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Does the Schwartz set violate anything that the Smith doesn't violate?<br><br><blockquote><hr>Date: Sat, 22 Sep 2007 17:45:17 -0500<br>From: rboundless@gmail.com<br>To: johnwong00@hotmail.com<br>Subject: Re: [Election-Methods] How important is the Schwartz criterion? Also, what is the Landau set, and ho<br><br>Hello, John.<br><br>I would say that it's not especially important. Smith is important, because a method that fails Smith will also fail Local Independence of Irrelevant Alternatives. However, as Markus Schulze once pointed out to me, requiring that the winner come from the Schwartz set won't offer a more robust sort of independence.<br><br>Raven<br><br><div><span class="EC_gmail_quote">On 9/22/07, <b class="EC_gmail_sendername">John Wong</b> <<a href="mailto:johnwong00@hotmail.com">johnwong00@hotmail.com</a>> wrote:</span><blockquote class="EC_gmail_quote" style="padding-left: 1ex;"><div>That sorta answers my question about Landau, but what about the Schwartz criterion? Is it important? <br><br></div></blockquote></div><br></blockquote><br /><hr />More photos; more messages; more whatever – Get MORE with Windows Live™ Hotmail®. NOW with 5GB storage. <a href='http://imagine-windowslive.com/hotmail/?locale=en-us&ocid=TXT_TAGHM_migration_HM_mini_5G_0907' target='_new'>Get more!</a></body>
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