Good comment, not sure what's happening; c stuff is pretty stable since > ~2.7/x

tried on 3.4.3 (default for this kata) - got a syntax error
works fine on 3.6.0

yep, word.lower() is executed for each char call, but who cares, especially if there is a lack of extra lines in oneliner's head? =)

I'm having trouble comprehenhing why this is here, why is a kata based around either a user doing trial and error or randoming using old formulas learnt and never hinted at?

Some of the katas of this series are great, ones like these in my opinion leave a lot to be desired and very little

The last digit of 7 ^ (6 ^ 21) != last digit of 7 ^ 6

Can you elaborate on how exactly does the algorithm work? For example ([7, 6, 21], 1) test case: as I understand it means 7 ^ (6 ^ 21). Either using my algorithm from "Last digit of a large number" kata or pow(n1, n2, 10) this transforms into 7 ^ 6, which gives the same last digit as 7 ^ 2, where the last digit of 49 is 9. Is this thing supposed to work in a different way or am I missing something?

mb

Not an issue, a question, especially when it's because your algorithm is not correct ;-)

That part means a(n) could be not equal to b(n)^2 or that b is shuffled.

That's correct but I found the descriptive language subpar because of the end tidbit of

"regardless of the order."

That would be against the instructions:

"Same" means, here, that the elements in b are the elements in a squared

Seem's the kata doesn't test for cases where a=b^2 aka reverse.
Making it a lot easier than needed be.

Read carefully the description and try to use the hint.