<div dir="auto"><div><br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Mon, Apr 24, 2023, 8:07 AM 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">For our purposes a chain is a list of alternatives on which every alternative is defeated by every higher listed alternative.<div dir="auto"><br></div><div dir="auto">If every alternative not in the chain is defeated by some member of the chain, it is said to be a "covering chain".</div><div dir="auto"><br></div><div dir="auto">The length of a chain is the number of alternatives comprising the chain.</div><div dir="auto"><br></div><div dir="auto">The head of the chain is the alternative that defeats all of the other members of the chain.</div><div dir="auto"><br></div><div dir="auto">The tail of the chain is the alternative defeated by every other chain member.</div><div dir="auto"><br></div><div dir="auto">Our proposed method is to elect the head of the covering chain of maximum strength among all those of minimum length.</div><div dir="auto"><br></div><div dir="auto">This method satisfies the ISDA criterion.</div><div dir="auto"><br></div><div dir="auto">In a public political election the MLMSCC chain will almost surely consist of fewer than three alternatives ... we can safely say never more than three.</div><div dir="auto"><br></div><div dir="auto">So the winner can be found by comparing the strengths of at most a few dozen chains.</div><div dir="auto"><br></div><div dir="auto">This would be the case, for example, if the top cycle were to be created by replacing each member of a rock-paper-scissors cycle with a different clone cycle of the same kind.</div><div dir="auto"><br></div><div dir="auto">In fact, in that case there would be exactly three dozen covering chains of length three to check.</div></div></blockquote></div></div><div dir="auto"><br></div><div dir="auto">Actually only 27 chains to check ... each of them isomorphic to (r,r)>(r,s)>(s,s).</div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto"><div dir="auto"><br></div><div dir="auto">To be continued (later today when I get another turn on the cell phone ...)</div><div dir="auto"><br></div><div dir="auto">-fws</div></div></blockquote></div></div><div dir="auto"><br></div><div dir="auto">Even the formidable "twisted prism" has only three chains to check ... the three defeats from the upper deck are the only minimum length covering chains.</div><div dir="auto"><br></div><div dir="auto">For the sake of completeness we now specify how to find the winner even in the statistically impossible case of more than 27 chains to check:</div><div dir="auto"><br></div><div dir="auto">Each faction may propose a covering chain. Elect the head of the strongest admissible proposal from among those tied for minimum length.</div><div dir="auto"><br></div><div dir="auto">Which brings us to the question of "chain strength" ... how do we gauge it?</div><div dir="auto"><br></div><div dir="auto">When the chain has only one link ... that is only one defeat ... the chain strength is simply that defeat strength.</div><div dir="auto"><br></div><div dir="auto">A chain of length three A>B>C has three defeats A>C, A>B, and B>C. To preserve reverse symmetry we define the defeat strength to be </div><div dir="auto"><br></div><div dir="auto">s(A>C) + min(s(A>B),s(B>C))*epsilon,</div><div dir="auto"><br></div><div dir="auto">where s is some gauge of pairwise defeat strength ... and epsilon is a positive infinitesimal that comes into play only when needed for tie breaking purposes.</div><div dir="auto"><br></div><div dir="auto">This definition of chain strength is extended indefinitely by recursion:</div><div dir="auto"><br></div><div dir="auto">s(A>B>...Y>Z) is given by s(A>Z) plus ...</div><div dir="auto">... min(s(A>...>Y),s(B> ...>Z))*epsilon</div><div dir="auto"><br></div><div dir="auto">That's it!</div><div dir="auto"><br></div><div dir="auto">For basic defeat strength of a sinle pair of alternatives, I highly recommend </div><div dir="auto"><br></div><div dir="auto">s(A>B)=approval(A)+disapproval(B),</div><div dir="auto"><br></div><div dir="auto">where approval X becomes disapproval X when all preferences are reversed.</div><div dir="auto"><br></div><div dir="auto">One such approval/disapproval pair is Equal Top Whole / Bottom Count Whole.</div><div dir="auto"><br></div><div dir="auto">EQTop(X) is the number of ballots on which X is not outranked. BotCount(X) is the number of ballots on which X outranks no candidate.</div><div dir="auto"><br></div><div dir="auto">Here's another example: </div><div dir="auto"><br></div><div dir="auto">approval(X) is the number of ballots on which some candidate outranked by X defeats every candidate that outranks X, </div><div dir="auto"><br></div><div dir="auto">... while disapproval(X) is the number of ballots on which some candidate that outranks X defeats every candidate outranked by X.</div><div dir="auto"><br></div><div dir="auto">If range/score ballots are used, then [disapproval, respectively, approval] of X is the number of ballots on which X is rated [below, resp, above] midrange.</div><div dir="auto"><br></div><div dir="auto">Now that you see how to respect the reverse symmetry, you can make up your own symmetric defeat strength gauge.</div><div dir="auto"><br></div><div dir="auto">Example:</div><div dir="auto"><br></div><div dir="auto">48 C</div><div dir="auto">28 A>B</div><div dir="auto">24 B</div><div dir="auto"><br></div><div dir="auto">The MinLentthMaxStrengthCoverinChain is C>A with strength EQTop(C)+BotCount(A)=48+72.</div><div dir="auto"><br></div><div dir="auto">Note also approvalC+disapprovalA is the same sum 120. </div><div dir="auto"><br></div><div dir="auto">By comparison, approvalB+disapprovalCis 52+52=104.</div><div dir="auto"><br></div><div dir="auto">Example 2</div><div dir="auto"><br></div><div dir="auto">a A>B(sincere A>C)</div><div dir="auto">b B>C</div><div dir="auto">c C</div><div dir="auto"><br></div><div dir="auto">s(A>B)=approvalA+disapprovalB</div><div dir="auto">=a+c</div><div dir="auto">s(B>C)=a+b+a</div><div dir="auto">s(C>A)=a+c+b+c</div><div dir="auto"><br></div><div dir="auto">So C is the head of the winning chain C>A.</div><div dir="auto"><br></div><div dir="auto">It's hard to think of another method that is half so simple and powerful.</div><div dir="auto"><br></div><div dir="auto">-fws</div><div dir="auto"><br></div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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