Talk:Ranked pairs

message: This article seems to define Ranked Pairs as using winning votes as the measure of defeat strength. Isn't that inaccurate, as far as what Tideman defined the method to be? KVenzke 17:42, May 22, 2005 (UTC)
* Probably. The winning-votes variant should be mentioned in a section. RSpeer 05:25, May 23, 2005 (UTC)
::It would be a tough choice since WV satisfies more criteria. The way the article is written now, anything that Schulze method satisfies is also satisfied by Ranked Pairs. But this will have to be qualified if this article defaults to margins. KVenzke 15:02, May 23, 2005 (UTC)
Tideman prefers margins at present, and has probably assumed that margins would be the defeat strength definition when publishing articles about ranked pairs, although he is open to the idea of WV. I think that it makes sense to present both ranked pairs/Tideman and beatpath/Schulze as base methods that can work with different definitions of defeat strength.Hermitage 23:23, 8 Jun 2005 (UTC)
replies: text: I disagree with Hermitage. I think it's much more important for the general public to treat MAM, Tideman's Ranked Pairs, Markus Schulze's beatpath, etc. as specific voting methods being advocated, which satisfy or fail specific criteria, than as abstract families of voting methods each having variants. MAM satisfies more criteria than either Tideman's Ranked Pairs or Schulze's beatpath, so it doesn't make sense to exclude MAM from Wikipedia or other discussions, or treat it as some unimportant variant. SEppley (talk) 16:29, 8 March 2010 (UTC)
thread_title:  Winning votes? 

message: This article lacks an evaluation based on criteria.  Since RP is a condorcet method, listing whether or not it satisfies the Smith criteria would be relevant.  I wonder if there are any volunteers for this?--Fahrenheit451 00:29, 15 Jun 2005 (UTC)
replies: text: RP and MAM etc satisfy the Smith criterion (a.k.a. the top cycle criterion). Assume x is in the Smith set (a.k.a. top cycle) and y is not. By definition, every candidate in the Smith set is ranked by a majority over every candidate outside the Smith set. Thus the "x over y" majority cannot be part of a majority cycle. (Proof: Suppose to the contrary that x over y is part of a cycle. This means there exists a sequence of three or more candidates a1, a2, ..., aN such that a1 is y, aN is x, a majority does not rank a2 over a1, a majority does not rank a3 over a2, etc., and a majority does not rank aN over aN-1. Since a1 is not in the Smith set, a2 can't be either, by the definition of the Smith set. Similarly, since a2 is not in the Smith set, a3 can't be either. By induction, aN can't be in the top cycle. But that is a contradiction since aN is x. Thus the lemma has been established.) Since x over y is not part of a cycle, it follows that RP (and MAM etc) will affirm that x finishes ahead of y. SEppley (talk) 18:28, 8 March 2010 (UTC)

text: KVenzke, do you mean Majority criterion? - McCart42 (talk) 22:57, 16 July 2005 (UTC)
replies: text: No, Plurality says that if A has more first preferences than B has votes of any rank, then B must not win with greater probability than A. Margins versions of RP and Schulze don't satisfy this. KVenzke 14:17, July 17, 2005 (UTC)
Kevin Venzke is mistaken above where he claims the Schulze method (a.k.a. beatpath) and the "winning votes" variation of Ranked Pairs satisfy the same criteria. Schulze's method fails Peyton Young's "local independence of irrelevant alternatives," which is satisfied by Ranked Pairs (regardless of whether it uses winning votes or margins) and by MAM. The winning votes variation fails Mike Ossipoff's Strong Defensive Strategy and Weak Defensive Strategy criteria (and Steve Eppley's Minimal Defense criterion, which is very similar to Ossipoff's Strong Defensive Strategy criterion) since according to Tideman each vote in Ranked Pairs is a strict ordering of the candidates; whereas MAM and Schulze allow weak orderings. Schulze's method fails Steve Eppley's "immunity from majority complaints," which is satisfied by the "winning votes" variation and by MAM. (I've distinguished MAM here because MAM differs in several ways from the winning votes variation of Ranked Pairs: 1. MAM allows each vote to be a weak ordering, allowing voters to express indifference and save time by leaving candidates unranked or ranked equally worst. 2. MAM includes only the majorities when constructing the "largest to smallest" order of precedence, whereas Ranked Pairs includes all N*N-1 ordered pairs, including ties and minorities; note that ties can be larger than majorities in the variation of RP most similar to MAM, which uses winning votes and allows votes to be weak orderings; note also that, by including ties in the order of precedence, RP always constructs a strict order of finish by the time it has considered the smallest pair, whereas MAM postpones tiebreaking involving pairwise ties until a final stage (if necessary). 3. MAM's tiebreaker differs subtly from Tideman-Zavist's 1989 tiebreaker, allowing MAM to completely satisfy the strong Pareto criterion, which is not completely satisfied by the winning votes variation of RP.) SEppley (talk) 18:28, 8 March 2010 (UTC)
thread_title:  Voting criteria pass and fail 

message: Do the Ranked Pairs variants, Maximize Affirmed Majorities and Maximum majority voting, deserve their own article (especially given the amount of overlap due to the Tennessee voting example)? My vote is no. -- Dissident (Talk) 23:21, 25 December 2005 (UTC)
replies: text: I agree. Maximize Affirmed Majorities (MAM) and Maximum majority voting (MMV) should be merged into Ranked Pairs, because MAM and MMV are only concreted formulations of Ranked Pairs. Markus Schulze 09:10, 26 December 2005 (UTC)

text: The merger is a good idea – especially if a comparison table of the variants is provided, similar to the one on Voting system. clacke 08:20, 21 April 2006 (UTC)

text: I disagree. Ranked Pairs is a specific method first described in 1987 by Tideman, refined in 1989 by Tideman & Zavist.  MAM and MMV are not "concreted formulations" of Ranked Pairs; they are significantly different from Tideman's method, which is what allows them to satisfy desirable criteria that Ranked Pairs fails. Several people in the Election-Methods maillist and in some websites have used the name Ranked Pairs to mean the same method as MAM, or to mean a family of methods that includes both MAM and Tideman's Ranked Pairs, but those usages have sown confusion. Not only was the MAM wikipedia article deleted, all information about MAM was deleted from the Ranked Pairs wikipedia article! THAT WAS NOT A MERGER; IT MAY EVEN HAVE BEEN VANDALISM, since MAM satisfies more desirable criteria than either Ranked Pairs or Schulze's method. (Ranked Pairs fails Mike Ossipoff's Weak and Strong Defensive Strategy criteria and fails my Immunity from Majority Complaints criterion. Schulze's method fails Peyton Young's Local Independence of Irrelevant Alternatives criterion and my Immunity from Majority Complaints criterion, and computer simulations show that majorities rank MAM winners over Schulze winners more often than vice versa.) For these reasons, MAM is the superior method and thus merits its own wikipedia article. (That article should mention the several ways in which it differs from Ranked Pairs). The Ranked Pairs article should say that Ranked Pairs is just Tideman's method and is not MAM, and that the name Ranked Pairs has been used in various websites and maillists, somewhat misleadingly, by some people when they mean the MAM method. SEppley (talk) 14:32, 8 March 2010 (UTC)
replies: text: I don't agree that "MAM satisfies more desirable criteria than Schulze's method". The Schulze method satisfies Woodall's CDTT criterion and guarantees that the winner is always chosen from the union of all sets with minimum worst defeat. See section 4.9 of my paper. Computer simulations by Norman Petry, Jobst Heitzig, and Barry Wright show that RP/MAM needlessly generates winners with strong worst defeats, while the winner of the Schulze method is almost always identical to the winner of the MinMax method. Anyway, WP:NOTAFORUM. Markus Schulze 18:35, 8 March 2010 (UTC)

text: Another problem of RP/MAM is that it has an exponential runtime when we want to calculate all winners. This fact makes it difficult to combine RP/MAM with some additional tiebreaker. Markus Schulze 18:57, 8 March 2010 (UTC)
replies: text: Isn't it a problem of the Schulze method too? Ranked Pairs has O(n4) complexity, Schulze method has O(n3) complexity. Wat 20 01:21, 8 July 2012 (UTC)
thread_title:  [[Ranked Pairs]] variants 

message: The link in the Criteria section to the voting criteria table appears to be broken. It links an anchor on the Voting system page, but there does not appear to be any voting criteria table on that page. Qutezuce 08:56, 16 January 2006 (UTC)
replies: text: I cleaned it up into the start of a more useful, self-contained section.  Thanks! Scott Ritchie 22:07, 16 January 2006 (UTC)
thread_title: Broken link

message: I ask this as a layman. -- nyenyec ☎ 16:11, 10 February 2007 (UTC)
thread_title:  Why the captial P in the title? 

message: I started a discussion recently on the election-methods mailing list about how to determine "second place" in a Ranked Pairs election.  An interesting fact I found is that in ranked pairs the second place winner can be determined either by looking at whomever is second in the (complete) graph OR by removing the winner from all ballots and rerunning the algorithm -- both will produce the same result. Scott Ritchie (talk) 22:03, 14 October 2008 (UTC)
replies: text: This criterion is called "decreasing sequential independence". Markus Schulze 22:25, 14 October 2008 (UTC)

text: Both Ranked Pairs and MAM satisfy Peyton Young's criterion "local independence of irrelevant alternatives" (LIIA). This means that deleting the winner from the ballots will not change the relative order of finish of the remaining alternatives. (It also means that deleting the alternative that finishes in last place will not change the relative order of finish of the remaining alternatives.) It follows by induction that deleting the top N finishers will not affect the relative order of finish of the remaining alternatives (for any N). The article should be revised regarding how the order of finish is constructed, since it is unnecessary to iteratively delete the "winner." SEppley (talk) 14:43, 8 March 2010 (UTC)

text: The Maximize Affirmed Majorities method (MAM) is also equivalent to finding the "best" order of finish, but MAM defines "best" in a significantly different way: the size of the opposing minority is not subtracted when measuring the size of a majority. (Minorities are not considered at all by MAM except in the rare case where two or more majorities are the same size, in which case MAM uses a form of tiebreaking that considers the opposing minorities.) In the example above, for MAM the size of the "x over y" majority is 51, and the "y over x" minority is not included in the order of precedence. (Ranked Pairs ranks all pairs—majorities, minorities and tied pairs—but MAM ranks only the majorities, and could perhaps have been named Ranked Majorities. There are several other subtle differences between MAM's tiebreaking and Ranked Pairs' tiebreaking, but those are beyond the scope of this note.) MAM allows votes to be "weak" orderings (that is, the voter can rank two or more candidates the same to express indifference between them) so it matters how majorities' sizes are measured, and MAM measures them in the way that permits satisfaction of several criteria failed by Ranked Pairs. For example, suppose 35% rank x over y, 20% rank y over x and 45% are indifferent between x & y, and suppose 55% rank y over z and 45% rank z over y. For Ranked Pairs 35%-20% is greater than 55%-45% so it gives the x over y pair greater precedence than the y over z pair. For MAM 55% is greater than 35%, so MAM gives the y over z majority greater precedence than the x over y majority. (Ranked Pairs as defined by Tideman requires votes to be strict orderings, which is another part of the reason it fails some of those criteria.) SEppley (talk) 16:01, 8 March 2010 (UTC)
thread_title:  Second place 

message: In the method of Ranked pairs candidates are divided into pairs & voters have to choose the winner of each pair. What if the number of candidates is odd? That must be a serious drawback of this voting method. Nikolay95 (talk) 20:28, 13 September 2012 (UTC)
replies: text: Candidates are not divided into pairs (as in a tournament?); pairing means that each candidate is compared to each other candidate.  For 3 candidates there are 3 pairings; for 5 candidates, 10 pairings; for 7 candidates, 21 pairings. —Tamfang (talk) 22:11, 13 September 2012 (UTC)
thread_title:  Odd number of candidates 

message: In the Example section, the subsection Summary follows subsection Ambiguity resolution example, yet the Summary subsection summarizes all subsection preceding Ambiguity resolution example without mention of the Ambiguity one. The Ambiguity resolution example subsection is effectively a second example, with separate example items (A>B>C>A as opposed to the Tennessee cities in the rest of the main section). It would seem appropriate to move the Summary subsection before the Ambiguity resolution example subsection. Anyone have a reason in opposition to this? — al-Shimoni (talk) 16:17, 17 November 2012 (UTC)
thread_title:  Subsection ordering in Example Section 

message: Is there any algorithm which calculates a Ranked Pairs ranking in parallel? The algorithm in Eppley's MAM procedure definition paper is sequential. Wat 20 18:05 16 February 2013 (UTC)
: The algorithm is fundamentally sequential, but parts of it can be parallelized. Counting ballots and computing the pairwise results can be done in parallel, provided that all of the results are collated in the end. Sorting the majorities can be parallelized using various divide-and-conquer sorting methods (merge is likely the best). Locking in majorities pretty much needs to be done sequentially. Computing transitive closure (to detect cycles) can be parallelized. None but the first really matters unless you have a large number of candidates though.
72.68.72.137 (talk) 04:38, 8 October 2014 (UTC)
thread_title:  Parallel computation of ranking 

message: The following line is in the "Lock" section of the algorithm description: "One way to resolve this issue is to allow cycles if they are needed to resolve ties (i.e., if a single new edge would not create a cycle, but multiple tied edges would), and then define the winners as the resulting Schwartz set."
Where does this come from? There seems to be no citation given. Can anyone give further description?  — Preceding unsigned comment added by TheShepherd7 (talk • contribs) 04:06, 18 November 2017 (UTC)
replies: text: This proposal has been added by Vectro in 2016. I am not aware that this proposal has ever been published somewhere. So it seems to be original research. Markus Schulze 08:38, 19 November 2017 (UTC)
replies: text: Markus, I am attempting to describe your method, so I am surprised to see you consider it OR! Perhaps I have misunderstood something. I added a citation to a secondary source. TheShepherd, for more information you may wish to look here. Vectro (talk) 00:12, 29 November 2017 (UTC)
replies: text: The Schulze method is not a concretization of ranked pairs. The Schulze method and ranked pairs are two different election methods in so far as they can lead to different winners even when both methods have decisive winners. Markus Schulze 19:22, 1 December 2017 (UTC)
thread_title:  Locking in Ties 

message: "Procedure > Sort" states that:
The pairs of winners, called the "majorities", are then sorted from the largest majority to the smallest majority. A majority for x over y precedes a majority for z over w if and only if one of the following conditions holds:
1. Vxy > Vzw. In other words, the majority having more support for its alternative is ranked first.
2. Vxy = Vzw and Vwz > Vyx. Where the majorities are equal, the majority with the smaller minority opposition is ranked first.
However, I'm having trouble seeing how the second criterion could ever be possible if there are no indifference or unstated candidates, as in the example, because surely Vxy = Vzw would imply that Vwz = Vyx.
Further, under "An example > Sort", the following is stated:
Nashville (68%) beats both Chattanooga and Knoxville by a score of 68% over 32% (a tie, unlikely in real life for this many voters). Since Chattanooga > Knoxville, and they are the losers, Nashville vs. Knoxville will be added first, followed by Nashville vs. Chattanooga.
but this appears to be in direct contradiction with the given procedure; abbreviating Chattanooga, Knoxville, and Nashville with c, k, and n respectively, and letting x = z = n, y = k, and w = c, we have that Vnk = Vnc = 68%, and Vkn = Vcn = 32%, so neither criterion is satisfied, and thus neither majority can precede the other.
It seems like to have the conditions agree with the reason given, we need to add a third condition similar to "Vxy = Vzw, Vyx = Vwz, and Vwy > Vyw". Or is the reason given incorrect?
I also can't seem to find any sources that back up this procedure, so I can't tell if there is indeed an error or if I'm just misunderstanding it. The article states in reference #1 that the original article uses a different approach to ranking the strength of a victory.
Edderiofer (talk) 14:54, 10 September 2018 (UTC)
replies: text: I fully agree – thanks for pointing this out. I’ve rewritten the procedure (and also deleted the tag complaining about paucity of references – Tideman’s paper is crystal clear and authoritative, so why would anyone need more?) Colin.champion (talk) 14:16, 10 January 2022 (UTC)
thread_title:  Is there a typo or error in the conditions in "Procedure > Sort"? 

message: The article now says
Alternatively, you can use the procedure above to pick the winner, make that candidate first place, drop them from the election (i.e. drop all comparisons/edges that include them), and then repeat the process.
Please supply a citation to a textbook or similar to support this.  Absent support, this is too easily confused with original research.  —Quantling (talk | contribs) 18:59, 25 February 2023 (UTC)
replies: text: I have removed the part about using the algorithm repeatedly.  If you want to re-insert it, please include a citation.  —Quantling (talk | contribs) 14:37, 28 February 2023 (UTC)
thread_title:  Repeated use of the algorithm? 

message: The river method is not a variant of ranked pairs. The river method and the ranked pairs method have different heuristics; the river method tries to find the "best" arborescence (according to its own definition for "better"), while the ranked pairs method tries to find the "best" complete digraph (according to its own definition for "better"). The river method and the ranked pairs method are not related; the ranked pairs method has been proposed by Nicolaus Tideman in 1987, while the river method has been proposed by Jobst Heitzig in 2004. The river method and the ranked pairs method find different winners (even when there are no pairwise defeats of equal strengths). The river method and the ranked pairs method have different properties; the river method satisfies independence of Pareto-dominated alternatives, while the ranked pairs method violates this criterion; the river method violates reversal symmetry and local independence of irrelevant alternatives, while the ranked pairs method satisfies these criteria. The current version of this article misleads the reader into believing that the river method satisfies all criteria that are satisfied by the ranked pairs method plus independence of Pareto-dominated alternatives. The fact that both methods might look very similar doesn't justify the claim that the one method is a variant of the other method. Markus Schulze 10:06, 16 May 2024 (UTC)
replies: text: I would agree with this. Further, I don't think River really belongs in any wikipedia article yet, as it is original research, so I do not think @Closed Limelike Curves should reinstate it. Jannikp97 (talk) 05:53, 17 May 2024 (UTC)
thread_title:  The river method is not a variant of ranked pairs.