<div dir="auto"><div>Congratulations!... a major tour de force ... discouraging the most common insincere strategies like never before, while retaining monotonicity and even inhancing Smith to Landau efficiency! </div><div dir="auto"><br><div class="gmail_quote" dir="auto"><div dir="ltr" class="gmail_attr">On Mon, Sep 12, 2022, 2:27 AM Kristofer Munsterhjelm <<a href="mailto:km_elmet@t-online.de">km_elmet@t-online.de</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Here's a method that generalizes fpA - sum fpC to pass:<br>
- Landau (with a certain provision)<br>
- monotonicity<br>
- independence of twins (a weaker clone independence)<br>
- output scores (not just rankings)<br>
and probably also passes DMTCBR and a weaker DMTBR (as yet unnamed <br>
criterion). My simulator indicates as such, but I don't have a proof <br>
yet. In addition, unlike IRV, it's summable.<br>
<br>
Let's call it Friendly Cover for now, for lack of a better name. Any <br>
name suggestions would be appreciated!<br>
<br>
(In addition, Friendly Cover passes all the three-candidate criteria <br>
that fpA-fpC does, like the chicken dilemma criterion.)<br>
<br>
This is quite a step towards getting strategy resistance, clone <br>
independence, and summability!<br>
<br>
I'll describe a simple version, then a more complex one, and then do <br>
some proofs for the criteria.<br>
<br>
First, the simple version. It has two problems, one of which a patch <br>
handles; I'll give an explicit version later that incorporates that <br>
patch more directly.<br>
<br>
For any candidate X:<br>
Let X's defeaters be the candidates who beat X pairwise,<br>
Let X's intermediates be the candidates who beat some defeater and are <br>
not defeaters themselves,<br>
and let X's friends be everybody else (including X himself).<br>
<br>
Then X's score is the sum of first preferences of X's friends, minus <br>
the sum of first preferences of X's defeaters.<br>
<br>
In a three candidate A>B>C>A cycle, this reduces to fpA - fpC because C <br>
is A's defeater and B is an intermediate, so the friend set is only A.<br>
<br>
However, there are two problems. First, we only get Landau if every <br>
candidate gets at least one first preference vote. Otherwise it's <br>
possible that A is the CW and has no first preferences, and B beats <br>
everybody but the CW and, despite being beaten by the CW, ties him by <br>
score. That happens because the only defeater of B is A, and A has zero <br>
first preferences, so B's score is A's score minus zero.<br>
<br>
I don't know how to robustly handle that yet, so I'm going to leave it <br>
for now. Perhaps funny is that people who make a big deal out of core <br>
support could see this as a feature and not a bug :-)<br>
<br>
Second, which *can* be fixed: if A ties C pairwise, C's defeat set <br>
contains B but not A, and A's friend set also contains B, then raising A <br>
on a B>A ballot could give C an additional point while not changing A's <br>
score, thus violating monotonicity. To fix that, we need this patch:<br>
<br>
Before calculating X's score, remove from X's friend set everybody <br>
who's a defeater of someone who X ties pairwise.<br>
<br>
But that's fairly inelegant, so instead I'll give an explicit version <br>
that's equivalent in the absence of pairwise ties. Thanks to Forest <br>
Simmons for discovering that equivalence:<br>
<br>
For any candidate X:<br>
Let X's defeaters be the candidates who beat X pairwise,<br>
Let X's friends be everybody X weakly covers,<br>
<br>
and the score is calculated as above.<br>
<br>
Ordinary covering is similar to "A defeats B", i.e. you can't be <br>
covering someone and that someone also be covering you. Weak covering is <br>
more like "A defeats or ties B": you can be weakly covering someone else <br>
who is also weakly covering you, and that means you, in the covering <br>
sense, tie that other candidate.<br>
<br>
The weak covering relation is defined like this: "A weakly covers B if A <br>
beats everybody B does, and B is beaten by everybody A is beaten by, <br>
pairwise". (Note that A covers himself.)<br>
<br>
Hence the "Cover" in "Friendly Cover". It can be shown that if there are <br>
no pairwise ties, X's friend set is exactly X and everybody X covers in <br>
the ordinary Landau sense, but I don't want to clutter up the post.<br>
<br>
This solves the problem we patched out because A can't cover B if B is a <br>
defeater of some C that A ties pairwise, because then A doesn't beat <br>
everybody B does (in particular, A doesn't beat C!)<br>
<br>
----<br>
<br>
Proofs:<br>
<br>
Proving Landau compliance involves a lot of nitty-gritty details, so <br>
I'll show it for a related strict covering that differs from <br>
Landau/Fishburn when there are ties. I think it can be shown for actual <br>
Landau too, but again, I don't want to clutter up the post; just ask me <br>
if you'd like the full thing.<br>
<br>
First assume that everybody has a first preference. That's the caveat I <br>
mentioned at the start.<br>
<br>
Let A strictly cover B if A weakly covers B, A beats B pairwise, and <br>
either A beats someone B doesn't, or B is beaten by someone A isn't. And <br>
let the uncovered set be everybody who isn't strictly covered by anyone.<br>
<br>
Suppose A strictly covers B. If B is beaten by one more candidate than <br>
A, then B's defeat set contains more candidates than A's. And since <br>
covering relations are transitive (if A covers B and B covers C, then A <br>
covers C), A's friend set is no smaller than B's. Hence B's sum over <br>
defeaters is larger than A's, and B's sum over friends is at most equal <br>
to A's. Thus A outscores B, as desired.<br>
<br>
If, on the other hand, A beats one more candidate than B (but both A and <br>
B are beaten by the same set), then since A beats B pairwise, A weakly <br>
covers B but B doesn't weakly cover A. So A's friend set contains A <br>
himself as well as everybody weakly covered by B, but B's friend set <br>
doesn't contain A. Hence B's sum over defeaters is the same as A's while <br>
A's sum over friends is greater than B's, so again A outscores B.<br>
<br>
<br>
Monotonicity is pretty easy: raising A can't make anyone appear in A's <br>
defeater set who wasn't there before, and can't remove someone from <br>
someone else's defeater set. Since fpA - fpC itself is monotone, that's <br>
all we need.<br>
<br>
<br>
Independence of twins: this is a weaker clone independence that minmax <br>
also passes. It says that if we replace a candidate X by two clones X1 <br>
and X2, then the outcome can't change.<br>
<br>
First, let's show that there's no teaming incentive. Suppose A wins and <br>
we clone A. Since everybody who beats A also beats all the clones, this <br>
can't remove any defeaters of A from the clones' defeater sets. And <br>
since the clones all beat candidates who the original A beats, it can't <br>
introduce new friends to any of the clones. Since the sum over defeaters <br>
can't decrease and the sum over friends can't increase, the A clones' <br>
scores can't increase.<br>
<br>
Now suppose that the winner is X and we clone Y. If Y is a defeater of <br>
X, then all the clones will be defeaters of X. If Y is covered by X, <br>
then every clone will also be covered by X. So cloning Y doesn't affect <br>
X's score. Since we know from the previous paragraph that it can't <br>
increase Y's score, Y can't go from loser to winner by being cloned. So <br>
there is no teaming incentive (and this is fully general, i.e. not just <br>
for twins).<br>
<br>
Now suppose A wins and we replace A with two candidates A1 and A2. <br>
Suppose also without loss of generality that A1 beats or ties A2 <br>
pairwise. Then if A1 ties A2, both weakly cover each other. Since they <br>
beat the same non-clone candidates and are beaten by the same non-clone <br>
candidates as A, all that happens is that their friend sets go from <br>
containing A to containing A1 and A2. And since the sum of clones' first <br>
preferences is the same as the original candidate's first preferences, <br>
the clones' scores don't change. If A won, both clones will still win; <br>
if A lost, both clones will still lose.<br>
<br>
If A1 strictly beats A2, then A1 weakly covers A2 but not vice versa. So <br>
A1's score is the same as A used to be while A2's score will at most be <br>
equal to A1 (and may be lower). In any case, if A won, A1 will win after <br>
the cloning, and if A lost, both clones will still lose afterwards.<br>
<br>
The problem is that in the more general case, it's possible that A could <br>
be cloned into A1, A2, and A3, who are in a cycle (i.e. A1>A2>A3>A1). <br>
Then each clone gets one of the other clones into his defeat set, which <br>
lowers all the clones' score and may make them lose. So if you have at <br>
least five candidates and a Landau set with three clones in it, there <br>
could be trouble.<br>
<br>
-km<br>
----<br>
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</blockquote></div></div></div>