<div dir="ltr"><div dir="ltr"><br></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Jun 12, 2019 at 1:37 AM C.Benham <<a href="mailto:cbenham@adam.com.au">cbenham@adam.com.au</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div bgcolor="#FFFFFF">
<p>They can't when there are three candidates. But plain MinMax
doesn't meet Clone-Winner<br>
<br>
You can have a trio of clones in a cycle who all narrowly
pairwise beat another single candidate, but<br>
whose strongest defeats (which are only to each other) are
stronger than any of that single (Condorcet Loser) candidate's<br>
defeats (to all of them).<br>
<br>
But Smith//MinMax doesn't have that problem and will nearly always
give the same winner as Ranked Pairs<br>
(and Schulze and River).<br>
<br>
A very small consolation prize for MinMax(Margins) is that unlike
those methods it meets Mono-add-Top (like IRV).<br>
<br></p></div></blockquote><div>How does all of this compare to Tideman's Alternative?<br></div><div><br></div><div>1. Identify Condorcet winner;</div><div>2. If no Condorcet winner, identify Top Cycle;</div><div>3. Eliminate all candidates not in Top Cycle;</div><div>4. Eliminate remaining candidate with fewest votes;</div><div>5. Go to 1.</div><div><br></div><div>I tend to use the Schwartz Set for the Condorcet Winner, and the Smith Set for the Top Cycle. I've found the Schwartz Set is nearly always one candidate. </div><div><br></div><div>I know TA is independent of smith-dominated alternatives, but I haven't determined if it's LIIA or reversal-symmetry (apparently Schulze isn't LIIA and ranked pairs is, but I don't see how). I doubt it satisfies later-no-harm and later-no-help, but the run-off step at 4 might allow it to.</div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div bgcolor="#FFFFFF"><p>
Chris Benham<br>
<br>
</p>
<div class="gmail-m_-1166335301478685373moz-cite-prefix">On 12/06/2019 2:25 pm, robert
bristow-johnson wrote:<br>
</div>
<blockquote type="cite">
<p> </p>
<p>I said some things wrong. First of all, if there are N
candidates, the number of Pairwise Defeats is <span style="color:rgb(0,0,0);font-size:9pt;white-space:pre-wrap">N⋅(N-1)/2, not double that value.</span></p>
<p><span style="color:rgb(0,0,0);font-size:9pt;white-space:pre-wrap">Secondly, I did the scenario with the cycle wrong. I wanted several N-3 values in the decreasing sequence, not just one. But I hope you have the idea.</span></p>
<p><span style="color:rgb(0,0,0);font-size:9pt;white-space:pre-wrap">r b-j</span></p>
<p><br>
<br>
---------------------------- Original Message
----------------------------<br>
Subject: [EM] Discounting ties, how can MinMax differ from
Ranked Pairs?<br>
From: "robert bristow-johnson" <a class="gmail-m_-1166335301478685373moz-txt-link-rfc2396E" href="mailto:rbj@audioimagination.com" target="_blank"><rbj@audioimagination.com></a><br>
Date: Tue, June 11, 2019 9:46 pm<br>
To: <a class="gmail-m_-1166335301478685373moz-txt-link-abbreviated" href="mailto:election-methods@electorama.com" target="_blank">election-methods@electorama.com</a><br>
--------------------------------------------------------------------------<br>
<br>
><br>
><br>
><br>
> Let's assume Margins, but I think this works for Winning
Votes also.<br>
> Begin with all candidates marked as "Plausible Winner"
(which means they are not yet marked as "beaten") and order all
N⋅(N-1) candidate Pairwise Defeats from strongest defeat
strength to weakest defeat strength. Call that ordered list the
"Original
List".<br>
> Now create another ordered list of Pairwise Defeats from
the above Original List.<br>
> <br>
> MinMax starts with<br>
> the entire Original List with N⋅(N-1) entries and
successively eliminates entries from the bottom up.<br>
> Referring to a quote in this Wikipedia
article: <a class="gmail-m_-1166335301478685373moz-txt-link-freetext" href="https://en.wikipedia.org/wiki/Minimax_Condorcet_method#Variants_of_the_pairwise_score" target="_blank">https://en.wikipedia.org/wiki/Minimax_Condorcet_method#Variants_of_the_pairwise_score</a> ,
MinMax starts at the bottom and:<br>
> "Disregards the weakest Pairwise Defeat until one
candidate is unbeaten."<br>
><br>
><br>
><br>
> Similarly Ranked Pairs starts with an empty list and
successively includes entries from the Original List from the
top down.<br>
> Cannot Ranked Pairs be concisely stated as:<br>
> "Include the strongest Pairwise Defeat until only one
candidate is unbeaten." ?<br>
> <br>
> Is that not an accurate description of both<br>
> methods?<br>
> Now, a decreasing sequence of numbers (*not* strictly
decreasing) of length N⋅(N-1)+1 that represents the number of
"unbeaten" candidates (of all Pairwise Defeats above it) having<br>
> value of N at the top, that can be thought of as having
each of the N⋅(N-1) Pairwise Defeats inserted between these
numbers of decreasing integer value.<br>
> Assuming no ties, the value of this sequence at the bottom<br>
> must be either 0 or 1. Also, assuming no ties, this
decreasing sequence can only be decremented by 1 or 0.<br>
> If the value of this sequence is 1 at the bottom, there is
a Condorcet Winner and all Pairwise Defeats are considered. But
regardless of whether MinMax or RP is used, this<br>
> ordered list is the same decreasing sequence of integer
values. If the value of this sequence of numbers is 0 at the
bottom, there is no Condorcet Winner. But, for all Pairwise
Defeats having "1" below them, would not the undefeated
candidate be the same<br>
> candidate?<br>
> This might look like:<br>
> <br>
> Plausible Winners ----- Pairwise Defeat<br>
> N<br>
> A>D<br>
> N-1<br>
> A>C<br>
> N-2<br>
> B>D<br>
> N-2<br>
> B>A<br>
> N-3<br>
> B>C<br>
> N-3<br>
> C>D<br>
> N-3<br>
> <br>
> Now if N=4 then the bottom N-3 is 1 and we have a Condorcet
Winner.<br>
> Now suppose we have a cycle:<br>
> Plausible Winners ----- Pairwise Defeat<br>
> N<br>
> A>D<br>
> N-1<br>
> A>B<br>
> N-2<br>
> B>D<br>
> N-2<br>
> C>D<br>
> N-2<br>
> B>C<br>
> N-3<br>
> C>A<br>
> N-4<br>
> <br>
> Now if N=4 then the bottom N-4 is 0 and we don't have a
Condorcet Winner. But the unbeaten candidate is A whether we
include the B>C Pairwise Defeats or not. Assuming no ties,
how can the unbeaten candidate be different?<br>
> You guys may have discussed<br>
> this before when I wasn't paying attention, but it seems to
me that if there are no ties and all of the Defeat strengths are
unequal values, the ordering of this list must be the same and
the number of Plausible Winners must decrease from N at the top
to 1 at the bottom if there is a CW (or N
to 0<br>
> if no CW). How can MinMax and Ranked Pairs elect a
different candidate?<br>
> Thank you for any attention and thought put to this.<br>
> --<br>
><br>
><br>
> r b-j <a class="gmail-m_-1166335301478685373moz-txt-link-abbreviated" href="mailto:rbj@audioimagination.com" target="_blank">rbj@audioimagination.com</a><br>
><br>
><br>
><br>
> "Imagination is more important than knowledge."<br>
><br>
> <br>
> <br>
> <br>
> <br>
> ----<br>
> Election-Methods mailing list - see
<a class="gmail-m_-1166335301478685373moz-txt-link-freetext" href="https://electorama.com/em" target="_blank">https://electorama.com/em</a> for list info<br>
></p>
<p> </p>
<p> </p>
<p> </p>
<p><br>
--<br>
<br>
r b-j <a class="gmail-m_-1166335301478685373moz-txt-link-abbreviated" href="mailto:rbj@audioimagination.com" target="_blank">rbj@audioimagination.com</a><br>
<br>
"Imagination is more important than knowledge."<br>
</p>
<p> </p>
<p> </p>
<p> </p>
<br>
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<pre class="gmail-m_-1166335301478685373moz-quote-pre">----
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