Yeah, that's a good point. I certainly don't claim that Michael is to blame for her actions.
the girl in question has publicly declared some of the psychological techniques she uses on people in order to induce altered states to be downstream of michael
Yeah I was initially going to dispute it and then I thought some more and realized it was probably correct.
...iirc you had LSD like a week or so before you had the cannabis? And you took the cannabis while fairly sleep deprived. And I definitely started getting worried about your mental state after the LSD, so even if you consider the psychotic break as starting a few days after taking cannabis I definitely think the psychedelics were a compounding factor.
Sapph is referring to @AprilSR (I'm involved in the situation, she's also commented down below confirming it to be her)
....is the second person me? You can say it is if it's me, I don't think it's an inaccurate description. Edit: thought about it a bit more and yeah it is probably me
I'm familiar with the events that Sapph refers to, and for the most part agree with the general description of them as well as the recommendations. If you don't want to become psychotic, don't do the things that are famously associated with becoming psychotic.
Well, perhaps, but due to global commerce I can just go to the store and buy a bar of soap much more easily.
And perhaps you are fond of that particular type of soap and it's a bit harder to find the specific type that you're looking for but it's still not really worth saving the old bathwater for it, instead of just looking for that specific type of soap?
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.
Wow, I came here fully expecting this post to have been downvoted to oblivion, and then realized this was not reddit and the community would not collectively downvote your post as a joke