Peter R [ARCHIVE] on Nostr: đź“… Original date posted:2016-05-09 đź“ť Original message:[9 May 16 @ 6:40 PDT] For ...
đź“… Original date posted:2016-05-09
đź“ť Original message:[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 at 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
>
> _______________________________________________
> bitcoin-dev mailing list
> bitcoin-dev at lists.linuxfoundation.org
> https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev
đź“ť Original message:[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 at 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
>
> _______________________________________________
> bitcoin-dev mailing list
> bitcoin-dev at lists.linuxfoundation.org
> https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev