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You get a specific Target value when you divide the maximum Target by the Difficulty.

But when you convert the Target in to the Bits format, you're only taking the first 3 bytes of the Target and losing some accuracy.

For example:

Difficulty: 166851513282.78

Original Target: 0x696f3ffffffe0c000000000000000000000000000000000
Target -> Bits: 0x180696f4
Bits -> Target:  0x696f4000000000000000000000000000000000000000000

So when a Miner is trying to get a low enough hash value for the block, are they trying to get below the Bits value or the Target value?

1 Answer 1

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Below the target value that corresponds to the bits field in the block.

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  • So the Bits field is ultimately a short (and slightly less accurate) representation of the Target, just for the purposes of having a representation of the Target in the block header?
    – inersha
    Commented Apr 14, 2016 at 12:01
  • Also, I'm assuming Bits is always rounded up?
    – inersha
    Commented Apr 14, 2016 at 13:30

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