# [EM] Teams

Susan Simmons suzerainsimmons at outlook.com
Sat Jul 10 09:15:45 PDT 2021

```Perhaps a traversal of the river basin tree, respecting at each node the order of the addition of the branches ... at each node, last added branch, first traversed ... or something like that.

It is possible to elevate or topologically "lift" the river tree to the form of a binary decision tree by that method ... and there are some possible uses for such a tree ...

Suppose sophisticated voters with complete information about each others' preferences were in possession of a binary tree whose leaves were the candidates.

At each node starting at the root, the voters choose which branch to follow. The leaf at the end of the path is the winning candidate.

This method gives zero incentive for insincere branch choices, given the tree. (unlike the related method that starts at the leaves and works towards the root).

How to get the tree without distortions of gamers?

Pollsters randomly polling one pairwise choice per "pollee?"

Sent from my MetroPCS 4G LTE Android Device

-------- Original message --------
From: Kristofer Munsterhjelm <km_elmet at t-online.de>
Date: 7/10/21 2:26 AM (GMT-08:00)
To: Susan Simmons <suzerainsimmons at outlook.com>, Daniel Carrera <dcarrera at gmail.com>
Cc: election-methods at lists.electorama.com
Subject: Re: [EM] Teams

On 7/10/21 9:23 AM, Susan Simmons wrote:
> The separate groups, teams, river basins exist to keep the snake from
> eating its tail ... that's why we consider only contests between the
> head of one group and the members of another (different) group. If the
> head is defeated by a member of its own group, then we essentially lock
> in a cycle.
>
> This doesn't need to be mentioned in the definition, but definitely
> included in the faq's, etc.
>
> The most natural place it comes up is in the recursive proof that every
> head (including the final one) has a beat path to each member of its group.
>
> BTW the fact that the final head has a beat path to every member of its
> group (i.e. to all of the other candidates) shows that the winner is in
> the Smith set.

Is there a way of extending River to producing a full social order in a
way that passes LIIA? If so, it'd seem to be strictly better than Ranked
Pairs because River also passes IPDA.

Well, perhaps River fails MAM's immunity to complaints criterion, but
that's rather specific.

-km
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