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Everything posted by harey

  1. It is not because of the gender. the Claytons indicates that Mrs Clayton is still alive and Jasper's mom is deceased. Therefore, Jasper is not a Clayton.
  2. In the same way, you exclude that Jasper is a Clayton: 2. In deference to an influential family member, the Claytons agreed that if they ever had a daughter they would name her Janice.
  3. There are 8 boxes, not 4. And there is another possible distribution you have to exclude.
  4. The way how the marbles were selected is not known, so you cannot do better then Bonanova. However, instead of grabbing the calculator:
  5. Tried to send you a private message, got an error. Tried admin@brainden, got suggestion you box is full.

  6. harey


    A big step forward. Now that we found the result, it remains to find the way to find the result.
  7. @rocdocmac Pas du tout. Mais c'est ma langue maternelle, je sais même écrire.
  8. I see now: plainglazed's formula is the simplification of my formula I was so hard looking for. On paper, I got a kind of unreadable proof, so I fed them into my computer. Up to n=30, no difference with 6 decimals.
  9. harey


    Nice try, but they do not share and they survive all. Hint:
  10. Though they are similar, I see at least one huge difference. In the traffic jam puzzle, you cut ANY part. In the marble problem, you cut the REMAINING part. Enough for different formulae. Remains me of http://brainden.com/forum/topic/18168-squirrel/
  11. harey


    A large colony of squirrels dug holes during the summer and in each hole, they put between 1 and 100 nuts (each quantity has the same probability). If each squirrel has to eat 100 nuts during the winter, how many holes must he find (in average)? Each hole contains 50.5 in average, so 2 should be enough, right?
  12. I see, I found the solution for 8 generals, confused by range() in Python. The distribution for 7 generals:
  13. The hard part is to generate the lock numbers from the generals, the formulas are quite complicated: There is a way to cheat: Both programs give the same list: The distribution: The solution sure is not unique, various combinations are possible. Not talking about the possibility that the lock 1 has to be unlocked to unlock locks 2 and 3.
  14. Hint 0: Hint 1: Hint 1.1: Hint 2: Hint 2.1:
  15. harey

    Dumb robot

    Congratulations, works. I checked it: I suspect a kind of recursivity, but it probably will not show up for small n. For n=2 and n=3, the robot stays where it should, which does not seem evident for larger n.
  16. harey

    Dumb robot

    I do not really understand what you mean by "simple path", but I think the answer is no. The figure H is OK, but you can very well have: * * * * * c * * * c * c c * * * c * * * * * * * * c * ... Your list must work for all these cases, as well as for all grids where there is/are one/three/four cement blocks. i.e. the solution Molly Mae proposed would not work for the first maze (so it does not matter anymore that it works for the 2nd and 3rd). Even calculating the number of possibilities for 2 blocks gives me headaches. 7 * 6 / 2? Wrong, the robot must be free to leave the corner and the lower right corner must remain accessible. To get insane...
  17. harey

    Dumb robot

    An interesting variant. The problem states "there is a path", but it does not state "there is a path for the robot". I think it is easier we stay with moving just to the next square.
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