On 12/14/05, <b class="gmail_sendername">Rob LeGrand</b> <<a href="mailto:honky1998@yahoo.com" target="_blank" onclick="return top.js.OpenExtLink(window,event,this)">honky1998@yahoo.com</a>> wrote:<div><span class="gmail_quote">
</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
Paul Kislanko wrote:<br>> The number of full ranked ballots is just the number of permutations of<br>> N alternatives = N!<br>><br>> If equal ranknigs are allowed, it's N! + 2^N - 1<br><br>I assume you mean that there are N! ways to arrange the candidates
<br>strictly, and then N - 1 spaces in between that could hold either a > or<br>a =, giving N! * 2^(N - 1). But I think that's an overestimate because<br>it would count both A=B>C and B=A>C. Did you mean something different?
</blockquote><div><br>I did a quick check with 3 candidates, and his formula appeared to give the correct answer (note that I eliminated redundant ones like your example):<br><br>a=b=c<br>a>b>c<br>a>b=c<br>a=b>c
<br><br>a>c>b<br>a=c>b<br><br>b>a>c<br>b>a=c<br><br>b>c>a<br>b=c>a<br><br>c>a>b<br>c>a=b<br><br>c>b>a<br><br>which is 13, which agrees with <a href="http://www.google.com/search?q=3%21%2B2%5E3-1" target="_blank" onclick="return top.js.OpenExtLink(window,event,this)">
http://www.google.com/search?q=3%21%2B2%5E3-1</a><br><br></div></div>Although i might not want to count a=b=c, which I would sorta consider "not voting"...<br><br><span class="gmail_quote"></span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
Paul Kislanko wrote:<br>And double yes to I'm glad someone's working on a way to model the different methods from a universal ballot format. It should make it easier to see the differences between Condorcet-based methods.
</blockquote><div><br></div>Cool. While I may not be 100% aligned with your philosophical objection to the matrix as an intermediate step, I am certainly interested in exploring some methods which don't use it. (I'm not trying to model the exisiting condorcet methods this way, but looking at other, new[?] methods that don't use the matrix)
<br><br>Unfortunately with 10 candidates, there are about 4 million possible unique ballots, so this non-lossy compression scheme may be less helpful than I hoped<br><br>rob<br>