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Jun 17, 2019 at 8:07 comment added user41014 I made some comment on your proof idea in my question. Maybe you can have a look?
Jun 17, 2019 at 7:41 comment added user41014 Represent those sequences in base $2$ and add the numbers. For example $\pi(\{1,\cdots,1\}) = 2^N-1$, where $N=|\{1,\cdots,1\}|$ is the length of the sequence.
Jun 17, 2019 at 7:38 comment added Bulat Can i ask you what is definition of your '+' operation? F.e. how it adds up two {1,1,1,...1} sequences? ;)
Jun 17, 2019 at 7:34 comment added user41014 I do not understand the argument with $2^N$. Maybe you can comment on that in your answer?
Jun 17, 2019 at 7:29 history edited Bulat CC BY-SA 4.0
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Jun 17, 2019 at 7:28 comment added Bulat yes, of course.
Jun 17, 2019 at 7:27 comment added user41014 what is N, the length of x in bits?
Jun 17, 2019 at 7:25 comment added Bulat @orgesleka compute it modulo 2^N
Jun 17, 2019 at 7:22 comment added user41014 but how do you guarantee that $g(x)>=0$?
Jun 17, 2019 at 7:20 vote accept CommunityBot
Jun 17, 2019 at 7:55
Jun 17, 2019 at 7:15 history answered Bulat CC BY-SA 4.0