Up to Elections: Results and Voting systems

Up to Mikes article on CONDORCET'S METHOD

Examples For CONDORCET'S METHOD

by Mike Ossipoff

To write to Mike write nkklrp before the "@" sign, and then write hotmail.com after the "@" sign. Or write ossipoff2002 before the "@" sign, and then write yahoo.com after the "@" sign.

Here's an example for the purpose of demonstrating the application of PC, RP, SD, SSD, CSSD, & BeatpathWinner to a set of rankings.

In this example, SSD, CSSD, and BeatpathWinner choose one winner; PC & SD choose a different winner; and RP returns a tie between those 2 winners.

If you're just interested in the examples themselves, there's no reason to not skip from here directly to the heading "Now the example".

As was mentioned at the Condorcet page, though RP, SD, SSD, CSSD, & BeatpathWinner choose from the Schwartz set in public elections, because there are no pairwise ties when there are so many voters, it's possible, in small committee voting, for RP & SD to choose outsidet the Schwartz set. This example was chosen to demonstrate that.

The Schwartz set was defined in the definition of SSD (Schwartz Sequential Dropping), at the Condorcet page.

Incidentally, even in small committee elections, all of the above-listed methods except for PC always choose from the Smith set. The Smith set is the smallest set of candidates such that every candidate in the set pairwise-beats every candidate outside of the set. X pairwise-beats Y if more voters rank X over Y than Y over X.

In public elections, where there are no pairwise-ties, the Smith & Schwartz sets are identical.

PC's ability to choose outside the Smith set, even in public elections, can lead to embarrassing, though not really important, transgression examples, which can be used against PC in a campaign for its adoption. On the other hand, PC is by far the simplest Condorcet version, which could be important in a campaign.

Now the example:

AB2 means that A beats B, with the defeat strength of 2, because 2 people have ranked A over B (and fewer have ranked B over A).

I don't yet have rankings for these pairwise defeats, but it's a sure thing that this example can happen, because it's been shown that, for any set of pairwise preference vote totals, there is some constant that we could add to them all that would give a set of pairwise vote totals that's consistent with some set of rankings. Since the order of the defeats is what counts in this example, of course the example would work just as well if some constant were added to all of the pairwise vote totals.

Of course if A beats B, I only list the number of people ranking A over B. Of course some smaller number ranked B over A, but it isn't necessary to specify that number. That's because, with these methods, we measure A's defeat of B by how many voters ranked A over B.

The best way to look at examples like this is to make a diagram in which the candidates are arranged in small circular arrangements. Since this example has two cycles, ABC & DEF, arrange those candidates in 2 triangular sets on the page. Then draw arrows to represent the defeats, so that if A beats B, there's an arrow from A to B. Then label the arrows with the defeat magnitudes.

In this example there are 2 cycles: the ABC cycle and the DEF cycle. Every candidate in {A,B,C} pairwise beats D & E, but ties F. It isn't necessary to specify those inter-cycle defeat magnitudes, because none of them are in a cycle, and so they'll all be kept in RP, and none will be dropped in SD or SSD or SSD, and none have a return beatpath in BeatpathWinner.

Here are the defeats in the 2 cycles:

AB5, BC6, CA4 DE3, EF1, FD2

What do PC, SSD, CSSD, BeatpathWinner, RP, & SD do in this example?

Again, if you're just interested in the examples themselves, skip directly to where, a few paragaraphs down, it says "So let's apply all 6 methods, in turn, to the example:"

First of all, since nothing in {A,B,C} is beaten by anything outside the set, {A,B,C} is an unbeaten set. Since it doesn't contain a smaller unbeaten set (every subset of {A,B,C} is beaten by someone in {A,B,C}), {A,B,C} is an innermost unbeaten set. It's the only one, and so it's the Schwartz set.

So we know that SSD, CSSD, & BeatpathWinner will choose from {A,B,C}.

RP and SD can choose outside the Schwartz set, and so they might choose outside {A,B,C}. They do in this example.

Since F pair-ties A, B, & C, and is in a cycle with D & E, all 6 candidates are the Smith set, because no subset of the candidates pairwise-beat everyone else.

So let's apply all 6 methods, in turn, to the example:

PC:

PC says: Drop the weakest defeat. Repeat till someone is unbeaten. This results in electing the candidate whose greatest pairwise defeat is the weakest.

EF1 is the weakest defeat. When it's dropped, F has no defeats, and therefore wins. We haven't specified the inter-cycle defeats, but their magnitudes don't matter, since all the candidates have bigger defeats than F does.

SSD:

The smallest defeat among the Schwartz set is CA4. When we drop it, A is unbeaten, and so A wins.

CSSD:

When we drop CA4, the Schwartz set now is {A}. Obviously there are now no defeats in the Schwartz set, and so its member, A, wins.

BeatpathWinner:

BeatpathWinner is equivalent to CSSD, and so we know it will pick A too. But that can also be shown from BeatpathWinner's own rule: A beatpath's strength is the strength of its weakest defeat. Following the beatpaths from A to B, and from A to C, and from B to A, and from C to A, it turns out that A has stronger beatpaths to B & C than they have to A.

Likewise, A has beatpaths to D,E, & F, but they have no beatpath to A. So, by BeatpathWinner's rule, A wins, because no one has a beatpath win against A. That can't be said for any other candidate.

SD:

Drop the weakest defeat that's in a cycle. That's EF1. Now F is undefeated, and F wins.

RP:

RP insists on solving all cycles, because any defeat that's contradicted by kept stronger defeats should be considered nullified by them.

First of all, all the defeats from the ABC cycle to the DEF cycle eventually get kept, since none of them are in cycles. Of course none of those intercycle defeats is to F.

So, just looking at the defeats in the cycles, we start by considering the strongest one, BC6. It doesn't cycle with kept defeats because there are none yet, since we haven't yet considered any other defeats. So we keep BC6.

Next we consider AB5. It too doesn't cycle with kept defeats, since we've only kept one defeat, and you can't make a cycle with fewer than 3 defeats. So we keep AB5. (In general, when applying RP's defining rule, we always keep the 2 strongest defeats for the reasons given above).

Considering CA4, it cycles with AB5 & BC6, which have been kept, and so we don't keep CA4. Since that's A's only defeat, that means that A will be a winner.

Next we similarly keep DE3 & FD2, but not EF1, because it would cycle with the 2 previously-kept defeats.

So when we've considered all the defeats, we have 2 candidates with no kept defeats: A & F.

It's a tie, and so we'd use a tiebreaker. I'd suggest, in organizations or small committees, Random Ballot. Whichever of {A,F} is ranked higher on a randomly- chosen ballot wins.

RP's choice makes a lot of sense, because any defeat that's constradicted by stronger defeats is understandably considered to be nullified by them, unless they themselves are nullified in that way.

But the better decisiveness of the other methods counts in their favor, for practical purposes, in a small committee vote. In a public election, of course, there's no decisiveness difference.

And it can also be reasonably argued that the winner should come from the initial Schwartz set. So RP's appeal, and the criterion of choosing from the initial Schwartz set can't be had in one method.

But we can define a method that says to use RP to choose from the initial Schwartz set. In other words, delete all the candidates who aren't in the initial Schwartz set, and then apply RP to the undeleted candidates. That would be my favorite for small committees, ideally, but, with RP being somewhat less decisive than the others, and subject sometimes to the need to solve ties during the middle of the count, when 2 defeats are equal. With the added wordiness of dealing with these things, RP doesn't look as practical for small committees as the other methods do. SD has much simplicity appeal in small committees and organizations, though, if we disregarded simplicity, I'd prefer the other methods to it. But simplicity can be important when proposing a voting system to an organization or a small committee, or to the public.

But, for public elections, RP is as briefly defined as SD, and RP is better. For a public proposal, it's a choice between RP's merit, and PC's especially extreme simplicity.

Mike Ossipoff ***

Mike Ossipoff

Up to Index of Ossipoff's Single Winner article