Cycles in transitive voting

[Michael Allan]

Welcome Peter,

> Basically each row of the transition matrix contains the shares of
> all the votes of a specific voter that he/she gives to each of the
> other voters (including him/herself).  From the transition matrix
> you get the stationary distribution by finding its eigenvalues and
> those eigenvalues are the final distribution of the votes. ...

I guess the transition matrix would have a single row in our case.
Our votes are atomic and cannot be split.  Instead of splitting, they
coalesce and cascade "like raindrops down the branches of a tree". *
This was a deliberate design choice for us because we wanted a
mechanism to enable and encourage consensus.

 * http://zelea.com/project/votorola/d/theory.xht#medium
   Cycles are described at figure 2.

-- 
Michael Allan

Toronto, +1 416-699-9528
http://zelea.com/


Peter Zbornik said:
> Hi Michael,
> 
> just a clarification:
> 
> I wrote:
> > By the way (probably an other topic), one way to allow cycles in
> > delegated votes and still being able to calculate the number of
> > votes of each person is to use the page rank algorithm which google
> > uses:
> >
> http://chato.cl/papers/boldi_bonchi_castillo_vigna_2011_viscous_democracy_social_networks.pdf
> 
> You answered:
> Boldi et al. use something similar to Brin and Page's "dampening".
> The vote loses strength as it goes from delegate to delegate.  We have
> a different aproach where the vote is stopped and "held" before it
> can cycle: http://zelea.com/project/votorola/d/theory.xht#cycle
> 
> My comment:
> You don't necessarily need to use "dampering".
> In theory you can use the stationary distribution of the Markov
> chain (which is equivalent to setting the dampering constant to one), see
> http://en.wikipedia.org/wiki/Markov_chain#Steady-state_analysis_and_limiting_distributions
> and http://en.wikipedia.org/wiki/Markov_chain#Internet_applications .
> 
> Basically each row of the transition matrix contains the shares of all the
> votes of a specific voter that he/she gives to each of the other voters
> (including him/herself).
> From the transition matrix you get the stationary distribution by finding
> its eigenvalues and those eigenvalues are the final distribution of the
> votes.
> 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.
> 
> Best regards
> Peter Zbotník