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Suppose below is target difficulty:

0x0000CDEF0A000000000000000000000000000000000000000000000000000000

So result value must lower than target to select block.

So, If Miner 'A' found:

0x000000000000000000000000076898778ABEFDC34798734590349843BDCAE854

Will miner 'A' can satisfy target difficulty or not?

And if not then please tell me the reason.

Thanks

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Yes. The test is that in order to be a valid proof of work, the hash found must be numerically less than or equal to the target. That is certainly true of the hash found by Miner A in your example.

  • Thanks, Still I am not clear. I referred blockexplorer.com some completed blocks,and found "Hash Value" has same number of Previous zeros as target has. I mean if Target has 4 previous zero(0x000043AbC..),then found hash has same number of zero(0x000024C...). Why all the Found hash same number of zeros? It may also generate more than 4 zeros and satisfy less than target. – Tushar Aug 14 '14 at 8:28
  • @tushar: It certainly could have 4 more zeros, but this will only happen 1/65536 of the time, so it's not surprising we rarely see it. – Nate Eldredge Aug 14 '14 at 15:35

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