[EM] Theorem equating Nanson AV-Borda with Condorcet: 3 candidates case shown
David Catchpole
s349436 at student.uq.edu.au
Sun Sep 10 15:43:46 PDT 2000
No, Nanson is _not_ a multi-winner method (there are Condorcet analogues
for multi-winner elections, but they have a general structure that their
Condorcet-analogue sets of winners, if they exist, are selected, using a
selected method, say STV, when compared to each of the other
candidates). It's explicitly a single-winner method. It's also an
eminently provable feature of Nanson that it elects a Condorcet winner
if she exists. All you have to demonstrate is that if there exists a
Condorcet winner, the Condorcet winner will have a higher Borda score
than some other candidate.
The reason I brought it up is because it seems apparent that in order for
truncation resistance to be satisfied, for every possible voting scenario,
e.g.
4 A>B>C
2 B>A>C
3 C>B>A
A method must be chosen such that one and only one of the consequent
"truncation scenarios" is held, e.g.-
(A) 4 A>[B,C]
2 B>A>C
3 C>B>A
elects A; or
(B) 4 A>B>C
2 B>[A,C]
3 C>B>A
elects B; or
(C) 4 A>B>C
2 B>A>C
3 C>[A,B]
elects C.
In other words, we have to find a unique property of one and only one
truncation scenario. Now, my first thought was "if there's a Condorcet
winner, she's still the Condorcet winner of 'her' truncation scenario
(e.g. (B)) and she remains in the majority against every other candidate
in 'their' truncation scenarios (e.g. B>C in (C))." Now- it's only
speculation, but I guessed the Nanson method 'cause it's more
"immediate" than other Condorcet completion methods.
***
The thing is, is it or is it not true, that for all "truncation
scenarios" the "Nanson winner" is not the candidate up to whom truncation
has occurred unless that candidate is the Nanson winner for the original
scenario?
***
In the case given,
(A) 4 A>[B,C]
2 B>A>C
3 C>B>A
"Negative scale" Borda scores- A B C
-17 -18 -19
C gets excluded. B wins.
(B) 4 A>B>C
2 B>[A,C]
3 C>B>A
"Negative scale" Borda scores- A B C
-18 -16 -20
C gets excluded. B wins (surprise).
(C) 4 A>B>C
2 B>A>C
3 C>[A,B]
"Negative scale" Borda scores- A B C
-15.5 -17.5 -21
C gets excluded. A wins.
In the "original scenario,"
4 A>B>C
2 B>A>C
3 C>B>A
"Negative scale" Borda scores- A B C
-17 -16 -21
C gets excluded. B wins (surprise).
So in this case, only (B) is satisfied (yay-that's what we wanted!). And
the result derived from that is the same as that using the explicit Nanson
method. Yay!
Thanks to Joe for reminding us what the Nanson method is called. Nanson
was a turn-of-the-century Australian mathematician who is remembered most
for being a poor administrator. But I couldn't remember his name. Might be
that as a member of the Australian Left I'm averse to names that end with
-anson!
On Mon, 11 Sep 2000, Craig Carey wrote:
> >From: Joe Malkevitch 'Email: joeyc at cunyvm.cuny.edu'
> >Date: Mon Sep 11, 2000 2:58am
> >Subject: [EM] Re: Sets of vertices leads nowhere; Mike Ossipoff
> >
> >
> >The election method which is an elimination method based on
> > the Borda count is usually known as Nanson's method. It has
> > the very nice property that if there is a condorcet winner
> > that the method chooses that candidate.
> >
> >Cheers, Joe
> >Joe Malkevitch Department of Mathematics,
> >York College (CUNY) Jamaica, NY 11451
> >Web Page: http://www.york.cuny.edu/~malk
>
>
> I aim to show the statement false, but the attempt fails, by the
> bottom of the message.
>
> This Alternative Vote method with Borda to calculate the 1st
> preference sums.
>
> a AB. (Sa,Sb,Sc) = (2, 1, 0 ).a = (4, 2, 0).a / 2
> b B.. (Sb,Sa,Sc) = (2, 1/2, 1/2).b = (4, 1, 1).b / 2
> c C.. (Sc,Sa,Sb) = (2, 1/2, 1/2).c = (4, 1, 1).c / 2
>
> Sum: (Sa,Sb,Sc) = (4a+b+c, 2a+4b+c, b+4c)
>
> (Sa<Sb) = (4a+b+c < 2a+4b+c) = (2a < 3b)
> (Sa<Sc) = (4a+b+c < b+4c) = (4a < 3c)
> (Sb<Sc) = (2a+4b+c < b+4c) = (2a+3b < 3c)
>
> /-------------------------------------------------------------------
> Check that the lines made by changing '<' into '=' meet:
> Elim a: 3c=2a+3b <--> 3c=6b, and 2=2a+2b+2c = 5b+2c
> Elim c: 6c=12b, 6=15b+6c, --> 6=15b+12b=27b <--> 2=9b, b=2/9
> --> c = (6/3)(2/9)=4/9,
> a = 3/9
> --> 9a = 3, 9b = 2, 9c = 4.
> (Sa,Sb,Sc) = (4a+b+c, 2a+4b+c, b+4c) / 9
> 9.(Sa,Sb,Sc) = (12+2+4, 6+8 +4, 2+16) = (18, 18, 18)
> \-------------------------------------------------------------------
> Case stage 1 & (A loses), (2a<3b)(4a<3c)
>
> a+b B. (c<a+b)
> c C. (a+b<c)
>
> Case stage 1 & (B loses), (3b<2a)(2a+3b<3c)
>
> a A. (c<a)
> c C. (a<c)
>
> Case stage 1 & (C loses), (3c<4a)(3c<2a+3b)
>
> a AB (b<a)
> b B. (a<b)
>
> +-------------------------------------------------------------------
> (A wins) = (3b<2a)(2a+3b<3c)(c<a) or (3c<4a)(3c<2a+3b)(b<a) = aW1
> (B wins) = (2a<3b)(4a<3c)(c<a+b) or (3c<4a)(3c<2a+3b)(a<b) = bW1
> (C wins) = (2a<3b)(4a<3c)(a+b<c) or (3b<2a)(2a+3b<3c)(a<c) = cW1
>
> aW1: (c<a) => (3c<4a)
> bW1: (a<b) => (2a<3b)
> cW1: (4a<3c) => (a<c)
>
> aW1 = (3c<4a).((3b<2a)(2a+3b<3c)(c<a) or (3c<2a+3b)(b<a))
> bW1 = (2a<3b).((4a<3c)(c<a+b) or (3c<4a)(3c<2a+3b)(a<b))
> cW1 = (a<c).((2a<3b)(4a<3c)(a+b<c) or (3b<2a)(2a+3b<3c))
>
> aW1: (2a+3b<3c)(c<a) => (2a+3b<3a) = (3b<a) => (3b<2a)
> bW1: (3c<4a)(a<b) => (3c<2a+2b) => (3c<2a+3b)
> cW1: (2a<3b)(a+b<c) => (5a<3c) => (4a<3c)
>
> aW1 = (3c<4a).((2a+3b<3c)(c<a) or (3c<2a+3b)(b<a))
> bW1 = (2a<3b).((4a<3c)(c<a+b) or (3c<4a)(a<b))
> cW1 = (a<c).((2a<3b)(a+b<c) or (3b<2a)(2a+3b<3c))
>
> I shan't complete the equations.
>
> The Condorcet method:
>
> Case of Condorcet:
>
> a AB.
> b B..
> c C..
>
> Internal pairwise comparisons:
> Case A:B = a:b
> a A
> b B
> c C..
>
> Case A:C = a:c
> a A
> b B..
> c C
>
> Case B:C = (a+b):c
> a AB
> b B
> c C
>
> aW2 = (A wins) = (b<a)(c<a) Condorcet 1-winner winners
> bW2 = (B wins) = (a<b)(c<a+b)
> cW2 = (C wins) = (a+b<c)
>
> goto X1
> (No winners) = (a<c).-aW2.-bW2.-cW2 or (c<a).-aW2.-bW2.-cW2
> = (a<c).T.-bW2.-[(a+b<c)] or (c<a).-[(b<a)].-bW2.T
> = (a<c).T.-bW2.-[(a+b<c)] or (c<a).-[(b<a)].-bW2.T
> = [(b<a) or (a+b<c)] . [(a<c)(c<a+b) or (c<a)(a<b)]
> = (b<a)(a<c)(c<a+b) or (a+b<c)(c<a)(a<b)
> = (b<a)(a<c)(c<a+b)
> = E say
>
> The presence of the (b<a) is a region where Condorcet may be getting
> the wrong answer.
>
> <<X1>>
> The statement is false if
>
> (not E) =>
> (aW1.-aW2 or -aW1.aW2 or bW1.-bW2 or -bW1.bW2 or cW1.-cW2 or -cW1.cW2)
>
> -----------------------------------------------------------------------
>
> Case #1: (a+b<c): Condorcet has C win:
>
> In the J. M. (Nanson) method:
>
> cW1 = (a<c) . ((2a<3b)(a+b<c) or (3b<2a)(2a+3b<3c))
> cW1 = (a<c) . ((2a<3b) or (3b<2a)(2a+3b<3c))
> cW1 = (a+b<c).((2a<3b) or (2a+3b<3c))
> cW1 = (a+b<c) = T.
>
> The two methods agree on Condorcet's C win region.
>
> Case #2: (b<a)(c<a): Condorcet has A win:
>
> aW1 = (b<a)(c<a) . (3c<4a).((2a+3b<3c)(c<a) or (3c<2a+3b)(b<a))
> aW1 = (b<a)(c<a) . ((2a+3b<3c) or (3c<2a+3b))
> aW1 = (b<a)(c<a)
>
> The two methods agree on Condorcet's A win region.
>
> Case #3: (a<b)(c<a+b): Condorcet has B win:
>
> bW1 = (a<b)(c<a+b) . (2a<3b).((4a<3c)(c<a+b) or (3c<4a)(a<b))
> bW1 = (a<b)(c<a+b) . ((4a<3c) or (3c<4a))
> bW1 = (a<b)(c<a+b)
>
> The two methods agree on Condorcet's B win region.
>
> -----------------------------------------------------------------------
>
> There the cases of 0 and 2 and 3 winners. Condorcet finds no
> winners so presumably the conjecture is not failed there too.
>
> I have lost access to my REDLOG code. Maybe the original poster could
> therefore inform [whomever] on the constraints that allow the statement
> happens to be true [e.g. an upper limit on the number of candidates].
>
> So a multiwinner Condorcet can be defined now, using STV instead of the
> Approval Vote.
>
>
> G. A. Craig Carey, Auckland
> http://www.egroups.com/messages/politicians-and-polytopes
>
>
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