[EM] two bit ratings

[Forest Simmons]

I would like to clarify two of my claims about my second method below.

On Wed, 19 Sep 2001, Forest Simmons wrote:

<snip>

> 
> II. The second method (my favorite two bit method) is similar to Ranked
> Pairs, only instead of winning votes or margin size determining the
> strength of each pair, it's the number of ballots on which the pair
> straddles the big gap, i.e. the number of ballots on which the two
> candidates in the pair have ratings that differ in the most significant
> bit.
> 
> If there is no pairwise unbeaten candidate, then weak pairs are zeroed out
> (starting at the weakest) until there is an unbeaten candidate. 

Clarification:

This description of which pairs to neutralize has to be changed to mimic
Ranked Pairs a little more closely in order to satisfy the Reverse
Symmetry Criterion. 

As in Ranked Pairs, lock in pairs, starting from the strongest, while
neutralizing pairs (i.e. over-riding their verdict) only when they
contradict the partial order established by the previously locked in
pairs. 


> 
> In this method it is useful for the voter to think of the two bit ratings
> as base ten expressions:
> 
> 00=zero < 01=one << 10=ten < 11=eleven
> 
> The gap sizes count for nothing if there is a pairwise unbeaten candidate.
> 
> Otherwise, the relative gap sizes (big vs small or zero) help determine
> the strength of the pairs.
> 
> It is necessary to compute two matrices: the pairwise margin matrix, as
> well as the pair strength matrix.
> 
> Both of these matrices are n by n matrices, where n is the number of
> candidates. 
> 
> The margin matrix is antisymmetric, and the strength matrix is symmetric
> with zeros down the main diagonal. So there are only n(n-1) numbers to be
> carried in the precinct summaries.
> 
> This method satisfies both the Condorcet Criterion and the FBC.
> 

Clarification:

It satisfies a version of the Condorcet Criterion relativized to the
ballot type:

If there is a candidate A such that for any other candidate B, it is the
case that candidate A is preferred over candidate B on more ballots than
candidate B is preferred over candidate A, then candidate A is the
method's designated winner. 

This is the closest we can get to the Condorcet Criterion for any method
that doesn't require complete rankings.  So, for example, if Ranked Pairs
allows truncation of ballots, this is the closest it can get to satisfying
the Condorcet Criterion.

> Because of the symmetry properties of the margin and strength matrices
> it should also satisfy the Reverse Symmetry Criterion.

Clarification:

In Ranked Pairs, the strength of a pair is determined by the absolute
value of the margin.  The Reverse Symmetry of Ranked Pairs derives from
the fact that when the order is reversed the margin matrix merely changes
sign, so the absolute values are the same, and the same pairs are locked
in. 

In our version, the strength matrix for reverse order is identical to the
strength matrix for the forward order because if two candidates straddle
the big gap in one order, then they also do in the other order. 

As in Ranked Pairs, the same pairs get locked in when the order is
reversed, so the Reverse Symmetry Criterion is indeed satisfied by this
method.

In terms of bits, reversing order is the same as subtracting all of the
two bit ratings from 11, which interchanges on with off, yes with no, zero
with one, blank with marked, or however you want to think of the bits.

So if two candidates differed in the most significant bit on some ballot
in the forward order, then in the reverse order on that same ballot they
would also differ in that bit. 


Forest