Topic: What kind of "puzzles" do bitcoin mining programs solve? (Read 704 times)
completely agree that this race))) and so it is faster, the more computing power of your equipment. the main thing you need to understand is that the earlier your equipment will make a profit, then you will not have questions, because you will know the answers to them.
thank you so much ^^
Just read this:
https://en.bitcoin.it/wiki/Difficulty
https://en.bitcoin.it/wiki/Target
People sometimes use "target" and "difficulty" interchangeably but they are different.
Here is the key formula that should bring it all together for you:
Anyhoo, just read the pages and you will understand it.
https://en.bitcoin.it/wiki/Difficulty
https://en.bitcoin.it/wiki/Target
People sometimes use "target" and "difficulty" interchangeably but they are different.
Here is the key formula that should bring it all together for you:
Quote
difficulty = difficulty_1_target / current_target
Anyhoo, just read the pages and you will understand it.
Can you post an example?
It is not a puzzle : it is a lottery.In simple words : Bitcoin mining protocol consist in a hash calculation of a block of transactions, and this hash (a hash of an other hash in act) has to be a number lower than the difficulty.
The difficulty number is very low and there very few chances to find a number lower than low.
This makes zero sense. If your explenation is right, that would mean that an infinite difficulty would lead to instantly block finding, which is the opposite of what is the case.
It would also mean that a difficulty of 2 would mean that it is extremely hard to find a block, which is also not true.
You're wrong, and the users who posted above you already explained it the right way.
Can you post an example?
It is not a puzzle : it is a lottery.In simple words : Bitcoin mining protocol consist in a hash calculation of a block of transactions, and this hash (a hash of an other hash in act) has to be a number lower than the difficulty.
The difficulty number is very low and there very few chances to find a number lower than low.
Here is a slightly simplified form of the puzzle:
Given a target value, T, and a block header, H, that includes the value N,
find a value for N such that:
SHA-256(SHA-256(H)) <= T
The only known way to solve the "puzzle" is brute-force guessing.
Given a target value, T, and a block header, H, that includes the value N,
find a value for N such that:
SHA-256(SHA-256(H)) <= T
The only known way to solve the "puzzle" is brute-force guessing.
Can you post an example?
It is not a puzzle. It is also not that hard to explain. Mining involves hashing the current block plus a changing nonce as fast as you can. First one to find the hash that meets the current difficulty wins. Not a puzzle, more of a race.
Can you post an example?
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