[EM] Rehabilitating Copeland

[Forest Simmons]

This example is based on three factions tightly distributed around their
respective candidates A, B, and C.

A and B are at the endpoints (vertices) of the longest side of the
triangle, while C is at the other vertex located closer to A than B.

Assuming voters' preferences favor closer candidates, the ballot profile
must be ...

a ACB
b BCA
c CAB

To keep things interesting we assume that no single faction has full
majority support.

The pairwise defeats are ...
A>B a+c
C>A b+c
C>B a+c

So C is the sincere Condorcet Winner, while B is the equally sincere
Condorcet Loser.

Could either of the losing factions get an advantage by voting insincerely?

Any unilateral change on the B faction ballot would either turn their
anti-favorite into the ballot Condorcet winner ... or simply reinforce C's
victory ... which might not be such a bad idea in view of the likelihood of
successful A faction shenanigans under many popular election methods ...
whether or not they satisfy the Condorcet Criterion.

For example, if the faction sizes were in the order a>b>c, then A would win
under IRV unless the B faction were to compromise in the form of favorite
Betrayal... C>B.

Suppose that the A faction were to "bury" the sincere CW by insincerely
reversing their CB preference to BC:

a AB
b BC
c CA

Then a "Condorcet Beat Cycle" would result ... with defeat strengths of

A>B (a+c), B>C (b+a), and C>A (b+c).

These respective defeat strengths are more revealing in the forms ...
n-b, n-c, and n-a, where n=a+b+c.

The classical Condorcet methods, including Ranked Pairs, Beatpath/CSSD,
MinMax, and River (no matter winning votes or margins in this context) all
elect the candidate of the largest faction because they all break the cycle
at the link with least strength n-a,n-b, or n-c, respectively, depending on
which of the respective factions A, B, or C is largest.

For example, if A is the largest faction the defeat C>A with strength n-a
is the weakest. When this defeat is annulled, the beatpath A>B>C gives the
finish order for all classical Condorcet Methods.

So back to our story ... does A's burial of the sincere CW pay?

Under classical Condorcet the answer is "yes", if the A  faction is
largest. If B is largest, it will backfire. If C is largest, it's a wash.

However the customary precaution of the CW faction truncating the other
candidates will thwart A's win.

How about under our refurbished Copeland "littlest help from friends"
method?

The respective designated helper friends for A, B; and C are B, C, and A,
respectively.

Their respective "helps" are their winning votes in befriending the
respective  faction's "enemies" C, A,and B.
The respectively helper victories over these enemies are B>C n-c, C>A n-a,
and A>B n-b.

So in this context, the help (meaning winning votes over enemy) is smallest
when the "enemy" faction is largest.

In our example, A wins when C is largest, B wins when A is largest, and C
wins when B is largest.

So A's burial of C pays off only if C is the largest faction. But the
sincere CW faction's customary policy of truncation below the CW will
thwart A's win, as in the classical case.

Our "least help" method has a built in opportunity for a sincerity check
... elect the sincere pairwise winner in a runoff  (instant or not) between
the help and the enemy (of the winner before this sincerity runnoff).

For example, if A wins by burying C ... it's because A's helper B, barely
beats C .... that is, with the weakest victory of any helper over enemy ...
so there is a perfectly good reason for confirming that win with a roll
call vote or a fresh set of ballots.

In our example, the fresh ballots or roll call present no temptation to
dissemble ... so C; the sincere CW must win the runoff contest between it
and B.

But what if (in a different context) the cycle was a natural cycle instead
of the result of conniving, unscrupulous buriers?

Consider this question: which is easier ... to create a cycle by
insincerely burying a sincere CW or to create a sincere cycle?

Take our geometric scenario for example ... no matter how you adjust the
faction sizes or the distances between them ... there will be no sincere
cycle.

In fact, with three factions it is impossible unless you separate the
candidate positions from their factions, and you carefully rotate all of
the candidates in the same sense (clockwise or widdershins) ... and by the
right amount ... and with the interior angles not too far from sixty
degrees ... not something that will happen very often at random or even by
natural bias.

In fact, sincere top cycles are much harder to contrive ... hence more rare
than insincere cycles.

And what if such a natural cycle did arise by chance, and you did get the
"wrong" cycle resolution because you were using a burial/ chicken resistant
Landau efficient method instead of Classical Condorcet?

Well, you still got a Landau winner (always a strong Smith candidate).

What you didn't get was rewarded-for-burial or rewarded for chicken
defection ... against which Classical Condorcet has no defense.

Just say'n ...

-Forest






On Wed, Mar 15, 2023, 10:30 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> First example ballot profile ... a case of Chicken Defection:
>
> 28 A>B
> 24 B (sincere B>A)
> 48 C
> The sincere CW, (namely A) has been subverted by the B faction's defection
> from a sincere  (A, B} clone coalition ... creating a ballot beat cycle of
> ABCA.
>
> A>B  28 to 72
> B>C 52 to 48, and
> C>A 48 to 28
>
> Classical Condorcet breaks the cycle at the weakest link A>B ... leaving
> the rest of the cycle BCA intact. So classical Condorcet elects B, thereby
> rewarding the defectors.
>
> How about our "littlest help from our friends" method?
>
> The the respectivevcovering pairs are the defeat pairs A>B, B>C, & C>A ...
> where the helpers are the defeat losers and the "enemies" befriended (i.e.
> covered) by the helpers are respectively C, A, & B, with respective help
> strengths of
> #[B>C]=52, #[C>A]=48, and #[A>B]=26.
>
> The last of these is the "least help" ... the help given by A to befriend
> C's  enemy B.
>
> So C is the candidate needing the least help from its friend to befriend
> its enemy.
>
> Therefore our least-help-from-friend(s) method elects C ... which
> constitutes a backfire against the defecting candidate B.
>
> Next time ... a geometrically generated family of examples...
>
> -Forest
>
>
> On Wed, Mar 15, 2023, 3:02 PM Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
>
>> Here's the Enhanced Copeland method we've been leading up to:
>>
>> Elect the candidate that needs the "littlest help from its friends" to
>> cover its enemies.
>>
>> Explanation:
>>
>> Last time we learned that if C is a candidate tied for Copeland winner,
>> then C's friends will cover its enemies ... in other words, every enemy of
>> C is a friend of a friend of C ... in other words C will be a Landau
>> candidate ... in other words,  our Enhanced Copeland Winner is the Landau
>> candidate that requires the least help from its friends to befriend its
>> enemies.
>>
>> Operationally we proceed as follows:
>>
>> 1. Each Landau candidate C (including all tied Copeland Winners)
>> designates a set of candidates H(C) as its "helpers".
>>
>> 2. A Deliberative Assembly checks the validity of these helper claims,
>> and eliminates any candidate C whose enemies are not covered by H(C).
>>
>> 3. Next the Deliberative Aassembly narrows down the candidate pool to
>> T=argmin #H(C) ... the set of candidates tied for the fewest (designated)
>> helpers needed to defeat all of its enemies.
>>
>> 4.From among the members of the  set T tied for fewest needed
>> (designated) helpers, elect the candidate C whose Max Pairwise Opposition
>> from an enemy of C to a designated helper of C, is minimal.
>>
>> This is the candidate C that requires the fewest helper friends (and the
>> least help from those so designated) in order to befriend all of its
>> enemies (especially the one hardest for a designated helper to befriend).
>>
>> In practice, no candidate in a public election will need more than one
>> helper, so the method simplifies to electing the candidate C whose
>> designated helper H most easily befriends the enemy E of C having the most
>> Pairwise Opposition against H.
>>
>> The less Pairwise Opposition that E has against H the less help (defined
>> as H's Pairwise Support against E) it takes to befriend E.
>>
>> So C is the candidate that needs the least help in befriending its
>> enemies.
>>
>>
>> Next time some examples ...
>>
>> On Tue, Mar 14, 2023, 10:11 PM Forest Simmons <forest.simmons21 at gmail.com>
>> wrote:
>>
>>> I've always had a soft spot in my heart for Copeland ... perhaps because
>>> it's such a friendly method. Let me explain ...
>>>
>>> Remember Kristofer's wry definition of friendship ... whoever doesn't
>>> beat you is your friend?
>>>
>>> By that definition the Copeland winner is simply the candidate with the
>>> most friends.
>>>
>>> And everybody is a friend of the Condorcet Winner (the unbeaten
>>> candidate if there is one) ... so Copeland is Condorcet efficient.
>>>
>>> Simple and compelling ... as far as it goes ... but it doesn't always go
>>> the whole distance ... what if there are two or three candidates tied for
>>> most friends?
>>>
>>> Various Copeland tie breaking proposals have been suggested over the
>>> years .... but none of them seem to naturally continue the basic friendship
>>> theme that inspired Copeland in the first place.
>>>
>>> Here's a friendship idea that might help:
>>>
>>> A set of candidates can have friends: you are a friend of a set if you
>>> don't beat every member of the set.
>>>
>>> Another way to say this ... is that you are a friend of a set iff the
>>> set covers you.
>>>
>>> This friendship/ covering relation leads to a natural hierarchy of
>>> candidates ...
>>>
>>> A Condorcet candidate has everybody as a friend ... and therefore covers
>>> everybody  ... without any help.
>>>
>>> A candidate  X that covers everybody with the help of only one friend F
>>> is the next level. Together X and F form a set X+F that cannot be beaten
>>> ... everybody is friendly to X+F just like everybody is friendly to a
>>> Condorcet Candidate, when there is one.
>>>
>>> So different candidates need different amounts of help from their
>>> friends to cover their "enemies", the candidates that are not their friends.
>>>
>>> The fewer friends you need to cover all of your enemies, the stronger
>>> you are.
>>>
>>> A Condorcet Winner has lots of friends and zero enemies.
>>>
>>> At the other extreme ... a Condorcet Loser has no friends to help cover
>>> his enemies .... which is unfortunate because all of the other candidates
>>> are his enemies.
>>>
>>> What about a Copeland candidate C tied for most friends?  Let F be the
>>> set of friends of C.  Then C and F together cover the enemies of C.
>>>
>>> Any time a candidate C with some of her friends F together cover all of
>>> her enemies, we say that C is a Landau candidate. So every max score
>>> Copeland candidate is a Landau candidate .... which makes Copeland a
>>> "Landau efficient" method.
>>>
>>> So the bigger the Copeland score of a candidate the fewer her enemies
>>> and the fewer friends she needs to help cover those enemies.
>>>
>>> This brings us to another imperfection of Copeland that we will take
>>> care of with the help of our friends:
>>>
>>> Suppose candidate F is a friend of candidate C and of nobody else. Then
>>> unscrupulous C supporters might be tempted to enter a whole "team" T of
>>> clones of F into the race to increase C's Copeland score and nobody else's.
>>>
>>> How can we effectively counter this kind of "Teaming" temptation?
>>>
>>> A natural way is to see how far we can pare down the friendly help while
>>> still covering the enemies of C .... just how many friends does C actually
>>> need to cover her enemies?
>>>
>>> This doesn't completely resolve the tie problem ... but it reduces the
>>> covering sets down to manageable size.
>>>
>>> In fact, the most amazing and useful thing I have learned in these
>>> friendly explorations, is that in public elections ... for all practical
>>> purposes ... it is 99.999... percent sure that at least one candidate C
>>> will have a friend F which is all the help she needs to cover her enemies.
>>>
>>> This fact hugely simplifies tie breaking and, at the same time,
>>> completely erases the clone teaming problem!
>>>
>>> To be continued ....
>>>
>>> -Forest
>>>
>>>
>>>
>>>
>>>
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