Timeline for What will a miner do when a high-fee transaction comes but he has constructed his block and is solving the puzzle?
Current License: CC BY-SA 4.0
4 events
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| Mar 7, 2020 at 21:37 | comment | added | Pieter Wuille | Computing a single hash takes time in the order of nanoseconds, and every attempt is independent. It's not because you've tried a billion failed hashes already that you're in any way "closer" to finding a solution. In other words: PoW is progress free, apart from the time it takes to compute a single hash, which is so short compared to the time scale at which transactions happen. | |
| Mar 7, 2020 at 18:45 | comment | added | MCCCS | SHA256, like other good cryptographic hash functions, has avalanche effect, which means even one bit changed in the input will cause the output to be entirely different. The new hashes found will be entirely different as the Merkle hash of the transactions (or the hash of all TXIDs in the block) can also act as a nonce to change output hash. @Better | |
| Mar 7, 2020 at 15:07 | comment | added | Better | Let's consider this situation: before the high-fee transaction comes, a miner didn't update his block and kept changing the "nonce" of the same block, the space of the useless results he got is S. Now the high-fee transaction comes, the miner reconstructs the block and restarts solving the puzzle, what if the results he gets fall into S again? Doesn't it mean that the work he has done is wasted? | |
| Mar 7, 2020 at 8:54 | history | answered | MCCCS | CC BY-SA 4.0 |