Re: Clarification on 3DES transforms

[Bill Sommerfeld <sommerfeld@apollo.hp.com>]

> DES_Key_I  = Truncate(MD5( D_Pad_I | K ),192)
> DES_KEY_I1 = first 64 bits of DES_KEY_I
> DES_KEY_I2 = second 64 bits of DES_KEY_I
> DES_KEY_I3 = third 64 bits of DES_KEY_I

MD5 only produces 128 bits of output, so this is not going to work
very well ...  you can't produce I3 because DES_KEY_I only has 128
bits, and you've also wasted 16 bits of entropy because of the @!#^%^&
DES key parity.

Note that if you use SHA instead, you get 160 bits out.  If you use
the bits efficiently, you only need 168 (not 192).  Where do you find
the remaining 8 bits?

I'm not endorsing the second one, mind you; I want to hear what the
Real Cryptographers(tm) have to say about it..

					- Bill

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Follow-Ups:

• Re: Clarification on 3DES transforms
• From: "C. Harald Koch" <chk@border.com>

References:

• Clarification on 3DES transforms
• From: Naganand Doraswamy <naganand@ftp.com>

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