# [EM] Proof idea that IRV can't be summable Classic List Threaded 6 messages Open this post in threaded view
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## [EM] Proof idea that IRV can't be summable

 Here's an idea that may be used to prove that any attempt to find a clever way of making IRV summable (in a linear sense) will fail: that it's not just the case that we haven't found such a solution, but that it can't be done. It's not a complete proof, but perhaps someone can complete it :-) And, of course, it could be wrong; I could be missing something. Consider a voting method as a union of convex polytopes expressed as sets of linear inequalities. The unknown variables are the ballot numbers: e.g. in the three-candidate no-truncation case, they're ABC, ACB, BAC, BCA, CAB, CBA: the number of people who expressed each preference. Then if the c!-dimensional vector is inside this union of convex polytopes, A wins (if the union is called a win region); or A wins or ties for first (if the union is called a win-or-tie region). As far as summability is concerned, we can pick whichever is more convenient as long as the difference between the win-and-tie region and the win region is not of full dimensionality (which means that as the number of voters approach infinity, the proportion of ties approach zero). This is, to my knowledge, the case for IRV. My strategy is to show that for (almost) any pair of ballot variables, there exists some win region with some polytope with some inequality that treats both ballot variables equally by scaling both equally; and another that treats them differently (by scaling them by different amounts). The inequalities need not be in the same polytopes, since the argument is that the inequality requires the method to distinguish between two variables (or treat them equally); and thus that any summary must also allow the method to do so. A violation of this will degrade at least one of the win regions. So, letting the two ballot variables be X and Y, the inequality that treats X and Y equally proves that if IRV is summable, the sums can't all have a term that amounts to (X-Y), because then the inequality in question could not treat X and Y equally. The second case proves that if IRV is summable, the sums can't all have a term that amounts to (X+Y), because then the inequality can't treat X and Y differently. If this can be proven for all pairs, then there must be c! different summed variables (since no pair can be treated differently nor equally), and since c! is not polynomial in c, we're done. Now, to familiarize, consider the win region for A. For IRV, the union is of the form (three candidate example given here): "B is eliminated, A wins pairwise against C" or "C is eliminated, A wins pairwise against B" If either of these is true, then A wins with certainty; otherwise, A does not. In this particular case, the inequalities of the first constituent polytope look something like: ABC + ACB - BAC - BCA > 0 (A outscores B in the first round) ABC + ACB - CAB - CBA > 0 (A outscores C in the first round) CAB + CBA - BAC - BCA > 0 (C outscores B in the first round) ABC + ACB + BAC + BCA - CAB - CBA > 0 (A beats C pairwise) So now take two ballot orders X and Y, but not so that Y is X reversed. Let the set S_1 be the set of candidates that must be removed so that the first candidate of X and Y match. Since Y isn't X reversed, this set is at most two candidates smaller than the set of all candidates. Then X and Y have a common factor in an inequality belonging to a polytope where all the candidates of S_1 have been eliminated; because both X and Y then count towards the first preferences of someone not in S_1, vs someone else not in S_1. Then let the set S_2 be the set of candidates that must be removed so that the first candidate listed differs. Since X is not Y, the set is at most two candidates smaller than the set of all candidates. Then, similarly, in the scenario where the candidates in S_2 are all eliminated, X counts towards someone's first preferences while Y counts towards someone else's. So if cX is the candidate listed first on X after the candidates in S_2 have been eliminated, then a polytope dealing with cX's win region after the candidates in S_2 have been eliminated will have a X + ... - Y - ... > 0 inequality. The cases where Y is the exact reverse of X can't be treated this way, but that's no problem as there are only O(c) of them anyway: hence these being summable won't make IRV itself polynomially summable. Caveat: I haven't shown that every pair of constituent polytopes for A's win region is nonconvex as a whole, and that there are no redundant inequalities (that each polytope's inequality matrix is full rank). That's necessary to make sure the win region can't be reconstructed in a way so that the tricky inequalities disappear. In other words, if I don't do this, then the following "proof" for the non-summability of Plurality would work: A wins if one of the following is true (one polytope per inequality):         for i = all 2^k subsets of ballots that begin in A                 for j = all 2^p subsets of ballots that don't                         e_i * x > e_j * x for some numbers k and p; where x is the ballot vector, and e_i and e_j are the indicator vectors for the respective sets. Then one could claim that due to the superpolynomial amount of inequalities, Plurality is non-summable. The flaw lies in that e.g. the inequalities         ABC > CBA         ACB > CBA are entirely subsumed by the inequality         ABC + ACB > CAB + CBA. So the method doesn't need to discriminate between two situations that it seems like it needs to at first glance. I would also, strictly speaking, need to show that (win or tie region) \ (win region) is not of full dimensionality; or deal with the win-or-tie region directly. Finally, I have a feeling there's some kind of more general matrix rank argument hiding in there somewhere. Perhaps it could be used to show just what kind of positional elimination systems are non-summable. Borda-elimination is summable because you can get the Borda score from the Condorcet matrix; why is it different? Are there any other systems like it? -km  Note that these methods are deterministic. If there are any random tiebreaks, they will have to be determined ahead of time. It's then clear that some random tiebreaks may make an otherwise summable method non-summable if they're "random enough". ---- Election-Methods mailing list - see https://electorama.com/em for list info
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## Re: [EM] Proof idea that IRV can't be summable

 Of course, all that learning won't alter what John Stuart Mill knew over 150 years ago that maiorocracy is not democracy. The fixation on single winner methods, (the monarchism hang-over) is candidate-centred not voter-centered or politician-based not people-based. Richard L. On 01/12/2020 22:25, Kristofer Munsterhjelm wrote: > Here's an idea that may be used to prove that any attempt to find a > clever way of making IRV summable (in a linear sense) will fail: that > it's not just the case that we haven't found such a solution, but that > it can't be done. > > It's not a complete proof, but perhaps someone can complete it :-) And, > of course, it could be wrong; I could be missing something. > > Consider a voting method as a union of convex polytopes expressed as > sets of linear inequalities. The unknown variables are the ballot > numbers: e.g. in the three-candidate no-truncation case, they're ABC, > ACB, BAC, BCA, CAB, CBA: the number of people who expressed each preference. > > Then if the c!-dimensional vector is inside this union of convex > polytopes, A wins (if the union is called a win region); or A wins or > ties for first (if the union is called a win-or-tie region). As far > as summability is concerned, we can pick whichever is more convenient as > long as the difference between the win-and-tie region and the win region > is not of full dimensionality (which means that as the number of voters > approach infinity, the proportion of ties approach zero). This is, to my > knowledge, the case for IRV. > > My strategy is to show that for (almost) any pair of ballot variables, > there exists some win region with some polytope with some inequality > that treats both ballot variables equally by scaling both equally; and > another that treats them differently (by scaling them by different > amounts). > > The inequalities need not be in the same polytopes, since the argument > is that the inequality requires the method to distinguish between two > variables (or treat them equally); and thus that any summary must also > allow the method to do so. A violation of this will degrade at least one > of the win regions. > > So, letting the two ballot variables be X and Y, the inequality that > treats X and Y equally proves that if IRV is summable, the sums can't > all have a term that amounts to (X-Y), because then the inequality in > question could not treat X and Y equally. The second case proves that if > IRV is summable, the sums can't all have a term that amounts to (X+Y), > because then the inequality can't treat X and Y differently. > > If this can be proven for all pairs, then there must be c! different > summed variables (since no pair can be treated differently nor equally), > and since c! is not polynomial in c, we're done. > > Now, to familiarize, consider the win region for A. For IRV, the union > is of the form (three candidate example given here): > > "B is eliminated, A wins pairwise against C" or > "C is eliminated, A wins pairwise against B" > > If either of these is true, then A wins with certainty; otherwise, A > does not. In this particular case, the inequalities of the first > constituent polytope look something like: > > ABC + ACB - BAC - BCA > 0 (A outscores B in the first round) > ABC + ACB - CAB - CBA > 0 (A outscores C in the first round) > CAB + CBA - BAC - BCA > 0 (C outscores B in the first round) > ABC + ACB + BAC + BCA - CAB - CBA > 0 (A beats C pairwise) > > So now take two ballot orders X and Y, but not so that Y is X reversed. > > Let the set S_1 be the set of candidates that must be removed so that > the first candidate of X and Y match. Since Y isn't X reversed, this set > is at most two candidates smaller than the set of all candidates. Then X > and Y have a common factor in an inequality belonging to a polytope > where all the candidates of S_1 have been eliminated; because both X and > Y then count towards the first preferences of someone not in S_1, vs > someone else not in S_1. > > Then let the set S_2 be the set of candidates that must be removed so > that the first candidate listed differs. Since X is not Y, the set is at > most two candidates smaller than the set of all candidates. Then, > similarly, in the scenario where the candidates in S_2 are all > eliminated, X counts towards someone's first preferences while Y counts > towards someone else's. So if cX is the candidate listed first on X > after the candidates in S_2 have been eliminated, then a polytope > dealing with cX's win region after the candidates in S_2 have been > eliminated will have a X + ... - Y - ... > 0 inequality. > > The cases where Y is the exact reverse of X can't be treated this way, > but that's no problem as there are only O(c) of them anyway: hence these > being summable won't make IRV itself polynomially summable. > > > > Caveat: I haven't shown that every pair of constituent polytopes for A's > win region is nonconvex as a whole, and that there are no redundant > inequalities (that each polytope's inequality matrix is full rank). > > That's necessary to make sure the win region can't be reconstructed in a > way so that the tricky inequalities disappear. In other words, if I > don't do this, then the following "proof" for the non-summability of > Plurality would work: > > A wins if one of the following is true (one polytope per inequality): > for i = all 2^k subsets of ballots that begin in A > for j = all 2^p subsets of ballots that don't > e_i * x > e_j * x > for some numbers k and p; where x is the ballot vector, and e_i and e_j > are the indicator vectors for the respective sets. > > Then one could claim that due to the superpolynomial amount of > inequalities, Plurality is non-summable. The flaw lies in that e.g. the > inequalities > ABC > CBA > ACB > CBA > are entirely subsumed by the inequality > ABC + ACB > CAB + CBA. > So the method doesn't need to discriminate between two situations that > it seems like it needs to at first glance. > > I would also, strictly speaking, need to show that (win or tie region) \ > (win region) is not of full dimensionality; or deal with the win-or-tie > region directly. > > Finally, I have a feeling there's some kind of more general matrix rank > argument hiding in there somewhere. Perhaps it could be used to show > just what kind of positional elimination systems are non-summable. > Borda-elimination is summable because you can get the Borda score from > the Condorcet matrix; why is it different? Are there any other systems > like it? > > -km > >  Note that these methods are deterministic. If there are any random > tiebreaks, they will have to be determined ahead of time. It's then > clear that some random tiebreaks may make an otherwise summable method > non-summable if they're "random enough". > ---- > Election-Methods mailing list - see https://electorama.com/em for list info ---- Election-Methods mailing list - see https://electorama.com/em for list info
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## Re: [EM] Proof idea that IRV can't be summable

 On 03/12/2020 08.31, Richard Lung wrote: > > Of course, all that learning won't alter what John Stuart Mill knew over > 150 years ago that maiorocracy is not democracy. The fixation on single > winner methods, (the monarchism hang-over) is candidate-centred not > voter-centered or politician-based not people-based. I have a hunch that anything that passes Droop proportionality must fail strong summability: that it's impossible to find a Droop-proportional multiwinner method where you can construct an array of numbers polynomial in the number of candidates, and then later use that array to find the outcome for any number of seats. The same approach could possibly be used to prove this, although it would be a lot harder. Proving multiwinner STV non-summable would not be too much harder than what I did in the post you replied to. There's nothing there that has to necessarily be single-winner. It's not clear whether the corresponding regions should be of candidates, or councils, though. (E.g. do we consider the region "A is on the council" as some analog of a win-or-tie region for A, or should the region be "the two-seat council is AB"?) If told to create something democratic without concern to current constraints, I'd probably just skip right to sortition. This would not just invalidate single-winner methods, but voting altogether; except, possibly, the method the assembly itself uses to decide. -km ---- Election-Methods mailing list - see https://electorama.com/em for list info
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## Re: [EM] Proof idea that IRV can't be summable

 Le jeudi 3 décembre 2020 à 04:20:54 UTC−6, Kristofer Munsterhjelm <[hidden email]> a écrit : >If told to create something democratic without concern to current >constraints, I'd probably just skip right to sortition. This would not >just invalidate single-winner methods, but voting altogether; except, >possibly, the method the assembly itself uses to decide. In this exchange "democratic" must mean that the assembly's seats are allocated proportionally. This leaves the issue of allocation of actual policy-making power as you suggest. Maybe there is a way to determine policy proportionally, and without using randomness. I don't think it can be based on decay of individual delegates' voting power (because if you use your power sub-optimally you may fail to influence policy) or on how many things you vote on (because proposals could be clones of each other etc.). So it might have to be based on time... A faction gets an amount of time in power. But realistically there is probably a minimum faction size you would want to allow to wield power. And to prevent whiplash you'd probably want a minimum amount of time that a faction could be in power. Could a faction representing 25% of the voters be allowed to set policy for even a year? If not, can we defend that without invoking the principle of majority rule? (I doubt it... And for me that is always the limitation, that no matter what, you have to implement majority rule somewhere in the process.) Kevin ---- Election-Methods mailing list - see https://electorama.com/em for list info