3 ms·
Let’s assume Alice is an antagonist coordinating with Carol to unmask bob. How many self to self tx does bob need to do to hide from Alice and carol?
by josh2600 8y ago
Let’s assume Alice is an antagonist coordinating with Carol to unmask bob.
How many self to self tx does bob need to do to hide from Alice and carol?
- rdl 8y agoI believe it depends on how much of the network the adversary can monitor, and high a confidence Alice/Carol need in Bob’s identity to take action. If the action is “subject him to additional off-chain scrutiny”, the bar is presumably lower than “criminal conviction or assassination solely on this evidence”. The more I learn about MW the less suitable it seems for my goals.
- josh2600 8y agoI think Alice and Carol need knowledge of 2 tx with Bob to coordinate a common ancestor; that's it. If you imagine the ledger like a deck of cards, there is probably a point where all txo's are shuffled (and we can likely make a conjecture that after N shuffles, the position of any txo is unknown, it's just hard to gauge how many shuffles we need to do for real privacy guarantees. If you want 6 sigma privacy, you might need to wait a very long time).