A big milestone - 2^70 hash operations

dree12

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Quote from: oakpacific on May 12, 2013, 09:20:56 PM

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

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.

oakpacific

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

Mysticsam_3579

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Quote from: BitshireHashaway on May 12, 2013, 07:34:38 PM

Quote from: nimda on May 12, 2013, 02:35:28 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:28:00 PM

Quote from: Scrat Acorns on May 12, 2013, 02:14:12 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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.

BitshireHashaway

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Quote from: nimda on May 12, 2013, 02:35:28 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:28:00 PM

Quote from: Scrat Acorns on May 12, 2013, 02:14:12 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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:)

nimda

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0xFB0D8D1534241423

Quote from: Remember remember the 5th of November on May 12, 2013, 02:28:00 PM

Quote from: Scrat Acorns on May 12, 2013, 02:14:12 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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

dree12

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Activity: 1246

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Quote from: Remember remember the 5th of November on May 12, 2013, 02:28:00 PM

Quote from: Scrat Acorns on May 12, 2013, 02:14:12 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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.

ShadowOfHarbringer

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Bringing Legendary Har® to you since 1952

Quote from: Remember remember the 5th of November on May 12, 2013, 02:28:00 PM

Quote from: Scrat Acorns on May 12, 2013, 02:14:12 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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.

Remember remember the 5th of November

legendary

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Reverse engineer from time to time

Quote from: Scrat Acorns on May 12, 2013, 02:14:12 PM

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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?

Scrat Acorns

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Activity: 293

Merit: 250

Quote from: Remember remember the 5th of November on May 12, 2013, 02:08:47 PM

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

legendary

Activity: 1862

Merit: 1011

Reverse engineer from time to time

Quote from: Scrat Acorns on May 12, 2013, 01:59:06 PM

Quote from: frozen on May 12, 2013, 01:49:44 PM

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

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Activity: 293

Merit: 250

Quote from: frozen on May 12, 2013, 01:49:44 PM

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.

frozen

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Can you explain how this is being calculated?

Scrat Acorns

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Activity: 293

Merit: 250

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.

Topic: A big milestone - 2^70 hash operations (Read 1499 times)