Proof-of-Work in Action

Block 550204 has two transactions in it. Below we can see the raw transaction data converted to the TXID through a double application of SHA256 and the Merkle root calculated from a double application of SHA256 upon the concatenated TXIDs that would make up the leaf nodes of the two-layer Merkle tree. Note that the output from the hash functions is displayed as little-endian notation so a reversal of the byte string of the hexadecimal character pairs of the Merkle root is necessary to get the same Merkle root that's displayed as big endian in block explorers. Similarly, if you are building a Merkle tree out of TXIDs that you have found on a block explorer, they will be listed in Big Endian meaning a byte reversal to Little Endian is necessary before concatenating as pairs.

TX	Raw TX Data	HASH256 (TXID in Little Endian)	HASH256 [1|2] (Little Endian)
1	01000000010000000000000000000000000000000000000000000000000000000000000000 ffffffff19033c65083838676a693166676c00966c47be0b09ffb3650300ffffffff016684814a00000 0001976a9145983d46a37697ef84bb0a0de00fcc169da06847588ac00000000	0f3691f02f24428e273f430c6a88410b2ad4f82150718c7a1c9ba50a417806e5	44049cf6ea2d2f283dc824cf7d47ca23b0dfe457e7496806a1162c74a32d6eaa
2	01000000011028cf06b9fd7eb47f53c0eb494faa73b703d179c4db8a82a4006e0c1d93178e01000 0006b483045022100af1bdcee3ebae916a6072e390f26cd4526741dd1ac79c8ad552253f24e4f9f 0d0220210722fb3900f246687de07dda8c66f48834041886c4bb640885a994e36351d7412103e 1d0ee3e2428df53ab81576b6e60631e97bf9a692ae57b311202a693196638ecfeffffff02e44e0f00 000000001976a9143fcf2e0815f62d7292867db34e0a48cebf5b25a788ac600bb1010000000019 76a914efdadcd32e8bedb8fa6eef66e7d6fb0b301f864288ac3a650800	e93ece6a702454aa3595349a89683f16ff651a14362e70bbaaa852eb5d70a455
Merkle Root or HASH256 [1|2] in Big Endian	aa6e2da3742c16a1066849e757e4dfb023ca477dcf24c83d282f2deaf69c0444

Next, we see the Merkle root from the above calculations input as one element into the data fields of the block header. These values were obtained from a block explorer so again conversions are necessary to get all the decimal elements to hexadecimal and then a byte string reversal of the hexadecimal character pairs before all the data elements can be concatenated into a single 80-byte string. Although confusing for the human interpreting the endianness conventions, it is much more efficient for hardware to process inputs in little endian, and although such gains would not be noticeable when only one operation is performed, BSV is designed to handle a transaction throughput of millions of transactions a second, so any optimisation processes have been implemented in the protocol design.

Data Field	Bytes	Value	Hexadecimal (Little Endian)
Version	4	20000000	00000020
Hash Previous Block	32	000000000000000001822fa70881fd8b3a26d85f347cca3fccbcd2a95b685792	9257685ba9d2bccc3fca7c345fd8263a8bfd8108a72f82010000000000000000
Merkle Root	32	aa6e2da3742c16a1066849e757e4dfb023ca477dcf24c83d282f2deaf69c0444	44049cf6ea2d2f283dc824cf7d47ca23b0dfe457e7496806a1162c74a32d6eaa
Timestamp	4	1538351826	d262b15b
nBits	4	402785011	f3020218
Nonce	4	1460619339	4b480f57
#550204 String	80	000000209257685ba9d2bccc3fca7c345fd8263a8bfd8108a72f8201000000000000000044049cf6ea2d2f283dc824cf7d47ca23b0dfe457e7496806a1162c74a32d6eaad262b15bf30202184b480f57

In order to evaluate whether an attempt at the hash puzzle was successful, the output is checked against a value that can be calculated from the 4-byte nBits value using the following formula.

= coefficient * 2 ^ ( 8 * (index — 3) )
= 0x0202f3 x 2^(8 x (0x18-3))
= 131827 x 2^(8(24-3))
= 4.93x10^55

0x denotes that this value is in hexadecimal

This calculation will yield an extremely large number, so by using the 4-byte value, the node software is able to parse the string for two pieces of information embedded within it. The first byte represents the index which will determine which power the following 3 bytes (coefficient) will be raised by. The nBits value is generally displayed in block explorers as a decimal number yet the calculation will be performed on the hexadecimal representation of that number by the node software. In the case of the block, we are working with the hexadecimal representation of nBits is 180202F3, where 18 (24 in decimal) will be the index and 0202F3 (131827 in decimal) the coefficient.

Hexadecimal	Decimal
1st Hash	eee9428aaf8117633011f5fc9f225641c239ee9839af775b48f67fd16512cd9e	108062592583125048371800204014412291248245351226987386384426656356890072763806
2nd Hash	2b68c4606e28ff6193f2734243eb0cddc660fd931b1636866dc80c6ec27bf7cf	1.96x1076
Difficulty Target	4.93x10^55
Difficulty minus Blockhash	-1.96x10^76

As we can see the first attempt at generating a successful proof of work solution was invalid as the output from the double hashing of the block header was much higher ($10^{21}$) than the target value specified from inputting the nBits value into the equation. Assuming that no new transactions have been appended to the Merkle tree and we are still within the same second of the UNIX time, a new attempt at the proof of work can be made by incrementing the nonce by 1. This yields an almost entirely similar 80-byte string as the first concatenated block header data inputs with only one hexadecimal character being incremented. Yet, when we check the output from the double application of the SHA256 hash function to the string we can see a significantly lower value with a large number of leading zeros.

Data Field	Bytes	Hexadecimal Value (Little Endian)
Version	4	00000020
Hash Previous Block	32	9257685ba9d2bccc3fca7c345fd8263a8bfd8108a72f82010000000000000000
Merkle Root	32	44049cf6ea2d2f283dc824cf7d47ca23b0dfe457e7496806a1162c74a32d6eaa
Timestamp	4	d262b15b
nBits	4	f3020218
Nonce	4	4c480f57
String	80	000000209257685ba9d2bccc3fca7c345fd8263a8bfd8108a72f8201000000000000000044049cf6ea2d2f283dc824cf7d47ca23b0dfe457e7496806a1162c74a32d6eaad262b15bf30202184c480f57
Block Hash (Little Endian)	32	2bbc7a5bfd73ab81e8ed273e7c0568ae9ff2aebb7e6657010000000000000000
Block Hash (Big Endian)		00000000000000000157667ebbaef29fae68057c3e27ede881ab73fd5b7abc2b
Block Hash (decimal)		3.28x10^55
Difficulty target		4.93x10^55
Difference (target-blockhash)		1.65x10^55

Converting the number to a decimal value and comparing it to the target value calculated by the 180202f3 used in the difficulty target equation, we can see that the output is indeed lower than the target value and a correct solution for the proof of work challenge has been found! In the next chapter we will look at how this solution is communicated to the nodes on the network and how they can check the proof of work before referencing this block hash in their next block header candidate.