[EM] Another multi-seat Condorcet method (PSC-CLE)

[Dan Bishop]

Here's an election method I thought of a few weeks ago, called 
Proportionality of Solid Coalitions by Condorcet Loser Elimination 
(PSC-CLE).  It has the advantage of simplicity, but the disadvantage of 
having a Borda-like "teaming" effect.


DEFINITION AND EXAMPLE

Consider an election for 2 seats with the ballots

33 A>D>B>C
33 B>D>A>C
32 C>D>A>B
2  D>A>B>C

**Step 1: Choose a Condorcet ranking method and use it to rank the 
candidates.  This ranking will be the inverse of the elimination order.**

For the example, I will use Ranked Pairs, which gives the order D>A>B>C, 
so the elimination order will be C, B, A, D.

**Step 2: Calculate the quota.**

quota = votes/(seats+1)+epsilon=33.333...+epsilon

**Step 3: Count the votes for each solid coalition, as in DSC.**

First preferences:

{A} has 33 votes
{B} has 33 votes
{C} has 32 votes
{D} has 2 votes

First 2 preferences:

{A,D} has 34 votes
{B,D} has 33 votes
{C,D} has 32 votes

First 3 preferences:

{A,B,D} has 68 votes
{A,C,D} has 32 votes

First 4 preferences:

{A,B,C,D} has 100 votes

**Step 4: Divide all the vote totals by the quota and discard the 
remainder**

{A} has 0.98999... quotas -> 0
{B} has 0.98999... quotas -> 0
{C} has 0.95999... quotas -> 0
{D} has 0.05999... quotas -> 0
{A,D} has 1.01999... quotas -> 1
{B,D} has 0.98999... quotas -> 0
{C,D} has 0.95999... quotas -> 0
{A,B,D} has 2.03999... quotas -> 2
{A,C,D} has 0.95999... quotas -> 0
{A,B,C,D} has 2.999... quotas -> 2

**Step 5: List the "proportionality rules".  Each set of C candidates 
with Q quotas is entitled to min(C, Q) seats.**

Elect at least 1 candidate from {A, D}.
Elect at least 2 candidates from {A, B, D}.
Elect at least 2 candidates from {A, B, C, D}.

**Step 6: If there are any proportionality rules of the form "Elect at 
least N candidates from this set of N candidates," then declare those N 
candidates elected, and remove them from the elimination order.  If the 
required number of candidates has been elected, then stop.**

So far, there aren't any.

**Step 7: Eliminate the next candidate in the elimination order.**

C is eliminated.  That leaves us with the candidates {A, B, D} and the 
proportionality rules:

Elect at least 1 candidate from {A, D}.
Elect at least 2 candidates from {A, B, D}.

**Step 8: Go back to Step 6.**

Nothing happens then, but at step 7, candidate B is eliminated.  Then we 
go back to Step 6 and declare A and D elected, and are done.

CRITERION COMPLIANCE

* (Single-winner) Condorcet Criterion: PASS

Suppose there is a Condorcet winner.

If there are no proportionality rules, then the CW wins by virtue of 
being last in the elimination order.

Suppose for contradiction that there is a proportionality rule requiring 
the election of a candidate other than the CW.  That is, there is a set 
of candidates S, not including the CW, that are ranked above the 
complement S' (including the CW) on at least one quota (i.e., a simple 
majority) of the ballots.  But this means that the candidates in S 
pairwise beat the CW, which is impossible by definition.  Therefore, if 
there is a proportionality rule, it must include the CW, who will win 
because the other members of the coalition would be eliminated first.

* Proportionality of Solid Coalitions: PASS.
This follows from the definition of the method.

* Independence of Clones: FAIL.

Consider the example election (in which A and D won) with D replaced by 
D1>D2>D3:

33 A>D1>D2>D3>B>C
33 B>D1>D2>D3>A>C
32 C>D1>D2>D3>A>B
  2 D1>D2>D3>A>B>C

The elimination order is C, B, A, D3, D2, D1.  The coalition {A, D1, D2, 
D3} is entitled to 1 seat, and {A, B, D1, D2, D3} is entitled to 2 
seats.  The winners are D1 and D2, which gives the D clone set an extra 
seat compared to the original election.

* (Polynomial) Summability Criterion: FAIL.
AFAICT, this method requires a summation array of size 2^n+n(n-1) = O(2^n).