[EM] IFNOP Method (was Re: Question about Condorcet methods)

[mrouse1 at mrouse.com]

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
Mrouse1 at mrouse.com