araver
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Everything posted by araver
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reality - XO zoology - X regally - OO dullish - O
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reality - XO zoology - X regally - OO
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You're such an independent Hacker ... prime material for Evil Mastermind-ness Don't mind octopuppy, the quest for the algorithm is still open.
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reality - XO zoology - X
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Correct answer again (although you did not use spoilers this time ). Yet I'm afraid we're not on the same page on what a proof means. I can say for example that xn+yn=zn has no solutions for natural numbers (n>2,x,y,z>0). And even if one would agree with me, one would still require proof, right? And simply saying that the best way to find a solution is considering x and y relatively close and then showing that for this case there is no z, is not a (complete) proof.
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Well TEXAS, IOWA, HAWAII, ARIZONA, ARKANSAS are grateful . Just figured that I was counting ABORT so actually only 5 remaining Idk really. I thought ARKANSAS after KANSAS would be a give-away so I refrained from that. But I was counting on the North/South pairs to help in some way but as no one was requesting them, I played them earlier than the end. Guess I wanted to be stopped So, care to explain this time?
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Password accepted. System shutting down. Awww, maaaan ... I knew those meddling kids will stop me from reaching a perfect score.
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By all means Anyway, a late response to one of your thoughts before. As the rules state, the Evil Mastermind accepts max 1 guess / player between 2 consecutive rounds of firing. So a big enough coalition of players can practically brute-force in between rounds (Assuming the Evil Mastermind is not trigger-happy, which I am not at the moment.)
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Even numbers that are also multiples of 4: Even numbers that are not multiples of 4 (m = 4*p+2) are not that easy. 2 is possible (a=1,b=1 1+1=2) I'm betting 6 isn't
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No particular reason. But I make a point in avoiding north/south pairs since they might contain a weakness I did not see. Here's the last of the pairs: NORTHCAROLINA TICBSQNK EDIT: I only got 6 states left. And I am indeed protecting one of them
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NORTHDAKOTA TIFAMOTC
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As I said, your answers to Q1, Q2, Q3 are correct and your proposed strategy works for that case. But there is a nice proof (besides trying it practically) for that in all cases, not just the one you mentioned (which may or may not be what player 1 choses to play). I'm adding a question to the OP, about a slightly modified version of the game, which hopefully might provide a hint on why the strategy works in all cases. Question 4) For what M and N does the second player always have a winning strategy, if the figure that needs to be covered is selected by the first player?
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reality - XO
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SOUTHDAKOTA PWTIFAMO
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i would :dance: if I could OK, 7 letters. Almost did a _ _ _ _ before I figured out it isn't ROLLO
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burden?
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Care to elaborate the reason? (as a bonus since its not asked directly in the OP?)
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I realized I haven't told why I expected my first hunch to work. It might be worth something so I'm sharing it. So 35->(25, 25). But further, I always got less lucky / unbalanced. E.g. if you split by asking #6 if #7 is Random: - the No branch has 20T and 5Rs for the last position which you can exploit by again halving the number of truthtellers (going backwards and asking #7 if #6 is Random). You get 2x15-combinations branches (5+20/2) - the Yes branch does not exhibit the same pattern. The most T's you can get on the same position is 16. (with 9 Rs in the rest of the combinations), so at the most, a same-type of question can only give you 2x17-combination branches (9+16/2). This inherent non-equilibrium has been puzzling me. (or flummoxing since I learned a new word the other day ) I based my initial calculations on an (inaccurate so far) estimate of the time/questions it takes to reduce it to a very small of combinations.
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Assume two players have an infinite amount of equal/identical regular polygon figures (N sides, N>=3) and play the following game: 1) First player places M>1 figures on the table in any position he wants. 2) Second player can move the figures on the table without rotating them (only translation is allowed). Figures can be made to overlap. Second player wins if he can completely cover one figure using some or all of the rest of the figures already on the table. Otherwise first player wins. Question 1) For what M and N does the second player always have a winning strategy? Question 2) For what M and N does the first player always have a winning strategy? Question 3) Is N important for the winning strategy? P.S. It's not original. I found a simpler version in an online puzzle competition (which is about to end anyway). But I liked it very much, so I'm sharing it
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Thank you. Appreciate it.
