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Smith and Schwartz
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Pharamond Curtis
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Sep 30, 2001 16:09 PDT
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First, thanks to Rob and Blake for answering my questions about the Ranked
Pairs tiebreaker.
I've been reading definitions of the Schwartz set, but I'm having trouble
understanding them. How is the Schwartz set different from the Smith set?
Are members of the Smith set always members of the Schwartz set? Does
anyone have any examples in which the Schwartz set is not identical to the
Smith set? Does anyone here see Ranked Pairs' failure to meet the Schwartz
criterion to be a serious flaw?
--
28.35 grams of prevention is worth 0.45 kilograms of cure.
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