[EM] Theorem equating Nanson AV-Borda with Condorcet: 3 candidates case shown

[Craig Carey]

 >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