[EM] IRV / RCv advances

Kristofer Munsterhjelm km_elmet at t-online.de
Fri Jul 13 09:37:51 PDT 2018

On 2018-07-13 08:53, Richard Lung wrote:
> It is ironic that the world seems to have this electoral reform battle 
> of the big-enders versus the little-enders (from Gullivers Travels). 
> That is to say the collectivist Europeans and their outliers, back a 
> proportional count with party lists - such as sabotages individual 
> choice, without a ranked choice for voters. While the good ole US of A 
> does not forget individual representation with a ranked choice, but 
> forgets equality of representation with a proportional count. The only 
> exception is Cambridge Mass. using STV, and one or two minor cases of 
> STV in Minnesota, I believe.
> One EM member says STV is BAD. As the ignored inventor of FAB STV, I 
> know the limitations of traditional STV but essentially it is on the 
> right lines, laid down by the original inventors, Carl Andrae and Thomas 
> Hare, namely the quota-preferential method, as the Aussies pithily 
> describe its essence.
> As for ranked pairing, my understanding is that it is not an independent 
> method at all, but a means of cross-referencing a ranked choice 
> electoral system. As previously mentioned to this email group, I wrote a 
> supplemetary chapter on this, in my book FAB STV: Four Averages Binomial 
> Single Transferable Vote.

If you mean Ranked Pairs, that is a method of its own. It goes like this:

1. Construct a pairwise matrix (call it d) from the set of ranked 
ballots. Usually this is defined as c[x, y] = number of voters who 
preferred x to y, and d[x, y] = c[x, y] if c[x, y] > c[y, x], otherwise 
d[x, y] = 0. (wv) Margins is also possible; see below.

2. Make a list of pairwise contests and sort it from greatest value of 
d, to least value of d. A pairwise contest entry is of the form "X over 
Y" and has strength equal to d[x, y].
E.g. if d[a, b] = 100 and d[b, c] = 80, then "A over B" comes before "B 
over C".

3. Go down the list and affirm a pairwise contest unless it would 
contradict what has already been affirmed. "Affirming X over Y" means 
that the final outcome must rank X over Y.

4. Return the candidate who has nobody affirmed over him; he is the winner.

Strictly speaking, what I've described above is MAM (without 
tiebreakers). Ranked Pairs "proper" uses margins (d[x, y] = c[x, y] if 
c[x, y] > c[y, x], otherwise d[x, y] = c[x, y] - c[y, x]).

An example instance of step 3 could go like:

If the list is:
	40: A over B
	34: B over C
	30: C over A
	 0: B over A
	 0: C over B
	 0: A over C

Then step three proceeds this way:
	Affirm A over B: A must be higher ranked than B in the outcome
	Affirm B over C: B must be higher ranked than C in the outcome
	C over A: Impossible because we've affirmed A over B and B over C, 
which means we indirectly have A over C, and C over A would contradict it.
	B over A: Contradicts A over B
	C over B: contradicts B over C
	Affirm A over C: A must be higher ranked than C in the outcome (though 
we already knew that).

And then the outcome is: A first, B second, C third. A is the winner.

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