<div dir="auto">First example ballot profile ... a case of Chicken Defection:<div dir="auto"><br></div><div dir="auto">28 A>B</div><div dir="auto">24 B (sincere B>A)</div><div dir="auto">48 C</div><div dir="auto">The sincere CW, (namely A) has been subverted by the B faction's defection from a sincere (A, B} clone coalition ... creating a ballot beat cycle of ABCA.</div><div dir="auto"><br></div><div dir="auto">A>B 28 to 72</div><div dir="auto">B>C 52 to 48, and</div><div dir="auto">C>A 48 to 28</div><div dir="auto"><br></div><div dir="auto">Classical Condorcet breaks the cycle at the weakest link A>B ... leaving the rest of the cycle BCA intact. So classical Condorcet elects B, thereby rewarding the defectors.</div><div dir="auto"><br></div><div dir="auto">How about our "littlest help from our friends" method?</div><div dir="auto"><br></div><div dir="auto">The the respectivevcovering pairs are the defeat pairs A>B, B>C, & C>A ... where the helpers are the defeat losers and the "enemies" befriended (i.e. covered) by the helpers are respectively C, A, & B, with respective help strengths of </div><div dir="auto">#[B>C]=52, #[C>A]=48, and #[A>B]=26.</div><div dir="auto"><br></div><div dir="auto">The last of these is the "least help" ... the help given by A to befriend C's enemy B.</div><div dir="auto"><br></div><div dir="auto">So C is the candidate needing the least help from its friend to befriend its enemy.</div><div dir="auto"><br></div><div dir="auto">Therefore our least-help-from-friend(s) method elects C ... which constitutes a backfire against the defecting candidate B.</div><div dir="auto"><br></div><div dir="auto">Next time ... a geometrically generated family of examples...</div><div dir="auto"><br></div><div dir="auto">-Forest</div><div dir="auto"><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Mar 15, 2023, 3:02 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com">forest.simmons21@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto">Here's the Enhanced Copeland method we've been leading up to:<div dir="auto"><br></div><div dir="auto">Elect the candidate that needs the "littlest help from its friends" to cover its enemies.</div><div dir="auto"><br></div><div dir="auto">Explanation:</div><div dir="auto"><br></div><div dir="auto">Last time we learned that if C is a candidate tied for Copeland winner, then C's friends will cover its enemies ... in other words, every enemy of C is a friend of a friend of C ... in other words C will be a Landau candidate ... in other words, our Enhanced Copeland Winner is the Landau candidate that requires the least help from its friends to befriend its enemies.</div><div dir="auto"><br></div><div dir="auto">Operationally we proceed as follows:</div><div dir="auto"><br></div><div dir="auto">1. Each Landau candidate C (including all tied Copeland Winners) designates a set of candidates H(C) as its "helpers".</div><div dir="auto"><br></div><div dir="auto">2. A Deliberative Assembly checks the validity of these helper claims, and eliminates any candidate C whose enemies are not covered by H(C).</div><div dir="auto"><br></div><div dir="auto">3. Next the Deliberative Aassembly narrows down the candidate pool to T=argmin #H(C) ... the set of candidates tied for the fewest (designated) helpers needed to defeat all of its enemies.</div><div dir="auto"><br></div><div dir="auto">4.From among the members of the set T tied for fewest needed (designated) helpers, elect the candidate C whose Max Pairwise Opposition from an enemy of C to a designated helper of C, is minimal. </div><div dir="auto"><br></div><div dir="auto">This is the candidate C that requires the fewest helper friends (and the least help from those so designated) in order to befriend all of its enemies (especially the one hardest for a designated helper to befriend).</div><div dir="auto"><br></div><div dir="auto">In practice, no candidate in a public election will need more than one helper, so the method simplifies to electing the candidate C whose designated helper H most easily befriends the enemy E of C having the most Pairwise Opposition against H.</div><div dir="auto"><br></div><div dir="auto">The less Pairwise Opposition that E has against H the less help (defined as H's Pairwise Support against E) it takes to befriend E. </div><div dir="auto"><br></div><div dir="auto">So C is the candidate that needs the least help in befriending its enemies.</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto">Next time some examples ...</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Mar 14, 2023, 10:11 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" target="_blank" rel="noreferrer">forest.simmons21@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto">I've always had a soft spot in my heart for Copeland ... perhaps because it's such a friendly method. Let me explain ...<div dir="auto"><br><div dir="auto">Remember Kristofer's wry definition of friendship ... whoever doesn't beat you is your friend?</div><div dir="auto"><br></div><div dir="auto">By that definition the Copeland winner is simply the candidate with the most friends.</div><div dir="auto"><br></div><div dir="auto">And everybody is a friend of the Condorcet Winner (the unbeaten candidate if there is one) ... so Copeland is Condorcet efficient.</div><div dir="auto"><br></div><div dir="auto">Simple and compelling ... as far as it goes ... but it doesn't always go the whole distance ... what if there are two or three candidates tied for most friends?</div><div dir="auto"><br></div><div dir="auto">Various Copeland tie breaking proposals have been suggested over the years .... but none of them seem to naturally continue the basic friendship theme that inspired Copeland in the first place.</div><div dir="auto"><br></div><div dir="auto">Here's a friendship idea that might help:</div><div dir="auto"><br></div><div dir="auto">A set of candidates can have friends: you are a friend of a set if you don't beat every member of the set.</div><div dir="auto"><br></div><div dir="auto">Another way to say this ... is that you are a friend of a set iff the set covers you. </div><div dir="auto"><br></div><div dir="auto">This friendship/ covering relation leads to a natural hierarchy of candidates ... </div><div dir="auto"><br></div><div dir="auto">A Condorcet candidate has everybody as a friend ... and therefore covers everybody ... without any help.</div><div dir="auto"><br></div><div dir="auto">A candidate X that covers everybody with the help of only one friend F is the next level. Together X and F form a set X+F that cannot be beaten ... everybody is friendly to X+F just like everybody is friendly to a Condorcet Candidate, when there is one.</div><div dir="auto"><br></div><div dir="auto">So different candidates need different amounts of help from their friends to cover their "enemies", the candidates that are not their friends.</div><div dir="auto"><br></div><div dir="auto">The fewer friends you need to cover all of your enemies, the stronger you are.</div><div dir="auto"><br></div><div dir="auto">A Condorcet Winner has lots of friends and zero enemies.</div><div dir="auto"><br></div><div dir="auto">At the other extreme ... a Condorcet Loser has no friends to help cover his enemies .... which is unfortunate because all of the other candidates are his enemies.</div><div dir="auto"><br></div><div dir="auto">What about a Copeland candidate C tied for most friends? Let F be the set of friends of C. Then C and F together cover the enemies of C.</div><div dir="auto"><br></div><div dir="auto">Any time a candidate C with some of her friends F together cover all of her enemies, we say that C is a Landau candidate. So every max score Copeland candidate is a Landau candidate .... which makes Copeland a "Landau efficient" method.</div><div dir="auto"><br></div><div dir="auto">So the bigger the Copeland score of a candidate the fewer her enemies and the fewer friends she needs to help cover those enemies. </div><div dir="auto"><br></div><div dir="auto">This brings us to another imperfection of Copeland that we will take care of with the help of our friends:</div><div dir="auto"><br></div><div dir="auto">Suppose candidate F is a friend of candidate C and of nobody else. Then unscrupulous C supporters might be tempted to enter a whole "team" T of clones of F into the race to increase C's Copeland score and nobody else's.</div><div dir="auto"><br></div><div dir="auto">How can we effectively counter this kind of "Teaming" temptation?</div><div dir="auto"><br></div><div dir="auto">A natural way is to see how far we can pare down the friendly help while still covering the enemies of C .... just how many friends does C actually need to cover her enemies?</div><div dir="auto"><br></div><div dir="auto">This doesn't completely resolve the tie problem ... but it reduces the covering sets down to manageable size.</div><div dir="auto"><br></div><div dir="auto">In fact, the most amazing and useful thing I have learned in these friendly explorations, is that in public elections ... for all practical purposes ... it is 99.999... percent sure that at least one candidate C will have a friend F which is all the help she needs to cover her enemies.</div><div dir="auto"><br></div><div dir="auto">This fact hugely simplifies tie breaking and, at the same time, completely erases the clone teaming problem!</div><div dir="auto"><br></div><div dir="auto">To be continued ....</div><div dir="auto"><br></div><div dir="auto">-Forest</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div></div></div>
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