Subject: Re: [Bitcoin-development] Optional "wallet-linkable" address format - Payment Protocol

From: Alan Reiner <[email-masked]>
Date: Wed, 19 Jun 2013 10:39:04 -0400

On 06/19/2013 10:25 AM, Timo Hanke wrote:
> Since you mention to use this in conjunction with the payment protocol,
> note the following subtlety. Suppose the payer has to paid this address
> called "destination": 
>>    Standard Address ~ Base58(0x00 || hash160(PubKeyParent * Multiplier[i]) ||
>> checksum)
> Also suppose the payee has spent the output, i.e. the pubkey
> corresponding to "destination", which is PubKeyParent * Multiplier[i],
> is publicly known. Then anybody can (in retrospect) create arbitrary
> many pairs {PublicKeyParent, Multiplier} (in particular different
> PublicKeyParent) that lead to the same "destination".
>
> Depending on what you have in mind that the transaction should "prove"
> regarding its actual receiver or regarding the receiver's PubKeyParent,
> this could be an unwanted feature (or it could be just fine). If it is
> unwanted then I suggest replacing
> PubKeyParent * Multiplier[i] by 
> PubKeyParent * HMAC(Multiplier[i],PubKeyParent)
> which eliminates from the destination all ambiguity about PubKeyParent.
>
> This modification would not be directly compatible with BIP32 anymore
> (unfortunately), but seems to be better suited for use in conjunction
> with a payment protocol. 
>
> Timo

It's an interesting observation, but it looks like the most-obvious
attack vector is discrete log problem:  spoofing a relationship between
a target public key and one that you control.   For instance, if you see
{PubA, Mult} produces PubB and you have PubC already in your control
that you want to "prove" [maliciously] is related to PubB, then you have
to find the multiplier, M that solves:  M*PubC = PubB.  That's a
discrete logarithm problem.

I'm not as familiar as you are, with the available operations on
elliptic curves, but it sounds like you can produce essentially-random
pairs of {PubX, Mult} pairs that give the same PubB, but you won't have
the private key associated with those public keys.  It's an interesting
point, and there may be a reason to be concerned about it.  Though, I
don't see it yet.

-Alan