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Private attack is to violate the safety of a block B by creating another chain of length equal to or longer than the longest chain containing B, after B has been confirmed.

As in my reference

If the adversary reveals its chain too early, i.e. before block B gets k-deep. It’ll still end up displacing B, but then it’s action do not lead to a safety violation.

Why won’t it pose a threat on safety? Isn’t displacing B in the future already means block mined by the malicious nodes be adopted as the longest chain and block containing malicious transaction might get confirmed once it become k-deep?

Ref: Private attack section in https://courses.grainger.illinois.edu/ece598pv/sp2021/lectureslides2021/ECE_598_PV_course_notes6.pdf

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The chances of a block being reorganized because of a new chaintip displacing the current one drop exponentially as the number of blocks in the current chaintip increases. Since in theory a chain of any depth could be reorganized, it is impossible to define a "safe" depth, but in practice we commonly see recommendations of up to 6 blocks before a transaction can reasonably assumed to be final. In other words, the safety violation depends on your assumption of how many confirmations are required.

It is my understanding that this is the kind of safety referred to in the document you linked. For example, say you're selling your car for Bitcoin but you agree with the buyer that you'll only give him the car keys as soon as the transaction has 6 confirmations. If the buyer tries to attack you by secretly mining a parallel chain and double spending their payment, the safety of your agreement is not violated if they decide to publish their private alternative chain too early. After all, you've agreed to wait to give them the keys before reaching the 6 confirmations.

Note of course that successfully mining a private chain in parallel would require the attacker to have a majority of hashrate (leaving aside selfish mining), which in the current environment is extremely difficult and costly.

See also Why is 6 the number of confirms that is considered secure?

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