Re: [bitcoin-dev] Compact Block Relay BIP

From: Peter R <peter_r@gmx•com>
To: Bitcoin Protocol Discussion <bitcoin-dev@lists•linuxfoundation.org>
Subject: Re: [bitcoin-dev] Compact Block Relay BIP
Date: Mon, 9 May 2016 18:42:45 -0700	[thread overview]
Message-ID: <6BF6388E-2D1F-4A3D-BA57-B1AA94E40F7A@gmx.com> (raw)
In-Reply-To: <5C2809F9-286D-49E4-89DB-7109B73F6076@gmx.com>

[9 May 16 @ 6:40 PDT]

For those interested in the hash collision attack discussion, it turns out there is a faster way to scan your set to find the collision:  you’d keep a sorted list of the hashes for each TX you generate and then use binary search to check that list for a collision for each new TX you randomly generate. Performing these operations can probably be reduced to N lg N complexity, which is doable for N ~2^32.   In other words, I now agree that the attack is feasible.  

Cheers,
Peter

hat tip to egs

> On May 9, 2016, at 4:37 PM, Peter R via bitcoin-dev <bitcoin-dev@lists•linuxfoundation.org> wrote:
> 
> Greg Maxwell wrote:
> 
>> What are you talking about? You seem profoundly confused here...
>> 
>> I obtain some txouts. I write a transaction spending them in malleable
>> form (e.g. sighash single and an op_return output).. then grind the
>> extra output to produce different hashes.  After doing this 2^32 times
>> I am likely to find two which share the same initial 8 bytes of txid.
> 
> [9 May 16 @ 4:30 PDT]
> 
> I’m trying to understand the collision attack that you're explaining to Tom Zander.  
> 
> Mathematica is telling me that if I generated 2^32 random transactions, that the chances that the initial 64-bits on one of the pairs of transactions is about 40%.  So I am following you up to this point.  Indeed, there is a good chance that a pair of transactions from a set of 2^32 will have a collision in the first 64 bits.  
> 
> But how do you actually find that pair from within your large set?  The only way I can think of is to check if the first 64-bits is equal for every possible pair until I find it.  How many possible pairs are there?  
> 
> It is a standard result that there are 
> 
>    m! / [n! (m-n)!] 
> 
> ways of picking n numbers from a set of m numbers, so there are
> 
>    (2^32)! / [2! (2^32 - 2)!] ~ 2^63
> 
> possible pairs in a set of 2^32 transactions.  So wouldn’t you have to perform approximately 2^63 comparisons in order to identify which pair of transactions are the two that collide?
> 
> Perhaps I made an error or there is a faster way to scan your set to find the collision.  Happy to be corrected…
> 
> Best regards,
> Peter
> 
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