The comments offering logical reasons to let the AI out really just makes me think that maybe keeping the AI in a box in the first place is a bad idea since we’re no longer starting from the assumption that letting the AI out is an unequivocally bad thing.
ToasterLightning
“If it is true that you would really do anything to see them perform, that implies that the performance is worth at least +100 utility to you, to make up for the loss of missing the essay. Therefore, I will allow you to turn it in, but only for 75% credit, disincentivizing lying about your true preferences but still preserving most of the mutual utility.”
I’m guessing they would have happily accepted a bet at 20:1 odds that my driver’s license would say “Mark Xu” on it.
Personally, I wouldn’t do it at 20:1 odds even if you said your name was “John Smith,” purely because of how many people go by a name different than the one on their driver’s license.
I’m just working my way through these problems in sequence.
1 is not particularly difficult to solve
Let’s imagine the base case: B-G. Obviously, there is 1 biochromatic edge. Adding either B or G to a biochromatic edge will turn it into B-B-G or B-G-G respectively, which means there is still 1 bichromatic edge.
If you add B to a B-B or G to a G-G it turns into B-B-B or G-G-G, which does not add or destroy any bichromatic edges.
The final case is adding G to B-B or B to G-G, which makes either B-G-B or G-B-G, adding two bichromatic edges. Since adding two to an odd number results in an odd number, and we begin with 1 bichromatic edge, we always have an odd number of edges.
For a formal proof, we’d have to prove the unspoken assumption that we can make any finite linear path made up of Blue/Green nodes where the start is a Blue node and the end is a Green node, by adding Blue/Green nodes in between a B-G path.
The proof is as follows: Besides the start and end nodes, every node has two connections. Thus, we can remove a node and connect its two adjacent nodes to each other in its place. Removing a node this way does not make it no longer a qualifying path under our definitions, and the removal of a node can be undone by adding it back in between the two nodes. Thus, since we can remove all the nodes until we’re left with a single B-G path, we can add them back until we’ve reached the original path, while still ensuring that there is an odd number of bichromatic edges.
I mean, to be completely fair, you can’t exactly phrase distracting you from your work as a good thing. Perhaps a less distracting lottery would be better?