Please use spoiler flag, comments are visible from the dashboard. I added it for you this time.

Of course, I was just commenting because for me, as a mathematician, it was on the first glance confusing that this actually works :D

To be fair, the type definition is Int -> Int, not (Num a) => a -> Int, so a Double would never be allowed into the function.

Note that this definition would not work for Double, e.g. 2.25=1.5^2 would be sent to 2^2=4, because 1.5 gets rounded to 2, although 2.25 is closer to 1 than to 4.

Very nice!

That's what all men see, lol

The boobs operator!

This is majestic.

This feels like a 8 kyu kata, not really a 7. Cheers.

Not hard to do by hand, just eta conversion and infix <-> prefix notation.

Definitely not a best practice in terms of code clarity though! :D

Fixed.

@mattmanj17 ..

• Thanks Matthew for naming the language making it more easier healing the issue
• It's really weird since 4 Solvers have done it in Haskell Without raising an issue concerning it , But Anyway I've mentioned the Haskell Translator to do the necssary fixing the problem
• Hope you all the best Matthew On/Off CW , Happy to follow you Bro .. Regards .. Zizou

@cliffstamp @MrZizoScream
I am solving in Haskell. The description says (1  ≤  a, b, c  ≤  10), but I am getting input < 1, which makes the problem harder.