Quote from: d5000 on May 15, 2019, 04:06:12 PM

@DDAsics: Monsterer is referring to the "nothing at stake" (N@S) problem, which affects mostly PoS algorithms (but in his opinion also affects PoC). If there is no value at risk if you mine more than one chain, this opens the way to some attack vectors. In PoW, attackers always have costs when they want to carry out a double spending attack, but in PoS they can try so continuously, practically "for free". Also in PoC there must be value at risk to mine more than one chain, if it doesn't want to suffer from N@S.

Thanks for that explanation, but I am still somewhat confused why that would be. In the case of mining multiple chains with PoC, there is not a "Nothing at Stake" problem that I can see. If you mine on 2 chains, you have to devote capacity resources to 2 chains, splitting how much you can earn. Only 1 of these chains will eventually win, and then all the value you earned on the second chain is lost. Had you devoted twice the capacity to the winning chain, you'd be much better off. It is completely different than the PoS model, where there is absolutely no resource usage required.

What I am trying to figure out is if there is any reason why resource usage in a non competitive environment such as PoC (or indeed an Proof of Useful Work algorithm) is any different than resource usage in a competitive environment, such as the existing SHA-256 PoW, that allocates rewards statistically rather than on a guaranteed basis. They seem to be identical to me from a mathematical point of view, but there is a group of people who insist they are not. The N@S problem is strictly a PoS issue, and does not enter into this particular question.