CaptainEd
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If you plot these as locations in a 4x4 grid, with rows labelled 1-4 and columns labelled a-d, the sequence Wolfgang has produced is symmetric about the center of the grid. That is, if you view the two characters as RC (row, column), and number the sequence items from 0 to 15, C(i) = 4-C(15-i), R(i) = 4-R(15-i) I think it's interesting that his sequence completely fills the 4x4 grid and extends no farther. I think it's interesting that his sequence exhibits this symmetry. I have NO CLUE how this relates to the location of 34. Sorry...
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Interesting thought, Bonanova, Since a non-edge square can be traversed a maximum of 2 units (all four diagonal legs), and the entry and exit to the grid take one unit each, an upper bound to the maximum must be 2 * 64 + 2 = 130. Probably the bound can be tightened by limiting it by the number of reflective faces that can be on the outside...
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Akram Abu, I'm having difficulty with the word "slice". Here are some questions that may help me: (1) Assumption: all the cutting has to be done on the grid lines, is that true? (2) Assumption: a cut can "turn a corner"; that is, the cut can go North for a while, then turn East, then turn North again, then possibly West, etc. Is that true? (3) Question: can we separate this piece into more than 2 pieces, and then reassemble them to form the two identical/symmetric shapes?
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Another question: if a square's diagonal length is 1 unit, then I guess that means a mirror square is significantly smaller than a grid square. Is that true, or did I misinterpret this statement? If so, then we have to fastidiously clear about exactly where inside a grid square we are placing a mirror square, as Princecharmthings points out.
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Actually, I think it's better. it means that, to evaluate a number's divisibility by 23, you can separate a long number string by cutting it every 22 digits, starting at the right end, and add the residues mod 23.
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Here's something that might help a little tiny bit. 10^22 = 1 (mod 23) That means for any number x =10^22 * y + z x (mod 23) = 10^22 * Y (mod 23) + z (mod 23) since we can replace 10^22 with 1, this means x (mod 23) = (y + z) (mod 23) <...edited to remove false conclusion about removing 22 consecutive zeroes...>
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And thank you, Sword. It was indeed an interesting challenge. As you see, it took us a bunch of tries to figure it out. I'd really like to know how Prof Templeton found his schedule. I can't keep track of 20 items myself...
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Yes, but I've simplified the even cases above. Also moved the fixed point to the SouthWest side.
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I've got a schedule that wastes one sitting, but is totally easy to create, for any N people. But right this minute, I can't create a spoiler...I'll keep trying.
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No, it's not that easy. The first half of my algorithm correctly matches each odd person with each even person. The second half matches people with some they've already seen and therefore missing some as well. Back to the drawing board.
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Thank you for the interesting problem--it seemed like we didn't know enough, but with the right sort of assumption, it became tractable.
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If we assume a (different) uniform deceleration as well, it is pretty heavily constrained. My first attempts at initial accelerations were way too high--if we decelerated uniformly, we traveled more than the allotted distance.
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I'm with curr3nt--it appears we don't know enough for a unique answer. Since it takes less time to decelerate than it did to accelerate, it appears we have to discover at least two values, the uniform acceleration and the separate uniform (or otherwise) deceleration. It feels like we have too much latitude (or, another way of stating it, too few constraints.
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Curr3nt, thanks for the hint (I was beginning to feel very dumb) AND thanks for the pointer to the new (to me) technology called "tineye". THere's so much to learn here in the Den...
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What a cute problem! Thank you, Superprismatic!
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That must have been a very scenic elevator, or you have a LOT of patience!
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What's funny to me is that the two different pictures with tables result in two different arrangements (clockwise around the table), but the same answer!
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Spoke too soon, indeed! Thanks. But look, Molly Mae believes P1 has the win after P1: 35 P2: 56 P1: 76 I wonder if he's right...
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4x5, yielding 5 2 6 1 3 4 7 8 looks like P2 wins in two more moves each. If I follow the thread right, we're demonstrating that P1 has no good counter to P1: 35 P2: 56 1 2 6 4 3 5 7 8 isn't this a surprise! I sure wish I could state a human-oriented strategy, a way for P2 to create each move in response to P1's, without exhausting the game tree. What a counterintuitive belief!
