> > To my knowledge, so far only two monotone, clone free,

> uncovered methods

> > have

> > been discovered. Both of them are ways of processing given

> monotone, clone

> > free

> > lists, such as a complete ordinal ballot or a list of

> alternatives in order

> > of

> > approval.

> >

> > The first method, "chain climbing," starts at the bottom of

> the list and

> > works

> > upward. It initializes a "chain" with the lowest alternative

> of the list

> > and

> > while there is any alternative that pairwise beats all the

> current members

> > of

> > the chain, the lowest such member of the list is added to the

> chain. The

> > last

> > alternative added to the chain is the chain climbing winner.

> >

> > The second method, the "covering chain" method, starts at the

> top of the

> > list

> > and works downward. A variable X is initialized as the

> alternative highest

> > on

> > the list. While some alternative covers X, the highest such

> alternative on

> > the

> > list becomes the new value of X. The final value of X is the

> covering chain

> > winner.

> >

> > Both methods pick from the uncovered set, so they are both Condorcet

> > efficient.

> >

> > Suppose, for example, that the alternatives A, B, C, D, and E

> are arranged

> > along

> > a line in alphabetical order with C at the voter median

> position. The

> > sincere,

> > rational ordinal ballot of a voter near alternative A would

> likely list the

> > alternatives in alphabetical order. If this ballot were taken

> as the list

> > L,

> > and the two methods were applied to L, the first method (chain

> climbing)> would

> > build up the chain in the order E,D, C, so C would be the

> chain climbing

> > winner. If the second method were used, the successive values

> of X would

> > be A,

> > B, and C, so C would also be the covering chain winner.

> >

> > The two methods approach the voter median alternative from

> opposite sides.

> >

> > If C were replaced with the cycle C1 beats C2 beats C3 beats

> C1, and the

> > list

> > order became

> >

> > A, B, C1, C2, C3, D, E,

> >

> > then the chain climbing chain would build up in the order E,

> D, C3, C2,

> > with C2

> > winning, while the successive covering chain values of X would

> be A, B, C1,

> > respectively, with C1 becoming the covering chain winner.

> >

> > This illustrates the general principle that when the Smith set

> is a cycle

> > of

> > three alternatives, chain climbing is more penetrating than

> the covering

> > chain

> > method. In this context the covering chain method will always

> stop at the

> > first

> > member of the top cycle that it encounters, thus pleasing

> those voters who

> > favor

> > the list order, while the chain climbing method always

> penetrates beyond

> > the

> > first member of the top cycle that it encounters, but it may

> or may not

> > make it

> > to the highest member of the top cycle on the list.

> >

> > In the above example, the A supporter who supplied the list

> would prefer

> > the

> > covering chain winner, but C2 might be better for the

> electorate as a

> > whole.

> >

> > Chain climbing has more penetrating power in another sense,

> too. It never

> > stops

> > short of the Banks set, a special subset of the uncovered set.

> When the

> > top

> > cycle has only three members, the Banks set contains all of

> the uncovered

> > set,

> > so this Banks set penetration is something beyond that shown

> in the above

> > example.

> >

> > The covering chain method is not guaranteed to pick from

> Banks, but it has

> > a

> > nice property that chain climbing lacks, namely it satisfies

> independence> from

> > pareto dominated alternatives (IPDA).

> >

> > So we cannot say that one of these methods is uniformly better

> than the

> > other.

> >

> > Here's another instructive example. Suppose that the list of

> alternatives> is

> >

> > L = [A, D, B, C] in that order, and that the pairwise "beats"

> relation is

> >

> > C beats B beats D beats A coupled with D beats C beats A beats B.

> >

> > Then, all of the alternatives are in the Smith set, but A is

> covered only

> > by C,

> > so C is the covering chain winner. On the other hand the

> chain climbing

> > sequence is C followed by D, so D is the chain climbing

> winner. This is

> > the

> > first example we have seen where the chain climbing winner

> comes out higher

> > on

> > the list than the covering chain winner. That can never

> happen when the

> > top

> > cycle (Smith set) has fewer than four members.

> >

> > Now suppose that we bubble sort the list L to get the list

> >

> > L' = [D, C, A, B]

> >

> > If the original list L were the approval order, then this list

> would be the

> > Majority Choice Approval order after the bubble sort, and we

> would discover

> > alternative D to be the MCA winner.

> >

> > The list L' is fair game for our two methods, since bubble sorting

> > preserves

> > monotonicity and clone independence, so let's see how they

> compare with MCA

> > in

> > this example:

> >

> > Chain climbing builds up its chain in the order B, A, C, and

> stops with C,

> > since

> > D does not beat B. So C is the chain climbing winner for the

> list L'. On

> > the

> > other hand, C at the top of list L' is uncovered, so C is the

> covering> chain

> > winner for list L'. So in this example, our two methods

> switch winners as

> > we go

> > from the unsorted list L to the bubble sorted list L'.

> >

> > So (under the assumption that L is the approval order) the MCA

> winner D is

> > the

> > chain climbing winner for the unsorted list L, as well as the

> covering> chain

> > winner for the sorted list L', while the MCA runner-up C is

> the covering

> > chain

> > winner for the unsorted list L and the chain climbing winner

> for the sorted

> > list

> > L' .

> >

> > Since MCA is not guaranteed to pick from the uncovered set, I

> am not

> > suggesting

> > that we consider it in the same class as our two uncovered

> methods. But

> > the

> > above comparison at least makes a connection with something

> more familiar

> > to

> > most of us.

> >

> > I do suggest the following:

> >

> > In any context where being as faithful as possible to the

> original list

> > order is

> > considered important, perhaps because the only reason for not

> automatically> electing the top of the list is a desire to

> satisfy Condorcet efficiency,

> > then

> > in this case I suggest computing both the chain climbing

> winner and the

> > covering

> > chain winner for the list L, and then going with which ever of

> the two

> > comes out

> > higher on L.

> >

> > On the other hand, when the list L is just considered a

> convenient starting

> > place, with no other special importance, then I suggest bubble

> sorting L to

> > get

> > L', and then flip a coin to decide which of the two methods to

> use for

> > processing this sorted list.

> >

> > This coin flip (after the ballots have been voted) will

> largely foil

> > attempts at

> > manipulation. But it can also be thought of in the following

> way: the two

> > different methods tend to approach the "center" of the voter

> distribution> from

> > opposite sides, so they can be thought of as bracketing the

> "center." The

> > coin

> > flip is an averaging mechanism that takes us closer (on

> average) to the

> > "center."

> >

> > If the coin flip is unpalatable, then apply only chain

> climbing to L',

> > since

> > chain climbing is more apt to penetrate to the Banks set in general.

> >

> > Thought and comments?

> >

> > Thanks,

> >

> > Forest

> > ----

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> >

>