[tor-dev] Proposition: Applying an AONT to Prop224 addresses?

[Ian Goldberg]

On Mon, Mar 27, 2017 at 01:59:42AM -0400, Ian Goldberg wrote:
> > To add an aside from a discussion with Teor: the entire "version" field
> > could be reduced to a single - probably "zero" - bit, in a manner perhaps
> > similar to the distinctions between Class-A, Class-B, Class-C... addresses
> > in old IPv4.
> > 
> > Thus: if the first bit in the address is zero, then there is no version,
> > and we are at version 0 of the format
> > 
> > If the first bit is one, we are using v1+ of the format and all bets are
> > off, except that the obvious thing then to do is count the number of 1-bits
> > (up to some limit) and declare that to be version number.  Once we're up to
> > 3 or 4 or 7 or 8 one-bits, then shift version encoding totally.
> > 
> > Teor will correct me if I misquote him, but the advantage here was:
> > 
> > a) the version number is 1 bit, ie: small, for the forseeable / if we get
> > it right
> > 
> > b) in pursuit of smallness, we could maybe dump the hash in favour of a
> > AONT + eyeballs, which would give back a bunch of extra bits
> > 
> > result: shorter addresses, happier users.
> 
> You indeed do not require a checksum under an AONT, but you do require
> redundancy if you want to catch typos.  Something like
> 
> base64( AONT( pubkey || 0x0000 ) || version)
> 
> is fine.  If you want "version" to be a single bit, then the AONT would
> have to operate on non-full bytes, which is a bit (ha!) annoying, but
> not terrible.  In that case, "0x0000" would actually be 15 bits of 0,
> and version would be 1 bit.  This would only save 1.4 base32 characters,
> though.  If you took off some more bits of the redundancy (down to 8
> bits?), you would be able to shave one more base32 char.  And indeed, if
> you make the redunancy just a single byte of 0x00, then the extra 0-bit
> for the "version" actually fits neatly in the one leftover bit of the
> base32 encoding, I think, so the AONT is back to working on full bytes.
> 
> But is a single byte of redundancy enough?  It will let through one out
> of every 256 typos.  (I thought we had spec'd 2 bytes for the checkcum
> now, but maybe I misremember?  I'm also assuming we're using a simple
> 256-bit encoding of the pubkey, rather than something more complex that
> saves ~3 bits.)
> 
> (Heading to the airport.)

OK, here are the details of this variant of the proposal.  Onion
addresses are 54 characters in this variant, and the typo-resistance is
13 bits (1/8192 typos are not caught).

Encoding:

raw is a 34-byte array.  Put the ed25519 key into raw[0..31] and 0x0000
into raw[32..33].  Note that there are really only 13 bits of 0's for
redundancy, plus the 0 bit for the version, plus 2 unused bits in
raw[32..33].

Do the AONT.  Here G is a hash function mapping 16-byte inputs to
18-byte outputs, and H is a hash function mapping 18-byte inputs to
16-byte outputs.  Reasonable implementations would be something like:

G(input) = SHA3-256("Prop224Gv0" || input)[0..17]
H(input) = SHA3-256("Prop224Hv0" || input)[0..15]

raw[16..33] ^= G(raw[0..15])
# Clear the last few bits, since we really only want 13 bits of redundancy
raw[33] &= 0xf8
raw[0..15] ^= H(raw[16..33])

Then base32-encode raw[0..33].  The 56-character result will always end
in "a=" (the two unused bits at the end of raw[33]), so just remove that
part.

Decoding:

Base32-decode the received address into raw[0..33].  Depending on your
base32 decoder, you may have to stick the "a=" at the end of the address
first.  The low two bits were unused; be sure the base32 decoder sets
them to 0.  The next lowest bit (raw[33] & 0x04) is the version bit.
Ensure that (raw[33] & 0x04 == 0); if not, this is a different address
format version you don't understand.

Undo the AONT:

raw[0..15] ^= H(raw[16..33])
raw[16..33] ^= G(raw[0..15])
# Clear the last few bits, as above
raw[33] &= 0xf8

Check the redundancy by ensuring that raw[32..33] = 0x0000.  If not,
there was a typo in the address.  (Note again that since we explicitly
cleared the low 3 bits of raw[33], there are really only 13 bits of
checking here.)

raw[0..31] is then the pubkey suitable for use in Ed25519.  As before
(and independently of the AONT stuff), you could sanity-check it to make
sure that (a) it is not the identity element, and (b) L times it *is*
the identity element.  (L is the order of the Ed25519 group.)  Checking
(a) is important; checking (b) isn't strictly necessary for the reasons
given before, but is still a sensible thing to do.  If you don't check
(b), you actually have to check in (a) that the pubkey isn't one of 8
bad values, not just the identity.  So just go ahead and check (b) to
rest easier. ;-)


This version contains two calls to SHA3, as opposed to the one such call
in the non-AONT (but including a checksum) version.  The benefit is
Alec's (and others') desire that there cannot be any bits an attacker
could twiddle that would leave both the key the same and the address
looking OK to somone who just spot-checks say the beginning and/or the
end.
-- 
Ian Goldberg
Professor and University Research Chair
Cheriton School of Computer Science
University of Waterloo