<p>The hardest possible target is all zeros plus a one (<code>0000000...1</code>). 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.</p>

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

CC BY-SA 3.0
2013-05-22T10:11:58.893Z
Do you mean that there's only one block with the hash zero?
mASOUD
CC BY-SA 3.0
2013-05-22T10:15:32.073Z
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.
Tom van der Woerdt
CC BY-SA 3.0
2013-05-27T16:25:27.777Z
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
CC BY-SA 3.0
2013-05-27T16:38:28.980Z
Also if block `0000000...1` was mined (by coincidence) today while the target is much higher, it would be harmless.
finnw
CC BY-SA 4.0
2024-01-19T14:51:23.983Z
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.
Pieter Wuille
CC BY-SA 3.0
2013-05-21T17:22:19.997Z
2013-05-21T17:22:19.997Z
Tom van der Woerdt
<p>According to <a href="https://en.bitcoin.it/wiki/Target" rel="nofollow">https://en.bitcoin.it/wiki/Target</a> 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)</p>

<hash><difficulty><hashpower>
What is the hash-rate of the Bitcoin network that results in the maximum difficulty?
mASOUD
# What is the hash-rate of the Bitcoin network that results in the maximum difficulty?

(Asked by: username_0 on 2013-05-21)

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)

**Tags:** hash, difficulty, hashpower


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## 1 answers

### Answer by username_1 on 2013-05-21. Score: 6 | Accepted answer

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.

> **Comments:**
> **username_0**: Do you mean that there's only one block with the hash zero?  
> **username_1**: 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.  
> **username_2**: 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.  
> **username_2**: Also if block `0000000...1` was mined (by coincidence) today while the target is much higher, it would be harmless.  
> **username_3**: 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.  

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