To answer the question, not quite. If you had two eggs and 14 tries, and observed the first egg break on the Nth try, you then have one egg and 14-N tries remaining. In other words, the tries are cumulative. Notice therefore that increasing n beyond m has no effect, since you can use up at most m eggs no matter your strategy.

Good kata, I learned something new today.

Nice kata for C++ training on parsers, error handling, and such.
With modern C++17 (variants and visitors, exception handling, string_view, and such) quite straightforward to
solve : just parse the way you describe the langage and... debug for errors.
Thus ... overranked IMHO

Note : would have needed a proper C++ translation (passing const std::string& and returning a std::string), pissed with those const char returning functions...*

Omg this is so hardcore.
I needed to dig into really advanced math hacks.
Finally i discovered good pure- math move to calculate this significally less then limited time - below 4 seconds, in Python and i am finally satisfied :D

I don't think Cubical.HITs.HitInt exists in the cubical library anymore. It seems to have been replaced with QuoInt. This makes it hard to complete this kata. Anyway this can be updated?

Please! don't use m and n that are not even defined in this kata (you have to go to the 3 kyu versions to understand what they mean). Use launches and eggs, instead.

Even better: copy the whole description of the kata!

Haskell noob here -- I wrote up something that seems to typecheck and pass the sample tests, but it times out on submission with logs "Generating tests...". Is this necessarily an issue with my solution module? Otherwise, how should I interpret this -- should I be looking for a simpler proof?

Great kata. Sounds simple, turns out to be very challenging.

Couldn't you copy and paste it here? Having to go to a different kata to see information you need in this one shouldn't be needed. Even more when this is the easier one, and the normal order for doing this should be easier first, harder later, not the other way around.

Thanks.
CONTIGUOUS. There is only one 2 next to 3.

Sorry, I don't understand the example {2,2,2,3}.
If subarray {2} is counted three times, why is {2,3} only counted once?

(unfortunately, the answer to nekoman's question is hidden)

"don't need to know neither"

"neither" => "either"

The description should clarify its comment on leading zeroes; should they be preserved in the transformation?

(if a b c) should be if a then b else c. If c is not given ((if a b)) and a is false, return Nul.

Well guess who has never realised this setting and went debugging for 1 hour straight

Commutative unital magma?