According to https://en.bitcoin.it/wiki/Target the target is a 256-bit number. As the total hash-rate of the Bitcoin network rises, this number decreases to increase the difficulty. How much hash-rate (in terms of GHash/s) would cause the target to reach the minimum value? will the target ever get to 0? Does that mean it will be impossible to find a new block? (the minimum of a SHA256 hash is 0 which is not less than 0)

1 Answer 1


The hardest possible target is all zeros plus a one (0000000...1). If such a block were ever mined (which is extremely unlikely: on average you'd need 2^255 hashes per 10 minutes, which is 9.6e73 hashes per second), it would definitely crash the network : there can only be one hash at that difficulty, which means that there can only be one block mined at that difficulty.

Of course, if the target is zero, no block can be mined at all.

  • Do you mean that there's only one block with the hash zero?
    – mASOUD
    Commented May 22, 2013 at 10:11
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    Yes. Block hashes must be unique, and the hash zero is already reserved as the prevblock for the genesis block, so a block with all zeroes is an impossible block. Commented May 22, 2013 at 10:15
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    The network would probably crash before that, because collisions would increase as the target approached the minimum, which would also cause unconfirmed transactions to pile up.
    – finnw
    Commented May 27, 2013 at 16:25
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    Also if block 0000000...1 was mined (by coincidence) today while the target is much higher, it would be harmless.
    – finnw
    Commented May 27, 2013 at 16:38
  • This is slightly off: a block is valid when its hash is less than or equal to the target. So if the target is 0, there is still one valid block hash possible. Commented Jan 19 at 14:51

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