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I'm reading https://en.bitcoin.it/wiki/Difficulty

And I don't get why max target is not (2^256 - 1) which is 0xFFFFFFFFFFFF... but 0x00000000FFFF...?

ycshao
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  • This is an interesting question. However I suggest you flesh it out a bit, citing the resource you mention on the relevant part. – nulldev Jan 06 '18 at 21:11
  • The main point is that the hash generated by miners needs to have a specific number of leading zeros, thus it needs to be lower than the hexadecimal target. – Paul Razvan Berg Jun 24 '18 at 09:31

1 Answers1

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If the maximum target was 2256-1, every candidate block would be a valid block.

As even the CPU miner in Bitcoin's first software release was capable of doing 100 kH/s or more, this would have led to a very rough start of the chain, with 1000s of blocks produced per second until the difficulty adjusted.

Because of that reason, the maximum target was probably set to a level that would guarantee not too frequent blocks after genesis.

Pieter Wuille
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