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Re: Smith and Schwartz
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Rob LeGrand
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Sep 30, 2001 18:45 PDT
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Pharamond Curtis wrote:
| | How is the Schwartz set different from the Smith set?
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Usually, it's not. But if there are pairwise ties, they might be
different. See Markus Schulze's post
http://groups.yahoo.com/group/election-methods-list/message/6493
for an algorithm to compute the Schwartz set. The same algorithm can
be used to calculate the Smith set by changing the line
if (d[i,j] > d[j,i]) then
to
if (d[i,j] >= d[j,i]) then
| | Are members of the Smith set always members of the Schwartz set?
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No, but the converse is true: The Schwartz set is always a subset of
the Smith set. So if a method passes the Schwartz criterion, it also
passes the Smith criterion (and thus the Condorcet criterion).
| | Does anyone have any examples in which the Schwartz set is not
identical to the Smith set?
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Sure. Here's one I generated randomly:
1:A>B>D>C
2:B>A>C>D
3:C>A>D>B
2:D>B>A>C
The Smith set is {A, B, C, D}, but the Schwartz set is just {A}. Note
that A has no pairwise losses. Ranked Pairs might choose B depending on
the tiebreaker, but Schulze, Minmax, Copeland, Dodgson and Borda always
pick A. IRV picks B.
| | Does anyone here see Ranked Pairs' failure to meet the Schwartz
criterion to be a serious flaw?
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Nah, not really. I would like a method to pass the Smith criterion,
but I think most agree that Schwartz is too strict. Besides, the two
sets will always be the same in large enough elections.
--
Rob LeGrand
honk-@aggies.org
http://www.aggies.org/honky98/
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