@koubiac,
You say that the only way for the attacker to try again
is to change the kernel, but if their attack fails
(chain is not accepted), then why can't they try
again with the same kernel?
Because if he tries again with the same kernel, he will produce exactly the same branch.
I'm not sure if this is clear or not. The hash being deterministic, the only way to try again (i.e. to try to obtain a different outcome) is to change the kernel.
No, you are not clear.
Look, an attacker can build any number
of DIFFERENT "branches" or chains very quickly.
Whether this so-called "kernel" changes
as a result of the various permutations of
transactions and blocks he's put together,
or whether it remains the same because the staking UTXOs
are the same, really doesn't matter.
Why doesn't it matter?
It doesn't matter because if the chain isn't accepted, the attacker
still has his UTXOs and can try again
Of course the attacker can try as many times as he wants I never said the contrary, what I'm saying is that he will never succeed
So either way, you do not need to change your UTXO set to
try more than once.
EDIT: The fact that the "hashes are deterministic" is really saying
nothing at all. That always is the case. How could they be random?
(Who would be generating the random numbers and how would they
be verified?) So yes, you would need to change the attacking
chain to get a different outcome against a different main chain,
but there's nothing stopping you from doing that.
I guess we're having a hard time understanding each other!
Let's do it differently, if you want give me some hypothesis: total UTXOs the attacker owns, what kind of attack he want to conduct (i.e. how far behind the attacker starts his fork) etc. and I will prove you mathematically that he will
never succeed if he doesn't own a very large portion of the mining coin.
The fact that he can try many times doesn't help him.
Otherwise, maybe you could describe how the attacker
tries many times and what he does to get different outcomes cause that's the part that's unclear to me in your explanation.