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EventHorizon

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Everything posted by EventHorizon

  1. n factorial (represented as n!) is the product of all positive integers up to n. So the number you are looking for is 1,000,000! ("one million factorial")
  2. I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points. How to find the arc length of a function: Restatement of problem using previous spoiler: One answer is simple, the other is less so.
  3. I looked at that for a while. Though it's not much, here's what I got...
  4. This puzzle is not quite solved yet. So here's my attempt
  5. I installed the plus4 app, and it is just a mastermind clone. +x means you have x numbers in the right spot, and -x means you have x numbers right, but they are in the wrong spots. It could be that a guess would result in +2-2, meaning that you have all the right numbers, but two are in the wrong spots.
  6. So the best you can do is 00 00 or 0 0 ... 0 0 000...00
  7. Looks like I misunderstood the initial configuration. I assumed the x's were already amoeba and not just space to vacate.
  8. Example 2 is not looking good. Perhaps it is unsolvable....
  9. Yay for integer sequences (found using your 19355 )... Time to look at 3-regular graphs... which aren't as simple and quite possibly don't have this property. So it looks like we basically just need to find how many 8-regular graphs there are with 26 labeled vertices. It'll be off by a little, but will be a good estimate for N.
  10. N is not going to be that simple to calculate. Any given adjacency you find could have multiple possible cycles that could be found within it (So it may incorrectly be counted multiple times). Adjacencies don't need to be as neatly organized as curr3nt's answer was.
  11. I think the bit debt idea will always have situations that produce contradictions, so I'm thinking it should be discarded. I looked over plainglazed's 38, and I think it works. Nice. Edit: I got another 38. It's slightly more complicated (1 more branch needed), but is based on yours, so I won't bother posting it. It uses 6of8's and 5of8's in combinations instead of groups of 5's in a bigger group of 20.
  12. I didn't really explain it too well. Here's a (hopefully) better attempt to show that with 3 bits of information, you can get 4 of the next 5 cards right and still end up with 2 bits after A couple examples Example 2 shows that this method does not work as I thought it did, since I reached a contradiction. So back to the drawing board... hopefully something from this can be salvaged. I think it may still work, but that I can't 'borrow' quite as many bits as I thought. I think there needs to be a 1 bit buffer at all times. Hopefully that solves the issue, but I'll need to lo
  13. So, after reading your post and looking at the fewest for 3 again. I thought of one configuration to try. Since testing a configuration with my code is fast, I threw it into my code and...
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