[EM] Binary dropping of candidates

[Kristofer Munsterhjelm]

Michael Rouse wrote:
> It's probably already been discussed before (most likely with a more 
> descriptive name), but the election methods list has been quiet, so...
> 
> Has anyone looked at making a ranked list  of candidates -- either by 
> number of first place votes, as in IRV, or Borda order, as in 
> Nanson/Baldwin -- and dropping candidates in a binary rather than unary 
> way?
> 
> By way of example, in Instant Runoff Voting, you see if a candidate has 
> a majority. If not, you drop the lowest candidate. If there is still no 
> candidate, you drop the next lowest candidate, and so on. With five 
> candidates, you have something like this:
> 
> .....   First round (no one dropped)
> ....1 Second round (one person dropped, ballots redistributed)
> ...11 Third round (two people dropped, ballots redistributed)
> ..111 Fourth round (three people dropped, ballots redistributed)
> (for five candidates, three drops are sufficient)
> 
> It's a method analogous to a unary counting system.
> 
> Under a binary dropping method, you have:
> (Note: If there is no winner after a round, you use the original 
> candidate ranking and proceed to the next round.)
> 
> 00000 First round (no one dropped)
> 00001 Second round (bottom candidate dropped, ballots redistributed)
> 00010 Third round (second-to-last candidate dropped, ballots redistributed)
> 00011 Fourth round (bottom two candidates dropped, ballots redistributed)
> 00100 Fifth round (middle candidate dropped, ballots redistributed)
> 00101 Sixth round (middle and last candidate dropped, ballots 
> redistributed)
> 00110 Seventh round (middle and second to last candidate dropped, 
> ballots redistributed)
> 00111 Eighth round (bottom three candidates dropped, ballots redistributed)

I'm not sure what you mean. The unary version works since the partial 
rankings are consistent. That is, you eliminate the last (which makes 
the last 1), then the last remaining (adding another 1) and so on.

However, in the binary case, you may need to "backtrack". Consider the 
first round giving a social ordering of A > B > C > D. You drop D, so

???D
0001

Then the next iteration gives A > C > B, so you get
??BD
0011

But what does 0010 mean? Does it mean to reinstate D? Say you do this 
and you get A > D > C. Then is 0011 equivalent to "eliminate D and B" or 
"eliminate D and C"? Because of the backtrack, you have two options, and 
it's not clear which is correct.

I suppose you could do it in a greedy manner:

First round gives A > B > C > D: last is D.
Second round gives A > C > B: next to last is B, so now you have ??BD
Third round gives A > D > C, so overwrite the next to last with C.
Fourth round then eliminates D and C...

But in that case, I don't see the logic. Single-elimination like IRV has 
the logic that you remove the "irrelevant" candidates until only one is 
left, but this removes and reinstates candidates with seemingly little 
pattern.