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These are followup questions from Transaction Fee in Confidential Transactions

In Confidential Transaction, Alice own 10btc sends 1 btc to Bob

Input : C1 = Commit (10btc) = 10 G + 10 H
Output: C2 = Commit (1 btc) = 1 G + 1 H
C3 = Commit (8.999 btc) = 8.999 G + 8.999H
C4 = Commit (stated as fee) (0.001 btc) = 0.001G + 0.001H
C1 = C2 + C3 + C4. (for simplicity reason, the r (blinding value rG) is the same as v (actual btc value vH).

My questions are
Q1. Is the winning miner is pre-determined already?
Q2. If Q1 = Yes? Then what is the coinbase address?
Q3. If Q1 = No? Then how does Alice know which address to send the fee to?
Q4. Suppose the fee is not yet sent to the winning miner yet? Then how does the winning miner know the blinding value of r = 0.001 since Alice don't know which miner to tell. Or the r and v values are not encrypted at all for this case.

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Q1. That depends on the blockchain's consensus rules. It has nothing to do with Confidential Transactions. In Blockstream Liquid, the miners (really, federation signers) rotate, and one by one propose a block which 2/3s of the other signers need to agree on. There are other systems that use some variant of Confidential Transactions that work completely differently.

Q2. Whatever the miner wants.

Q3. Alice doesn't send the fee to any address. Her transaction simply lists inputs, outputs, and fee (the fee is not an output, it has no address; it's just a field that says "the fee is 0.001", in cleartext - no Pedersen commitment). In your example, the transaction would list {inputs: [utxo with output C1,witness], outputs:[(C2,address1),(C3,address2)], fee:0.001}, and the validity rule would be C1 = C2 + C3 + 0.001*H).

Q4. There is no blinding for the fee, so there is nothing the miner needs to know.

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  • Thank You very much Pieter. Now, it very super clear.
    – Cisco Mmu
    Commented Jul 10, 2021 at 4:27

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