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Jun 28, 2013 at 17:00 answer added CCat timeline score: 1
S Jun 27, 2013 at 20:27 history suggested Kaveh CC BY-SA 3.0
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Jun 27, 2013 at 20:23 review Suggested edits
S Jun 27, 2013 at 20:27
Jun 22, 2013 at 14:09 comment added gowers I think I've now found a construction that does what I want.
Jun 20, 2013 at 9:36 comment added gowers Actually, scratch that -- I miscalculated the information-theoretic bound, which gives that exponentially many would be needed.
Jun 19, 2013 at 8:45 comment added gowers I now think it may be possible to do something by composing polynomially many Feistel permutations.
Jun 19, 2013 at 6:55 comment added gowers Yes. I was vague about it, but the precise requirement I would like is that $k$ should be at most a polynomial function of $n$ (or perhaps a very slightly superpolynomial function).
Jun 19, 2013 at 1:08 comment added Ryan O'Donnell You must want some upper bound on $k$, right? Otherwise you could take $k = \log(2^n!)$ and then define $\phi(x,y) = \pi_x(y)$ where $\pi_x$ is the $x$th permutation on $\{0,1\}^n$ in lexicographic order. Then $\phi$ is completely indistinguishible from truly random. (I'm ignoring here the issue that $\log(2^n!)$ is not an integer, but one could hack around this.)
Jun 18, 2013 at 17:13 comment added Timothy Chow If you don't get a good answer here on MO, you might try asking your question at cstheory.stackexchange.com
Jun 18, 2013 at 17:00 comment added gowers I have now found a source that seems to suggest that the Luby-Rackoff construction won't give hardness greater than $2^n$. So it looks as though a different idea would be needed. But maybe there are some different ideas out there.
Jun 18, 2013 at 14:14 history asked gowers CC BY-SA 3.0