at the burial temptation scenario:
>
> 49  A>B (sincere is A>C)
> 27  B>C
> 24  C>A
>
> The sincere CW is C.
>
> Now suppose that the A faction buries C as indicated above:
>
> TACC will elect B. whether or not the A faction approves B.
>
> Forest
>

I like it a lot Forest and believe it would make most voters happy
with the outcome. Great that it's monotonic and strategy resistant so
encourages sincere voting.

What does "clone free" mean again please?

It is a little tricky to count, not very intuitive, so I wonder how to
explain it so most people could understand its logic?  The logic seems
solid.

It is precinct-summable with an n x n matrix (n = # candidates) using
the diagonal for the # approval votes for each candidate.  Not sure
how to do the sampling mathematics for post-election audits - might be
a challenge to figure out how to limit the risk of certifying an
incorrect candidate.

Do you have a multi-winner version or a proportional representation version?

Regards,
-- 

Kathy Dopp
http://electionmathematics.org
Town of Colonie, NY 12304
"One of the best ways to keep any conversation civil is to support the
discussion with true facts."

Fundamentals of Verifiable Elections
http://kathydopp.com/wordpress/?p=174

Realities Mar Instant Runoff Voting
http://electionmathematics.org/ucvAnalysis/US/RCV-IRV/InstantRunoffVotingFlaws.pdf

View some of my research on my SSRN Author page:
http://ssrn.com/author=1451051

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Kevin Venzke]

Hi Forest,

--- En date de : Mar 9.11.10, fsimm...@pcc.edu  a écrit :
> As we know, in the above scenario the result that is best
> assuming sincere votes
> is not the same as the result that is best assuming that
> the Gore faction
> truncated in order to create the cycle.
> 
> Assuming sincere votes giving Bush the win encourges
> favorite betrayal. 
> Assuming insincere truncation of Nader in the second
> faction and giving Gore the
> win rewards that truncation.  Nader cannot be given
> the win because of the
> Plurality criterion.
> 
> So there is no completely satisfactory deterministic
> solution.  

I agree technically, but on these ballots I don't think it's likely that
any special incentive is operating on the Gore voters, or that one can
be found that would make them completely stop their truncation.

In my opinion this unpleasant situation will tend to be solved prior to
election day. As I've said before, there is too little to gain and 
everything to lose in one "side" running two strong candidates. So, 
nomination disincentive is actually unavoidable, I think.

I don't usually regard truncation as insincere in the same way that
burial or compromise is insincere, but it seems particularly unusual to
call it insincere in the context of an approval-based method.

> However, on the whole, if we want determinism and Condorcet
> efficiency,  it
> seems more important to prevent insincere cycle producing
> truncation than
> worrying about lack of FBC compliance, since (1) the FBC
> and Condorcet
> efficiency are incompatible, any way, and (2) sincere
> Condorcet cycles are rare
> (in my opinion).

Well for (1), also burial resistance and LNHarm are incompatible with
Condorcet. I'm also not talking about strict FBC compliance, but just
minimal defense, which is basically a weak form of FBC for majorities.

For (2) I am not sure what all you include in "sincere." For strictly
ranked sincere ballots, I agree that cycles should be rare. If truncation
(as under an approval-based method) isn't considered insincere then I
think sincere cycles need not be uncommon. If they are, I would guess it's
because in practice not enough viable candidates are nominated to see the
problem.

> Now consider the following propositions (relative to the
> above scenario):
> 
> (1) Woodall's Plurality Criterion says that Nader should
> have less probability
> than Bush.  
> 
> (2) It seems that Gore and Nader together should have more
> probability than Bush.
> 
> (3) Gore should not have more probability than Bush,
> because that would
> encourage truncation in another case where the sincere
> ballots of the middle
> faction turn out to be Gore>Nader .
> 
> No deterministic method can satisfy these three
> conditions.

The way I phrase this problem with this scenario is that no method can
simultaneously satisfy Plurality, Minimal Defense, and LNHarm. Which one
to lose (with 3 candidates) depends on what you're going for...

Personally, the methods I'm interested in all either satisfy MD or LNHarm.
I don't really care about Condorcet except as it can service MD or other
criteria that protect literal majorities.

Kevin


  

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Kristofer Munsterhjelm]

Kathy Dopp wrote:


What does "clone free" mean again please?


It means that if people always vote a certain set of candidates next to 
each other, but not necessarily in the same order, the probability that 
the win comes from that set is independent of how many are in that set. 
It's intended to model resistance against vote-splitting and teaming: if 
a party splits in two or two parties always voted next to each other 
join, and nothing else happens, it won't affect the outcome.



It is a little tricky to count, not very intuitive, so I wonder how to
explain it so most people could understand its logic?  The logic seems
solid.


You might do it by saying that the method admits stronger and stronger 
candidates into the set, according to the base method (Approval in this 
case), until it finds the strongest candidate that is also strong (beats 
the other members) in the pairwise sense.


The problem, however, would be to explain why having a candidate that is 
strong in both respects is desirable, and better than a candidate that 
is stronger in one respect but not strong at all in the other.



It is precinct-summable with an n x n matrix (n = # candidates) using
the diagonal for the # approval votes for each candidate.  Not sure
how to do the sampling mathematics for post-election audits - might be
a challenge to figure out how to limit the risk of certifying an
incorrect candidate.

Do you have a multi-winner version or a proportional representation version?


I imagine you could run CPO-STV on this - as you could with any other 
summable Condorcet method - but it would lose its summability properties 
and would not be polynomial in space or time, nor would it be monotone.


Making a PR method that reduces to a Condorcet method in the 
singlewinner case, yet is polytime, seems to be very hard. For that 
matter, making a PR method that's polytime and monotone (and better than 
SNTV) seems very hard as well.


Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[Jameson Quinn]

2010/11/10 C.Benham 

>  Chris wrote ...
>
> >* 31: A>B*>* 32: B>C*>* 37: C>A*>**>* Approvals: B63, A68, C69. 
> >A>B>C>A.*>**>* TACC elects A, but C is positionally the dominant candidate 
> >and*>* pairwise beats A.*>**>* For a Condorcet method with pretension to 
> >mathematical elegance,*>* I don't* *see how that* *can be justified.*>**>* 
> >Chris Benham*>**>* PS: Could someone please refresh our memories: What is 
> >the*>* "Banks Set"?*
> Forest Replies:
>
>
> As you know C is the DMC winner, and would be a slightly better winner, given
> that the ballots are sincere.  But DMC is not as burial resistant and 
> truncation
> resistant as TACC.
>
> It is interesting that DMC and TACC have opposite rules for which of the top 
> two
> approval members of the top cycle (of three) wins.  DMC awards the win to the
> one (of these two) that beats the other.  TACC awards the win to the one that 
> is
> beaten by the other.
>
> Chris: I have long since abandoned the "Definite Majority Choice" (DMC)
> method in favour of Smith//Approval (as my preferred Condorcet method),
> which also elects C here. I still like the Definite Majority criterion,
> which says that no candidate that is pairwise beaten by a more approved
> candidate is allowed to win. I think that (in isolation) meeting the
> Condorcet criterion is desirable, but not so holy that on discovering there
> is no voted CW the method should proceed on the assumption that there is
> really a "sincere CW" that has been victimised by strategists the method
> should try to frustrate or punish. Condorcet methods are vulnerable to
> Burial, period. Futile attempts to address this should not be at the expense
> of producing winners that can have no philosophical justification on the
> assumption that all the votes are sincere (or are all equally likely to be
> sincere). The TACC winner A simply has no shred of justification versus the
> Smith//Approval winner C. Forest:
>
> I've come around to the belief that most Condorcet cycles in ordinary 
> elections
> are artificial, so chances are that this cycle was created from the burial of 
> B
> by the C faction.  Giving C the win only rewards this manipulation.
>
> Chris: I can't see any remotely rational justification for assuming that
> this is the case rather than, say, the cycle was created by the A voters
> burying C.
>
>
The rational conclusion would be that the most probable burial, is the one
that favors the actual winner. That is, you can't assume that the ballots
exist independently of the system being used. If that is so, then Chris is
right: it is essentially futile for a Condorcet method to try to
de-incentivize burial. That's not true, though, for truncation; and
remember: in the real world, strategy will not be unanimous, so attempted
burial will tend to look similar to probabilistic truncation.

So, for its truncation resistance, TACC seems to me to be a
better-than-average, but not outstanding, Condorcet method. So far, though,
that's just based on arbitrary examples; I'd certainly be happier if there
were some rigorous truncation-resistance property which TACC could be shown
to [probabilistically?] meet.

JQ

Election-Methods mailing list - see http://electorama.com/em for list info

Re: [EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[fsimmons]

- Original Message -
From: "C.Benham" 
Date: Saturday, November 13, 2010 11:20 am
Subject: My Favorite Deterministic Condorcet Efficient Method: TACC
To: em 
Cc: Forest W Simmons 

> Regarding my example
> 
> 31: A>B
> 32: B>C
> 37: C>A
> 
> Forest:
> >
> > >I've come around to the belief that most Condorcet cycles in
> > ordinary elections
> > >are artificial, so chances are that this cycle was created from
> > the burial of B
> > >by the C faction. Giving C the win only rewards this manipulation.
> >
> >

[big snip]

> It is nothing like a sufficient counter to my original central 
> point, 
> that on the ballots as voted C is the strongest
> candidate and solidly dominates the TACC winner A.
> 
> Interpreting the ballots as 3-slot ratings, C has the highest 
> Approval 
> score and the the highest Top-Ratings
> score. And C pairwise beats A

I agree that if the ballots as voted are assumed to be sincere, then C has a 
better case than A.  And, of 
course, C can say, "We have to give them the benefit of a doubt that they are 
sincere."

I've just become more and more skeptical about Condorcet cycles being sincere. 
If my skepticism is 
justified, then, given a choice, I would rather have a method that prevents 
artificial cycles than a method 
that takes them seriously.  If the method prevents them, then how it would deal 
with them becomes 
completely moot.

Of course it would be nice to have a method that assumes sincerity and makes 
artificial cycles back-
fire, like Mr. Magoo coming out smelling like a rose while oblivious to all of 
the danger he has survived.

I am willing to admit that my skepticism could be misguided.  I would just like 
to see some evidence of 
that.  Kris M. makes a valid point; since all of our current methods in use 
tend to polarize the electorate, 
we any current cycles would be artificial, but once a Range or Condorcet method 
is adopted, this could 
change.

I'm not a die hard Condorcet supporter. In fact my truly favorite methods are 
neither Condorcet efficient 
nor deterministic; hence the title of this thread is intended to connote a 
deliberate restriction of attention 
to lesser evil methods that might be acceptable to Condorcet enthusiasts.  So 
far most Condorcet 
supporters seem to think that we have to have cycles, and therefore.the 
important thing is how to deal 
with them rather than how to prevent them.


Election-Methods mailing list - see http://electorama.com/em for list info