<html><head><style type="text/css"><!-- DIV {margin:0px;} --></style></head><body><div style="font-family:times new roman, new york, times, serif;font-size:12pt"><DIV>I'm not sure I've followed everything about how to determine the pecking order and how to calculate the distance between candidates, and why it's good to base the challenge order on this, but I'll go along with it!</DIV>
<DIV> </DIV>
<DIV>Could we work on a similar distance basis for STV? There are S candidates to be elected, so we could start with the bottom S in the pecking order. For the challenger (single candidate), can distance be measured between a set of candidates and a single candidate? I'm sure it can, so we could calculate the most distant candidate from the current "champion set".</DIV>
<DIV> </DIV>
<DIV>Then we need to see which of the S+1 candidates is to be eliminated. We now need to determine the order of comparison here as well. Essentially there's two orders of comparison we need to worry about. The first, as already described, is which candidate is to be pitted against the current "champion set". Then, once this is determined, we need to determine the order of comparison of the S+1 sets that contain all members of the "champion set" and the challenger candidate.</DIV>
<DIV> </DIV>
<DIV>Each set here will have just one of the S+1 candidates missing so in terms of starting at the bottom of the pecking order, we probably want to start off with the set that excludes the candidate at the top of the pecking order (the top out of the S+1 that are being considered here). Then on the basis that we can probably calculate distance between a set and a candidate, we can also probably calculate the distance between two sets. So we take our "champion set" containing S of the S+1 candidates, and compare it against the most distant other remaining set that still remains in this group. When only one of these sets remains, this becomes the new "champion set", and we find the most distant challenger and continue as before. Also, the winner in the set comparisons can be done as in Schulze STV (or another one if you prefer) as Schulze STV compares sets that differ by only one candidate anyway, so it seems fit for purpose.</DIV>
<DIV> </DIV>
<DIV>So, Forest, is this of any potential use?<BR></DIV>
<DIV style="FONT-FAMILY: times new roman, new york, times, serif; FONT-SIZE: 12pt"><BR>
<DIV style="FONT-FAMILY: times new roman, new york, times, serif; FONT-SIZE: 12pt"><FONT size=2 face=Tahoma>
<HR SIZE=1>
<B><SPAN style="FONT-WEIGHT: bold">From:</SPAN></B> Toby Pereira <tdp201b@yahoo.co.uk><BR><B><SPAN style="FONT-WEIGHT: bold">To:</SPAN></B> fsimmons@pcc.edu<BR><B><SPAN style="FONT-WEIGHT: bold">Cc:</SPAN></B> election-methods@lists.electorama.com<BR><B><SPAN style="FONT-WEIGHT: bold">Sent:</SPAN></B> Thu, 21 July, 2011 12:19:36<BR><B><SPAN style="FONT-WEIGHT: bold">Subject:</SPAN></B> Re: [EM] HBH<BR></FONT><BR>
<DIV style="FONT-FAMILY: times new roman, new york, times, serif; FONT-SIZE: 12pt">
<DIV>Excellent - if you think it's a good idea it must be!</DIV>
<DIV> </DIV>
<DIV>I don't think it would be as simple as checking one possible result against one other that differs by one candidate. How would we decide which of the "current champion set" to remove to put the new candidate in for our comparison? We'd have to check the merits of each possible result. There wouldn't be that many though. So if S candidates are to be elected, the "current champion set" would obviously have S members. Then we introduce a challenger candidate giving us S+1 candidates. There will then only be S+1 possible sets of S members involving these candidates (one for each candidate's absence). Instead of comparing each of these sets against each other, we can stick to the principle of HBH and its winner-stays-on nature. So when we determine the winning set of S out of the S+1 candidates, we eliminate the candidate not in the winning set. Then we pick another challenger and so on, one at a time, until S candidates remain. These S are
elected.</DIV>
<DIV> </DIV>
<DIV>As for deciding the order of comparison, I'm not sure I'm as well qualified as you!<BR></DIV>
<DIV style="FONT-FAMILY: times new roman, new york, times, serif; FONT-SIZE: 12pt">
<DIV style="FONT-FAMILY: arial, helvetica, sans-serif; FONT-SIZE: 13px"><FONT size=2 face=Tahoma>
<HR SIZE=1>
<B><SPAN style="FONT-WEIGHT: bold">From:</SPAN></B> "fsimmons@pcc.edu" <fsimmons@pcc.edu><BR><B><SPAN style="FONT-WEIGHT: bold">To:</SPAN></B> Toby Pereira <tdp201b@yahoo.co.uk><BR><B><SPAN style="FONT-WEIGHT: bold">Cc:</SPAN></B> election-methods@lists.electorama.com<BR><B><SPAN style="FONT-WEIGHT: bold">Sent:</SPAN></B> Thu, 21 July, 2011 2:22:03<BR><B><SPAN style="FONT-WEIGHT: bold">Subject:</SPAN></B> Re: [EM] HBH<BR></FONT><BR>Good idea. Let's play with it.<BR><BR>----- Original Message -----<BR>From: Toby Pereira <BR>Date: Wednesday, July 20, 2011 4:44 pm<BR>Subject: Re: [EM] HBH<BR>To: <A href="mailto:fsimmons@pcc.edu" rel=nofollow target=_blank ymailto="mailto:fsimmons@pcc.edu">fsimmons@pcc.edu</A><BR>Cc: <A href="mailto:election-methods@lists.electorama.com" rel=nofollow target=_blank ymailto="mailto:election-methods@lists.electorama.com">election-methods@lists.electorama.com</A><BR><BR>> I was thinking - Schulze STV
compares every result against every <BR>> other result <BR>> that differs by just one candidate, which could be a lot of work <BR>> for a computer! <BR>> So could your HBH system be used for STV elections? Determine <BR>> the order of <BR>> comparison and compare two results that differ by one candidate <BR>> and the "losing <BR>> candidate" is eliminated. So each pairwise comparison eliminates <BR>> a candidate and <BR>> it's all done much more quickly.<BR>> <BR>> <BR>> <BR>> <BR>> ________________________________<BR>> From: "<A href="mailto:fsimmons@pcc.edu" rel=nofollow target=_blank ymailto="mailto:fsimmons@pcc.edu">fsimmons@pcc.edu</A>" <BR>> To: <A href="mailto:fsimmons@pcc.edu" rel=nofollow target=_blank ymailto="mailto:fsimmons@pcc.edu">fsimmons@pcc.edu</A><BR>> Cc: <A href="mailto:election-methods@lists.electorama.com" rel=nofollow target=_blank
ymailto="mailto:election-methods@lists.electorama.com">election-methods@lists.electorama.com</A><BR>> Sent: Mon, 18 July, 2011 19:25:01<BR>> Subject: [EM] HBH<BR>> <BR>> HBH stands for Hog Belly Honey, the name of an inerrant <BR>> "nullifier" invented by <BR>> a couple of R.A. Lafferty <BR>> <BR>> characters. The HBH is the only known nullifier that can "posit <BR>> moral and <BR>> ethical judgments, set up and <BR>> <BR>> enforce categories, discern and make full philosophical <BR>> pronouncements," in <BR>> other words eliminate the <BR>> <BR>> garbage and keep what's valuable. The main character, the "flat <BR>> footed genius," <BR>> Joe Spade, picks the <BR>> <BR>> name "Hog Belly Honey," for it "on account it's so sweet."<BR>> <BR>> The whole idea of HBH is just starting at the bottom of a <BR>> pecking order and <BR>> pitting (for elimination) the <BR>> <BR>>
current champ against the most distant challenger. I hope you <BR>> will keep that in <BR>> mind as we introduce <BR>> <BR>> the necessary technical details.<BR>> <BR>> HBH is based on range style ballots that allow the voters to <BR>> rate each <BR>> alternative on a range of zero to <BR>> <BR>> some maximum value M. [Keep this M in mind; we will make <BR>> explicit use of it <BR>> presently.]<BR>> <BR>> Once the ballots are voted and submitted, the first order of <BR>> business is to set <BR>> up a "pecking order" for <BR>> <BR>> the purpose of resolving ties, etc. Alternative X is higher in <BR>> the pecking <BR>> order than alternative Y if <BR>> <BR>> alternative X is rated above zero on more ballots than Y is <BR>> rated above zero. <BR>> If both have the same <BR>> <BR>> number of positive ratings, then the alternative with the most <BR>>
ratings greater <BR>> than one is higher in the <BR>> <BR>> pecking order. If that doesn't resolve the tie, then the <BR>> alternative with the <BR>> greatest number of ratings <BR>> <BR>> above two is higher, etc.<BR>> <BR>> In the practically impossible case that two alternatives have <BR>> exactly the same <BR>> number of ratings at each <BR>> <BR>> level, ties should be broken randomly.<BR>> <BR>> The next order of business is to establish a proximity relation <BR>> between <BR>> alternatives. For our purposes <BR>> <BR>> closeness or proximity between two alternatives X and Y is given <BR>> by the number<BR>> <BR>> Sum over all ballots b, min( M*(M-1), b(X)*b(Y) ).<BR>> <BR>> [The minimization with M*(M-1) clinches the method's resistance <BR>> to compromise, <BR>> as explained below.]<BR>> <BR>> This proximity value is a useful measure of a certain kind
of <BR>> closeness of the <BR>> two alternatives: the larger <BR>> <BR>> the proximity number the closer the alternatives in this limited <BR>> sense, while <BR>> the smaller the number the <BR>> <BR>> more distant the alternatives from each other (again, in this <BR>> limited sense).<BR>> <BR>> For the purposes of this method, if two alternatives Y and Z <BR>> have equal <BR>> proximity to X, then the one that <BR>> <BR>> is higher in the pecking order is considered to be closer than <BR>> the other. In <BR>> other words, the pecking <BR>> <BR>> order is used to break proximity ties.<BR>> <BR>> Next we compute the majority pairwise victories among the <BR>> alternatives. <BR>> Alternative X beats alternative <BR>> <BR>> Y majority-pairwise if X is rated above Y on more than half of <BR>> the ballots.<BR>> <BR>> For the purposes of this method, the "victor"
of a pair of <BR>> alternatives is the <BR>> one that beats the other <BR>> <BR>> majority pairwise, or in the case where neither beats the other <BR>> majority-pairwise it is the one that is <BR>> <BR>> higher in the pecking order. Of the two, the non-victor <BR>> alternative is called <BR>> the "loser." In other words, <BR>> <BR>> the pecking order decides pairwise victors and losers when there <BR>> is no majority <BR>> defeat. [This convention <BR>> <BR>> on victor and loser is what makes the method plurality <BR>> compliant, as explained <BR>> below.]<BR>> <BR>> Next we initialize an alphanumeric variable V with the name of <BR>> the lowest <BR>> alternative in the pecking <BR>> <BR>> order, and execute the following loop:<BR>> <BR>> While there remain two or more discarded alternatives<BR>> discard the loser between V and the alternative most distant
<BR>> from V,<BR>> and replace V with the name of the victor of the two.<BR>> EndWhile<BR>> <BR>> Finally, elect the alternative represented by the final value of V.<BR>> <BR>> This HBH method is clone free, monotone, Plurality compliant, <BR>> compromise <BR>> resistant, and burial <BR>> <BR>> resistant.<BR>> <BR>> Furthermore, it is obviously the case that if some alternative <BR>> beats each of the <BR>> other alternatives majority <BR>> <BR>> pairwise, then that alternative will be elected.<BR>> <BR>> Let's see why the method is plurality compliant:<BR>> <BR>> If there is even one majority defeat in the sequence of <BR>> eliminations, every <BR>> value of V after that will be the <BR>> <BR>> name of an alternative that is rated positively on more than <BR>> half of the <BR>> ballots. If none of the victories are <BR>> <BR>> by majority defeat, then the
winner is the alternative highest <BR>> on the pecking <BR>> order, i.e. the one with the <BR>> <BR>> greatest number of positive ratings.<BR>> <BR>> Let's see why the method is monotone:<BR>> <BR>> Suppose that the winner is moved up in the ratings. Then its <BR>> defeat strengths <BR>> will only be increased, and <BR>> <BR>> any proximity change can only delay its introduction into the <BR>> fray, so it will <BR>> only face alternatives that <BR>> <BR>> lost to it before.<BR>> <BR>> Let's see why it is compromise resistant:<BR>> <BR>> Since Favorite and Compromise are apt to be in relatively close <BR>> proximity, and <BR>> pairwise contests are <BR>> <BR>> always between distant alternatives, if Compromise gets <BR>> eliminated, it will <BR>> almost certainly be by <BR>> <BR>> someone besides Favorite, so there can hardly be any incentive <BR>> for rating <BR>>
Favorite below Compromise.<BR>> <BR>> Furthermore, there is no likely advantage of rating Compromise <BR>> equal to <BR>> Favorite, because rating <BR>> <BR>> compromise just below Favorite already makes the maximum <BR>> possible contribution <BR>> M*(M-1) to their <BR>> <BR>> proximity sum, i.e. the best you can do to make sure they are <BR>> pitted against <BR>> each other only after all of <BR>> <BR>> the other alternaties have been eliminated (if at all).<BR>> <BR>> How about burial?<BR>> <BR>> I don't have such an easy argument for burial resistance, but <BR>> the experiments I <BR>> have conducted show <BR>> <BR>> that more likely than not it won't pay off. I hope that Kevin <BR>> will run his <BR>> simulations on the method for <BR>> <BR>> (hopefully) more support on that account.<BR>> <BR>> I realize that the method sounds complicated from the <BR>>
description above, but all <BR>> of the complication is <BR>> <BR>> from the details of tie breaking, including what to do when <BR>> defeats are not <BR>> majority-pairwise.<BR>> <BR>> Other than that, as mentioned at the beginning, it is just <BR>> starting at the <BR>> bottom of the pecking order and <BR>> <BR>> pitting (for elimination) the current champ against the most <BR>> distant challenger.<BR>> <BR>> Aint that sweet?<BR>> <BR>> ----<BR>> Election-Methods mailing list - see <A href="http://electorama.com/em" rel=nofollow target=_blank>http://electorama.com/em</A> for <BR>> list info<BR>> <BR></DIV></DIV></DIV></DIV></DIV></div></body></html>