chalst: unfortunately the electronic voting systems are far less accurate than the paper-based ones they're replacing. Not only do they not have a paper trail, making any sort of double-checking impossible, but they run on un-audited proprietary software. The software has been known to crash in deployment. The solution? Reboot it.
There is one thing which could be done to increase the accuracy of vote counting and the difficulty of vote fraud. Instead of creating a single paper ballot, create two, which go into two different bins, and are sent to two different vote counting agencies, which are run by two different political organisations.
Sadly, the current trend is in the exact opposite direction, with paper trails getting eliminated entirely. They're replaced by unpublished electronic protocols which we're simply supposed to trust. There is a very real possibility that in the near future every major election in the united states will be rigged.
I've been thinking about how to apply ciphergoth's new trust metric ideas, discussed in my last few entries, with negative certs for spam resistance. The two match up very well. When a bucket is full, in addition to dribbling out more positive water, it also dribbles out some anti-water to all of its negative certs. Anti-water can filter backwards until it's negated an entire bucket then flow backwards from there. This approach needs more polish, but I like its game-resistance properties.
I figured out an in some ways better way to do simple peer ordering. For each new inclusion, figure which peer has the greatest total sum weight pointing at it, then include that peer and move weight of all the peers which certify it onto that one, setting the weight to 1/numcerts * peerweight for each certifier. Unfortunately this removes the disincentive for certifying peers by replacing it with a strong incentive for certifying peers, since the more peers you certify the less your strength gets diluted.
More thought is required. I've figured out one nice straightforward criterion though. If the seed certifies two peers, and the first one certs x peers and the second one x+y peers, then they should get to alternately appoint includes until they've both selected x, then the y extras should get added at the end.
It's impossible to position three solid tori in 3-space in such a way that they're borromean linked without deforming them. Is it possible to position a larger number of solid torii such that the entire set of them is linked but no pair is?
How can you find the midpoint of two points using a compass only, no straightedge?