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Everything posted by plasmid

  1. I also come up with there not being a solution, although admittedly with a bit of hand waving toward the end. Without a computer (or at least not much): Edit: SMH I just realized that the stipulation of i being in the range from 1 to 2j doesn't hold for all cases (specifically I listed examples of i=3, j=1 that I overlooked) so I would need to rethink if there are more possibilities, or trudge through everything by hand.
  2. Giving it a shot at writing with different notation
  3. Having to actually write it out in such a way that others can read and follow it does wonders for imposing clarity...
  4. I think I've got a solution, but not everyone ends up being restricted to one exact age. Does that mean I messed up somewhere?
  5. Rats, I was going to say that ( Sqrt(7!) - 7^0 ) / 7 is close enough that no one's likely to notice the difference.
  6. I agree with harey. As was said earlier, you can rule out the Drummer from being the Truth teller because he said he tells truth and lies, and you can rule out the Piper as being the Truth teller because he says someone else always tells the truth and there's only one Truth teller. So there are three possible Truth tellers. (1) If the Jester is the Truth teller then the Piper must be the Liar, as Babysnoot said, but that's not enough to rule out other possible truth tellers. (2) If the Juggler is the Truth teller then you know the Drummer must be the Liar. Since the Drummer can't be the Truth teller, and the Truth teller said his statement that he tells both truths and lies is false, the only thing the Drummer could be is the Liar. This kind of works against Babysnoot's argument. (3) If the Bear is the Truth teller, then we know the Juggler isn't the Liar but I can't narrow it down more than that. I also considered the possibility that the statement that the other three "tell a mixture of truth and lies" means that there must be both some true statements and some false statements among the three, but that still doesn't help. If the Drummer and Juggler are mixes then the Drummer's statement is True and the Juggler's statement is False, so any solution excluding them as Truth teller and Liar would work. And for the Drummer = Liar / Juggler = Truth teller scenario, you would have the Jester's first statement be True and the Piper's statement be False, so that doesn't rule anything out.
  7. I'm not sure I agree with harey. While it's true that if a plane's power is lost it will drop more slowly than a ballistic due to the drag of air against the wings keeping it from accelerating too quickly, while it's flying straight and level and the net vertical velocity is zero I would imagine that it shouldn't have an impact. I'm also puzzled by the fact that a plane can stay aloft when the force of gravity is greater than the thrust of the engine. It probably has something to do with the fact that a plane can only stay aloft when the airflow across its wings is laminar, and that if the airflow becomes turbulent (for example if the pilot pulls the nose up too much and the plane goes into a stall) then it can no longer maintain enough lift and starts falling. But I don't know how to account for that with equations and account for balancing out the force of gravity.
  8. Building on Thalia's answer...
  9. The first thing that jumps out at me is that when I do Einstein puzzles by hand I usually need to do many iterations of going through the rules and filling out as much as I can, and then going through the rules again to narrow things down further based on what I’ve eliminated so far. In this case there’s a single loop of For cr = 1 to cntRules that gets executed once, as if a human were to just go through all the rules once instead of iteratively. If you as a human are able to solve the puzzle by going over each of the rules once in the order that they're presented, then the program should work. Otherwise maybe it would help if you wrap the for/next loop of cr in a loop similar to the one you currently have with bDoAgain checking for any changes since the last iteration before letting the program end. If you can go through this particular Einstein puzzle by hand and complete it in a single iteration (so the above isn't applicable), then could it be an issue with how the rules are encoded? I'm afraid I can't get the app working, but I’m used to seeing things like “The plumber lives in a house to the left of the guy who goes birdwatching on Thursdays” which aren’t so easily encodable with a plus/minus system.
  10. We'll have to see the algorithm and what it's doing in order to say much. Could you share it and walk through how it would handle an example Einstein puzzle that folks without the game could follow (say the one at https://web.stanford.edu/~laurik/fsmbook/examples/Einstein'sPuzzle.html)? If you solve the puzzle the old fashioned way and keep track of what you eliminate and why, is your algorithm able to recreate the path to the solution? If not, then what parts is it missing?
  11. Ah, I see. So in that spirit you could present your solution #2 this way (not particularly efficient, but crystal clear that it works)
  12. I'm afraid I might be missing something, because it seems like whenever you switch parity you end up losing all the ground you previously covered. Consider xp2008's solution... Similarly for EventHorizon's first solution You could argue that if you start off covering a finite area, then cover a slightly larger area, then cover a slightly larger area, etc. then you will eventually catch the groundhog because the groundhog must be at some finitely numbered hole which you will eventually reach. My counterargument to that is:
  13. Only if they're imaginary stonenibblers.
  14. Just a reference to something commonly associated with it...
  15. I'm pretty sure this isn't optimal, but it's a start.
  16. I agree completely with EventHorizon, and will try to summarize in a way that addresses the OP and deconvolutes the paradox at its heart:
  17. The part that I turned red in the quote bugs me a little bit.
  18. It looks like that will indeed work. And I know I just said this about the previous solution, but now I really don't think there's a way to do it with any fewer breadcrumbs. As for ever finding a general solution to the breadcrumbing problem,
  19. I would have to disagree with that since I just posted a solution with only one breadcrumb going from C to E. But I do agree that there probably isn't a better solution for the pentagon case than adding five new breadcrumbs.
  20. That's the best solution so far, and matches the number of breadcrumbs I had even though it seems substantially different. I strongly suspect that there aren't any better solutions with fewer breadcrumbs, although I don't know of a way of proving it, let alone finding a general solution for determining the fewest number of breadcrumbs it would take to "teach" a dumb but hungry mouse to travel any arbitrary path. And I sort of doubt that anyone has figured it out since this sort of puzzle isn't a common thing. But I'll go ahead and share my perspective on the problem and the solution I reached.
  21. That's correct. The mouse can take whatever path you want it to take, even if it's not straight lines from A to C to E to B to E, as long as it reaches the specified destinations in the specified order. And it took me more than a day or two, so I certainly wouldn't fault you if it takes more than that.
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