<div dir="auto">River is actually really easy to explain:</div><div dir="auto">1. List all pairwise matches from biggest to smallest margin of victory.</div><div dir="auto">2. If a candidate loses a match, cross them out (declare them to be a loser). Cross out any redundant matches that involve them (anything that would make them get eliminated twice).</div><div dir="auto">3. Cross out any elections that would create a cycle.</div><div dir="auto"><br></div><div dir="auto"><br></div><div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Mar 9, 2024 at 2:46 PM Michael Ossipoff <<a href="mailto:email9648742@gmail.com">email9648742@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)"><div dir="auto">Wow. River doesn’t need the exhaustive pairwise-count? How does its count-time compare to that of RP?</div><div dir="auto"><br></div><div dir="auto">I didn’t know that about River. I believed that only Sequential-Pairwise was the only exception to the need for the exhaustive pairwise-count.</div><div dir="auto"><br></div><div dir="auto">The exhaustive count requires, per voter, counting one pairwise-vote for each possible pair of candidates.</div><div dir="auto"><br></div><div dir="auto">How many votes need to be counted per voter in River?</div><div dir="auto"><br></div><div dir="auto">If one only cares about finding the winner, rather than an output-ranking, could the count-instruction be written more briefly?</div><div dir="auto"><br></div><div dir="auto">As written, it’s much too complicated for a public-proposal.</div><div dir="auto"><br></div><div dir="auto">Someone said that River is better at deterring burial. I disagree. It seems to me that skipping a defeat if its defeated is defeated in an already-kept defeat undermines autodeterence.</div><div dir="auto"><br></div><div dir="auto">Only one of the CW’s defeats is kept. That means that every Bus but one can’t have its defeat dropped, so only one Bus survives.</div><div dir="auto"><br></div><div dir="auto">I like it if the exhaustive pairwise-count isn’t needed, but can the count-instructions be written more briefly, if only the winner is needed?</div><div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Mar 9, 2024 at 07:02 Kevin Venzke <<a href="mailto:stepjak@yahoo.fr" target="_blank">stepjak@yahoo.fr</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)">Hi Mike and everyone,<br>
<br>
First off, if anyone was missing my site, it is back up. I had to find different<br>
hosting (a bit abruptly).<br>
<br>
Where I was trying to link right before the site went down:<br>
<a href="http://votingmethods.net/cond" rel="noreferrer" target="_blank">votingmethods.net/cond</a><br>
works out a given (or random) scenario for Schulze, RP, or River. You just have to<br>
expand sections at the bottom of the result. So it could be worth a look.<br>
<br>
Mike wrote:<br>
> Is River as easy to define &. explain as RP?.<br>
<br>
I see I should try to write out clearly how I suggest to understand River.<br>
<br>
There is no "final ranking" in River. Instead every candidate begins "below no one"<br>
or "subordinated to no one." This is sort of a ranking but the "trees" we make go<br>
only one level down: you will never be able to ascend two positions from a given<br>
candidate.<br>
<br>
1. Initially each candidate is subordinated to no one.<br>
2. Consider each pairwise defeat from strongest to weakest.<br>
3. When you consider a defeat, ask whether the loser is subordinated to anyone?<br>
If so: Ignore the defeat and proceed to the next.<br>
If not, then ask:<br>
4. Is the defeat winner subordinated to the defeat loser? If so, ignore the defeat<br>
and go to the next.<br>
5. Is the defeat winner subordinated to someone else? If so, the defeat loser, along<br>
with everyone subordinated to them, becomes subordinated to the candidate that the<br>
defeat winner is subordinated to.<br>
6. Otherwise it must be that the defeat winner is subordinated to no one. So here<br>
the defeat loser, along with everyone subordinated to them, becomes subordinated to<br>
the defeat winner.<br>
7. End loop. Go to the next defeat.<br>
8. In the end, the candidates subordinated to no one are the winners.<br>
<br>
Alternatively instead of talking about subordination, you can say that each<br>
candidate has their own "bin" and starts in their own and may move to another.<br>
This would allow you to merge steps 4 and 5:<br>
"4. The defeat loser, along with everyone *in the loser's bin*, moves to whichever<br>
bin the defeat winner is currently located in."<br>
And if the latter bin happens to be the loser's bin, in effect nothing happens. We<br>
don't need a rule saying to ignore the defeat, because the bin movement doesn't<br>
change anything either.<br>
<br>
I can understand if a reader eyeballs all that and says this looks like a mess and<br>
it's not clearer than RP.<br>
<br>
But hear me out on the *ease* of it:<br>
<br>
1. If you are programming River, you never actually check for a cycle, whether a<br>
proposed defeat would create one. And comparing to Schulze, you never trace a<br>
beatpath or find its strength, or (by its other algorithm) have to find the Schwartz<br>
set repeatedly.<br>
2. If you are solving it by hand, it would be enough to have a fridge magnet for<br>
each candidate, start them out in imaginary bins, and push the magnets around in a<br>
straightforward way to track who is subordinated to whom.<br>
<br>
It may be possible to define RP more concisely, but it takes some work to figure out<br>
what it is actually saying to do to solve it.<br>
<br>
Hopefully the above explains it better than I have before.<br>
<br>
Kevin<br>
<a href="http://votingmethods.net" rel="noreferrer" target="_blank">votingmethods.net</a><br>
</blockquote></div></div>
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