[EM] Method X, Landau

[Richard Lung]

On 03/08/2023 08:01, Richard Lung wrote:
>
> My feeling is that "Method X" is doomed to failure, while based on 
> competing elimination methods. Eliminations being based on rules 
> outside the power of the voters, are bound to be arbitrary, and 
> arbitrary rules cannot provide a definitive method X.
>
> Tho I think it is good that you look for a definitive method.
>
>
> Regards,
>
>  Richard Lung.
>
>
> On 03/08/2023 01:54, Kevin Venzke wrote:
>> I plan to implement Method X when I get a moment. I'm not completely 
>> skeptical about its monotonicity since it doesn't seem like there's 
>> an obvious argument why it shouldn't be monotone.
>>
>> For Landau, I use some code like this, which I base on the 
>> explanation on Wikipedia:
>>
>> let landau = [];
>> for(x in candidates) {
>> let i_am_covered = false;
>> for(z in candidates) {
>> if( x != z ) { // X can't cover themselves
>> let found_y = false;
>> for(y in candidates) {
>> // Note that beats_or_ties should evaluate as true if the two 
>> operands are the same candidate
>> if( x beats_or_ties y && y beats_or_ties z ) {
>> // alternative to the above line which should be right:
>> // if( not( y beats x || z beats y ) ) {
>> found_y = true; // found a counterexample to Z covering X
>> break; // cease checking for Ys for this Z
>> } //if x...
>> } //for y
>> if( not found_y ) {
>>   i_am_covered = true; // Z evidently covers X
>> break; // cease checking for Zs for this X
>> } // if not found_y
>> } // if x!=z
>> } // for z
>> if( not i_am_covered ) landau.append( x );
>> } // for x
>> return landau;
>>
>> Kevin
>> (end)
>>
>>
>> Le mercredi 2 août 2023 à 09:54:45 UTC−5, Forest Simmons 
>> <forest.simmons21 at gmail.com> a écrit :
>> On Wed, Aug 2, 2023, 4:15 AM Kristofer Munsterhjelm 
>> <km_elmet at t-online.de> wrote:
>>> I tried to check Landau//X. It had plenty of Other strats, which is not
>>> unexpected given your proof about independence of covered alternatives
>>> plus first preference tiebreak implying nonmonotonicity... but I think
>>> my implementation of Landau is suspect, so I'm not including it here.
>> Let M be the matrix whose (j,k) element is one (else zero) if 
>> candidate j defeats candidate k.
>>
>> Then candidate j is uncovered iff the j_th row of (I+M)^2 is devoid 
>> of zeros  ... because the (j,k) element of this matrix is the number 
>> of distinct short beatpaths (i.e. of length 0,1, or 2) from candidate 
>> j to candidate k.
>> ----
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