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$f: \{0,1\}^n -> \{0,1\}$

The function "depends on i" if there exists two $o/1$ strings (A and B) where A and B differ only at position i and $f(A) \not= f(B)$.

How many n-bit Boolean functions do not depend on i?

I have no clue how to start this one. If someone could explain me through the logic of it, I would be extremely grateful, and maybe I could figure out a similar problem. Thank you.

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  • $\begingroup$ Can you start by determining how many $n$-bit Boolean functions there are? $\endgroup$ – MartianInvader Apr 2 '14 at 23:32
  • $\begingroup$ @MartianInvader How would I do that? $\endgroup$ – Jessica Apr 3 '14 at 0:16
  • $\begingroup$ Well, what are the choices you make when defining an $n$-bit Boolean function? $\endgroup$ – MartianInvader Apr 3 '14 at 0:45
  • $\begingroup$ Would the total number of functions be $2^{2^n}$? $\endgroup$ – Jessica Apr 3 '14 at 1:08
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The answer is $2^{2^{n - 1}}$ n-bit Boolean functions that do not depend on i.

When determining the number of possible string combinations, you don't need to consider the i-th bit so there are $2^{n - 1}$ possibilities. And there are still 2 outputs to think about so the number of n-bit Boolean functions is $2^{2^{n - 1}}$.

Thanks MartianInvader :)

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