Do Miners have to get below the Target or the Bits value?

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?

inersha

Posted 2016-04-14T11:16:41.197

Reputation: 2 236

Answers

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

Pieter Wuille

Posted 2016-04-14T11:16:41.197

Reputation: 54 032

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 – 2016-04-14T12:01:27.353

Also, I'm assuming Bits is always rounded up? – inersha – 2016-04-14T13:30:09.040