bushindo
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Everything posted by bushindo
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I think it is important to remember that we are limited to the availability of english words and the corresponding idiosyncratic distribution of word length. For instance,
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I think this paradox is a lot deeper than it seems. The mathematical community, at least, doesn't seem to agree on a single answer to this problem.
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1 more time
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This seems fun.
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Some clarification please. In the bolded passage, what do you mean by "improve the value of the hand as much as possible"? Does it mean discard the hand so as to maximize the chance of drawing a better hand, and that all better hands count equally? E.g. if we have a pair, then an improvement to 2 pairs counts the same as an improvement to a royal flush? Does it mean discard some cards so that we maximize the expected 'value' of the new hand for some payoff scale? If so, what is that payoff scale?
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How about 21?
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The logic seems good to me. The first part of step III is particularly nice as it extracts maximum information out of the 3rd weighing. Great work, plainglazed!
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I now can do 1 better than that
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Minor improvement
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Mea culpa. The puzzle is flawed and wasn't completely thought through. I apologize. Congratulations to hettieann for the counter example.
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The bolding in the quote is mine. I assume by Rubik's cube you mean the specially-modified cube in my OP. The question in the OP is not whether such a cube is solvable, but rather why is it not solvable. In other words, I'm asking for a proof that no sequence of rotations or pivots will return the cube to the traditional solved state.
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Let's consider the following 4x4x4 Rubik's Cube, The above cube is in the solved state (each of the 6 faces has identical color). I now take some paint and paint over two squares so that the resulting cube now looks the image below, Notice that essentially we swapped the position of a red with that of a blue square. The top face and the three faces hidden from view are unchanged. Show that, using the normal pivoting and rotating operations of Rubik's cube, it is not possible to return the cube to the solved state like that of the first image.
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Here's the computation of the winning percentage. It isn't as elegant as I would like, so if there's a better way, I'd love to hear about it.
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That's a marvelous approach! Thanks for sharing this great puzzle and the elegant solution.
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Thanks for the interest, howardl1963. I describe my solution below
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Sorry for the late response. I didn't internet access for the last few days. Here is how I approached the problem,
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Here is a rough stab at the strategy,
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Here's a strategy to fully compensate both men
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Assuming that 'falling into the black hole' means crossing the event horizon,
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This is indeed a cool problem
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I'm intrigued with KlueMaster's solution with 49 cells as well. If KlueMaster could elaborate on it, that would be great.
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Let's say we have 343 biological cells stacked into a 7x7x7 cube. In this arrangement, each cell is surrounded by a maximum of 6 neighbors. The cells can be infected with a disease. In particular, if a healthy cell is surrounded by 3 or more infected cells, it will become infected as well. What is the minimum number of infected cells that you need in the beginning that is guaranteed to eventually infect the entire 7x7x7 cube?
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I was hoping to save the ah-ha trick for the next puzzle, but it's Christmas, no point being a Scrooge . Here it is
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Good work, superprismatic. Looks like the Easter Bunny is going to file for bankruptcy very soon. This puzzle didn't last that long. I'll have to think of a harder one next time.
