@hidden@grips Just do what I do and put fried eggs in a tortilla! Rn I’m having 3 easy over eggs and a slice of pepper jack in a ... um .... err.... whole wheat tortilla :reimufacepalm:
@Moon the fact that they can’t acknowledge their own intern without having gibberish wrapping onto another line should tell them everything they need to know about user retention
Suppose that there are N instances. Steve Jail creates alt accounts while the American Society for the Prevention of Cruelty to Animals (ASPCA) purges his alts. They take turns as follows: - First, Steve Jail creates exactly two alt accounts anywhere he likes. (They may be on the same instance or on different instances.) - Second, the ASPCA chooses *one* instance and purges *all* of Steve's alts from that instance.
By definition, an instance reaches the *boiling point* if Steve Jail manages to have M alts on that instance existing at the same time.
Steve Jail wants to ensure that *some* instance reaches the boiling point, while the ASPCA wants to prevent it. Who wins? (The answer depends on N and M.)
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Example:
Suppose there are four instances (N = 4) and the boiling point is M = 3. Initially, Steve Jail's alt count is (0, 0, 0, 0).
Steve Jail creates alts on instances 1 and 3. Now the alt count is (1, 0, 1, 0).
The ASPCA purges instance 3. Now the alt count is (1, 0, 0, 0).
Steve Jail creates alts on instances 1 and 4. Now the alt count is (2, 0, 0, 1).
The ASPCA purges instance 1. Now the alt count is (0, 0, 0, 1).
Steve Jail creates two alts on instance 4. Now the alt count is (0, 0, 0, 3), so instance 4 has reached the boiling point. Steve Jail wins but his dog loses.
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Extra credit: How does the answer change if, on each turn, Steve Jail can create three alts?
@Arcana@lain nonono I meant I agree with Lain’s sentiment which I believe agrees with yours. Any government, sufficiently large, collapses into a black hole from which the money supply cannot escape no matter its velocity. I was just quibbling with Lain’s statement that nobody can have a taxable income of 1B.
@allison@sun_eater@mrsaturday@meso Capital G and I are pretty funny, but my biggest complaint is that the lowercase "r" looks nothing like it. The older system (Palmer Method) on which D'Nealian was based doesn't have this problem...
I decided to put them on a line because it makes the analysis more uniform. (Starting from a ring, Eliza would destroy any instance and then it becomes a line anyways.) But CMD and udongle's simpler solution works just as well for a ring. All that is needed is that Eliza can totally disconnect the network after destroying 50 instances.
@hidden@MercurialBlack@ceo_of_monoeye_dating@jeffcliff@roboneko@scenesbycolleen I just played through it. One of the best interactive math explanations I've seen, *despite* its quadruple-vaxxed tone. I actually learned something new because I *incorrectly* guessed that "always cheat" would win because they can exploit "always cooperate."
It slightly annoys me that they changed all the standard terminology. For example, "copycat" is usually called "tit for tat," and "copykitten" is "tit for two tats." I find the phrase "tit for tat" to be funnier too.
Here's a funny bit of lore about the prisoner's dilemma tournament which the book is based on. It was a real public tournament which anybody could enter their own bot into. Lots of professors submitted incredibly sophisticated strategies, and they were all BTFO by the humble "copycat" which was the winner.
There's a little-known way to gank prisoner's dilemma tournaments. What you do is submit one "main" bot and hundreds of "feeder" bots. Your bots start each round by executing a secret "handshake" move sequence; if they recognize each other as both being your bot, then the "feeder" will purposely let the "main" bot exploit it. If the opponent is not recognized as one of your own bots, then your bot will just play "copycat." As long as you can enter enough "feeder" bots into the tournament, your "main" bot will be guaranteed to win. The implication for cult behavior is clear.
Lastly, there's an error in one of the explanations (see pic). None of the games considered in this slideshow are "zero-sum." Zero sum would mean that the payoffs in every case are exactly opposite: if you get +X, then I get -X. Since the "cooperate" case gives positive payout to both players, this game is not zero sum.