Re: Time to exhaustively break 40-bit RC4?

["Ian Farquhar" <ianf@sydney.sgi.com>]

On Dec 12,  3:30pm, Raph Levien wrote:
> The key schedule operation in RC4 does 256 "swap" operations. Let's
> say it takes four instructions to do each swap. So, it's 2000
> instructions per key. A one-MIPS processor can search 500 keys a
> second. There are about 30 million seconds in a year, so that's 15
> billion keys a year. 40 bits is a trillion keys, so it works out to 66
> years, which is well within the Pentium-style accuracy of the
> calculations I've done.

No, because as you're doing an exhaustive keysearch, you can "pipeline"
the key generation process in software.  Each key requires 256 swaps,
certainly, but there are only two swaps difference between the key
for "0000000000" and "0000000001" (assuming a 40 bit key).  If you
recursively generate keys, then you can generate successive keys
like this:

1. Copy the keystate from iteration n-1 (keep the partial keystates
   on a stack).
2. Do the swap for this portion of the key, and for 255 out of 256
   keys, you will have a new one in 2 swaps.

(In reality, it would be faster to undo the last swap rather than copying
the key, and keeping the swaps on a stack rather than the keystate on
a stack.  These are implementation issues I haven't given a huge amount
of thought to as yet.)

Unless there is some hidden complexity which I have overlooked - in which
case I will be delighted to stand corrected - this will produce a key
fast enough to allow an average workstation to search the 40-bit keyspace
using a known plaintext attack in a couple of hours or less.  If this is
the case, 40-bit RC4 might as well be crypt(1), and 48-bit RC4 looks
pretty shakey too.

I was planning to code this over the xmas break, dependent on whatever
other commitments fall on me during that period.  I realised it was possible
a couple of months ago after pondering ways of parallelising the RC4 key
generation process in hardware.

						Ian.