Comment
Author: Admin | 2025-04-28
Two competing public chains the portion of the network that picks up on the attacker’s chain is Z.State 0: If the attacker’s private chain is the same as the public chain, mine on the private chain. With probability X, the attacker discovers a block and advances to state 1 (private chain 1 block ahead). With probability 1-X, the public network discovers a block, and the attacker resets his private chain to the public chain.State 1: If the attacker’s private chain is 1 longer than the public chain, mine on the private chain. With probability X, the attacker advances to state 2 (private chain 2 blocks ahread). With probability 1-X, the public network discovers a block, setting the system to state 0′.State 0′: The attacker publishes his block. There are now two competing chains, both one block long. With probability X, the attacker will discover another block, causing the network to switch over to the private chain. The attacker gains a revenue of 2, and the system resets to state 0. With probability (1-X)Z, the network finds a block on top of the attacker’s block. The attacker and the network gain a revenue of 1, and the system resets to state 0. With probability (1-X)(1-Z), the network finds a block on top of its own block, the network gains a revenue of 2 and the system resets to state 0.State 2: With probability X, the attacker advances to state 3 and earns a revenue of 1 (technically, the attacker will earn the revenue later, but it’s easier to account for it here). With probability 1-X, the network finds a block, so the attacker publishes his 2-block private chain, which is still one block longer than the public chain, so the network will switch to the attacker’s chain. The attacker earns a revenue of 2.State n (n > 2): with probability X, the attacker advances to state n+1 and earns a revenue of 1. With probability 1-X, the attacker falls back to state n-1.To see why this strategy works, suppose that Z is close to one. In this case, there is never any chance that the attacker has to discard a block; the only time that might happen is from state 0′, and if Z ~= 1 almost all of the network, attacker and other nodes included, is mining on the attacker’s block so the attacker’s block will not be discarded. Thus,
Add Comment