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This appears to be Elian Script. I'm not sure I can read the writer's handwriting entirely (and they seem to have added some nonstandard things like a zigzag for T), but the first few lines read: PAR?NRE RENR ?HE C PROGRAMMING LANGUAGE! I REALLY WANT MY PEN! Edit by OP: I spend some time to fully decipher it but I think Deusovi deserves the ...

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I'm not entirely sure this is exactly a puzzle (but also not sure enough to suggest closing the question or anything). Anyway, I guess the reason is that

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Was it Reasoning: and and

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Here's my guess Take the first 100 digits of pi: 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 STEP 1: Based on whether a digit is odd and even, convert it to AB format. Result: ABAAA BBAAA BAAAA BABBB BBBAA BABAA ABBBB BAAAA BAAAA AAAAB ABBBA ABABB AABAB ABABB BBBBB BBBAA BBBBB ...

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Perhaps

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I think the history is something like this. Start with 8 colours because 8 is a small power of 2. These are simple: bit 0 for red, bit 1 for green, bit 2 for blue, but 000 means grey rather than black because you're only controlling the foreground and not the background. Then add another 8 colours. Actually, in the image here it looks as if 8-15 may be the ...

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Did you eat Because

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A couple of possibilities come to mind: 1: A trusted third party. Have someone else toss the coin. 2: "Geohashing" type approach: toss the coin by unpredictable means that everyone can independently observe, even over the internet. 3: Crowd-seeded pseudorandom number generator: have each participant send you a number. Add them up, and seed a PRNG with the ...

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I was able to find one of the solutions using "paper and pencil". I stopped searching for more solutions after that. It is certainly very very time consuming. In my explanation I name rows as A,B,C,D,E - columns as 1, 2, 3, 4, 5. My search strategy is based on an observation that As you can see this gave me only 4 possible values for $B1$ Page 2 ...

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Not sure if I'm going at this the right way, but Because

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Looks like this is It says

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Looking at this: CoinBrothers it is rarely true. For example, Australia 2019: https://coin-brothers.com/catalog/coin3771

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To a real-life problem I had to give a real-life answer: But you asked for an actual tiling, without gaps, so here it is. PS: there is a simpler pattern where pairs disassemble with a single translation:

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Agree on a future public event that all parties can observe, and a means to generate a bitstream from it. Perhaps the parity of the last digit of the Dow Jones Industrial Average at a predetermined set of times? (I was going to suggest daily high temperatures from a given source for a list of N cities, but you'd probably end up with much higher correlation ...

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This is just for the teaser question (do not know the answer for the real one). I think you can pull the puzzle apart by

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Counting the first male as Generation 0 (and his mother as Generation 1) the male will have

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I think the messages read: The code is ... How did I decode it?

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Partial answer: Assumption: Assumption #2: Also, here's the transctipted text from both (don't think there's any need to hide it as a spoiler): Left (since it's handwritten I may have misinterpreted certain characters): 659CK4MG3659XTG39C - MG/AG/GMAH B5659XTV8CKXT6599CV8CKW2G3CK W2A34M4MA3 MSA3659G3CK CK K9CK9CK9 9CX16599C854A3 9C G3A3P2A34M F78544M, ...

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Note that It turns out that For example, Indeed, The answer is obviously

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Although I was too lazy to get the angles exact, the idea should hold in principle if the picture isn't quite right. They interlock.

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The answer is: Have a look at the following diagram: If you: Here is one possible roster:

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You should The intuition behind this decision: Calculation Generalization calculation

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EDIT: Just to let you know before you read this, this actually isn't the solution at all. I've made lots of assumptions and serious mistakes! :) The way to think about this problem is My algorithm is really simple: Notice that This algorithm is generalizable From this it's easy to see that And that's it.

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Very partial solution Suppose there are an even number of people: say 2n. (Call them A1..An and B1..Bn.) Then as per Brandon_J's conjecture However, We can get a lower bound on the number of meeting periods needed from Here's a construction that's not too bad, though in general it's far from optimal. [EDITED to add:] No, wait, I'm not sure $m-1$ is ...

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Here is a good solution for 8 people: I don't have a general solution for larger numbers yet.

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Disclaimer: This is a proof of p_sutherland's answer. If you consider this correct, please accept their answer. Spoiler-splitter Spoiler-splitter Spoiler-splitter

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[EDIT: After this was answered, the puzzle was restricted to $8$ players. In this case, $8+3+3-3=11$ games are required in the worst case.] The following method requires up to games. Suppose $n=2^k$. First, Then, Finally, In total, at most games are played. Suppose we add Alice so that $n=2^k+1$. Modifying the above method:

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How about the Any two adjacent pieces can be slid apart, but I don't think the whole thing can be split by sliding from any direction.

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Florian F's 2nd pattern is far and away my favorite, but if anyone was curious, I'll post my answers. First I wanted to show an example of something that doesn't work but really seems like it should: It's just like the third example from the question, but it uses joinery such that the pieces slide together at an angle. It comes apart in the same way, ...

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