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Adding onto @chytrik answer, the "proof" that it's impossible to find such a hash is purely probabilistic; perhaps there is some transaction hash which by sheer mind-boggling coincidence would not be hashable with all the others or any nonce to find a block with the necessary difficulty. If we supposed that this were the case, the next question would be how ...


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I think that the answer to your question is no, a transaction which makes every block it is included in invalid does not exist. An important core property of a cryptographic hash function is that there is no discernable relationship between the input data, and output data. If such a relationship existed, then miners could 'cheat' by only creating blocks (...


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