Problem 5 (Bitcoin Selfish Mining, 20 Points)

Consider the selfish mining problem in Bitcoin. Notations are given in Lecture 3, Slide 35.

1. [10 pts] Describe all events that cause each transition in the state transition diagram in Lecture 3, Slide 36.
2. [10 pts] At which transitions in the state transition diagram in Lecture 3, Slide 36, the honest miners would earn the block reward, and how many?

slide 35:

slide 36:

Analysis Notations: - a: fraction of the selfish miner's computing power - y: probability that an honest miner chooses to mine on the selfish miner's block - State 0: there is only one longest chain all nodes are mining on (no branch) - State O': there are two public branches of length one: the main branch, and the branch that was private to the selfish miner - State i (i = 1,2, ...): the selfish miner's private branch is i blocks longer than the main branch Analysis (Cont.) Markov chain: selfish miner mines a block honest miner mines a block on the main branch 21-a 1- Y(1- honest miner mines a block on the selfish miner's block 0,24V DC c9-a 2 (3 (4 1-a 1-a 1-a . de 1- a 1-a honest miner mines a block A selfish miner can obtain a revenue larger than its relative size a if 1-1 a > 3 2y a > 25% for y = 0.5 Analysis Notations: - a: fraction of the selfish miner's computing power - y: probability that an honest miner chooses to mine on the selfish miner's block - State 0: there is only one longest chain all nodes are mining on (no branch) - State O': there are two public branches of length one: the main branch, and the branch that was private to the selfish miner - State i (i = 1,2, ...): the selfish miner's private branch is i blocks longer than the main branch Analysis (Cont.) Markov chain: selfish miner mines a block honest miner mines a block on the main branch 21-a 1- Y(1- honest miner mines a block on the selfish miner's block 0,24V DC c9-a 2 (3 (4 1-a 1-a 1-a . de 1- a 1-a honest miner mines a block A selfish miner can obtain a revenue larger than its relative size a if 1-1 a > 3 2y a > 25% for y = 0.5