Quote from: achow101 on August 30, 2017, 02:14:25 PM

The only problem with this setup is the trusted third party. Everyone would have to trust that you don't give them the generated private key and that they don't give you their private key. But everyone is going to know their private key anyways, so there is a lot of security risk there.

Instead of using a trusted third party, what you could do is have each participant generate their own new key pair. When you generate addresses, you add all participants' public keys along with the one you generated in order to find a vanity address. This would be multi-party split key generation. Once an address is found, all participants send their private keys to the person who got their vanity address. Then they generate new key pairs and send the public keys back to you for the next vanity address to be generate. In this way, no one but the receiver of the vanity address can actually know the full private key to it.

This problem can be solved as follows: third parties who will provide public keys for generating a large number of vanity addresses should work according to some rules, for example:
1. function "get public key" - always returns a new public key to generate a large number of vanity addresses;
2. function "get private key" - based on a public key from function 1 the private key is given out. You can request a private key ONLY 1 time, then the public key is considered inactive and can not be used. When you repeatedly request the same private key, the service should give an error and say that the key has already been issued (compromised).

The service for generating vanity addresses will work according to the following algorithm: a request for a public key from a third party, a private key, it will not be requested, because the private key will then be compromised. Next, the service looks for many vanity addresses, if found, then sends the customer a vanity address and a third-party public key. And when the customer wants to get a private key from a third party, he will already know whether the key is compromised or not. And then the algorithm repeats.

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