I am reading page 39 in this "Post Quantum Cryptography" book. Why does equation 15 hold? There is no further knowledge about f and you definitely cannot use any power laws. So, why is $f^{2^w-1-b_i}(f^{b_i}(x_i)) = f^{2^w-1} $ valid?

I have my doubts, because $f(x) = x+1$ is already a counterexample.

  • $\begingroup$ Could you elaborate on why you think x+1 is a counterexample? $\endgroup$ – Maeher May 29 '15 at 15:10
  • $\begingroup$ Because I took $f^k(x)$ as $f(x)f(x)f(x)$ and not as $f(f(f(x)))$. $\endgroup$ – null May 29 '15 at 15:19
  • $\begingroup$ Yes, with poncho's answer, my comment is moot. I did not think of that possible confusion. $\endgroup$ – Maeher May 29 '15 at 15:20

In this notation, $f^k(x)$ means "apply $f$ $k$ times in succession". For example, $f^3(x)$ is defined to be $f(f(f(x)))$.

Because of this definition $f^a(f^b(x)) = f^{a+b}(x)$ holds trivially (even though we known nothing else about $f$), as the the left side means "do $f$ $b$ times, and then do it $a$ times", while the right means "do $f$ $a+b$ times".

$f(x) = x+1$ is not a counterexample. In this case, we have the simplification $f^k(x) = x+k$ (for example, $f^3(x) = f(f(f(x))) = (((x + 1) + 1) + 1) = x+3$, and so $f^{2^w-1-b_i}(f^{b_i}(x)) = 2^w - 1 - b_i + (b_i + x) = 2^w-1 + x = f^{2^w-1}(x)$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.