Klye @khird@qoto.org
Mod
Joined Sep 2018
@freemo I don't think three flips solves six boxes. Your patterns are 1-2-3, 1-2-5, 1-3-5, 2-3-5, unless I'm forgetting one. If box four doesn't match box six you're out of luck.
@freemo jumping right to the general case!
I think it's going to be exponential difficulty in at least flips-per-round, and probably also in boxes.
My algorithm is this:
Go through all known cases: if I were told the state of the boxes, for each pattern I could choose, what possible combinations would I find?
For each state-choice-combination triplet, and each possible combination I could set the coins to, what state (or superposition of states) could I leave it in?
Go through the unknown states the same way, by superposing the known ones, building all the way up to completely unknown (all states possible).
Make the decision tree by minmaxing, like in two-player game theory. If the starting state of "completely unknown" has a path to either all-heads or all-tails, count the steps in the path. If there isn't a path linking the starting state to the winning states, it's unsolvable.
@freemo only opening two boxes per turn, you can't do five or six. I think there might be a solution if you allow the player to open three per turn, at least for five.
For five boxes, choosing two: whenever you choose adjacent, suppose you choose boxes one and two; whenever you choose skip-one, suppose you choose one and three. Box four contains a heads, and box five contains a tails. You'll never be able to get all five the same.
For six, choosing two: suppose your adjacent choices are one and two, your skip-one are one and three, and your opposite are one and four. You'll never be able to get five to match six.
@freemo exactly!
For similar reasons, even under the original instant-win rules, a version with five boxes is unsolvable. No matter how many times you open any arrangement of two boxes, fate can always allow there to be two boxes that don't match and you never see.
I think five boxes, opening three per turn, should work, but I haven't figured out what the target number of moves should be.
@freemo proof of impossibility could be a solution to the puzzle 😎
@freemo unless you can logically deduce from the the coins you've seen that it must be solved or must not be solved, you cannot make the assumption in either direction.
@freemo I don't believe you can guarantee a win in five without the instant-win condition in the original problem.
@freemo No. If you know they're all the same, you can just leave them for the remaining turns. Or if you got them all the same without realising it, you can flip them out, flip them back, whatever you like until your N turns are up. I'm just removing the assumption that the fact the game hasn't ended implies all four are not the same.
@freemo Let's extend the riddle by saying you have to set the coins the same after exactly N moves. You're allowed to have all four coins the same before then, but you won't win unless they're still the same (or again the same) at exactly N moves.
What's the smallest N for which there's a guaranteed winning strategy?
Solution
@freemo Here's the strategy I'd follow, compliant with the hard rules.
1st spin: Open two boxes diagonally opposite one another. Set both coins to heads.
2nd spin: Open two adjacent boxes. Set both coins to heads.
3rd spin: Open two diagonally opposite boxes. If one coin is tails, set it to heads and you win. If both coins are heads, set exactly one to tails and leave the other one heads.
4th spin: Open two adjacent boxes and flip both coins. If they were the same, you win.
5th spin: Open two diagonally opposite boxes. Flip both coins. You win.
Reasoning:
The first two moves guarantee that at least three of four coins are heads. If all four coins are heads, you've won, so we only have to worry about the case where exactly three are heads.
The third move either turns over the last remaining tails, in which case you win, or it leaves you with two adjacent heads and two adjacent tails.
The fourth move either turns over all the remaining heads, all the remaining tails, or one of each. In the first and second cases, you win; in the third case you are left with alternating heads and tails where you formerly had adjacent heads and adjacent tails.
The fifth move turns over all the remaining heads or all the remaining tails, depending on whether you happen to open the odds or the evens, so you win.
@lefarfadet it says "randomly" which i took to mean it can stay the same or turn 90 degrees clockwise, 90 degrees anticlockwise, or 180 degrees around, but you don't know because you were blindfolded while it was spinning.
@diresock sound, nothing. Have you seen how they *look* beneath their disguises?
https://www.boredpanda.com/blog/wp-content/uploads/2017/01/owls-without-feathers-fb-png__700.jpg
@valleyforge What happens when the management no longer thinks the company's viable, or the proprietor wants to retire, etc.? Selling to a competitor allows the business's customers to keep receiving whatever service it had been providing.
For example, in aviation: when US Airways bought out the bankrupt American Airlines, it inherited the obligations to transport AA passengers who already bought tickets. When Thomas Cook went bankrupt without a buyer, hundreds of thousands of passengers just lost their money (technically, they held unsecured debt for whatever refund they were due, but that was effectively worthless, given that the airports secured their debt by impounding TC's assets), and many were stranded abroad.
@mandlebro In the context of the specific problem (mixture of two Gaussian distributions) he's discussing, the likelihood is nonzero. He's set only μ̂₁=yₓ, σ̂₁=0, but μ̂₂ and σ̂₂ can be anything; so he's effectively mixing a finite Gaussian with a Dirac delta function as an edge case. If you say Δₓ=0 and all other Δ=1, you're using the Dirac delta to explain yₓ and the finite Gaussian to explain all other y. The Dirac delta doesn't have to explain the other points, so the fact that it'd be zero at those values of y doesn't force the likelihood to zero.
That said, I don't see how he concludes this gives *infinite* likelihood (or even necessarily the maximum). It seems to me it just collapses to the likelihood of the finite Gaussian - one point goes to certainty (i.e. 1) and the others go to their likelihood under the finite Gaussian, which is between zero and one. But this isn't the sort of math I do much of, and intuition isn't always reliable when terms are going to infinity.
Hopefully my reasoning makes sense as to why the likelihood isn't zero in general. If it becomes clear to you why it goes to infinity and/or represents a maximum, please share your insights - you've got me curious! Alternatively, if you disagree with my reasoning, and you still think it should be zero, I'm happy to reconsider my position on it.
@realcaseyrollins They don't usually call it converting unless it involves a baptism, so Christians of other denominations are simply "received" into the church, but yeah it happens. And the rules for baptism are pretty permissive, so most Christian denominations (with a few exceptions, notably the Mormons) are considered validly baptised.
Source: went to Catholic school as a non-Christian, they make sure you know how you could join up if you want
It would make sense, then, for replies to appear as comments, so let's test that hypothesis.
@freemo I blame mystery novels. The clever detective finds some clues, and announces he's got a theory but he needs more evidence to confirm it. This has led to the public using the word "theory" for what scientists would call a "hypothesis" and thereby devaluing the term.
[Here](https://git.sr.ht/~vpzom/lotide/tree/master/item/openapi/openapi.json#L281)'s what they have - I don't see any human-readable documentation, and all the paths are marked "unstable", but it seems to be what there is for now.
Yes* - I just thought it would be a nice addition to the comparison for completeness's sake as a fourth example (which is why I numbered it as 4).
* actually, ß is from ſʒ, not ſs, but functionally that makes little difference
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