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Old 2020-06-28, 11:49   #1
tgan
 
Jul 2015

2×5 Posts
Default IBM July 2020

http://www.research.ibm.com/haifa/po.../July2020.html
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Old 2020-07-21, 07:48   #2
Dieter
 
Oct 2017

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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.
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Old 2020-07-22, 07:33   #3
tgan
 
Jul 2015

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I got an answer after few days that my answer is correct
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Old 2020-07-30, 05:33   #4
0scar
 
Jan 2020

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Default 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)?
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Old 2020-07-30, 05:57   #5
0scar
 
Jan 2020

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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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