This is the letter that I said I'd send about demonstrations of compliances with the defensive strategy criteria.
*** First let me point out that SD, DCD, & Tideman all have one thing in common, any defeat that they drop is the weakest defeat in some cycle. When DCD drops the weakest defeat in each cycle, that's obvious. When SD drops the weakest defeat that's in a cycle, that defeat must also be the weakest defeat in any cycle that it's in. SD, DCD & Tideman have a symmetrical relation. DCD just simultaneously drops each cycle's weakest defeat, which means that it drops every defeat that's the weakest defeat in a cycle. SD could be worded as saying: Drop the weakest defeat that is the weakest defeat in some cycle that it's in. Repeat till there's an unbeaten candidate. Tideman iterates down from the top, saying: Drop the strongest defeat that is the weakest defeat in some cycle. Repeat till all cycles are solved. [It goes without saying that, when the iteration is from the top, "the weakest defeat in some cycle" means "the weakest defeat in some cycle with undropped defeats]. *** I haven't recommended Tideman because it's motivation & aren't, to me, as obvious & natural as those of DCD & SD & SSD. *** So all three of those methods, DCD, SD, & Tideman, drop defeats that are the weakest in some cycle to which they belong. SD starts with the weakest such defeat. Tideman starts with the strongest such defeat. And DSC simply drops all such defeats simultaneously. *** These symmetrically-related methods I call the cycle Condorcet versions. I'll later show that SSD can be counted among them, but that isn't important now. First, I'd like to show that any method that drops a defeat only' if it's the weakest defeat in some cycle automatically meets all of the 5 defensive strategy criteria. *** 1. SFC: If no unfelt preferences are voted, then there's no way that a CW can have a majority defeat. By assumption the CW beats B by a majority. If B isn't in a cycle with the CW, then its majority defeat by the CW can never be dropped, since the method in use won't drop a defeat that isn't the weakest defeat in some cycle. If B is in a cycle with the CW, then, because the CW can't have a majority defeat, then B's defeat by the CW can't be the weakest defeat in that cycle; the CW's defeat in that cycle is weaker. So B's defeat won't be the one that is dropped, and B can't win. The CW will become unbeaten before B could. 2. GSFC (This demonstration is similar to the above). If no unfelt preferences are voted, then there's no way that a member of the sincere Smith set can have a majority defeat against him (because, compared to any other candidate, more people prefer a sincere Smith set candidate to him than vice versa). A member of the sincere Smith set has a majority against B, who isn't in the sincere Smith set. Give the name Z to the sincere Smith set member who beats B by majority. As before, if B isn't in a cycle with Z, then Z's majority defeat of B isn't in a cycle and can't be dropped. If B is in a cycle with Z, then all of the defeats of sincere Smith set members by nonmembers that are part of that cycle are weaker parts of that cycle than B's defeat by Z. If we only drop defeats that are the weakest in some cycle, then Z some member of the sincere Smith set will become unbeaten before B does. *** 3. WDSC, SDSC, & SrDSC: A majority of all the voters prefer A to B. Say they all rank A over B. B has a majority defeat. Now, say that the members of that majority refuse to include B in their ranking. That means that B can't have a majority defeat against anyone. That means that if A & B are in a cycle, then B's defeat by A can't be the weakest defeat in that cycle, because B doesn't beat anything by majority. Some other member of the cycle will become unbeaten before B can. This argument also applies to SDSC & SrDSC, since the members of that majority achieved their goal merely by not ranking B. They didn't have to vote anyone equal to or over a more-liked candidate, or violate the terms of SrDSC. [The fact that they have refused to vote for B and also for those less-liked than B doesn't count as voting a less-liked candidate equal to a more-liked one; instead, it could be said that these voters have equally _not_ voted for B and those less-liked candidates]. *** So the Cycle Condorcet methods meet all 5 defensive strategy criteria. *** Before I show that SSD meets them too, let me introduce another criterion, a generalization of the 5 defensive strategy criteria: Beatpath Criterion (BC): If X beats Y, and if Y's strongest beatpath to X is weaker than X's defeat of Y, then Y shouldn't win. *** Definition of a beatpath: X has a beatpath to Y if either X beats Y, or if X beats someone who has a beatpath to Y. The strength of a beatpath is the strength of its weakest defeat. *** Since Y doesn't have any beatpath to X that is as strong as X's defeat of Y, that means that if Y is in a cycle with X, then Y's defeat by X can't be the weakest defeat in that cycle. That means that any method that only drops a defeat that is the weakest defeat in some cycle will never drop Y's defeat by X before it breaks the cycle elsewhere, making someone else unbeaten. Therefore, the Cycle Condorcet versions meet BC. To show that methods that meet BC meet those 5 criteria that the Cycle Condorcet methods meet: The Cycle Condorcet methods meet the defensive strategy criteria because, in each case, B's majority defeat isn't the weakest defeat in a cycle with the candidate who beats B (call him Z). If any cycle including B & Z has a weaker defeat than B's defeat by Z, then B doesn't have a beatpath to Z as strong as Z's defeat of B, and so B can't win using any method that meets BC. *** Why SSD meets BC: If Z beats B, and Z has no beatpath to B as strong as B's defeat by B, then every cycle including B & Z has a defeat that's weaker than B's defeat by Z. If B isn't in the Schwartz set then his defeat can't be dropped & he can't win. If B is in the Schwartz set, then anything that beats him must be in the Schwartz set too, since the Schwartz set is an unbeaten set, not beaten from without. If some candidate is in the Schwartz set, then anyone who beats him must also be in the Schwartz set. So, anyone in a cycle containing B & Z must be in the Schwartz set. According to BC's premise, any cycle containing B & Z has a defeat weaker than B's defeat by Z (from the paragraph before last). That means that B's defeat by Z isn't the weakest defeat in the Schwartz set. Any cycle containing those 2 candidates has a weaker defeat elsewhere. When those defeats have all been dropped, as SSD repeatedly drops the Schwartz set's weakest defeat, some candidate other than B will have been made unbeaten. That could happen the first time a defeat is dropped, but it has to happen by the time that the weakest defeat is dropped from each cycle containing B & Z. So someone other than B will be the undefeated before B is, and B can't win. *** So SSD meets BC, and therefore meets all 5 of the defensive strategy criteria. *** Schulze meets BC for obvious reasons. If there's no beatpath from B to Z as strong as Z's defeat of B, that means that Z has a (1-defeat) beatpath to B stronger than any beatpath from B to Z, meaning that B can't be the candidate with a Schulze win over each one of the other candidates, and that B can't win. *** I've shown that the Cycle Condorcet methods and SSD & Schulze meet BC, and therefore meet all 5 of the defensive strategy criteria. Now, in my last of these messages, I'd like to show, as I said I would, that SSD can be counted as one of the Cycle Condorcet versions, because if a defeat is among the Schwartz set members, it is in a cycle. That will be the next & last of this series of messages. Mike Ossipoff ______________________________________________________ Get Your Private, Free Email at http://www.hotmail.com
