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  4. For every group of 6 people put on a lock that none of them have the key to. Then give the remaining 7 people each a key to the lock. This ensures no minority can open it and ensures that any group larger than 6 can open the chest. In total there are 13 choose 7 locks. And 7 * (13 choose 6) keys. The keys are distributed equally so each pirate (7 * (13 choose 6))/13 keys
  5. Bab

    Going to a Party

    Women and wine and wickedly wild Worcestershire! Rule: starts w “W”
  6. A robot is placed on the square A1 of a standard chessboard and has to reach H8. It understands orders Up, Down, Left, Right. On some squares, there might be a cement block; is such a case, the robot does not execute the order and continues with the next one on your list. There always is at least one possible path. The list of directions is finite. The robot might reach the destination somewhere in the middle of your list. Give the answer in the form UURRU. (This list will work i.e. on an empty 3x3 chessboard or with a single block on A3 but will fail with a single block on B3.)
  7. There are 144 distinct solutions times 8 rotations and reflections for a total of 1152 solutions for the 4x4 Magic Court Card Puzzle. Solution # 1: AS KH QD JC QC JD AH KS JH QS KC AD KD AC JS QH. Solutions # 2: AS KH QD JC JD QC KS AH KC AD JH QS QH JS AC KD. Plus there's 1150 more solutions that satisfies the conditions that each of the four card ranks and four card suits appears only once in each of the four rows, four columns, and the two diagonals. This also works if the cards came from a four color deck.
  8. phaze

    Tank Vandalism

    After two years of war the Russians still haven't found Zorro
  9. Yeah, that's not a very interesting definition of the state. Try the one I described. So if the angel and devil are approaching lantern 5, the state would be the state of lantern 5, followed by the state of lantern 6,..., ending with the state of lantern 4. This would be equivalent to cyclically bit shifting the binary number so the next lantern the angel and devil will visit is at the front. Say there are 5 lanterns: 1 is on, 2 is on, 3 is off, 4 is off, 5 is off. If the angel and devil are approaching lantern 4, the state is 00110. This is because looking forward from the pair's position they see off,off,on,on,off.
  10. OK, let's say that a repeated pattern of lit/unlit lanterns is a win for the devil, regardless of the position of the angel and devil. Let's also say that if the angel and devil make a complete circuit of the path without either of them changing any lanterns, the devil wins (otherwise the angel can just stall indefinitely without ever getting any closer to his goal). In this scenario, the solution is: To explain the winning strategy, I'll first point out the following: So, to win the game:
  11. My intention was to define repetition (Devil's win) as a repeated state of the lanterns and a repeated position of Devil and Angel, but looks like I failed to do it properly. My apologies.
  12. I had this solution in mind, unfortunately, it fails because of repeated situations. Starting situation: 6 lanterns, only lantern 6 is lit. First round: lantern 1: no action lantern 2: devil turns it on remaining lanterns: angel does not do anything (If the devil turns a lantern on before the angel has turned any off, then the angel should just do nothing on that circuit. ) Situation: Lanterns 2 and 6 are lit 2nd round: lantern 1: no action lantern 2: angel turns it off (on each circuit, turn off all lit lanterns until the devil turns an unlit lantern on) Now, only the lantern 6 is lit: the game ends if the lanterns return to a previously encountered state What am I missing?
  13. Evilhubert, your solution is exactly the solution I had in mind, congrats on solving! I'm happy that you found the puzzle enjoyable.
  14. There's a solution that I find simple and elegant enough to be satisfying In my explanation, I'm going to think of the lanterns as numbered. The first lantern our celestial beings come to after starting the game will be Lantern 0. The next will be Lantern 1, then L2, L3 and so on. The last lantern can be referred to as LT, with T = Total number of lanterns minus 1. Each time they walk round the lanterns and return to L0 will be referred to as a circuit. So who wins? So what's the winning stragegy? But how can we prove the strategy will always work for any number of lanterns? I just want to give a big thank you to witzar for posting the problem - I really enjoyed this one!
  15. actually this seems pretty straight forward to me, the angel simply turns off all lit lanterns and wins.
  16. Was there an easier method? Concave
  17. Good. Yeah, I overthink everything and can get rather paranoid.
  18. Here's an example from my code's 6 lantern output: 111110 <one lit lantern (interpret 0 as "on" and 1 as "off") 111101 <one lit lantern You suppose the devil turns the lantern 5 on. But what if he turns the lantern 4 and not 5 on? 111110 <one lit lantern (interpret 0 as "on" and 1 as "off") 111010 your turn P.S. I am not touchy (did not even think it could be interpreted this way), but I think things over. In the last time, I wrote a lot of nonsense, i.e. my first answer.
  19. And by "put the devil in his place", I didn't mean you @harey. I reread that and remembered you said "As I am a nice guy, I let you the role of the angel." So if you found that belittling, I apologize, as that was not my intent. I am sure you see things I don't and vice-versa.
  20. First, let's nail down the definition of configuration so we know exactly when the devil wins. It cannot simply be the state of the lanterns, since not changing a lantern on a given step would then repeat the configuration. So the location of the devil and angel need to be part of it. There are two ways I see of incorporating this. Include which lantern number the devil and angel are at and the string of lantern states. Which could be notated by a location marked in a string of lantern states (e.g., "01101;001") or a number and the string (6,01101001). The lantern numbers don't matter, just look forward from the lantern the angel and devil are at (e.g., ";00101101"). And then you don't need the location marker ";" because it is always at the front. Option 1 has more states, lanterns x 2^lanterns compared to just 2^lanterns. Let's choose option 2 since less states means more of a chance of repeat and gives the devil a better chance. That is the notation I used for my code. I notice that you use the word "turn" to sometimes mean one whole stroll around all the lanterns. That is something that threw me off a little. I'd call that a revolution or a full rotation. Alright, time to put the devil in his place.
  21. @EventHorizonMy post is unrelated to yours. The number of lamps is not important, just a little bit more that those I quote. "Let's start with one lantern lit, i.e. 33.": This is not a binary number, it is the 33rd lantern assuming the starting position is the 1st lantern (conveniently chosen), following the lantern 32 and followed by the lantern 34.
  22. I couldn't follow that, harey.
  23. recreating the results of the code...
  24. Hello everyone, I’ve created a little puzzle that follows the cryptographic principle of zero-knowledge proof. Let P = xx, the age of Peter To find xx, I will provide you with means to verify the statements of the puzzle, without giving you any direct informations about the ages of the characters. The ages of the characters are not given but can be found. Although there are an infinite number of answers that could verify the information I provide, there is one answer that can be verified to 99% assuming the puzzle is honest and verifiable, and that Peter has a realistic age and life. How old is Peter ? - Peter has 5 children, Matthew, Nancy, Phil, Quinlan and Ryan - Peter’s age is the sum of the ages of all of his children - The concatenation of his children’s ages forms a palindrom - Peter’s age is a semi-prime number - 2 of his children are the same age - One of his children is half the age of one of his older siblings - Quinlan is younger than Phil - Only two of his children have a job - At least 2 of his children have a palindrome age - Matthew can’t read - Peter didn’t have a child before the age of 30 - If x is the age of the child < 10, then we’ll write 0x, such that a 1 year-old child = 01
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