[8439] in cryptography@c2.net mail archive
Re: NONSTOP Crypto Query
daemon@ATHENA.MIT.EDU (Pat Farrell)
Sat Jan 13 14:19:22 2001
Message-Id: <5.0.2.1.0.20010113122719.034eea70@mail.goibsdsl.com>
Date: Sat, 13 Jan 2001 12:40:05 -0500
To: Ben Laurie <ben@algroup.co.uk>, Ray Dillinger <bear@sonic.net>,
cryptography@c2.net
From: Pat Farrell <pfarrell@pfarrell.com>
Cc: John Young <jya@pipeline.com>, pfarrell@pfarrell.com
In-Reply-To: <3A5FAFB1.33FF89E2@algroup.co.uk>
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At 01:30 AM 1/13/2001 +0000, Ben Laurie wrote:
>Hmm. 6 kHz has a wavelength of 5 cm. I would guess you can easily get
>resolution to 1/10 of a wavelength under ideal conditions. Which is .5
>cm, which is half the size of a key, more or less.
You don't have to locate the exact key to save a lot of complexity.
A standard PC keyboard has 47 keys on the main section.
Ignoring shifts, control, alt, combinations, etc. you have to deal with
47^N easy options per secret key of length N.
Lets assume you don't get the key as a fact from the sound inference,
but rather you get a probability density function that is weighted heavily
arround a single key, and then arround the keys "one key away" and
with decreasing probability for "two keys away" and so on until you get
to the maximum of 14 or so keys away.
If Ben's estimate is close to accurate, you should see a two standard deviation
circle of only 9 or so keys.
Since 47^6 is 229,345,008 and
9^6 is only 531,441
this technique can whack out a factor of 500 in the "likely" exhaustive
search of
a six character passphrase. Obviously it saves more on longer passphrases.
It also saves more if the user enters control/alt/shift combinations.
Interesting.
Pat
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