<div dir="ltr">Hi Forest,<br><br>thanks for correcting my error. I agree that B is not able to successfully win this chicken dilemma scenario by defecting and using burial.<br><br>Overall, this has the flavor of a Smith//Approval variant, replacing Approval with "Highest implicitly approved [Smith] candidate [HIASC] defeating lowest implicitly approved Smith candidate [LIASC]".<br><br>To flesh it out, I'd like more details on the following:<br><ul><li>What is the motivation/rationale for excluding Smith candidates that tie or are defeated by the LIASC? What makes the HIASC "better"? In the most recent example, it is not clear to me that C is better than A, and letting C win is merely supposed to provide motivation for B voters to not defect and bury.<br></li><li>What happens in the event of ties in lowest implicit approval among Smith Candidates?</li><li>Are there other tie situations to handle?<br><br></li></ul>Other examples:<br>-----------------<br>(Due to Chris Benham)<br><pre style="white-space:pre-wrap;color:rgb(0,0,0)">25 A>B
26 B>C
23 C>A
26 C
Basic scores:
A: 49, B: 51, C:75
A lowest basic score (implicit approval / above bottom) and defeats B; </pre><pre style="white-space:pre-wrap;color:rgb(0,0,0)">C > A, therefore C wins Q&CBRS</pre><pre style="white-space:pre-wrap;color:rgb(0,0,0)">C is the most top ranked and the most above-bottom ranked candidate. </pre><pre style="white-space:pre-wrap;color:rgb(0,0,0)">WV, MMPO, IRV, Benham elect B
-----------------
</pre><pre style="white-space:pre-wrap;color:rgb(0,0,0)">35 A
10 A=B
30 B>C
25 C
Basic scores: A:45, B40, C55. B is lowest basic score.
C55 > A45, A35>B30, B40>C25
C is defeated by B. A is remaining candidate defeating B, thus wins Q&CBRS.
A both pairwise-beats and positionally dominates B, but WV, Margins, MMPO all elect B.<font face="arial, sans-serif">
This is somewhat encouraging, however, I'd still like to get better motivation for the earlier example as in the bullet point questions above.</font></pre></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Jan 5, 2022 at 9:15 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com">forest.simmons21@gmail.com</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 dir="auto">Ted,<div dir="auto"><br></div><div dir="auto">We may need a different example ... let's go through this one slowly and carefully ...<div dir="auto"><span style="font-family:sans-serif">46: A > B</span><div style="font-family:sans-serif" dir="auto">44: B > C (sincere B or B > A)</div><div style="font-family:sans-serif" dir="auto">05: C > A</div><div style="font-family:sans-serif" dir="auto">05: C > B</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">A gets 46+5=51 basic points from the first and third factions, resp.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">B gets 46+44+5=95 points from the first, second, and fourth factions, resp.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">C gets 44+5+5=54 points from the last three factions.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">If this is right, A has the lowest score, and C wins as the only candidate pairwise undefeated by A.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">I don't doubt that this method could fail Chicken Defense, but it would have to be a marginal failure that the attacked faction could easily counter with a partial counter defection.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">For a range of examples let's look at</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">p: C</div><div style="font-family:sans-serif" dir="auto">q: A>B</div><div style="font-family:sans-serif" dir="auto">r: B (sincere B>A)</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">Where r+q>p>q>r</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">A beats B because q>r</div><div style="font-family:sans-serif" dir="auto">B beats C because r+q>p</div><div style="font-family:sans-serif" dir="auto">C beats A because p>q</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">The respective basic scores for A, B, & C are q, p+q, & p.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">The lowest of these is q which is A's score. So C wins, as the only candidate pairwise undefeated by A.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">So the method robustly punishes Chicken defection over a range of standard examples.</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto">Not bad for Quick and Dirty/Clean!</div><div style="font-family:sans-serif" dir="auto"><br></div><div style="font-family:sans-serif" dir="auto"><br></div></div><br><br><div class="gmail_quote" dir="auto"><div dir="ltr" class="gmail_attr">El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern <<a href="mailto:dodecatheon@gmail.com" target="_blank">dodecatheon@gmail.com</a>> escribió:<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 dir="ltr">Hi Forest,<br><br>Unfortunately, your new method does not handle Chicken-Dilemma types of burial, which I am interested in. See for example Chris Benham's example:<br><br>46: A > B<div>44: B > C (sincere B or B > A)</div><div>05: C > A</div><div>05: C > B<br><br></div><div>A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B are both lowest basic score. If C is taken as lower, then C defeats A and B wins, rewarding burial.<br><br>With approval cutoff at second rank, even Smith//Approval does better.</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer" target="_blank">forest.simmons21@gmail.com</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 dir="auto"><div>Ted, <div dir="auto"><br></div><div dir="auto">Thanks for providing a great example that illustrates how the burial defense works.</div><div dir="auto"><br></div><div dir="auto">I agree that it's probably better, especially when there are many candidates, to eliminate the non-Smith candidates before counting the basic scores.</div><div dir="auto"><br></div><div dir="auto">In the case of public elections for political office we expect the Smith set to be small ... usually a singleton, and occasionally a triplet ... unless factions think they can get away with burial!</div><div dir="auto"><br></div><div dir="auto">In general, I like ratings information (as in your version of Approval Sorted Margins) better than rankings ... the Q&C and Q&D methods were created to see what a minimal acceptable rankings based burial resistant method would look like.</div><div dir="auto"><br></div><div dir="auto">With ratings and lots of candidates I would go back to Range Based (or Total Approval) Chain Climbing for a burial resistant method:</div><div dir="auto"><br></div><div dir="auto">While there is no pairwise undefeated candidate among the remaining candidates ... eliminate from the remaining candidates all of those that do not pairwise defeat the remaining candidate X that has the lowest Range score (or alternately .. lowest below midrange approval score ... with or without renormalization as candidates are eliminated).</div><div dir="auto"><br></div><div dir="auto">In the case of a three candidate Smith Set this method first eats away all of the non-Smith candidates ... then the lowest score Smith candidate X (the one that got buried) and finally Y, the one responsible for burying X, leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as the sole survivor... someone not preferred over X by Y.</div><div dir="auto"><br></div><div dir="auto">It would be interesting to see how this method works on Colin Champion's five candidate example... especially to see if the renormalizations are worth the trouble ... and if perhaps the below midrange approval scores (or the ASM approval scores) work better than range scores.</div><div dir="auto"><br></div><div dir="auto">That's a lot of work! Do you have any students that need a project?</div><div dir="auto"><br></div><div dir="auto">My best,</div><div dir="auto"><br></div><div dir="auto">Forest</div><br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <<a href="mailto:dodecatheon@gmail.com" rel="noreferrer" target="_blank">dodecatheon@gmail.com</a>> escribió:<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 dir="ltr">Your Q&CBRS works well in a situation in which Approval Sorted Margins does not: (due to Colin Champion):<br><br>Sincere:<div><br><font face="monospace">1: B > D > A > E > C<br>1: B > D > E > C > A<br>5: C > A > D > B > E<br>1: D > A > C > B > E<br>1: D > C > B > A > E<br>2: E > B > D > C > A<br></font><div style="font-family:arial,sans-serif"><br style="color:rgb(0,0,0)"></div><div style="font-family:arial,sans-serif">D is the beats-all Condorcet voter.<br><br></div><div><font face="arial, sans-serif">5 C-first voters bury D:</font><br><br><font face="monospace" style="font-family:arial,sans-serif">1: B > D > A > E > C<br>1: B > D > E > C > A<br>5: C > A > B > E > D # Was C > A > D > B > E<br>1: D > A > C > B > E<br>1: D > C > B > A > E<br>2: E > B > D > C > A<br><br>Pairwise array:<br></font><font face="monospace">[-- 6. 2. 5. 8.]</font><br><font face="monospace">[5. -- 4. 9. 9.]</font><br><font face="monospace">[9. 7. -- 5. 7.]</font><br><font face="monospace">[6. 2. 6. -- 4.]</font><br><font face="monospace">[3. 2. 4. 7. --]</font><br><br><font face="arial, sans-serif">Now, if I understand your method correctly, we first find the Smith Set. In the burial case, it is all 5 candidates, A, B, C, D, E.<br></font><br><div style="font-family:arial,sans-serif">Then we find the basic scores for each Smith candidate:<br><br>A: 8</div><div style="font-family:arial,sans-serif">B: 11</div><div style="font-family:arial,sans-serif">C: 10</div><div style="font-family:arial,sans-serif">D: 6</div><div style="font-family:arial,sans-serif">E: 9<br><br>The Smith candidate with the smallest basic score is D (our previous CW).<br><br>Smith candidates that defeat D are <b>B</b> and <b>E</b>. B has highest basic score, therefore is the winner. C's strategy did not help C.<br><br>With Approval Sorted Margins, C is able to win using Burial.<br><br>The Basic Score needs to be tabulated separately from pairwise, and depends on the other rankings on the ballot: each candidate gets a point for any rating above minimum candidate rating, if using a ratings ballot. In the case of many candidates, this could lead to basic score ties, since each ballot would certainly have many candidates rated 0, so in those cases I would recommend recounting after eliminating all candidates outside the Smith Set.<br><br>I've been thinking about the impracticality of computing pairwise arrays in "jungle" elections, those with, say, >9 candidates. If a ratings ballot were used, with rankings inferred, I would recommend a floating score threshold, starting at 1% of maximum approval, but rising until at most 9 distinct candidate scores are above the threshold (allows score-tie clusters), then recounting to get the reduced pairwise array and basic scores. If the lowest basic score is tied, eliminate non-Smith candidates and recount basic scores.<br><br><u>For elections with 9 or fewer candidates and no lowest-basic score ties in the Smith set, this is summable, but requires recounts in the event of more candidates or lowest basic score ties.</u><br><br>Looking back at Colin's burial example, what happens if basic score is recalculated using ballots scoring candidate X above D?<br><br>A: 5</div><div style="font-family:arial,sans-serif">B: 9</div><div style="font-family:arial,sans-serif">C: 5<br>E: 7<br><br>B still wins. So your overall basic score as a proxy for ballots scoring above the lowest Smith basic score candidate is a good proxy in this case. </div></div></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</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 dir="auto"><div dir="auto">Let's call it Q&CBRS.<div dir="auto"><br></div><div dir="auto">Pre-requisite background:</div><div dir="auto"><br></div><div dir="auto">Candidate X pairwise defeats candidate Y iff candidate X is ranked above/before/ahead of candidate Y on more ballots than not.</div><div dir="auto"><br></div><div dir="auto">A defeat chain is a sequence of candidates in which each candidate pairwise defeats the subsequent member of the chain.</div><div dir="auto"><br></div><div dir="auto">Using a "bubble sort" procedure to sort a list of candidates into pairwise order produces a defeat chain of the listed candidates.</div><div dir="auto"><br></div><div dir="auto">In this way we can easily find a defeat chain that includes all of the candidates. The first candidate in such a chain is an example of a Smith candidate. More generally, any candidate who has a defeat chain to any other candidate is a member of the Smith Set.</div><div dir="auto"><br></div><div dir="auto">Q&CBRS:</div><div dir="auto"><br></div><div dir="auto">First, find the "basic score" for each candidate defined as the number of ballots on which it is ranked above one or more candidates.<br></div><div dir="auto"><br></div><div dir="auto">Then let X be the Smith candidate with the smallest basic score.</div><div dir="auto"><br></div><div dir="auto">Finally, among the candidates not defeated by X, elect the one with the greatest basic score.</div><div dir="auto"><br></div><div dir="auto">That's it! Quick & Clean!</div><div dir="auto"><br></div><div dir="auto">Note that if there is only one Smith candidate X, then X will be the only candidate not defeated by X, and therefore the one elected .</div><div dir="auto"><br></div><div dir="auto">In general a candidate not defeated by X will defeat X, and thereby find itself at the head of a defeat chain to every other candidate, i.e. it will be a Smith candidate. </div><div dir="auto"><br></div><div dir="auto">When there is only one Smith candidate, that candidate will not be pairwise defeated by any other candidate.</div><div dir="auto"><br></div><div dir="auto">I repeat the entire method procedure here:</div><div dir="auto"><br></div><div dir="auto"><div dir="auto" style="font-family:sans-serif">First, find the "basic score" for each candidate defined as the number of ballots on which it is ranked above one or more candidates.<br></div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">Then let X be the Smith candidate with the smallest basic score.</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">Finally, among the candidates not defeated by X, elect the one with the greatest basic score.</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">Where else can you find such a simple, quick, and clean election method?</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif"><br></div></div><div dir="auto"><br></div><div dir="auto"><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</a>> escribió:<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 dir="auto">Good idea!<div dir="auto"><br></div><div dir="auto">Although it seems to me that the highest approval candidate would have to pairwise beat or tie the approval cutoff candidate X pairwise (which would be impossible for a non-Smith candidate to do) ...I could be wrong ... and in any case redundancy reinforces communication and understanding.</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <<a href="mailto:dodecatheon@gmail.com" rel="noreferrer noreferrer noreferrer noreferrer" target="_blank">dodecatheon@gmail.com</a>> escribió:<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 dir="ltr">Hi Forest,<br><br>This sounds like an interesting method to me!<br><br>However, I would change the winning criteria to "Elect the most approved member of the Smith Set".</div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer noreferrer noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</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 dir="auto"><div dir="auto">I apologize for the defiant tone at the end of the previous message ... I must have gotten carried away with the "Dirty Dozen"' theme.</div><div dir="auto"><br></div><div dir="auto">But isn't it frustrating to you when people use the 2nd law of thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to justify their stubborn resistance to any kind of engineering progress?</div><div dir="auto"><br></div><div dir="auto">In the previous message Q&D Burial Resistant Condorcet was formulate in the typical "stitched together" form ... "Elect the CW if there is one, Else ..."</div><div dir="auto"><br></div><div dir="auto">In this message I would like to formulate a seamless version:</div><div dir="auto"><br></div><div dir="auto">Let X be the Smith candidate who on the fewest ballots is ranked ahead of any other Smith candidate. On each ballot approve all candidates down to X, but include X only when no Smith candidate is ranked ahead of X.</div><div dir="auto"><br></div><div dir="auto">Elect the candidate approved on the most ballots.</div><div dir="auto"><br></div><div dir="auto">This method can be described as electing the approval winner when the approval cutoff is (at the rank of) the weakest of the Smith candidates, which itself is approved on (and only on) those ballots which do not approve any other Smith candidate.</div><div dir="auto"><br></div><div dir="auto">In other words, the approval cutoff is inclusive only when necessary to ensure approval of at least one member of Smith.</div><div dir="auto"><br></div><div dir="auto">Since a Smith member is approved on every ballot, the method satisfies the Condorcet Criterion, i.e. it elects the only Smith member when Smith is a singleton.</div><div dir="auto"><br></div><div dir="auto">How does that grab you?</div><div dir="auto"><br></div><div dir="auto">-FWS</div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer noreferrer noreferrer noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</a>> escribió:<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 dir="auto"><div dir="auto">Is there any burial resistant Condorcet method simpler than this?</div><div dir="auto"><br></div><div dir="auto">The basic pre-requisite is to understand that whenever there is no Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of candidates whose members are not defeated by any candidates outside of the cycle, just as a Condorcet Winner is a candidate undefeated by any other candidate.</div><div dir="auto"><br></div><div dir="auto">Here's the Q&D burial resistant method:</div><div dir="auto"><br></div><div dir="auto">Lacking a Condorcet Winner elect the candidate X having the greatest pairwise victory over the top-cycle member Y that has the smallest ratio of first to last place votes within the top cycle.</div><div dir="auto"><br></div><div dir="auto">Two examples illustrate the method:</div><div dir="auto"><br></div><div dir="auto">Example 1.</div><div dir="auto"><br></div><div dir="auto">49 C</div><div dir="auto">26 A>B</div><div dir="auto">25 B (sincere B>A)</div><div dir="auto"><br></div><div dir="auto">The top cycle is ABCA</div><div dir="auto"><br></div><div dir="auto">Candidate A has the smallest ratio 26/74 of first to last place votes.</div><div dir="auto"><br></div><div dir="auto">Candidate C is the only candidate with a pairwise victory over it, so C wins.</div><div dir="auto"><br></div><div dir="auto">Notice how our rule does not reward B for insincerely lowering A to (equal) last?</div><div dir="auto"><br></div><div dir="auto">Example 2.</div><div dir="auto"><br></div><div dir="auto">45 A>B (sincere A>C)</div><div dir="auto">35 B>C</div><div dir="auto">25 C>A</div><div dir="auto"><br></div><div dir="auto">Candidate C has the smallest ratio 25/45 of first to last.</div><div dir="auto"><br></div><div dir="auto">Candidate B wins as the only candidate with a pairwise victory over C. So A's burial of C backfires.</div><div dir="auto"><br></div><div dir="auto">Typically, the faction A that buries or truncates a Condorcet Winner C to create a top-cycle cannot by so doing become a pairwise victor over the buried Condorcet Winner ... but must (in order to create a cycle) help some other candidate B defeat C by insincerely voting B>C.</div><div dir="auto"><br></div><div dir="auto">Our Quick and Dirty method insures that if the sincere CW's rightful victory is subverted, it goes to B, not to A.</div><div dir="auto"><br></div><div dir="auto">Is this method quick enough and dirty enough for the FairVote IRV promoters?</div><div dir="auto"><br></div><div dir="auto">What objections/criticisms might they have?</div><div dir="auto"><br></div><div dir="auto">Do they have any counter proposal that rivals this one in any way?</div><div dir="auto"><br></div><div dir="auto">If so, let them educate us ... we'll gladly join them if they can show us a better way!</div><div dir="auto"><br></div><div dir="auto">If not, then they should join us to educate the politicians, public, and last but not least, the academics still stuck in the pre-EM era!</div></div>
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