A big milestone - 2^70 hash operations

[Scrat Acorns]

The Bitcoin network reached a big milestone yesterday: it has performed over 2⁷⁰ (double) hashes since its conception.

Current count is:
1,191,253,457,820,256,820,938 = 2^70.01297 hashes

making it by far the most powerful distributed network computing system ever to have existed.

[frozen]

Can you explain how this is being calculated?

[Scrat Acorns]

Can you explain how this is being calculated?

You can get the hash count every time there's a SetBestChain in debug.log.

Then you log2 that number.

[Remember remember the 5th of November]

Can you explain how this is being calculated?

You can get the hash count every time there's a SetBestChain in debug.log.

Then you log2 that number.

But it's still probably just an estimate.

[Scrat Acorns]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

[Remember remember the 5th of November]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

And what percent is that number, from all the possible 2^256 SHA256 hashes?

[ShadowOfHarbringer]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

And what percent is that number, from all the possible 2^256 SHA256 hashes?

I join the question.

[dree12]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

And what percent is that number, from all the possible 2^256 SHA256 hashes?

That depends on whether or not double-SHA256 is reversible. For our purposes, we can assume it is. If it is, then the chances of a collision is near null and 2⁷⁰ is a very miniscule percentage of all possible double-SHA256 results.

To be exact, 2⁻¹⁸⁶.

NB: If a function f (x → y) is reversible over a domain D, that means that ∀y∈D ∃x such that f(x) = y. Here we define D to be all numbers between 0 and 2²⁵⁶−1.

[nimda]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

And what percent is that number, from all the possible 2^256 SHA256 hashes?

2^70 / 2^256 = 2^-186 = 1.02 x 10^-56 = 1.02 * 10^-54 %
0.00000000000000000000000000000000000000000000000000000102 %

Very small. Exponents are powerful.

Edit: fixed # of 0's

[BitshireHashaway]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

And what percent is that number, from all the possible 2^256 SHA256 hashes?

2^70 / 2^256 = 2^-186 = 1.02 x 10^-56 = 1.02 * 10^-54 %
0.00000000000000000000000000000000000000000000000000000102 %

Very small. Exponents are powerful.

Edit: fixed # of 0's

And just wondering, what happens when it reaches 100% or 2^256. And yes I agree, exponents are amazingly powerful:)

[Mysticsam_3579]

But it's still probably just an estimate.

My understanding is that it may overshoot or undershoot between blocks (because it doesn't know the exact number, only the average) but the average over all these years should be very close to the real number.

And what percent is that number, from all the possible 2^256 SHA256 hashes?

2^70 / 2^256 = 2^-186 = 1.02 x 10^-56 = 1.02 * 10^-54 %
0.00000000000000000000000000000000000000000000000000000102 %

Very small. Exponents are powerful.

Edit: fixed # of 0's

And just wondering, what happens when it reaches 100% or 2^256. And yes I agree, exponents are amazingly powerful:)

Precisely when we have done 2^255 hashes, when we have 50% left.   

[oakpacific]

With this kind of hashing capability, we can already birthday attack MD5/SHA1 fairly easily.

[dree12]

With this kind of hashing capability, we can already birthday attack MD5/SHA1 fairly easily.

That's rather interesting, actually. Since MD5 is only 128-bit, a hypothetical Bitcoin network on MD5 would have likely found over 2000 collisions by now. Obviously, collisions are not a problem in this context. But it sheds light into just how powerful the Bitcoin network is.

That hypothetical Bitcoin network would probably die in the near future from lack of space to adjust difficulty into. Bitcoin is lucky to be using 128-bit cryptography.