абидна да?

ну да получаеца

для богини не сделали аниму адаптацию

капес он чмоха

как я

Он просто дома сычует

Граф дизайнер

он типа сказал шо у него огромный член а потом сказал что у него маленький член?

чикак

хай привет

кичхош

ребят нужна помощь

в каком порядке надо смотреть второй сезон харухи?

зашушли

загуглил

4 permutations are contained within each n-cycle. So the goal of this algorithm is to systematically generate the sequence, and show that any other method would give a less efficient method. In the long run, I'd like to create some axioms/theorems for the proof, such as more overlap=more efficiency. I think this will call for some modular arithmetic to generalize it for all n but I'm not sure how to do so.

*by follow, I mean use it as a "rule" to tell you the next number in the sequence*

Start with 1 and follow (1 2 3 4) 6 times. We do this by convention. 1,[2,3,4,1,2,3]

Follow (1 4 2 3) 5 times. This group must be chosen, as (23) is in it. 1,2,3,4,1,2,3,[1,4,2,3,1]

Follow (1 2 4 3) 5 times. This is the last group with (3 1) in it. 1,2,3,4,1,2,3,1,4,2,3,1,[2,4,3,1,2]

Since no remaining groups have (1 2) in them, we have to choose one

A): Following (1 4 3 2) 6 times would end the sequence in wrong: 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2[1,4,3,2,1,4] No group left has (1 4) in it, so this option looses efficiency

B): Following (1 3 2 4) 6 times would end the sequence in wrong: 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2[1,3,2,4,3,2] No group left has (3 2) in it, so this option looses efficiency

C): Following (1 3 4 2) 6 times works: 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2[1,3,4,2,1,3]

Follow (1 3 2 4) 5 times. This is the last group with (1 3) in it. 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2,1,3,4,2,1,3,[2,4,1,3,2] Follow the last group (1 4 3 2) 5 times, to complete the sequence. 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2,1,3,4,2,1,3,2,4,1,3,2,1,4,3,2,1

но я в математике не силен