Re: How can I secure a Debian installation?
- To: debian-user@lists.debian.org
- Subject: Re: How can I secure a Debian installation?
- From: Florian Kulzer <debian-lists@florian-kulzer.eu>
- Date: Sat, 1 Feb 2014 15:41:38 +0100
- Message-id: <20140201144138.GA15324@isar.localhost>
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- In-reply-to: <1391263230.16341.60.camel@tagesuhu-pc>
- References: <52E76165.7020903@q.com> <20140128094643.6ee293e7@jretrading.com> <CAD4guxPnaDnwBgjEms8PnMT22myXJj7esT01ug-P-feR4+_rQw@mail.gmail.com> <20140128184234.GL3888@copernicus.demon.co.uk> <20140130185311.51a2a863@X200> <20140130202637.GN3888@copernicus.demon.co.uk> <CAD4guxODr4vFdqyyg20uJdiK-8zk-dDQX9poa5asX8HuG1hkBg@mail.gmail.com> <52EB405C.5070909@gmail.com> <20140201092121.GA31378@sid.nuvreauspam> <1391263230.16341.60.camel@tagesuhu-pc>
On Sat, Feb 01, 2014 at 12:00:30 -0200, André Nunes Batista wrote:
>
> Isn't it the case where the randomness of the key/password composes the
> overall quality of the crypto substitutions in such a way that 4096bit
> keys would necessarily provide better protection against cryptanalysis
> when compared to dozens of random, valid characters?
As far as I understand it, that is correct: A 4096bit key gives you
2^4096 possibilities, while a string of n random characters selected
from a set of, let's say, 50 members (letters, numbers, special
characters) has 50^n possible values. To break even with the 4096bit
key, such a random-string password would therefore have to have a length
of n=4096*ln(2)/ln(50) characters, which is about 725.
--
Regards, |
Florian | http://www.florian-kulzer.eu
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