In an earlier post I defined the

Any such strategy $\sigma$ induces a randomized strategy $\overline\sigma$ that acts like $\sigma$ in average, but is memoryless. In each state, when averaged over all runs, the strategies $\sigma$ and $\overline\sigma$ take the same actions with the same probabilities:

One can show that $\sigma$ and $\overline\sigma$ visit red transitions equally often in average; in particular, $\overline\sigma$ visits at least one red transition in average. The

*price of memory*, $u$, and asked:
Is $u < 1$? Worth €100.

Determine $u$. Worth another €100, even if $u=1$.

I'm happy to announce that there is now a solution:
Determine $u$. Worth another €100, even if $u=1$.

The price of memory, $u$, is 1.

I don't want to repeat the precise definition of $u$, but here is a short recap:
Suppose you have an acyclic finite MDP and a strategy $\sigma$ that guarantees to visit a red transition with probability 1.
This strategy $\sigma$ might use memory, i.e., might base its behaviour in a state on the history of the run, as indicated in yellow:
Any such strategy $\sigma$ induces a randomized strategy $\overline\sigma$ that acts like $\sigma$ in average, but is memoryless. In each state, when averaged over all runs, the strategies $\sigma$ and $\overline\sigma$ take the same actions with the same probabilities:

One can show that $\sigma$ and $\overline\sigma$ visit red transitions equally often in average; in particular, $\overline\sigma$ visits at least one red transition in average. The

*price of memory*is defined as the smallest global (independent of the MDP or any strategy) constant $u$ such that the probability that $\overline\sigma$ does**not**visit a red transition is at most $u$. Trivially, $u \le 1$, and examples as above show that $u \ge \frac12$. Let us now prove that $u=1$:
Consider the following MDP, $M_1$:

In the controlled state in the middle, the strategy $\sigma$ goes left if and only if no red transition has been visited yet (which is the case if and only if the initial random state has chosen the right transition). In this way, $\sigma$ visits exactly one red transition with probability 1. It also follows that in the controlled state in the middle, the randomized strategy $\overline\sigma$ picks either outgoing transition with probability $\frac12$. Overall, the probability in $M_1$ that $\overline\sigma$ does not visit a red transition is $u_1 := \frac12 \cdot \frac12 = \frac14$. It follows that $u \ge u_1 = \frac14$, which is not a strong statement. Consider now the following MDP, $M_2$, which contains two copies of $M_1$:

As before, there is a strategy $\sigma$ that ensures to visit exactly one red transition: in any controlled state, $\sigma$ goes left if and only if no red transition has been visited yet. As before, the resulting randomized strategy $\overline\sigma$ picks, in any controlled state, either outgoing transition with probability $\frac12$. The probability in $M_2$ that $\overline\sigma$ does not visit a red transition is $u_2 := (\frac12 + \frac{1}2u_1)^2 = \frac{25}{64} \approx 0.39$. It follows that $u \ge u_2$.

One can define a sequence of MDPs $M_1, M_2, M_3, \ldots$ in this way:

Defining $u_{i+1} := (\frac12 + \frac{1}2u_i)^2$ and generalizing $\sigma$ in the natural way, we obtain that in $M_i$ the probability that $\overline\sigma$ does not visit a red transition is $u_i$. It is straightforward to check that the sequence $u_1, u_2, u_3, \ldots$ is non-decreasing and converges to the only fixed point of the function $f(x) = (\frac12 + \frac{1}2x)^2$, which is $1$. Since $u \ge u_i$, it follows that $u=1$.

I should award the prize of €200 to the person who found this solution.
The only issue is: that would be myself.
Was the problem difficult to solve?
For me, yes.
Could a Maths or CS undergrad have found this solution?
Absolutely.
Does this result, $u=1$, lead to a paper?
It might, but my collaborators and I would have preferred $u \lt 1$. Oh well.
In the controlled state in the middle, the strategy $\sigma$ goes left if and only if no red transition has been visited yet (which is the case if and only if the initial random state has chosen the right transition). In this way, $\sigma$ visits exactly one red transition with probability 1. It also follows that in the controlled state in the middle, the randomized strategy $\overline\sigma$ picks either outgoing transition with probability $\frac12$. Overall, the probability in $M_1$ that $\overline\sigma$ does not visit a red transition is $u_1 := \frac12 \cdot \frac12 = \frac14$. It follows that $u \ge u_1 = \frac14$, which is not a strong statement. Consider now the following MDP, $M_2$, which contains two copies of $M_1$:

As before, there is a strategy $\sigma$ that ensures to visit exactly one red transition: in any controlled state, $\sigma$ goes left if and only if no red transition has been visited yet. As before, the resulting randomized strategy $\overline\sigma$ picks, in any controlled state, either outgoing transition with probability $\frac12$. The probability in $M_2$ that $\overline\sigma$ does not visit a red transition is $u_2 := (\frac12 + \frac{1}2u_1)^2 = \frac{25}{64} \approx 0.39$. It follows that $u \ge u_2$.

One can define a sequence of MDPs $M_1, M_2, M_3, \ldots$ in this way:

Defining $u_{i+1} := (\frac12 + \frac{1}2u_i)^2$ and generalizing $\sigma$ in the natural way, we obtain that in $M_i$ the probability that $\overline\sigma$ does not visit a red transition is $u_i$. It is straightforward to check that the sequence $u_1, u_2, u_3, \ldots$ is non-decreasing and converges to the only fixed point of the function $f(x) = (\frac12 + \frac{1}2x)^2$, which is $1$. Since $u \ge u_i$, it follows that $u=1$.

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