Perfect representation


  A winner set W provides perfect representation for a group of n voters and a total seat size k if n = ks for some positive integer s and the voters can be split into k pairwise disjoint groups N1, . . . , Nk of size s each in such a way that there is a one-to-one mapping μ : W → {N1, . . . , Nk} such that for each candidate a ∈ W all voters in μ(a) approve a.

In other words, in a multi-winner approval-voting election, for a given set of ballots cast, if there is a possible election result where candidates could each be assigned an equal number of voters where each voter has approved their assigned candidate and no voter is left without a candidate, then for a method to pass the perfect representation criterion, such a result must be the actual result.


Perfect representation is not compatible with strong monotonicity. Consider the following election with two winners, where A, B, C and D are candidates, and the number of voters approving each candidate are as follows:

100 voters: A, B, C

100 voters: A, B, D

1 voter: C

1 voter: D

A method passing the perfect representation criterion must elect candidates C and D despite near universal support for candidates A and B. This demonstrates that methods passing perfect representation can at best be only weakly monotonic, and could therefore be seen as an argument against perfect representation as a useful criterion.


Every winner set that provides perfect representation also provides Proportional Justified Representation [1]. In contrast, Extended Justified Representation may rule out all winner sets that provide perfect representation. [2]

Perfect Representation In the Limit

While perfect representation arguably has flaws as a proportionality criterion, the Perfect Representation In the Limit criterion (PRIL) states that the result of a voting method must approach perfect representation as the number of elected candidates increases, as long as the ballots make this possible. This criterion does not demand that a method passes lower quota so does not disqualify the Sainte-Laguë method and is compatible with strong monotonicity. [3]