3 ms·

It doesn't bother you that your method counts only unique hash values tested? Or that this value is being used on the site to talk about power expenditure (whic

It doesn't bother you that your method counts only unique hash values tested? Or that this value is being used on the site to talk about power expenditure (which is clearly related to total hashes per second, not unique hashes)? You're right that you can compute the N-second average rate easily. But nobody reports the "network hash rate" as such. And since the denominator matters a lot in determining what the number means, pretty much all the numbers you find out in the world are (1) misreported and (2) bogus.

gibybo 14y ago
It does count only unique hash values, but why should that bother me? People don't randomly generate the same 256 bit number very often (in fact, the probability is so low that it has almost certainly never happened) When I search for "bitcoin network hashrate" on google, the first link is http://bitcoin.sipa.be/ http://bitcoin.sipa.be/, which does in fact report the N-second average rate. It reports the 7-days, 14-day, and 30-day average rate.
- Danieru 14y ago
  Duplicate hashes are an issue because of the birthday problem: http://en.wikipedia.org/wiki/Birthday_problem http://en.wikipedia.org/wiki/Birthday_problem The chance of two hashes being equal is small but the number of two hash pairs grows exponential to the number of hashes.
  - gibybo 14y ago
    No, it's still not an issue. Check out the table for 256 bit numbers in the article you linked. To get a 0.1% probability of even a single collision, you need to generate 1.5e37 hashes. At 10 TH/s, that would take 3e17 years. 10 million times longer than the age of the universe.
  - mikeash 14y ago
    The birthday paradox basically says that you can expect a collision when you've explored the square root of the problem space. For birthdays, you expect a collision in a group of sqrt(365) ~= 19 people. For a 256-bit hash, you expec a collision in a group of 2^128 hashes. That is still huge. Absent a weakness in the hash being found, it's unlikely anyone will ever generate two colliding hashes.

gibybo 14y ago

It does count only unique hash values, but why should that bother me? People don't randomly generate the same 256 bit number very often (in fact, the probability is so low that it has almost certainly never happened) When I search for "bitcoin network hashrate" on google, the first link is http://bitcoin.sipa.be/ http://bitcoin.sipa.be/, which does in fact report the N-second average rate. It reports the 7-days, 14-day, and 30-day average rate.

Danieru 14y ago
Duplicate hashes are an issue because of the birthday problem: http://en.wikipedia.org/wiki/Birthday_problem http://en.wikipedia.org/wiki/Birthday_problem The chance of two hashes being equal is small but the number of two hash pairs grows exponential to the number of hashes.
- gibybo 14y ago
  No, it's still not an issue. Check out the table for 256 bit numbers in the article you linked. To get a 0.1% probability of even a single collision, you need to generate 1.5e37 hashes. At 10 TH/s, that would take 3e17 years. 10 million times longer than the age of the universe.
- mikeash 14y ago
  The birthday paradox basically says that you can expect a collision when you've explored the square root of the problem space. For birthdays, you expect a collision in a group of sqrt(365) ~= 19 people. For a 256-bit hash, you expec a collision in a group of 2^128 hashes. That is still huge. Absent a weakness in the hash being found, it's unlikely anyone will ever generate two colliding hashes.