Recursive IRV

Jack Waugh

The recursive method alternates between looking for a good candidate and looking for the worst candidate.

I have not checked @rob's assertion that just one level of recursion will make the system comply with Condorcet.

spelunker

Now I am fully confused. Would you be able to give a complete (and self-contained) description of the rule? (Pseudocode would also work)

Jack Waugh

@rob , this is kind of the key section of your code for this, right?

doEliminationLoop = (candidates, ballotsExp, isNegative) => {
  var results = {
    rounds: []    
  };
  var numCandidates = candidates.length;
  
  for(var i=0; i<numCandidates; i++) {
    var candidatesSorted = // (depth>0) ? 
        //sortedCandidatesByEliminationRounds(candidates, ballotsExp, true, 1) :
        sortedCandidatesByFirstPlaceVotes(candidates, ballotsExp);
    console.log(depth + " " + isNegative)
    console.log(JSON.stringify(candidatesSorted,0,2))
    var toBeEliminated = candidatesSorted[(isNegative)?0:candidatesSorted.length-1];
    results.rounds.push(candidatesSorted);
    //setValue('output_2', JSON.stringify(candidatesSorted,0,2))
    // maybe use Array.filter?
    candidates = []; // rebuild array without winner
    for(var j=0; j<candidatesSorted.length; j++) {
      if(candidatesSorted[j].name != toBeEliminated.name) {
        candidates.push(candidatesSorted[j].name);
      } 
    }
  }
  results.winner = results.rounds[results.rounds.length-1][0]
  return results
}

His condition on isNegative is key here. He's sorting the candidates the same way, based on top rankings, but he's picking whom to eliminate from the list from either the top or the bottom of the list.

Jack Waugh

@spelunker I could be vague on some details and we might need @rob to clarify, unless one of us wants to fully reverse engineer his code and see why it produced different results for the Burlington, Vt. election than what you would predict, @spelunker.

spelunker

This still very much seems like IRV and not Condorcet compliant though. Just imagine you have two voters and three candidates with preferences a>b>c and c>b>a. Wouldn't this method delete b in the first step which was the condorcet winner?

Jack Waugh

Thanks for giving an example to work.

Hare to the zeroth power is choose-one plurality.

Hare to the first power is Hare.

Hare squared does about (n - 1)(n - 1)/2 rounds, concerned with promotion of candidates rather than elimination of them.

I have three candidates and want to get down to two. Whom do I promote first? It's a tie between a and c. I have to appeal to @rob for tiebreaking rules, or make up my own.

Can you give an example that doesn't tie at any stage?

rob

@spelunker Hey sorry if my explanation wasn't clear. I actually started to develop a much better version, where you could choose any depth, and (the hard part) it had the ability to show output so it was understandable, but then I messed up and lost my code or something and dropped it for a while. (I'm sure I'll come back to it)

Look at it this way. What IRV attempts to do, prior to doing the final calculation (i.e. "determine who is preferred by more people"), is to remove irrelevant alternatives so the calculation is not affected by them. But it does the elimination rather crudely, based on "negative plurality". (least first choice votes)

This just takes it to another level, using the same logic for the elimination as it does for the final determination. As you go to further and further recursion depths, this should get more and more accurate, as the "crude part" is less and less relevant.

My general hypothesis is that Condorcet compliance is simply an easy to define step on the way to our goal of "immune to vote splitting." IRV logic is obviously closer to this goal than plurality, but since it can result in missing the Condorcet winner, it isn't very far toward the goal. Applying the logic twice appears to make it Condorcet compliant. Applying it 3 or more times moves it ever closer to the goal.

Since it is technically impossible to apply it an infinite number of times, this doesn't violate Arrow's theorem (and the like).

Jack Waugh

@rob This is sort of like engineered chess programs. They look ahead a certain number of moves and then resort to crude heuristics.

The ones that are not "engineered" use connectionism and/or Darwinism, but techniques like that are not relevant here. Before people tried those, they were engineering the solution, i. e. designing it top to bottom by hand.

rob

@spelunker said in Recursive IRV:

Just imagine you have two voters and three candidates with preferences a>b>c and c>b>a. Wouldn't this method delete b in the first step which was the condorcet winner?

It might, but I don't consider that a good example since to me it is (or should be) a three way tie.

If you have a case where there are no pairwise ties, and there is a Condorcet winner which it misses, I'd be interested in seeing it.

Jack Waugh

I believe it's a Copeland tie and a Dasgupta-Maskin tie.

a vs. b: no winner.

a vs. c: no winner.
c vs. b: no winner.

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