So in the case of N divisible by 3, there are impossible end states. But can you still go ahead and declare that you will sort your library within the possible end states? If so how would one prove that one did not step outside of possible states without showing the cards themselves?
Maybe one could first divide the library into 3 piles and sort within each piles first.
Well, due to it being divisible by 3, possible end states just mean that you can just make an arbitrary permutation within each triplet of cards. But you can do that simply in exactly N/3 casts, at which point it's probably just easier to simulate them.
Roukanken|1 year ago