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A comment on the previous post has me rethinking or wanting to clarify some things.

I was comparing the input space to the output space; this wasn't meant as a direct measure of the work done, though that was probably unclear. Ranking 25 candidates *does* have an space of 25! possibilities, vs. a smaller output set. Making 9 votes *does* cover a space of 512 possible votes, with 512 possible election outcomes. But it's also true that voting STV is nothing like a linear search of 25! possibilities, and for the work we'd better look at the computations involved.

Closed party list: your decision is a linear O(N) search through the number of parties, finding the maximum. The physical act of voting is probably trivial. I was going to say that proportionality can increase freely for the voter, but that's not true: lowering the threshold to get in means you can have smaller and thus more viable parties. 12% means no more than 8 parties can get in, 6.25% means 16, 3% means 32, 1% means 100 parties could get in. (In reality some will be big and taking up much of the vote, but still.) Whatever the threshold, though, learning of a new party is constant time: you compare them to your current favorite. Proxy voting would be exactly the same, with fewer warm bodies in seats.

Open party list: similar, except now N is the number of candidates. If proportionality is high, there may be lots of candidates, and physically voting might mean a log(N) search to find yours -- or linear, if the ballots are randomly unordered. A huge district with 100 members could have very fine-grained proportionality but also mean each party running up to 100 candidates -- big ballot. Still, pretty simple to do and understand.

STV: ranking N candidates is a sorting problem, O(N log(N)) in the ideal case, though possibly O(N^2) in practical naive sorting. Learning of a new candidate means comparing them to on average half the other candidates if you're simple, or to log(N) of them if you're clever. The physical act of voting... well, depends on the machine probably; Cambridge has you filling out a wide array of scantron bubbles, and I've needed a second ballot in both elections due to messing up the first one. I'm sure there are better ways.

Re-weighted score voting (RSV): O(N), you go down the list of candidates and rate each one. A new candidate simply means rating that one. Much simpler, cognitively and physically.

Referendums: lj:notthebuddha pointed out a twist. Naively, voting on N proposed laws is simply O(N), like score voting: go down the list voting up or down. 9 laws would mean 512 possibilities, but only 9 decision points. New law, new decision point. But it's possible for proposals to interact, so that in a worst case you are having to consider all the different possibilities, with exponential explosion: 10 laws meaning 1024 possibilities!

In practice, they don't interact that much; even more important, you don't get to vote on that many items at once, and pruning is enforced by time and temporal ordering. The Swiss vote on 3-4 referendums at a time. In a high-frequency legislature, you vote on one law at a time. US state ballots I don't know; voting every 2-4 years may allow them to pile up, vs. the Swiss every 3 months.

On the flip side, as I said before, here any increase in work is matched by an increase in power over the outcomes, whereas it's unclear that the higher workload of STV compared to proportional score voting has any benefit whatsoever, and the benefit of either of those compared to open party list depends on how much your grouping of the candidates cuts across the parties they've grouped themselves into. (That is, open party list means your vote for a successful candidate can spillover into another party member, based on the party list; STV/RSV lets you spillover to an unrelated candidate of your choice. And Cambridge elections are ostensibly non-partisan.)


Also on the information theory front: picking among from say 8 candidates or parties means expressing 3 bits of choice, every 2 to 6 years based on standard practice these days. 32 candidates, 5 bits. An American would be very lucky to have 8 choices, say if both main parties were running 4 candidates in their primaries. (Though California now has a top-two "open primary" system which can mean lots of choices up front... I think this is a terrible system, but another time for that.) Commonly we have like 1 bit: incumbent or some obscure challenger, so it's basically "keep or toss?" Bit rate from 3/2 years (8 candidates, House) to 1/6 year (incumbency, Senate).

By contrast, 9 referendums a year means 9 bits of voter input a year. The Swiss actually seem to be average between 12 and 16. Plus, any law or treaty could be subjected to referendum, and anything could be an initiative, so so there's some harder to measure aura of voter input as I imagine the legislature tries to avoid anything obviously unpopular, while Congress could do lots of unpopular (or not do lots of popular) things as long as those weren't more important than key issues of crime and the economy.

See the comment count unavailable DW comments at http://mindstalk.dreamwidth.org/394019.html#comments


Damien Sullivan

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