Confidential Transactions(people.xiph.org)
people.xiph.org
Confidential Transactions
https://people.xiph.org/~greg/confidential_values.txt
7 comments
Related, this was just published today: http://voxelsoft.com/dev/sumcoin.pdf
> A commitment scheme lets you keep a piece of data secret but commit to it so that you can not change it later.
> commitment = SHA256( binding_factor || data )
> Tell someone the commitment, then [later] reveal both the data and blinding factor.
It looks like I can change my data, then generate a binding factor that will combine to produce the original hash input.
> commitment = SHA256( binding_factor || data )
> Tell someone the commitment, then [later] reveal both the data and blinding factor.
It looks like I can change my data, then generate a binding factor that will combine to produce the original hash input.
>> A commitment scheme lets you keep a piece of data secret but commit to it so that you can not change it later.
>> commitment = SHA256( binding_factor || data )
>> Tell someone the commitment, then [later] reveal both the data and blinding factor.
>It looks like I can change my data, then generate a binding factor that will combine to produce the original hash input.
If you can find SHA256 collisions on demand. But if you can do that, you should probably be writing a paper about it and advancing the state of the art.
>It looks like I can change my data, then generate a binding factor that will combine to produce the original hash input.
If you can find SHA256 collisions on demand. But if you can do that, you should probably be writing a paper about it and advancing the state of the art.
Assume for a moment that no hashing is performed.
I compute C = B || D.
I reveal C.
I later choose new data D'.
I compute C = B' || D'.
I reveal B' and D'.
Since both B and D were secret, B' and D' are accepted.
Secretly masking data lends to malleability. (EDIT: Not a mask)
EDIT: As CJefferson points out the operation is not a mask, but concatenation of a fixed length random value which invalidates this example. Exploiting this secrecy would require a weakness in SHA256 that allows input prefixes to produce colliding hash states (hard).
I compute C = B || D.
I reveal C.
I later choose new data D'.
I compute C = B' || D'.
I reveal B' and D'.
Since both B and D were secret, B' and D' are accepted.
Secretly masking data lends to malleability. (EDIT: Not a mask)
EDIT: As CJefferson points out the operation is not a mask, but concatenation of a fixed length random value which invalidates this example. Exploiting this secrecy would require a weakness in SHA256 that allows input prefixes to produce colliding hash states (hard).
Here || denotes concatenation. Therefore your only options are to change where you split B and D into two two strings. If you (as is common) either fix the length of B, or make sure the splitter marker characters can't occur in B, then given C, B and D are fixed.
Only if you can break SHA. But still, good point.
This is related to:
https://news.ycombinator.com/item?id=9684066
https://news.ycombinator.com/item?id=9684066