[EM] Manual Construction of Smith Set

Kristofer Munsterhjelm km_elmet at t-online.de
Sun Jan 16 10:31:38 PST 2022


On 15.01.2022 21:50, Richard, the VoteFair guy wrote:

> Yet I believe that lots of audience members would give more trust to a
> method that eliminates candidates one at a time. This is the simplicity
> advantage that instant-runoff voting (IRV) has. In contrast, starting
> with the Copeland winner might seem like magic to people who fear math.
> 
> Would eliminating from the bottom always work?
> 
> In the Wikipedia example below, it does work. It identifies B and E as
> outside the Smith set, so the remaining candidates are in the Smith set.
> 
> 5 wins for A
> 
> 5 wins for D
> 
> 4 wins for G
> 
> 3.5 wins for C
> 
> 2.5 wins for F
> 
> 1 win for B
> 
> 0 wins for E
> 
> (Half indicates a tie.)
> 
> So, my question is: Does elimination from the bottom always work?
> 
> My guess is the answer is yes.

I think so too, in theory. Suppose that A is in the Smith set and B is
not. Then A beats everybody who B beats and then some. So B will be
ranked below A by Copeland score. Hence the only way someone in the
Smith set will at the bottom of the list is if everybody is in it.

But you still need to decide when to stop, which can be pretty
difficult. At first it'd seem to be obvious: if the bottom-most
candidate beats anyone else pairwise, he'd be in the Smith set, no? But
that doesn't quite work. Consider something like this:

The candidates are ABCD. There's a Condorcet winner, A, who beats
everybody. But there's also a loser three-cycle, B>C>D>B. Thus whoever
you choose to eliminate first of B, C, and D, that candidate beats
someone else despite being outside the Smith set. And batch-eliminating
everybody with an identical least score fails in more complicated scenarios.

To justify the Copeland winner being the initial candidate, I would just
point to the reasoning above: any non-Smith set member beats or ties
fewer people than every Smith set member (almost by definition). So the
topmost by number of candidates beaten can't be outside the Smith set.

If the voters are expected to know what a Smith set is, that leap
shouldn't be too large. And if not, then every Smith//X method will look
like magic to some degree. (Better use Benham or BTR-STV.)

-km


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