BB(n) isn't computable, even given unlimited computational resources.
Running all n-state machines won't work, because you may have some machines that continue indefinitely, but without repeating. (Remember that while the number of states is finite, the tape is infinite.) No matter how long you run them for, you can't be sure whether they are going to terminate at some point in the future, or if they'll continue forever without halting.
This is why computing BB(n) for arbitrary n is equivalent to solving the halting problem.
Princeton's COS 226 covered this topic quite well when I took it. It was a good intro to both algorithms and DS content, as well as practical skills for leetcode style questions.
You can check out the Syllabus for the course as taught at Princeton here[0], or take the Algorithms I and II courses on Coursera[1] which cover ~2/3 of the content but have video lectures.
The point of the tax is not that the revenue will be used to reduce waste. The point is that it will increase the cost of producing waste for the polluter, ideally exactly to the point where it matches the cost to society of that waste.
The purpose isn't to make people stop polluting entirely, but to make it so they will only do so when it brings them a benefit larger than the societal cost of the pollution.
Not married, but in a long-term relationship living with my partner. I listen to them while commuting to work or while doing household chores, and consistently get through 4+ hours/week of podcasts.
I don't find my listening habits change that much when I'm alone, as either way I tend to only listen to them while I'm engaged in some other activity. (When I have completely free time I prefer more interactive activities like games or music.)
Running all n-state machines won't work, because you may have some machines that continue indefinitely, but without repeating. (Remember that while the number of states is finite, the tape is infinite.) No matter how long you run them for, you can't be sure whether they are going to terminate at some point in the future, or if they'll continue forever without halting.
This is why computing BB(n) for arbitrary n is equivalent to solving the halting problem.