[EM] Re: Approval/Condorcet Hybrids

[Forest Simmons]

> From: Ted Stern <tedstern at u.washington.edu>
> Subject: Re: [EM] Re: Approval/Condorcet Hybrids
> To: Jobst Heitzig <heitzig-j at web.de>
> Cc: election-methods-electorama.com at electorama.com,	Ted Stern
> 	<tedstern at u.washington.edu>
>
> On 11 Jan 2005 at 14:40 PST, Jobst Heitzig wrote:
>>>
>>> But ... your argument that, if W differs from A, this implies "that W
>>> beat every candidate that A beats head to head" does not follow.  It
>>> only implies that W has highest approval in U(A).
>>
>> No, Forest is right, he defined:
>>>> Let U(A) be the set of uncovered candidates that cover the approval
>>>> winner A.
>>
>> Hence every member of U(A) not only defeats A but covers A (or is equal
>> to A), so it defeats all candidates A defeats, by definition!
>
> Unless I'm confusing something, "cover" doesn't mean direct defeat.  It means
> there is a beatpath of length 1 or 2.

A beat path of length 1 or 2 to each other candidate means "uncovered."

If A covers B then A beats (in one step) every candidate that B beats (in 
one step).


>
> Maybe I'm arguing myself in circles.  U(A) is the set of candidates that cover
> A.  If this set is of size > 1 and includes at least one candidate besides A,
> they don't cover each other.  If W has highest approval in that set, there
> might be another candidate that defeats W.  I guess I wanted to say that W
> doesn't have a particularly strong property other than the highest approval
> one.
>
> BTW, how does one choose the Dutta minimal covering set from the set of
> uncovered candidates?
>

I'll let Jobst answer this one.


Forest