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 2020-06-28, 11:49 #1 tgan   Jul 2015 128 Posts IBM July 2020
 2020-07-21, 07:48 #2 Dieter   Oct 2017 10100002 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 2·5 Posts I got an answer after few days that my answer is correct
 2020-07-30, 05:33 #4 0scar   Jan 2020 2×5 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 2×5 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

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