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

Question

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"?

Answer 1

How to calculate the target from bits

Let's start with a block-header, always 80-bytes that looks like this:

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
header = "04000000b9e2784a84e5d2468cee60ad14e08d0fee5dda49a37148040000000000000000e9dd2b13157508891880ef68729a1e5ecdde58062ebfa214a89f0141e5a4717faefd2b577627061880564bec".decode('hex')
print sha256(sha256(header).digest()).digest()[::-1].encode('hex')

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.