 mersenneforum.org July 2020
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 Register FAQ Search Today's Posts Mark Forums Read 2020-06-28, 11:49 #1 tgan   Jul 2015 318 Posts July 2020   2020-07-21, 07:48 #2 Dieter   Oct 2017 6E16 Posts Does anyone of you know, how the new puzzlemaster reacts to submissions that aren‘t correct? Oded has answered every time very fast - that was a good way to help. But I guess that he wasn‘t obliged to do so.   2020-07-22, 07:33 #3 tgan   Jul 2015 318 Posts I got an answer after few days that my answer is correct   2020-07-30, 05:33 #4 0scar   Jan 2020 31 Posts Spin-off A friend of mine suggested a nice "spin-off" puzzle. If you solved July 2020 main problem by using three letters or more, and your dictionary includes letters "I", "B", and "M", then how many times does substring "IBM" appear in your solution? Let IBM(n) be the number of occurrences within n-th string of your solution. Does this sequence follow some nice pattern? A trivial upper bound is IBM(n) <= LENGTH(n)/3 = FIBONACCI(n)*FIBONACCI(n+1)/3. A slightly sharper bound is given by RARE(n), the minimum among the occurrences of letters "I", "B", and "M". What about your ratio IBM(n) / LENGTH(n)? What about your ratio IBM(n) / RARE(n)?   2020-07-30, 05:57 #5 0scar   Jan 2020 31 Posts As LENGTH(n) grows exponentially, I think that it makes sense to compute the ratios for n up to 10, or their limits as n grows to infinity (only for the brave) For my currently best solution, IBM / LENGTH > 0.2   2020-08-28, 14:28 #6 0scar   Jan 2020 31 Posts Spin-off solution The target sequence G(n) = F(n)*F(n+1) satisfies an order-3 homogeneous linear recurrence: G(n) = 2*G(n-1) + 2*G(n-2) - G(n-3), so we need a homogeneous linear system of order at least three to generate it, which means a dictionary of at least three letters. Three letters actually suffice; I found three such solutions. Dictionary 1: {"A": "BC", "B": "AAB", "C": "ABC"}. Dictionary 2: {"A": "AB", "B": "AAAC", "C": "AAC"}. Dictionary 3: {"A": "AB", "B": "AABC", "C": "B"}. In each case, the game starts with letter "A"; letter "C" is always the rarest one, at least for index n>2. Dieter proved a nice closed-form representation for letter-counting functions. Dictionary 1: A(n) = F(n+1)*F(n-2); B(n) = F(n)*F(n-1); C(n) = F(n-1)^2. Dictionary 2: A(n) = F(n)^2; B(n) = F(n-1)^2; C(n) = F(n-1)*F(n-2). Dictionary 3: A(n) = F(n)*F(n-1) + F(n-2)^2; B(n) = F(n)*F(n-1); C(n) = F(n-1)*F(n-2). Dictionary 1 seems the most promising one, as letter "C" occurs slightly more often (by a factor F(n-1)/F(n-2), which converges to the golden ratio phi=1.618...). We can change the three spellings in 2*3*6 = 36 ways. We can replace letters "A","B","C" with letters "I","B","M" in 6 ways. Among the 216 possible variations, I chose the following one: {"M": "IB", "I": "MIM", "B": "IBM"} The game starts with letter "M"; letter "B" is the rarest one for n>2, so the upper bound IBM(n) <= B(n) is sharp. This solution almost reaches such bound: IBM(n) = B(n) if n is odd, IBM(n) = B(n)-1 if n is even. Proof. In each string of the sequence, we can only find the two-letter substrings "BM", "IB", "IM", and "MI". The sequence starts with a one-letter string, so the claim is trivially true for n=1. The given list contains all the two-letter substrings of the chosen spellings, so the claim is also true for n=2. Then, only substrings "BM" and "MI" can arise from a permitted concatenation of spellings: "BM": "IBMIB", "IB": "MIMIBM", "IM": "MIMIB", "MI": "IBMIM", so the claim is also true for every index n>2 by induction. In particular, letter "B" is always preceded by letter "I" and followed by letter "M", unless it occurs at the beginning or at the end of a string, so the lower bound IBM(n) >= B(n)-2 holds. For index n>1, it's easy to check that the first few string characters follow a regular pattern with period 2: IB... -> MIM... -> IB... -> MIM... and the same holds for the last few string characters: ...IB -> ...IBM -> ...IB -> ...IBM so letter "B" never occurs at the begin of a string, but it occurs at the end of even-indexed strings. For the given solution, the ratio IBM(n)/G(n) converges to the value 1/phi^3 ~ 0.236 Last fiddled with by 0scar on 2020-08-28 at 14:32   2020-08-31, 11:30 #7 tgan   Jul 2015 110012 Posts Very impressive Oscar     Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post a1call Puzzles 36 2018-08-06 14:02 Xyzzy Puzzles 4 2016-08-06 22:51 Xyzzy Puzzles 16 2015-08-19 16:13 Xyzzy Puzzles 6 2014-11-02 19:05 LaurV Lounge 8 2012-07-06 00:13

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