[Cryptography] BitCoin Question - This may not be the best place to ask

[Robert Christian]

On Dec 22, 2013, at 9:44 PM, Robert Christian <robertjchristian at gmail.com> wrote:

> 
> On Dec 22, 2013, at 6:31 PM, Robert Christian <robertjchristian at gmail.com> wrote:
> 
>> Exactly my point.  What's the collision resolution strategy and why isn't this a scary proposition?
>> 
> 
> I did the math on this and it starts to make sense without a collision strategy.  For ID's, it's 34 characters that can be 0..9, a..z, and A..Z.  That's 64^34.  If you could do 5,000 wallet generations per second, it's =(64^34)/(5000*60*60*24*365*1E+51) to get 16 percent all possible addresses within a year with all systems working full time - being a total of 100,000,000,000,000,016,384,608,344,632,472,552,568,168,984,184,560 machines on the task.  I think that settles it as a non-efficient means of hacking, at least for the next decade or so until quantum computing comes into play.
> 
> Can someone please check my math?  And my premise in general?  I feel like I might be missing something fundamental here…. and I think at this point it’s established that there is no collision strategy at all regardless of the fact it’s unlikely?

>>

Edit:  There are only 58 possible characters.  0OlI are excluded from the set to avoid being misread.  and there are a few characters within the address used for version and checksum.

> 
> 
>> On Sunday, December 22, 2013, Steve Weis wrote:
>> On Sun, Dec 22, 2013 at 4:30 PM, Robert Christian
>> <robertjchristian at gmail.com> wrote:
>> > 2) I am pointing out that addresses are finite, and 34 chars long... They
>> > can only be upper or lower case, or 0..9.  So at the end of the day, after
>> > all the fancy stuff, the number of all possible bitcoin addresses is
>> > (26*2+10)^34 possible unique ids.
>> >
>> > So the number of possible unique addresses is actually relatively smalll.
>> > Right?
>> 
>> The address has 20-bytes of hash, a network ID byte prefix, and a
>> 4-byte checksum. So, there are 2^160 possible unique addresses. This
>> is converted into a 34 character base-58 string.
>> 
>> You do bring up one point that many key pairs will collide for a
>> particular address. That's why the hash function must be assumed to be
>> collision resistant.
>> 
>> As for when we might see collisions, with a birthday attack you'd
>> expect there to be a 50% chance of some collision existing when there
>> are roughly 2^80 addresses.
> 

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