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<TITLE>election-methods Digest, Vol 23, Issue 4</TITLE>
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<DIV id=idOWAReplyText942 dir=ltr>
<DIV dir=ltr><FONT face=Arial color=#000000 size=2>Paul,</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>Actually, I did not assume that there was
any linear or two dimensional relationship. You can use any measure of
closeness that you want, linear, non-linear, ten dimensional, or infinite
dimensional (for example the norm of the difference of the quantuum mechanical
wave functions of the candidates, if you like). </FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>My only assumption is that there is
some measure d of distance that (for each pair of candidates A and
B) satisfies </FONT><FONT face=Arial size=2>d(A,B)=d(B,A)
, </FONT><FONT face=Arial size=2>and that when d(A,B)<d(A,C),
consistency requires that candidate A rank candidate B ahead of candidate C
.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>These distances are numbers, so they can be
linearly ordered. </FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>In fact, when there are n candidates, there
are C(n,2) of these pairwise distances, which can be ordered in C(n,2)!
ways. Each of these orderings is in an equivalence class of size n!
obtained by permuting the candidates. So there are at most C(n,2)!/n!
distinct geometrically consistent cases.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>If n=3, then C(n,2)=3, and C(n,2)!=n!, so
there is only one distinct geometrically consistent case.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>If n=4, then C(n,2)=6, and C(n,2)!=720, and
n!=24, so that leaves 720/24=30 geometrically distinct cases. If you
will write out all thirty of these cases (as I have), you can verify
for yourself that only one of them fails to always have a Condorcet
Winner.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>In the process you will see that only one
of them (a different one) fails to always have a Condorcet Loser, as
well.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>So most of the time, in the context of
Candidate Published Orderings, Concorcet will yield an unambiguous social
ordering of the candidates, with no cycles to resolve.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>More importantly (and precisely) it
means that if the four candidates accurately judge their relative closeness
to each other (and vote sincerely) then there is only one chance in
240 that there will not be a Condorcet Winner.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>Why 1/240 instead of 1/30? Because,
as I mentioned in my previous message, even in the one case where there is not
always a Condorcet Winner, only one eighth of the distributions of the
voters among the four factions will result in a cycle.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>I would say that's amazing, and extremely
relevant to the topic of this thread.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>Forest</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV></DIV>
<DIV dir=ltr><BR>
<HR tabIndex=-1>
<FONT face=Tahoma size=2><B>From:</B> Paul Kislanko
[mailto:kislanko@airmail.net]<BR><B>Sent:</B> Wed 5/10/2006 7:35
PM<BR><B>To:</B> Simmons, Forest<BR><B>Subject:</B> RE: [EM] Voting by selecting
a published ordering<BR></FONT><BR></DIV>
<DIV>
<DIV dir=ltr align=left><SPAN class=966333302-11052006><FONT face=Arial
color=#0000ff size=2>The problem is you just translated a multidemionsal problem
into a 2-dimensional one by assuming there was a linear relationship among the
candidates, when in fact the social choice problem is that assigning a metric
that relates A & B is impossible if there's more than 1 issue and more than
3 voters.</FONT></SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=966333302-11052006><FONT face=Arial
color=#0000ff size=2></FONT></SPAN> </DIV>
<BLOCKQUOTE dir=ltr
style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #0000ff 2px solid; MARGIN-RIGHT: 0px">
<DIV class=OutlookMessageHeader lang=en-us dir=ltr align=left>
<HR tabIndex=-1>
<FONT face=Tahoma size=2><B>From:</B> election-methods-bounces@electorama.com
[mailto:election-methods-bounces@electorama.com] <B>On Behalf Of </B>Simmons,
Forest<BR><B>Sent:</B> Wednesday, May 10, 2006 5:19 PM<BR><B>To:</B>
election-methods@electorama.com<BR><B>Subject:</B> Re: [EM] Voting by
selecting a published ordering<BR></FONT><BR></DIV>
<DIV></DIV>
<DIV id=idOWAReplyText46705 dir=ltr>
<DIV dir=ltr><FONT size=2>I've been exploring certain questions in relation to
this topic of Candidate Published Orderings.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>First of all, which combinations of candidate
orderings make sense geometrically?</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>This question is not hard to answer in the case of
three candidates, A, B, and C.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>Suppose that candidates A and C are the furthest
apart from each other as measured by some symmetric metric (Euclidean or
not).</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>Then geometric consistency requires that the A and C
orderings are </FONT><FONT size=2>ABC and CBA, respectively.
</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>The B ordering could be either BAC or
BCA. </FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>Either of these can be obtained from the other by a
permutation of the letters.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>In summary, there is essentially only one case (up
to permutations of the letters) that is geometrically consistent, when there
are three candidates:</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>x:ABC, y:BAC, z:CBA ,</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>and no matter the sizes of the factions x, y, and z,
there will always be a Condorcet Candidate.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>There are many more cases to consider when there are
four candidates A, B, C, and D.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>However, it turns out that (up to permutations of
the letters) there are only thirty (30) cases that are geometrically
consistent, and 29 (all but one) of these are like the above case in that
there will always be a Condorcet Candidate, no matter how the faction sizes w,
x, y, and z are distributed.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>The geometrically consistent case that does not
always have a Condorcet Winner is the following:</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>w:ACDB, x:BCDA, y:CBDA, z:DBCA.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>Even this case has a Condorcet Winner
for precisely 87.5 percent of the distributions of faction sizes w, x, y,
and z. The exceptions occur when all of the following relations hold
simultaneously:</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>x+y<50%, z<50%, x+z>50%</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>which describes precisely one eighth of the volume
of the tetrahedron given by</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>x+y+z < 100% </FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>in the first octant of the x, y, z, coordinate
system. (Note that w is determined from x, y, and z by w+x+y+z=100%
.)</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>To see that this case is geometrically consistent,
locate candidates A, B, C, and D respectively, at the points</FONT></DIV>
<DIV dir=ltr><FONT size=2>(4,2), (0,0), (1,0), and (0,2) in a
Cartesian Coordinate Plane. </FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>The important thing about these positions is
that</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT
size=2>d(A,B)>d(A,D)>d(A,C)>d(C,D)>d(B,D)>d(B,C) .</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>I'll stop here for now, to give you all a chance to
digest these interesting facts.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>To me they are rather amazing facts, and potentially
useful, IF we can figure out how to incorporate them
appropriately into this candidate published ordering setting, or more
generally, into a one faction per candidate setting.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>More on that later.</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2>Forest</FONT></DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT size=2></FONT> </DIV>
<DIV dir=ltr><FONT
size=2></FONT> </DIV></DIV></BLOCKQUOTE></DIV></BODY></HTML>