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May 5, 2020 at 23:28 vote accept Evgeny Kuznetsov
May 5, 2020 at 23:28 comment added Evgeny Kuznetsov Oh, I misread the "corresponding bit". I understood it as "first bit", I don't know why. Now I see what happens there completely. Sorry it was already clear that we can set $p=1$ without lose of generality.
Apr 29, 2020 at 21:52 history edited Ilya Bogdanov CC BY-SA 4.0
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Apr 29, 2020 at 21:51 comment added Ilya Bogdanov If you speak on part 1; I somehow forgot to mention that wlog $p=1$. Is it clearer now? (will change the answer right now)
Apr 25, 2020 at 17:35 comment added Evgeny Kuznetsov Thank you for the answer and sorry, I somehow missed it. I read it now and, I can not see yet why should $f(z)$ have $0$ on first bit, in case $z$ has first bit determined.
Sep 12, 2019 at 19:33 history edited Ilya Bogdanov CC BY-SA 4.0
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Sep 11, 2019 at 20:25 history answered Ilya Bogdanov CC BY-SA 4.0