1997-06-25 - Re: Comparing Cryptographic Key Sizes II

Header Data

From: “Peter Trei” <trei@process.com>
To: cypherpunks@toad.com
Message Hash: b036289b07dc6a55f2e9618c5edb86ce62914abed39c913b2951a9cf3cde07c1
Message ID: <199706251502.IAA17998@toad.com>
Reply To: N/A
UTC Datetime: 1997-06-25 15:19:26 UTC
Raw Date: Wed, 25 Jun 1997 23:19:26 +0800

Raw message

From: "Peter Trei" <trei@process.com>
Date: Wed, 25 Jun 1997 23:19:26 +0800
To: cypherpunks@toad.com
Subject: Re: Comparing Cryptographic Key Sizes II
Message-ID: <199706251502.IAA17998@toad.com>
MIME-Version: 1.0
Content-Type: text/plain



There is still quite a bit of confusion about the difficulty
of factoring 512 bit moduli. I'll try to clear that up.

Adam writes:

> 
> For example, 40 bit RC4 (a symmetric key cipher) can be broken in a
> few hours with a few hundred workstations but 40 bit RSA (a public key
> cipher) can be broken in a fraction of a second on one PC. A 350 bit
> RSA key is roughly equivalent in strength to 40 bit RC4, and 512 bit
> RSA key is thought to cost roughly the same to break as a 56 bit DES
> key.
> 
> (That last comparison (DES vs 512 bit RSA) glosses over many issues.
> Here's a summary if you're interested: no one's broken a 512 RSA bit
> key yet, and you need lots of memory to break RSA at that key size, a
> PC with 128 Mb would be required to participate.  In contrast, you
> need hardly any memory to break DES, any PC will do.  The internet
> based breaking of a DES key in answer to RSA DSI's challenge involved
> many participants, and included some participants with low end 486 PCs
> with 1Mb of memory.  Theoretically 512 bit RSA could be broken more
> quickly than DES, but, as you need more memory than typical
> workstations have, a distributed internet attack with the same group
> of participants as for the DES break would clearly take longer.  There
> are a number of other factors also.)

Here's the sequence of events:

 We cracked a DES key, searching about 1/2 the keyspace. I used the
published speeds of the clients, and my own knowledge of the Pentium
processor, to estimate that this search used about 457,000 MIPS 
years.

 In his usenet note announcing the factoring of RSA-130 using GNFS,
(General Number Field Sieve) Lenstra estimates that the effort was 
about 500 MIPS years, '1/10th of the effort for RSA-129, using QS
(Quadratic Seive)'.

Bruce Schneier, in AC2, has a table giving 28,000 MIPS years as the
time to factor a 512 bit number using GNFS.

 I wrote to Lenstra, asking him how big a modulus he thought could
be factored with 1000x the power he used for RSA-130.

 Lenstra responded that that power should be able to factor a 600
bit modulus using GNFS, but that the sieving clients would need about
128 M of memory, and that the matrix problem would be 'very big' (he
did NOT say impossible).

 RSA-129 was factored using QS, on clients with as low as 8M of 
memory. In the paper "The Magic Words Are Sqeamish Ossifrage", in
which the factorization of RSA-129 using QS is described, the authors
(including Lenstra) estimate the QS could factor a 512 bit modulus
in about 500,000 MIPS years.

 I wrote to Lenstra, asking if QS could be used instead, to deal with
the memory and matrix problems. Lenstra responded to the effect that
it could.

So, to summarize:

Using GNFS, on clients with 128M of memory, you could factor a 512
bit modulus in 28,000 MIPS years. With 500,000 MIPS years, you could
factor a 600 bit modulus.

Using QS, in 500,000 MIPS years you could factor a 512 bit modulus on
machines with modest memory requirements.

The effects of memory speed and bandwidth would slow things down
somewhat.

-----------------------------
One of the things that is bothering me about the reports I've been 
seeing is 'they took 6 months to break the key using thousands of
machines.' This is inaccurate - heavy work on the DES challenge did
not start for several months after the challenge was announced, and
the number of machines grew gradually as the work was done. 

Lets crunch some numbers:

Suppose an average client can test 500,000 keys/sec (this is 
a conservative figure).

Suppose we had 18,000 clients, 24x7, 100% utilized for this work.
(there were 14,000 in DESCHALL, 3000 in SolNet, 1000 misc(last is a 
conservative guess)

That's 9x10^9 keys/sec

To search 1/2 the keyspace, we need to test 2^55 keys = 3.6x10^16 keys.

->about 4,000,000 seconds

= 46 days.
-----------------------

Peter Trei
trei@Process.com



 






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