EventHorizon
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Everything posted by EventHorizon
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Oops....ignored that the area still needs to be integral. If that was removed and I could use many straight edges, I could get a polygon with as small an area as I want. My solution of 3 still works, but it's going to be harder to get lower than I previously thought.
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if corner to corner is good enough, you could offset each coin by half it's width to get rid of the center square.
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I have a solution with an area of 3 (which I won't give away yet), but I have another question... Do I have an eraser? Or does everything I write need to be part of the polygon? (i.e., can I cross lines drawn with the straight edge to find a point, then erase the lines?) Or, equivalently, can I use more than one (infinitely long, infinitesimally thin) straight edge?
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Way too high. From my java code (posted a little earlier) I have some numbers on this. We now just need a good equation to find them. Of course, if you find a problem in my code let me know. I'm pretty sure it works though.
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Something didn't seem right about my answer, so I thought I would run some numbers.
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It was a pentomino (area of 5). But I've made another interesting discovery...
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There's a large amount more....I've found 49 so far....and that's just in about a minute of looking.
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Ahh...not only that, but you can simulate unfair coins/dice by adding in multiple copies of the same outcome. The probabilities for the unfair coins need to be rational numbers though.
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We ask a slightly different question: How many people must be in a room for chance to favor every day on the calendar being the birthday of someone in the room? For simplicity you can ignore leap years. Oh, and thanks for yet another amusing mental diversion.
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I like my solution....it extends seamlessly to any number of sides for a die very easily. In fact, even with an unfair coin it could simulate a fair die of any number of sides (just let heads=1 and tails=0).
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Notice that if there is an H in either the first or second position, you need to get two T's in a row and then something else. Prof. Templeton simply needs to wait for the two T's in a row.
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I had some explanation in my post, but I guess it was insufficient or switched between player 1 and player 2 too much. Here's the same thing with better/more explanation. Hope that helps.
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That didn't take long...
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Another good one from Bonanova.
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Oops, didn't see Primes last post before posting my last post. Guess that's the problem with opening tabs and ignoring web pages for a while. So wikipedia also only proved that it works, and didn't show why xor works. I guess it's good to know that wikipedia didn't have a significantly better proof (at least one with a derivation of the magic formula).
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Finally read post #2. That's awesome. As for the proof. I suppose I will let wikipedia handle it. Though if I come up with a surprisingly simple/intuitive reason for xor, I'll post it.
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Hey tarunark, You got question 1, but missed 2 and 3. Three, as originally written, may be unsolvable....and definitely too hard and/or time consuming for most braindenners... let alone career mathematicians (it would require significant research and/or mathematical knowledge (it is an unsolved problem in Ramsey Theory)). Good job on question 1 though.
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I'm a bit late to the game, but here's a proof (for the modified game (take last stone to win))... See any flaws/holes?
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Yup, cliques are what we are looking for. Your answers are right. The method you used to get the answer to (2) is what I used as well. Though it has been proven for 3 relationships, I think it may just be an upper bound for more. What do you think? If it has been proven to be right, could the same method be used for (new 3)?
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Yup, that's the equation to start with.
