Strictly speaking, only a Condorcet winner is a stable approval winner; only he can win an approval election he’s the presumptive winner of. Elsewhere, I proposed generalizing the Condorcet winner to what I’ll go ahead and call the p-winner: the winner approval voting converges on when voters approve all candidates they prefer to their weighted average (score or utility) candidate, with weight equal to approval rate to the power of p. A Condorcet winner is the p=∞-winner, there is always a Greatest-p-winner (and a p-winner for p=0), and any method that always ranks Greatest-p-winners above all other candidates is resolvable. I could not determine whether Greatest-p is compatible with Smith or even Condorcet loser.
I now propose an alternative, obviously Smith-efficient generalization: the q-winner, which, rather than letting p vary, lets vary the quota (as an approval rate) a winner must have to be locked in (i.e. immune to defeat). This is based on the idea of costly counteroffers (or re-votes, in this case). The Condorcet winner is the unique q=1-winner, there is always a Greatest-q-winner (and a q=.5-winner), and any method that always ranks Greatest-q-winners above all other candidates is resolvable. For example:
The approval top cycle is: Cb, Bc, Ab, Ca, Cb…C has the biggest win (9 approvals against Ca) and is thus the Greatest-q-winner.
To find the Greatest p-winner, you could, beginning with p=2^(x=0), repeatedly find the p-winners and then increase (if there are more than 1) or decrease (if there are 0) x by 1 until the integer part of Greatest-p’s x has been found, after which you begin splitting the difference between the x’s Greatest-p’s x is between. Finding the Greatest-q-winner would be easier. You would just find the approval top cycle (i.e. the cycle of approval distributions repeated approval balloting converges on) and then find the biggest win (in terms of the winner’s approval rate) in that cycle. Greatest-p requires cardinal ballots, but ranked ballots would be sufficient for Greatest-q.