Regarding direction 17: There might be some potential drawbacks to ADAM. I think its possible that some very agentic programs have relatively low $g$ score. This is due to explicit optimization algorithms being low complexity.

(Disclaimer: the following argument is not a proof, and appeals to some heuristics/etc. We fix $M=M_0$ for these considerations too.) Consider an utility function $\hat{U}$. Further, consider a computable approximation of the optimal policy (AIXI that explicitly optimizes for $\hat{U}$) and has an approximation parameter n (this could be AIXI-tl, plus some approximation of $\hat{U}$; higher $n$ is better approximation). We will call this approximation of the optimal policy ${\pi }_n^{\hat{U}}$. This approximation algorithm has complexity $K\left({\pi }_n^{\hat{U}}\right)=C+K\left(\hat{U}\right)+K\left(n\right)$, where $C>0$ is a constant needed to describe the general algorithm (this should not be too large).

We can get better approximation by using a quickly growing function, such as the Ackermann function with $n=A(k,k)$. Then we have $K\left({\pi }_{A(k,k)}^{\hat{U}}\right)=C+K\left(\hat{U}\right)+K\left(A\right(k,k\left)\right)\le C+K\left(\hat{U}\right)+\log \left(k\right)$.

What is the $g$ score of this policy? We have $g\left({\pi }_{A(k,k)}^{\hat{U}}\right)={max}_U\left({min}_{{\pi }^{\prime }:\dots }K\right({\pi }^{\prime })-K(U\left)\right)$. Let $\overline{U}$ be maximal in this expression. If $K\left(\overline{U}\right)\ge K\left(\hat{U}\right)-C$, then $$g\left({\pi }_{A(k,k)}^{\hat{U}}\right)=\underset{{\pi }^{\prime }:E_{{\zeta }_{M_0}{\pi }^{\prime }}\left(\overline{U}\right)\ge E_{{\zeta }_{M_0}{\pi }_{A(k,k)}^{\hat{U}}}\left(\overline{U}\right)}{min}K\left({\pi }^{\prime }\right)-K\left(\overline{U}\right)\le K\left({\pi }_{A(k,k)}^{\hat{U}}\right)-K\left(\hat{U}\right)+C\le 2C\log \left(k\right)$$.

For the other case, let us assume that if $K\left(\overline{U}\right)<K\left(\hat{U}\right)-C$, the policy ${\pi }_{A(k,k)}^{\overline{U}}$ is at least as good at maximizing $\overline{U}$ than ${\pi }_{A(k,k)}^{\hat{U})}$. Then, we have $$g\left({\pi }_{A(k,k)}^{\hat{U}}\right)=\underset{{\pi }^{\prime }:E_{{\zeta }_{M_0}{\pi }^{\prime }}\left(\overline{U}\right)\ge E_{{\zeta }_{M_0}{\pi }_{A(k,k)}^{\hat{U}}}\left(\overline{U}\right)}{min}K\left({\pi }^{\prime }\right)-K\left(\overline{U}\right)\le K\left({\pi }_{A(k,k)}^{\overline{U}}\right)-K\left(\overline{U}\right))\le C+\log (k)$$.

I don't think that the assumption ($({\pi }_{A(k,k)}^{\overline{U}}$ maximizes $barU$ better than $({\pi }_{A(k,k)}^{\hat{U}}$) is true for all $\hat{U}$ and $k$, but plausibly we can select $\hat{U}$ such that this is the case (exceptions, if they exist, would be a bit weird, and if ADAM working well due to these weird exceptions feels a bit disappointing to me). A thing that is not captured by approximations such as AIXI-tl are programs that halt but have insane runtime (longer than $A(k,k)$). Again, it would feel weird to me if ADAM sort of works because of low-complexity extremely-long-running halting programs.

To summarize, maybe there exist policies ${\pi }_{A(k,k)}^{\hat{U}}$ which strongly optimize a non-trivial utility function $\hat{U}$ with approximation parameter $n=A(k,k)$, but where $g\left({\pi }_{A(k,k)}^{\hat{U}}\right)\le 2C+\log \left(k\right)$ is relatively small.

I am going to assume that in the code, when calculating p_alice_win_given_not_caught, we do not divide the ${\lambda }_2$ term by two (since this is not that consistent with the description. I am also assuming that $0.25-t_2$ is a typo and $0.5-t_2$ is meant, which would also be more consistent with other stuff). So I am going to assume assume a symmetrical version.

Here, P(Alice wins) is $e^{-{\lambda }_1{t}_1-{\lambda }_2{t}_2}-e^{-{\lambda }_1/2-{\lambda }_2/2}=:f({\lambda }_1,{\lambda }_2,t_1,t_2)$. Wlog we can assume $0<t<1$ (otherwise Bob will run everything or nothing in shielded mode).

We claim that $\left(\right({\lambda }_1^{\ast },{\lambda }_2^{\ast }),(t_1^{\ast },t_2^{\ast }\left)\right):=\left(\right({\mu }^{\ast },{\mu }^{\ast }),(t/2,t/2)$ is a (pure) Nash equilibrium, where ${\mu }^{\ast }=\log t/(t-1)$.

To verify, lets first show that Alice cannot make a better choice if Bob plays $(t_1^{\ast },t_2^{\ast })=(t/2,t/2)$. We have $f({\lambda }_1,{\lambda }_2,t_1^{\ast },t_2^{\ast })=e^{-({\lambda }_1+{\lambda }_2)t/2}-e^{-({\lambda }_1+{\lambda }_2)/2}$. Since this only depends on the sum, we can make the substitution $\mu =({\lambda }_1+{\lambda }_2)/2$. Thus, we want to maximize $\tilde{f}(\mu ,t)=e^{-\mu t}-e^{-\mu }$. We have ${\tilde{f}}^{\prime }(\mu ,t)=-t{e}^{-\mu t}+e^{-\mu }=0$. Rearranging, we get $t{e}^{-\mu t}=e^{-\mu }$. Taking logs, we get $\log t-\mu t=-\mu$. Rearranging, we get $\mu =\log t/(t-1)$. Thus, $\mu ={\mu }^{\ast }$ is the optimal choice. This means, that if Bob sticks to his strategy, Alice cannot do better than $f({\mu }^{\ast },{\mu }^{\ast },t/2,t/2)$.

Now, lets show that Bob cannot do better. We have $f({\lambda }_1^{\ast },{\lambda }_2^{\ast },t_1,t_2)=e^{-{\mu }^{\ast }{t}_1-{\mu }^{\ast }{t}_2}-e^{-{\mu }^{\ast }/2-{\mu }^{\ast }/2}=e^{-{\mu }^{\ast }t}-e^{-{\mu }^{\ast }}$. This does not depend on $t_1$ and $t_2$ anymore, so any choice of $t_1$ and $t_2$ is optimal if Alice plays $({\mu }^{\ast },{\mu }^{\ast })$.

(If I picked the wrong version of the question, and you actually want some symmetry: I suspect that the solution will have similarities, or that in some cases the solution can be obtained by rescaling the problem back into a more symmetric form.)