Smith//Condorcet tally shortcut? (was Re: Condorcet tally shortc

[Steve Eppley]

Mike O wrote:
>For the Smith set, my procedure included a shortcut, intended mostly
>to save labor in hand-counts: Not only is something admitted to the
>Smith set if it beats or ties something in the Smith set, but also if
>it is beaten by as few as (or beats as many as) something in the
>Smith set. 
-snip-

That Smith Set shortcut assumes it's known how many pairwins and/or
pairlosses each candidate has, which means first calculating ALL the
pairs.  But calculating all the pairs is far more time-consuming than
calculating the Smith set so there may be a more effective *overall* 
shortcut for Smith//Condorcet.

It appears that this can be done by finding the Copeland winner, the
"seed" of the Smith set, without finding *every* candidate's win/loss
record.  This would use a convergent process analogous to the 
Condorcet's Method shortcut, but with "fewer losses" substituted
for "smaller largest opposition" when keeping the BestYet of two 
candidates.

Then, given the Copeland seed, the Smith set is determined.  Some of 
the pair counts for this will already have been done during the Smith 
Seed process, but some others will need to be calculated since every
candidate needs to be paired against every member of Smith.

Then, given the Smith set, the Smith//Condorcet winner can be found
using the same shortcut which works for Condorcet, but with the
candidates limited to those in Smith.  Any rows and columns of 
the pair matrix calculated during the previous processes can be 
discarded if they involve candidates not in the Smith set.  Most
of the pair counts needed to complete the //Condorcet tally will 
already have been done during the Smith Seed and Smith Set processes.

---Steve     (Steve Eppley    seppley at alumni.caltech.edu)