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tgan 2020-06-28 11:49

July 2020

Dieter 2020-07-21 07:48

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.

tgan 2020-07-22 07:33

I got an answer after few days that my answer is correct

0scar 2020-07-30 05:33

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)?

0scar 2020-07-30 05:57

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

0scar 2020-08-28 14:28

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.


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:
"IM": "MI[COLOR="red"]MI[/COLOR]B",
"MI": "I[COLOR="red"]BM[/COLOR]IM",
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

tgan 2020-08-31 11:30

Very impressive Oscar:smile::smile::smile:

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