I’ve been doing some checking of the IFNOP method (I should probably rename it, since it involves ignoring the lowest preferences, not simply the fewest in number). Here is the method again, for reference:
1. Each voter ranks as many candidates as desired. 2. If there is a Condorcet order (or a Condorcet winner in a single-winner race), congratulations! No more work is necessary. 3. If there is one or more circular ties, create an NxN matrix, where N is the number of candidates. 4. For every pair that is part of a circular tie, place the number of preferences in the corresponding box. For example, if 50 people put A in the 7th position and B in the 12th position on their ballot, then put (50 A>B) in box (7,12). 5. If a person ranks two or more candidates the same, they are assumed to be as close as possible to any candidate they are compared to. For example, in the order A>B>C=D>E, C and D are both ranked 4th when compared with E (you would have C>E and D>E in the box (4,5), and both are ranked 3rd when compared with A and B (A>C, A>D in box (1,3), and B>C, B>D in box (2,3) respectively). For this ballot, there is no score added for C>D or D>C. 6. Once all the ballots have had their preference pairs counted in the appropriate box, you are ready to begin counting. Look at each pair in the lowest-ranked box. If ignoring part or all of the votes for a pair resolves a tie, then do so, otherwise go on to the next box on the list and add these “ignore” votes to the appropriate pair in the previous box. The order of addition is bottom ranking to top, closest ranking to farthest (see the examples later in this email). The one that breaks the circular tie first determines the order. To give an example of how this would work, here are two elections that yield 5-way circular ties. The first election appears rather simple to resolve to a "best fit" order, though in fact it has a complication with the top candidate (A) in the largest order also being the lowest candidate in the second-largest order. This makes it yield odd results in many tie-breaking methods: Election #1 Ballots “Simple” 5-way cycle 22 ABCDE 21 BCDEA 20 CDEAB 19 DEABC 18 EABCD 100 Total Putting the pairs in order in the appropriate boxes, using the order: (4,5)(3,4)(3,5)(2,3)(2,4)(2,5)(1,2)(1,3)(1,4)(1,5) (4,5) DE(22) EA(21) AB(20) BC(19) CD(18) (3,4) CD(22) DE(21) EA(20) AB(19) BC(18) (3,5) CE(22) DA(21) EB(20) AC(19) BD(18) (2,3) BC(22) CD(21) DE(20) EA(19) AB(18) (2,4) BD(22) CE(21) DA(20) EB(19) AC(18) (2,5) BE(22) CA(21) DB(20) EC(19) AD(18) (1,2) AB(22) BC(21) CD(20) DE(19) EA(18) (1,3) AC(22) BD(21) CE(20) DA(19) EB(18) (1,4) AD(22) BE(21) CA(20) DB(19) EC(18) (1,5) AE(22) BA(21) CB(20) DC(19) ED(18) Arranging the pairings high to low for clarity: 82 DE 18 (64) 81 CD 19 (62) 80 BC 20 (60) 79 AB 21 (58) 78 EA 22 (56) 63 CE 37 (26) 61 BD 39 (22) 60 DA 40 (20) 59 BC 41 (18) 57 EB 43 (14) Note, the number in parenthesis is the total of votes that need to be ignored to break the pair comparison. Finding out how far you have to go to break the circular ties (using the order of boxes above): DE (64) 22+21+0+20+0+0+1 CD (62) 18+22+0+21+0+0+1 BC (60) 19+18+0+22+0+0+1 AB (58) 20+19+0+18+0+0+1 EA (56) 21+20+15* CE (26) 0+0+22+0+4 BD (22) 0+0+18+0+4 DA (20) 0+0+20* AC (18) 0+0+18 EB (14) 0+0+14 Note, since each candidate has two other candidates that beat it, two separate pairs of “ignored” votes are needed to break a particular cycle. With this method, the single top winner is A -- EA is the only pair that can be ignored by (2,3), while all others take until (1,2). In the bottom group, three of them -- including DA, which breaks the other tie involving A – are resolved by (3,5). Removing A leaves two adjacent cycles that share two candidates: BCE and BDE. The link between E and B is the weakest, and breaking this creates the relation B>C>D>E. This makes the complete profile A>B>C>D>E. As a comparison, Borda would find the following: A = 4*22+3*18+2*19+20 = 200 B = 4*21+3*22+2*18+19 = 205 C = 4*20+3*21+2*22+18 = 205 D = 4*19+3*20+2*21+22 = 200 E = 4*18+3*19+2*20+21 = 190 This creates the ranking B=C>A=D>E, which doesn't appear to be as reasonable as the order given above. ** The second circular tie is a bit less obvious, though as easy to calculate: Election #2 Ballots Complicated 5-way cycle 18 EABCD 19 CDEAB 20 ABCDE 21 BCDEA 22 DEABC 100 Total Putting the pairs in order in the appropriate boxes, using the order: (4,5)(3,4)(3,5)(2,3)(2,4)(2,5)(1,2)(1,3)(1,4)(1,5) (4,5) CD(18) AB(19) DE(20) EA(21) BC(22) (3,4) BC(18) EA(19) CD(20) DE(21) AB(22) (3,5) BD(18) EB(19) CE(20) DA(21) AC(22) (2,3) AB(18) DE(19) BC(20) CD(21) EA(22) (2,4) AC(18) DA(19) BD(20) CE(21) EB(22) (2,5) AD(18) DB(19) BE(20) CA(21) EC(22) (1,2) EA(18) CD(19) AB(20) BC(21) DE(22) (1,3) EB(18) CE(19) AC(20) BD(21) DA(22) (1,4) EC(18) CA(19) AD(20) BE(21) DB(22) (1,5) ED(18) CB(19) AE(20) BA(21) DC(22) Arranging the pairings high to low (parentheses indicate difference): 82 DE 18 (64) 81 BC 19 (62) 80 EA 20 (60) 79 AB 21 (58) 78 CD 22 (56) 62 DA 38 (24) 60 AC 40 (20) 60 CE 40 (20) 59 BD 41 (18) 59 EB 41 (18) Finding out how far you have to go to break the circular ties (using the order of boxes above): DE (64) 20+21+0+19+0+0+4 BC (62) 22+18+0+20+0+0+2 EA (60) 21+19+0+20 AB (58) 19+22+0+17* CD (56) 18+20+0+18 DA (24) 0+0+21+0+3 AC (20) 0+0+20 CE (20) 0+0+20 BD (18) 0+0+18 EB (18) 0+0+18* AB is the weakest top cycle with this method, though the result is very close – AB requires 17 votes ignored in (2,3), while CD requires just one more (and in fact requires fewer ignored votes in lower cycles). EB is resolved by (3,5), which ties for the lowest, so this makes B the winner. D is the next one to resolve (CD and BD are next in line), and removing B and D leaves the circular tie ACE. Note that CD would be considered a weaker cycle than AB under many other Condorcet tie-breaking methods. This method considers higher-ranked comparisons to be more important (and more likely correct) than lower-ranked ones. For ACE, we notice that both AC and CE require 20 points to ignore (box 3,5), while EA requires 60 points (box 2,3). Minimizing "ignores," we must have either EAC or CEA -- both have 20 ignores -- which requires a tiebreaker. I give C the advantage, since AC still has two preferences remaining in (3,5) -- to flip CE we would need to go up to (2.4) -- but there might be another method that can be used. The final result would be B>D>C>E>A For a comparison, in Borda you have A = 4*20+3*18+2*22+19 = 197 B = 4*21+3*20+2*18+22 = 202 C = 4*19+3*21+2*20+18 = 197 D = 4*22+3*19+2*21+20 = 207 E = 4*18+3*22+2*19+21 = 197 This is actually fairly close to the order given above. B and D are flipped, while A, C, and E are in a three-way tie. Borda gives you D>B>A=C=E I’m going to try to compare the results this method gives with other Condorcet-completion methods, though I’ll probably be slow at it (I’m not a programmer, so I have to do things by hand). I’m especially curious about which methods of strategic voting would be effective against it, since that is one of the best ways to see if a method will work in an election. If anyone else has any thoughts or examples where this method would fail, I’d be happy to see them, and if anyone has any questions (which is quite likely, since my examples are not always the clearest, or even the most coherent :) ) I’d be happy to try to answer them. Thanks! Michael Rouse [EMAIL PROTECTED] ---- election-methods mailing list - see http://electorama.com/em for list info
