[EM] Discounting ties, how can MinMax differ from Ranked Pairs? ooops!
John
john.r.moser at gmail.com
Wed Jun 12 05:36:27 PDT 2019
On Wed, Jun 12, 2019 at 1:37 AM C.Benham <cbenham at adam.com.au> wrote:
> They can't when there are three candidates. But plain MinMax doesn't meet
> Clone-Winner
>
> You can have a trio of clones in a cycle who all narrowly pairwise beat
> another single candidate, but
> whose strongest defeats (which are only to each other) are stronger than
> any of that single (Condorcet Loser) candidate's
> defeats (to all of them).
>
> But Smith//MinMax doesn't have that problem and will nearly always give
> the same winner as Ranked Pairs
> (and Schulze and River).
>
> A very small consolation prize for MinMax(Margins) is that unlike those
> methods it meets Mono-add-Top (like IRV).
>
> How does all of this compare to Tideman's Alternative?
1. Identify Condorcet winner;
2. If no Condorcet winner, identify Top Cycle;
3. Eliminate all candidates not in Top Cycle;
4. Eliminate remaining candidate with fewest votes;
5. Go to 1.
I tend to use the Schwartz Set for the Condorcet Winner, and the Smith Set
for the Top Cycle. I've found the Schwartz Set is nearly always one
candidate.
I know TA is independent of smith-dominated alternatives, but I haven't
determined if it's LIIA or reversal-symmetry (apparently Schulze isn't LIIA
and ranked pairs is, but I don't see how). I doubt it satisfies
later-no-harm and later-no-help, but the run-off step at 4 might allow it
to.
> Chris Benham
>
> On 12/06/2019 2:25 pm, robert bristow-johnson wrote:
>
>
>
> I said some things wrong. First of all, if there are N candidates, the
> number of Pairwise Defeats is N⋅(N-1)/2, not double that value.
>
> Secondly, I did the scenario with the cycle wrong. I wanted several N-3
> values in the decreasing sequence, not just one. But I hope you have the
> idea.
>
> r b-j
>
>
>
> ---------------------------- Original Message ----------------------------
> Subject: [EM] Discounting ties, how can MinMax differ from Ranked Pairs?
> From: "robert bristow-johnson" <rbj at audioimagination.com>
> <rbj at audioimagination.com>
> Date: Tue, June 11, 2019 9:46 pm
> To: election-methods at electorama.com
> --------------------------------------------------------------------------
>
> >
> >
> >
> > Let's assume Margins, but I think this works for Winning Votes also.
> > Begin with all candidates marked as "Plausible Winner" (which means they
> are not yet marked as "beaten") and order all N⋅(N-1) candidate Pairwise
> Defeats from strongest defeat strength to weakest defeat strength. Call
> that ordered list the "Original List".
> > Now create another ordered list of Pairwise Defeats from the above
> Original List.
> >
> > MinMax starts with
> > the entire Original List with N⋅(N-1) entries and
> successively eliminates entries from the bottom up.
> > Referring to a quote in this Wikipedia article:
> https://en.wikipedia.org/wiki/Minimax_Condorcet_method#Variants_of_the_pairwise_score ,
> MinMax starts at the bottom and:
> > "Disregards the weakest Pairwise Defeat until one candidate is
> unbeaten."
> >
> >
> >
> > Similarly Ranked Pairs starts with an empty list and successively
> includes entries from the Original List from the top down.
> > Cannot Ranked Pairs be concisely stated as:
> > "Include the strongest Pairwise Defeat until only one candidate is
> unbeaten." ?
> >
> > Is that not an accurate description of both
> > methods?
> > Now, a decreasing sequence of numbers (*not* strictly decreasing) of
> length N⋅(N-1)+1 that represents the number of "unbeaten" candidates (of
> all Pairwise Defeats above it) having
> > value of N at the top, that can be thought of as having each of the
> N⋅(N-1) Pairwise Defeats inserted between these numbers of decreasing
> integer value.
> > Assuming no ties, the value of this sequence at the bottom
> > must be either 0 or 1. Also, assuming no ties, this decreasing sequence
> can only be decremented by 1 or 0.
> > If the value of this sequence is 1 at the bottom, there is a Condorcet
> Winner and all Pairwise Defeats are considered. But regardless of whether
> MinMax or RP is used, this
> > ordered list is the same decreasing sequence of integer values. If the
> value of this sequence of numbers is 0 at the bottom, there is no Condorcet
> Winner. But, for all Pairwise Defeats having "1" below them, would not the
> undefeated candidate be the same
> > candidate?
> > This might look like:
> >
> > Plausible Winners ----- Pairwise Defeat
> > N
> > A>D
> > N-1
> > A>C
> > N-2
> > B>D
> > N-2
> > B>A
> > N-3
> > B>C
> > N-3
> > C>D
> > N-3
> >
> > Now if N=4 then the bottom N-3 is 1 and we have a Condorcet Winner.
> > Now suppose we have a cycle:
> > Plausible Winners ----- Pairwise Defeat
> > N
> > A>D
> > N-1
> > A>B
> > N-2
> > B>D
> > N-2
> > C>D
> > N-2
> > B>C
> > N-3
> > C>A
> > N-4
> >
> > Now if N=4 then the bottom N-4 is 0 and we don't have a Condorcet
> Winner. But the unbeaten candidate is A whether we include the B>C
> Pairwise Defeats or not. Assuming no ties, how can the unbeaten candidate
> be different?
> > You guys may have discussed
> > this before when I wasn't paying attention, but it seems to me that if
> there are no ties and all of the Defeat strengths are unequal values, the
> ordering of this list must be the same and the number of Plausible Winners
> must decrease from N at the top to 1 at the bottom if there is a CW (or N
> to 0
> > if no CW). How can MinMax and Ranked Pairs elect a different candidate?
> > Thank you for any attention and thought put to this.
> > --
> >
> >
> > r b-j rbj at audioimagination.com
> >
> >
> >
> > "Imagination is more important than knowledge."
> >
> >
> >
> >
> >
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> >
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>
> --
>
> r b-j rbj at audioimagination.com
>
> "Imagination is more important than knowledge."
>
>
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