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<div dir="ltr" data-setdir="false">Hi Kristofer,</div><div><br></div>
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Le jeudi 23 janvier 2020 à 17:19:01 UTC−6, Kristofer Munsterhjelm <km_elmet@t-online.de> a écrit :
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<div style="font-size: 13px;">>My idea is that if A and B are in the same set, then either A beats<br></div><div style=""><div dir="ltr" style="font-size: 13px;">>someone who beats B, or B beats someone who beats A. If it's the former,<br></div><div dir="ltr" style="font-size: 13px;">>then admitting A>B would mean make B twice defeated, which is not<br></div><div dir="ltr" style="font-size: 13px;">>allowed. If it's the latter, then admitting A>B would create a cycle,<br></div><div dir="ltr" style="font-size: 13px;">>which is not allowed either. So, unlike Ranked Pairs, we don't need to<br></div><div dir="ltr" style="font-size: 13px;">>know which is the case, because either situation implies A>B should not<br></div><div dir="ltr" style="font-size: 13px;">>be admitted. So a disjoint set structure suffices.<br></div><div dir="ltr" style="font-size: 13px;"><br></div><div dir="ltr" data-setdir="false" style=""><font size="3">This seems right. Every candidate in a set, except one, has a path of losses back to</font></div><div dir="ltr" data-setdir="false" style=""><font size="3">the one exception. So no defeat between any of these candidates can be admitted.</font></div><div dir="ltr" data-setdir="false" style=""><font size="3"><br></font></div><div dir="ltr" data-setdir="false" style=""><font size="3">I can't speak to the O notation...</font></div><div dir="ltr" style="font-size: 13px;"><br></div><div dir="ltr" style="font-size: 13px;">>(Here I thought I had found a cloneproof O(n^2) Condorcet method. But I<br></div><div dir="ltr" style="font-size: 13px;">>found out it's *just* in excess of that, because of that log factor...)<br></div><div dir="ltr" style="font-size: 13px;">></div><div dir="ltr" style="font-size: 13px;">>... actually, isn't this just Kruskal's algorithm?<br></div><div dir="ltr" style="font-size: 13px;"><br></div><div dir="ltr" data-setdir="false" style=""><font size="3">It seems similar (if we say there are two edges between each pair of vertices, to</font></div><div dir="ltr" data-setdir="false" style=""><font size="3">represent the magnitude of the vote on each side) but </font><span style="font-size: medium;">given that the River graph will</span></div><div dir="ltr" data-setdir="false" style=""><span style="font-size: medium;">be directed (with a clear root) I wonder if it must be different,</span><span style="font-size: medium;"> or if that is just </span><span style="font-size: medium;">a cosmetic</span></div><div dir="ltr" data-setdir="false" style=""><span style="font-size: medium;">issue.</span></div><div dir="ltr" data-setdir="false" style=""><font size="3"><br></font></div><div dir="ltr" data-setdir="false" style=""><font size="3">Kevin</font></div><div dir="ltr" data-setdir="false" style="font-size: 13px;"><br></div><div dir="ltr" data-setdir="false" style="font-size: 13px;"><br></div></div>
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