>From: Joe Malkevitch 'Email: [EMAIL PROTECTED]' >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
