Re: [Ag Meinungsfindungstool] Delegated Voting

[Peter Zbornik <pzbornik AT gmail.com>]

Hi,

just a comment:
I wrote:
"And the solution is:
A gets 0.4 of the votes
B gets 0.2 of the votes
C gets 0.4 of the votes"

What this solution means is that if you imput the "stationary
distribution" above into the matrix, then you get the same
distribution in the next round, i.e. nothing changes.

So let's do that to check out that it works:

The matrix was:
                 A          B          C
> A          0.5        0          0.5
> B           1          0           0
> C           0          0.5       0.5

A has 0.4 votes - A gets half and C half - i.e. 0.2 each.
B has 0.2 votes - A gets all of them
C has 0.4 votes - B gets half and C half - i.e. 0.2 each.

So after delegating the votes, A has 0.2+0.2=0.4 votes
B has 0.2 votes and C has 0.2+0.2=0.4 votes

That is - the distribution of votes between the voters after the
delegation is the same as before the delegation.
That is why the distribution is called stationary.

Best regards
Peter Zborník

2012/10/12 Peter Zbornik <pzbornik AT gmail.com>:
> Hi,
>
> I'll try to explain the "pageRank"-method of vote delegation.
>
> Imagine each voter as a row in a matrix.
> The part of the vote(s) of the voter, that the voter delegates to the
> other voters are the columns in the matrix.
> So each cell specifies how great a part of the vote(s) of the voter
> that the voter gives to an other voter each delegation.
>
> Example: Say we have three voters, A, B, C and the following matrix:
>              A          B          C
> A          0.5        0          0.5
> B           1          0           0
> C           0          0.5       0.5
>
> OK, so A gives half of his votes to himself and half to C
> B gives all votes to A
> C gives half of the votes to B and half to himself.
>
> Lets say we start with A having one (or 100) votes.
> Then A gives half to himself and half to C.
> In the next round A has half of the votes.
> A gives half of the half to himself and half of the half to C
> C gives half of his half to B and half of his half to himself
> In the next round B has some votes, which he gives to A.
>
> You might imagine, that this distribution of votes might settle down
> to some stable state (which it does after a while)
>
> Let's let these people give the votes to eachother for a very long
> time and we have something called a stationary distribution of a
> markov chain.
> (see here 
> http://en.wikipedia.org/wiki/Markov_chain#Steady-state_analysis_and_limiting_distributions
> and http://en.wikipedia.org/wiki/Markov_chain#Internet_applications).
>
> The matrix above is called a "translation matrix".
>
> I guess the stationary distribution might be thought of as the average
> share of all votes that a voter receives in each turn when the votes
> are delegated after a very long time.
>
> From the transition matrix you get the stationary distribution by
> solving a matrix equation.
>
> So lets find the stationary distribution to the matrix above.
> This page seems to do the trick (I have forgotten how to do it by hand
> - that was like ten years ago):
> http://math.plussed.net/markov/index.php
>
> And the solution is:
> A gets 0.4 of the votes
> B gets 0.2 of the votes
> C gets 0.4 of the votes
>
> No vetos, all fair and square, no limitations.
>
> You can introduce damping factors which limit the strength of the vote
> every time it is delegated (bringing it closer and closer to zero at
> every transfer (this is what google do).
> By this you limit the number of times each vote can be delegated.
>
> OK, I hope I made things a bit clearer.
> I guess you need a mathematical statistitian to program this solution.
> But it is really not that difficult.
>
> Best regards
> Peter Zborník
>
>
> 2012/10/12 Alexander Praetorius <citizen AT serapath.de>:
>> 2012/10/12 Peter Zbornik <pzbornik AT gmail.com>
>>>
>>> Hi there,
>>>
>>> I am new to this discussion.
>>> Nevertheless, talking about circles in delegation (A delegates to B
>>> and B delegates to A), the best solution seems to be to use Googles
>>> pagerank algorithm (modified of course).
>>> Here is the paper:
>>>
>>> http://chato.cl/papers/boldi_bonchi_castillo_vigna_2011_viscous_democracy_social_networks.pdf
>>
>>
>> Could you elaborate on it?
>>
>>>
>>>
>>>
>>> Seems to be the right solution.
>>>
>>> Best regards
>>> Peter Zborník
>>> Prague, Czech republic
>>>
>>> 2012/10/12 Alexander Praetorius <citizen AT serapath.de>:
>>> > Tatsächlich ist das ein Schwachpunkt.
>>> > Alle würden mit dem abstimmen bis zum letztmöglichen Zeitpunkt warten.
>>> > Vielleicht kann man hier weiterdenken.
>>> >
>>> > Was wenn die NICHTABSTIMMENDEN zu gleichen Anteilen gemäß der Zeit für
>>> > die
>>> > sie im Vorfeld zur Deadline für einen Antrag aufgrund von delegation
>>> > gestimmt haben verteilt werden?
>>> >
>>> > BEISPIEL: A,B,C,D,E (Cliquendelegation)
>>> > Abstimmungszeitraum nach Einstellung bis Deadline: 7 Tage.
>>> >
>>> > Tag 1: A stimmt für X (=> B,C,D,E stimmen für X)
>>> > Tag 2: B stimmt gegen X (=> C,D,E stimmen gegen X)
>>> > Tag 4: C stimmt für Y als Alternative zu X (=> D, E stimmen für Y)
>>> >
>>> > Deadline läuft ab.
>>> >
>>> > pro X: 1 Tag * 5 Stimmen + 6 Tage * 1 Stimme = 11 Tagesstimmen
>>> > kontra X: 2 Tage * 4 Stimmen + 4 Tage * 1 Stimme = 12 Tagesstimmen
>>> > pro Y: 4 Tage * 3 Stimmen = 12 Tagesstimmen
>>> >
>>> > Hierdurch würde man den Anreiz mit der Stimmvergabe zu warten
>>> > invertieren.
>>> > Ein möglichst frühes abstimmen lohnt, weil man "Tagesstimmen" ansammeln
>>> > kann.
>>> > Sobald jemand weiteres abstimmt, nimmt der alle Stimmen mit, aber jemand
>>> > drittes kann die Stimmen wieder zurück zum initialvorschlag mitnehmen.
>>> > Das motiviert politische Teilhabe.
>>> >
>>> > Gruß
>>> > Alex
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > 2012/10/12 Slash <pirate_slash AT yahoo.com>
>>> >>
>>> >>
>>> >> Also, jetzt mal meine 2 cents:
>>> >> Is' wirklich nur was für Konsens-Gruppen, aber für den Fall kann man
>>> >> sich
>>> >> den Aufwand dann auch gleich sparen und auf Delegationskette setzen.
>>> >>
>>> >> In heterogenen Gruppen wär' es ein fragwürdiges, sehr gewichtiges
>>> >> Machtinstrument; wer zeitlich zuletzt stimmt, hat das meiste Gewicht.
>>> >> Und da der Mensch nun mal egoistisch ist, würde diese Möglichkeit
>>> >> nicht ungenutzt bleiben, was nicht im Sinne der Sache ist.
>>> >>
>>> >>
>>> >> Viele Grüße,
>>> >> / aka Oliver
>>> >>
>>> >> --
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>>> >
>>> >
>>> >
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>>> > ***********************************************
>>> > Alexander Praetorius
>>> > Rappstraße 13
>>> > D - 60318 Frankfurt am Main
>>> > Germany
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>>> > [mail] citizen AT serapath.de
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>>
>>
>>
>>
>> --
>>
>> Best Regards / Mit freundlichen Grüßen
>> ***********************************************
>> Alexander Praetorius
>> Rappstraße 13
>> D - 60318 Frankfurt am Main
>> Germany
>> [skype] alexander.praetorius
>> [mail] citizen AT serapath.de
>> [web] http://wiki.piratenpartei.de/Benutzer:Serapath
>> ***********************************************
>>
>>
>>
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