[EM] IFNOP method (continued)

[mrouse1 at mrouse.com]

Well, it's a little late, but I did a test of the IFNOP (Ignore Fewest
Number of Preferences) method using 4 examples from the Wikipedia
article on the Schulze method. I picked Schulze because it is fairly
well-behaved with fewer voting paradoxes than many other methods.

Example 1 (45 voters; 5 candidates):
5 ACBED
5 ADECB
8 BEDAC
3 CABED
7 CAEBD
2 CBADE
7 DCEBA
8 EBADC

Putting the pairs in the appropriate boxes:

(4,5) ED(8) CB(5) AC(8) BD(7) DE(2) BA(7) DC(8)
(3,4) BE(8) EC(5) DA(8) EB(14) AD(10)
(3,5) BD(8) EB(5) DC(8) ED(7) AE(2) EA(7) AC(8)
(2,3) CB(5) DE(5) ED(8) AB(3) AE(7) BA(10) CE(7)
(2,4) CE(5) DC(5) EA(8) AE(3) AB(7) BD(10) CB(7)
(2,5) CD(5) DB(5) EC(8) AD(10) BE(2) CA(7) BC(8)
(1,2) AC(5) AD(5) BE(8) CA(10) CB(2) DC(7) EB(8)
(1,3) AB(5) AE(5) BD(8) CB(3) CE(7) CA(2) DE(7) EA(8)
(1,4) AE(5) AC(5) BA(8) CE(3) CB(7) CD(2) DB(7) ED(8)
(1,5) AD(5) AB(5) BC(8) CD(10) CE(2) DA(7) EC(8)

The first number is the number of votes for the pair, the second is the
number of votes for the reverse pair, and the number in parenthesis is
the difference:

30 AD 15 (15)
25 BA 20 (5)
33 BD 12 (21)
27 CA 18 (9)
29 CB 16 (13)
24 CE 21 (3)
28 DC 17 (9)
23 EA 22 (1)
27 EB 18 (9)
31 ED 14 (17)

Now, let's see how far we have to go (by ignoring votes) before each
pair is broken. The order of comparison is
(4,5)(3,4)(3,5)(2,3)(2,4)(2,5)(1,2)(1,3)(1,4)(1,5). Zeroes mean there
are no corresponding pairs in that box.

BD (21) 7+0+8+0+6
ED (17) 8+0+7+2
AD (15) 0+10+0+0+0+5
CB (13) 5+0+0+5+3
CA (9) 0+0+0+0+0+7+2
DC (9) 8+0+1*
EB (9) 0+9
BA (5) 5
CE (3) 0+0+0+3
EA (1) 0+0+1

Pair winners are on the left of the arrow, pair losers on the right
BCE-->A
CE-->B
D-->C
ABE-->D
C-->E

DC is the first link broken -- at (3,5) -- leaving C the winner. Since
C>E is now redundant, E is in second place. Since both C and E have been
removed, B is in third place, which puts A in fourth place and D in
fifth place. The completed order is:

C>E>B>A>D.

Just for comparison, Schulze selects E as the winner. Borda gives the
order  E>A>B>C>D, partly because it considers comparing a first and
second place vote equivalent to comparing a fourth and fifth place vote.

************

Example 2 (30 voters; 4 candidates):
2 ACDB
3 ADCB
4 BACD
3 CBDA
3 CDBA
1 DACB
5 DBAC
4 DCBA

Putting the pairs in the appropriate boxes:

(3,4) DB(2) CB(4) CD(4) DA(3) BA(7) AC(5)
(2,3) CD(2) DC(3) AC(5) BD(3) DB(3) BA(5) CB(4)
(2,4) CB(2) DB(3) AD(4) BA(3) DA(3) AB(1) BC(5) CA(4)
(1,2) AC(2) AD(3) BA(4) CB(3) CD(3) DA(1) DB(5) DC(4)
(1,3) AD(2) AC(3) BC(4) CD(3) CB(3) DC(1) DA(5) DB(4)
(1,4) AB(5) BD(4) CA(6) DB(1) DC(5) DA(4)

The first number is the number of votes for the pair, the second is the
number of votes for the reverse pair, and the number in parenthesis is
the difference:

15 AC 10 (5)
19 BA 6 (13)
16 CB 9 (7)
16 DA 9 (7)
18 DB 7 (11)
13 DC 12 (1)

D is the unique winner, while the three remaining candidates are in the
circular tie C>B>A>C. The link A>C is the easiest one to break,
resulting in the final order of D>C>B>A. For comparison, the Schulze
method also gives D as the winner, and Borda gives the complete order
D>C>B>A as well.

********************

Example 3 (30 voters; 5 candidates):
3 ABDEC
5 ADEBC
1 ADECB
2 BADEC
2 BDECA
4 CABDE
6 CBADE
2 DBECA
5 DECAB

Putting the pairs in the appropriate boxes:

(4,5) EC(5) BC(5) CB(1) CA(4) DE(10) AB(5)
(3,4) DE(5) EB(5) EC(5) BD(4) AD(6) CA(5)
(3,5) DC(5) EC(5) EB(1) EA(4) BE(4) AE(6) CB(5)
(2,3) BD(3) DE(8) AD(2) AB(4) BA(6) BE(2) EC(5)
(2,4) BE(3) DB(5) DC(3) AE(2) AD(4) BD(6) BC(2) EA(5)
(2,5) BC(3) DC(5) DB(1) AC(2) DA(2) AE(4) BE(6) BA(2) EB(5)
(1,2) AB(3) AD(6) BA(2) BD(2) CA(4) CB(6) DB(2) DE(5)
(1,3) AD(3) AE(6) BD(2) BE(2) CB(4) CA(6) DE(2) DC(5)
(1,4) AE(3) AB(5) AC(1) BE(2) BC(2) CD(6) DC(2) DA(5)
(1,5) AC(8) AB(1) BC(2) BA(2) CE(10) DA(2) DB(5)

The first number is the number of votes for the pair, the second is the
number of votes for the reverse pair, and the number in parenthesis is
the difference:

30 DE 0 (30)
21 AD 9 (12)
21 AE 9 (12)
20 DC 10 (10)
20 EC 10 (10)
19 BE 11 (8)
19 CA 11 (8)
18 AB 12 (6)
17 BD 13 (4)
16 CB 14 (2)

DE (30) 10+5+0+8+0+0+5+2
AD (12) 0+6+0+2+4
AE (12) 0+0+6+0+2+4
DC (10) 0+0+5+0+0+5
EC (10) 5+5
BE (8)  0+0+4+2+2
CA (8)  4+4
AB (6)  5+0+0+1
BD (4)  0+4
CB (2)  1+0+1

C-->A
AC-->B
DE-->C
AB-->D
ABD-->E

The link that breaks first is CA, at (3,4), making A the winner. BD
breaks next, so D is second place. EC breaks next, so E is in third
place. This leaves C>B, so the final order is: A>D>E>C>B

For comparison, the Schulze method gives B as the winner, or exactly the
opposite. Using Borda, we get D>A>B>C>E

***********

Example 4 (9 voters; 4 candidates):
3 ABCD
2 DABC
2 DBCA
2 CBDA

Putting the pairs in the appropriate boxes:

(3,4) CD(3) BC(2) CA(2) DA(2)
(2,3) BC(5) AB(2) BD(2)
(2,4) BD(3) AC(2) BA(4)
(1,2) AB(3) DA(2) DB(2) CB(2)
(1,3) AC(3) DB(2) DC(2) CD(2)
(1,4) AD(3) DC(2) DA(2) CA(2)

The first number is the number of votes for the pair, the second is the
number of votes for the reverse pair, and the number in parenthesis is
the difference:

7 BC 2 (5) 2+3
6 DA 3 (3) 2+0+0+1
5 AB 4 (1) 0+1
5 AC 4 (1) 0+0+1
5 BD 4 (1) 0+1
5 CD 4 (1) 1

D-->A
A-->B
AB-->C
BC-->D

The link A>B is the easiest one to break, taking one point, at (2,3).
This leaves B as the overall winner. D is next, with the link C>D broken
at (3,4) combined with the winner B. A resolves to third place, and C to
4th place. The completed order is: B>D>A>C

For comparison, the Schulze method gives a tie win to B and D, and Borda
gives the order B>D>A>C, which is the same as for IFNOP.

Some thoughts:

The Schulze and IFNOP methods can give give the same winners (in example
2 and a tie win in 4), different winners (example 1), and completely
opposite winners (e.g. example 3). IFNOP places a higher priority on top
choices than on bottom choices, with the justification being that while
top choices are generally what the voter honestly prefers (unless
compromising), bottom choices are a mix of most hated, most unknown, and
most strategic votes.

Looking at the votes, the winners given by IFNOP seem reasonable -- even
in the example where it gave the exact opposite of result from Schulze,
the candidate it chose defeated the Schulze winner.

If anyone can come up with a good voting paradox, or an obvious winner
that loses (or obvious loser that wins), please let me know. It's more fun
seeing where voting systems fail than it is trying all the examples that
work well.

Michael Rouse
mrouse1 at mrouse.com