Great series! Here's the generic Bellman equation for (V(a,b)) in any state where the chains have diverged ((a \geq 1)):

Explanation:

• Flow cost $(-h_{A}c)$: Alice is continuously mining at hash rate $h_A$, paying cost $c$ per EH. This is a constant flow cost regardless of state.
• Alice finds the next block (Poisson rate ${\lambda }_A$): The state would transition to $(a+1,b)$. Alice then decides optimally: should she publish now (earning $P(a+1,b)$) or continue mining privately (earning $V(a+1,b)$)? She picks the maximum.
• Bob finds the next block (Poisson rate ${\lambda }_B$): The state would transition to $(a,b+1)$. Again, Alice decides optimally: publish ($P(a,b+1)$) or continue privately ($V(a,b+1)$)?

This equation uses the standard Bellman hint you provided, with two Poisson events (Alice vs. Bob finding the next block), both with Alice's optimal publish-or-continue decision embedded.

Two boundary cases worth noting:

1. When $a=0$ and $b=0$ (the case already solved in Problem 1), Alice hasn't diverged yet, so her decision on finding a block is whether to publish immediately ($P(1,0)$) or go private ($V(1,0)$). Bob finding a block doesn't change the strategic state — both just mine on $C-B$ and reset: $V(0,0)$ on the right-hand side.
2. When $b\ge a$, Alice's (P(a,b)) yields only $V(0,0)$ (since Bob's chain is at least as long), meaning publishing is never better than waiting in the hope of pulling ahead — unless $a+1>b$ after her next block.

The key recursive insight: Alice's value in any state depends on the future states she can reach, and in each future state she again makes the optimal publish-or-continue choice. This is what makes it a genuine Bellman equation.