# [EM] Simmons cloneproof method is not cloneproof

Chris Benham chrisjbenham at optusnet.com.au
Mon Jan 1 15:39:26 PST 2007

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Warren Smith wrote:

>see http://groups.yahoo.com/group/RangeVoting/message/2934
>for counterexample  (plus linear program explainign how I found the counterexample)
>wds
>
> Ballots:
> 6: A>B>C
> 3: C>A>B
> 4: B>C>A
>
> A wins under Simmons voting since
> A beats B pairwise ==> 6 ballots count against B
> C beats A pairwise ==> 3 ballots count against A
> B beats C pairwise ==> 4 ballots count against C
>
> Now add two clones of A in a Condorcet cycle.
>
> Then A1 is beat pairwise by A2 with 1/3 of the 6 of the
> A-top ballots, i.e. 3, and ditto A2 and A3, all have 2
> A-top ballots against them.
> Plus, all the Ak have got C's toprank votes
> against them, which is 3. So in total, each A-clone
> has 5 ballots against it, while C has only 4
> ballots against it.
>
> Hence C is now the winner thanks to A's cloning.
>
> So SIMMONS IS NOT CLONEPROOF!!

> If we agree only to clone non-winners,
> or if, when cloning a winner, all voters agree to rank the clones
> EQUALLY, THEN Simmons is cloneproof.
>
> Proof:
> After cloning,
> the A-beats-B relations are unaffected under these constraints,
> and the number of top-rank-votes-against X are either unaltered -
> or increased (increase is only possible for nonwinner X since
> winning-X clones never pairwise-beat each other).
> QED
>
> However... this weakened kind of cloneproofness is a good
> deal less impressive than genuine cloneproofness.

So  Simmons meets Clone-Loser, but can fail Clone-Winner when there are
three or more factions
in a top cycle and the candidates in one of those factions are in a
sub-cycle. That is a very very mild
failure of  Clone-Independence and arguably not a practical worry.

If that is the full extent of the bad news (and maybe even if it isn't)
then I think this method remains a
great contender (for "best practical Condorcet method") because of its
tremendous Burial resistance
and simplicity.

Chris Benham

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