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# Voting matters - Issue 4, August 1995

## Trying to find a winning set of candidates

#### I D Hill

In *Voting matters* issue 2, I introduced the idea of Sequential STV and came to the conclusion that
it should not be recommended for general use. But there remains something
very attractive in trying to find a set of candidates, of the right size for
the number to be elected, such that if an STV election were conducted with
that set plus any other one candidate, all other candidates being treated
as withdrawn, that set would always be the winners.

We know from Condorcet's paradox that in the one-seat case, where the set is
of size 1, there may not be any winner who fulfils the criterion, but at
least if we can find such a winner, the result is unique.

In the multi-seat case, we can still get results where no set satisfies the
criterion. For an example, consider 4 candidates for 2 seats and votes 1
AB, 1 BC, 1 CD, 1 DA. If we choose AB to test we find that ABD leads to AD
as winners; testing AD we find that ACD leads to CD; testing CD we find BCD
leads to BC; testing BC we find that ABC leads to AB. So round in circles
we go.

But now things are far worse for, even where a set to satisfy the criterion
is found, it may not be unique. Again consider 4 candidates for 2 seats and
votes 6 A, 6 B, 5 C, 5 D, 4 DA, 4 DB, 4 CA, 4 CB, 4 BC, 4 BD, 4 AC, 4 AD.
If we choose AB as potential winners, we find that ABC elects AB and ABD
elects AB, which would seem to confirm the choice; but if we choose CD we
find that ACD elects CD and BCD elects CD, so that choice is also confirmed.
Looking at the votes we can see that AB is, in fact, the better choice, but
merely to find any set that fulfils the criterion is not adequate.

Can we then say that, having found a potential winning set, we need only
look at disjoint sets to see if there are any others? Again things are not
as easy as that. Consider 6 candidates for 4 seats, with the same votes as
in the last example, with the addition of 20 E, 20 F. Then if we choose
ABEF as potential winners, we find that ABCEF elects ABEF and ABDEF elects
ABEF, seeming to confirm the choice; but if we choose CDEF we find that
ACDEF elects CDEF and BCDEF elects CDEF, so that choice is also confirmed,
and the sets are not disjoint as they both contain E and F.

It is clear therefore that there cannot be a universally best algorithm.
For everyday practical use, I believe that simple STV by Meek's method
should remain the algorithm of choice.

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