Up: Issue 4 Next: Paper 4 Previous: Paper 2

# 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if an STV election were conducted with that set plus any other one candidate, all other candidates being treated as withdrawn, that set would always be the winners.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.

Up: Issue 4 Next: Paper 4 Previous: Paper 2