Subject: [Cfrg] MASH Hash Function

From: John Wilkinson <[email-masked]>
Date: Wed, 16 Mar 2005 20:28:02 -0500

I read the intent of Phillip's message to be:

hash of message m = HMAC-SHA1 of m under key SHA1(m)

More explicitly, this is:

MASH(m) = SHA1(opad^SHA1(m) | SHA1(ipad^SHA1(m) | m))

If this is correct (and it would be nice to hear from Phillip to 
confirm), then I don't see any obvious way of breaking this 
construction unless one can generate collisions for SHA1 with an 
arbitrary, unknown IV.

Ferguson and Schneier have suggested a similar construction in their 
book "Practical Cryptography" (p. 93). They suggest:

h_DBL(m) := h(h(m) || m)

This is much like the inner-hash of HMAC, without the padding of the 
"key" out to one block-length.
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