Maximum Difficulty? | Bitcointalksearch.org

Rannasha

hero member

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Merit: 500

Quote from: sgk on December 19, 2013, 12:58:29 AM

I understand that the mining difficulty of Bitcoin is controlled by the number of leading zeros in the 24-digit block solution. Now if one assumes no leading zeros and all 24 digits have to be brute-forced in order to get the solution, it would be considered maximum mining difficulty.

This assumption is incorrect. Miners attempt to find a hash that is smaller than a target value, so having at least a given amount of leading zeroes. The higher the difficulty, the lower the target value and the higher the number of leading zeroes.

t3a

full member

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I don't think there is enough computing power in the universe to get to that point.

Anyway, to get the difficulty from a target, you divide 0x00000000FFFF0000000000000000000000000000000000000000000000000000 by the target.

If the target was 0x000000...0001, your difficulty would be 0x00000000FFFF0000000000000000000000000000000000000000000000000000 which in decimal is approximately 2.695953529101131*10^67

You can read more about this here: https://en.bitcoin.it/wiki/Difficulty

sgk

legendary

Activity: 1470

Merit: 1002

!! HODL !!

I understand that the mining difficulty of Bitcoin is controlled by the number of leading zeros in the 24-digit block solution. Now if one assumes no leading zeros and all 24 digits have to be brute-forced in order to get the solution, it would be considered maximum mining difficulty.

My question is what is this maximum difficulty value? Also, in other words, how many leading zeros are there in current difficulty?

Topic: Maximum Difficulty? (Read 1035 times)