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MinLV(erw) Sorted Margins Elimination

C
C.Benham
Sun, Aug 6, 2023 3:11 PM

MinLV(erw) Sorted Margins Elimination  is no longer my favourite (or
even one of my several favourite)
Condorcet voting methods.

For its complexity and and clunkiness it doesn't give enough, not enough
"bang for buck" as Forest would put
it.

This change of mind has been partially motivated by Kevin Venzke
surprising me with his finding that it fails
mono-raise even with only three candidates.

(I already expected that the eliminations would cause it to fail with
more than three candidates).

Quoting Kevin:

For mono-raise, let me run this one by you:
486: A>C>B
263: B>C>A
143: C>B>A
108: B>A>C

We have a A>C>B>A cycle with losing votes:
A: 486, B 371, C 406.
So they should be ordered
A: 486, C 406, B 371.
A does beat C and C does beat B, so the order is correct.
Eliminate B, and then A is elected.

Now change:
263: B>C>A  -->  B>A>C

Same cycle but losing votes are
A: 486, B 371, C 143.
That's the right initial order.
Both adjacent orders are wrong. A-B is a smaller difference.
So swap that:
B 371, A: 486, C 143.
Eliminate C, and now B wins.
That means raising A made A lose.

My new favourite Condorcet methods are mostly based on some of Forest's
ideas. More on that in
another post.

Chris Benham

MinLV(erw) Sorted Margins Elimination  is no longer my favourite (or even one of my several favourite) Condorcet voting methods. For its complexity and and clunkiness it doesn't give enough, not enough "bang for buck" as Forest would put it. This change of mind has been partially motivated by Kevin Venzke surprising me with his finding that it fails mono-raise even with only three candidates. (I already expected that the eliminations would cause it to fail with more than three candidates). Quoting Kevin: > For mono-raise, let me run this one by you: > 486: A>C>B > 263: B>C>A > 143: C>B>A > 108: B>A>C > > We have a A>C>B>A cycle with losing votes: > A: 486, B 371, C 406. > So they should be ordered > A: 486, C 406, B 371. > A does beat C and C does beat B, so the order is correct. > Eliminate B, and then A is elected. > > Now change: > 263: B>C>A  -->  B>A>C > > Same cycle but losing votes are > A: 486, B 371, C 143. > That's the right initial order. > Both adjacent orders are wrong. A-B is a smaller difference. > So swap that: > B 371, A: 486, C 143. > Eliminate C, and now B wins. > That means raising A made A lose. My new favourite Condorcet methods are mostly based on some of Forest's ideas. More on that in another post. Chris Benham