# How is a block header hash compared to the target (bits)?

I'm trying to wrap my head around the mining process by doing a small example of block hashing.

According to the Wiki entry about difficulty, the target for a block hash can be read from the "bits" part of the header as follows: In this example, the bits part is `535f0119`.

``````535f0119 * 2**(8*(0x1b - 3))
``````

My resulting target would be:

``````535f0119000000000000000000000000000000000000000000000000
``````

The target in decimal would be:

``````8780002705592212783085671453687210878315895819816253650256038723584
``````

Let's say the hash I got with my current nonce is

``````4d47599dd86834282a8ae6f20ba454704ddbe6eb23aa31b9fdec97fc7679b559
``````

How can I now compare if the hash is smaller than the target? What do I have to do with the hash to be able to say "hash < target"?

## How to calculate the target from bits

`04000000b9e2784a84e5d2468cee60ad14e08d0fee5dda49a37148040000000000000000e9dd2b13157508891880ef68729a1e5ecdde58062ebfa214a89f0141e5a4717faefd2b577627061880564bec`

From the 80-bytes, the bits are actually the 72nd to 76th byte:

`04000000b9e2784a84e5d2468cee60ad14e08d0fee5dda49a37148040000000000000000e9dd2b13157508891880ef68729a1e5ecdde58062ebfa214a89f0141e5a4717faefd2b57**76270618**80564bec`

or

`76270618`

This number, however, is in little-endian, so we have to reverse the bytes:

`18062776`

The first byte is the "exponent"

`e = 0x18`

The next 3 bytes are the "coefficient"

`c = 0x062776`

You plug this into a formula:

`target = c * 2**(8*(e - 3))`

In our case, that is:

`target = 0x062776 * 2**(8*(0x18 - 3))`

Which turns out to be:

`0000000000000000062776000000000000000000000000000000000000000000`

Let's calculate the hash of this block header using Python 2:

``````from hashlib import sha256
``````

The output is

``````0000000000000000040199a6c7b922f711ee7e98cd58863b8b981b02d2b83e13
``````

You can compare this to the target

``````>>> 0x0000000000000000040199a6c7b922f711ee7e98cd58863b8b981b02d2b83e13 < 0x0000000000000000062776000000000000000000000000000000000000000000
True
``````

That's how we know a block satisfies the proof-of-work.

• Could difficulty be argued to be a percent then? i.e. d = (target hash)/(most difficult target hash)
– Josh
Commented Jun 25, 2018 at 17:09
• In the last line of code the comparison shouldn't be `<=` instead of `<`? Commented Oct 4, 2022 at 22:28