TL;DR: I define a three-player game (Arthur, Nimue, Merlin) where Nimue is shown a hidden bit $b$ chosen by Merlin and tries to communicate it to her ally Arthur, but Arthur must act computably while Merlin acts adversarially, replacing Nimue's chosen symbols $\mathtt{P},\mathtt{Q}$ by actual numbers from fixed sets $P,Q\subseteq\mathbb{N}$. When do Arthur+Nimue have a winning strategy in this game?
In other words, for which pairs $(P,Q)$ can the information contained in a single bit can be transmitted through an adversarial interpreter to a computably bounded receiver?
THE GAME:
Let $P,Q\subseteq\mathbb{N}$ (we will soon assume $P\cap Q = \varnothing$). Consider the following game played between three players, “Arthur”, “Nimue” and “Merlin”, with Arthur and Nimue allied against Merlin.
(The game is described in the format used by Takayuki Kihara's paper “Lawvere-Tierney topologies for computability theorists”; however I do not assume familiarity with this paper, and I suspect it may not even be particularly relevant for answering the question. But see remarks below for what the question “means” in the context of Kihara's paper.)
Informal description: Merlin initially chooses a bit $b \in \{0,1\}$ and shows it to Nimue but not to Arthur. Nimue's goal is to communicate $b$ to Arthur (i.e., Arthur's goal is to guess the value of $b$), but Nimue isn't allowed to communicate directly to Arthur (they did agree a common strategy beforehand). At each turn, Nimue chooses an element of $\{\mathtt{P},\mathtt{Q}\}$ and shows it to Merlin, who then chooses an element of the corresponding set ($P$ or $Q$) and shows that element to Arthur. (Arthur only sees the elements of $\mathbb{N}$ that Merlin shows him; Merlin and Nimue see everything.) If Arthur can guess $b$ then he declares its value and the game ends. Otherwise, the game continues for a new turn. The crucial limitation is that Arthur must play following a computable strategy (Merlin and Nimue are not limited in this way). Arthur and Nimue win if Arthur makes a correct guess; otherwise, Merlin wins.
More precise description: On turn $0$, Merlin chooses $b \in \{0,1\}$. On each subsequent turn $i\geq 1$:
Nimue chooses an element of $\{\mathtt{P},\mathtt{Q}\}$.
Merlin chooses an element of $P$ if Nimue chose $\mathtt{P}$, and an element of $Q$ if Nimue chose $\mathtt{Q}$.
Arthur (seeing only Merlin's moves on turns $i\geq 1$) either chooses to end the game by declaring a value $a \in \{0,1\}$ or to continue the game.
Arthur+Nimue win the game if Arthur ends the game by declaring $a=b$ (or if Merlin fails to abide the requirement of playing in $P$ or $Q$). Merlin wins if $a\neq b$ or if the game lasts forever.
A Nimue strategy is a function $\sigma \colon \{0,1\} \times \bigcup_{i\geq 0}\mathbb{N}^i \to \{\mathtt{P},\mathtt{Q}\}$ that picks an element in $\{\mathtt{P},\mathtt{Q}\}$ in function of Merlin's previous moves ($b$ and the finite sequence with values in $\mathbb{N}$). An Arthur strategy is a partial computable function $\tau \colon \bigcup_{i\geq 1}\mathbb{N}^i \dashrightarrow \{0,1,\texttt{continue}\}$ that picks an element of $\{0,1,\texttt{continue}\}$ in function of Merlin's visible moves (the finite sequence of length $i$ with values in $\mathbb{N}$, but obviously not $b$). The function is not required to be defined on all possible input values, but if $\tau$ is undefined when it is Arthur's turn to play, this is considered a forfeit by Arthur and Merlin wins (just as when the game lasts indefinitely).
We say that Arthur+Nimue have a winning strategy if there is a (computable!) Arthur strategy $\tau$ and a Nimue strategy $\sigma$ such that, if they play according to them, they win against any valid play by Merlin.
THE MAIN QUESTION: Can we characterize pairs $(P,Q)$ such that Arthur+Nimue have a winning strategy?
Since the above main question might be too difficult or too general to admit a satisfactory answer (there might not be a characterization meaningfully different from the game itself), I am interested in partial answers as well, or even just interesting examples. To avoid posing a problem that is too open-ended, here are two specific questions that I don't know the answer to:
Specific question #1: Does there exist $(P,Q)$ such that Arthur+Nimue have a winning strategy but for which it does not hold that there exists a partial computable function defined at least on $(P\times Q) \cup (Q\times P)$ with value $0$ on $P\times Q$ and $1$ on $Q\times P$? (See examples below for the significance of this condition, which gives Nimue a simple strategy.)
Specific question #2: Do Arthur+Nimue have a winning strategy if $P$ is the set $\{e \in \mathbb{N} : \varphi_e\text{ is total}\}$ of codes of total computable functions, and $Q = \mathbb{N} \setminus P$ is its complement? (A positive answer here would imply a positive one to specific question #1.)
SPECIAL CASES AND EXAMPLES:
If $P$ resp. $Q$ is empty, then Arthur and Nimue trivially win by having Nimue play $\mathtt{P}$ resp. $\mathtt{Q}$ so that Merlin cannot make a choice. This is uninteresting so we now assume $P$ and $Q$ are inhabited.
If $P$ and $Q$ have an element $n$ in common, then Merlin can play $n$ on every turn $i\geq 1$ and clearly Arthur+Nimue cannot have a winning strategy. So we might as well assume $P\cap Q = \varnothing$.
If ($P\cap Q = \varnothing$ and) $P$ is computable, then Arthur+Nimue have a winning strategy: Nimue chooses $\mathtt{P}$ if $b=0$ and $\mathtt{Q}$ if $b=1$, and Arthur computably tests whether the integer $n$ communicated to him by Merlin is in $P$: if so, he ends the game by declaring $0$, and if not, he ends the game by declaring $1$. The same strategy works, mutatis mutandis, if $Q$ is computable.
If ($P\cap Q = \varnothing$ and) $P$ is merely computably enumerable, then Arthur+Nimue also have a winning strategy, it is just slightly more complicated. On turn $1$, Nimue chooses $\mathtt{P}$ if $b=0$ and $\mathtt{Q}$ if $b=1$ (and Arthur continues the game), and on turn $2$, Nimue chooses the opposite of what she chose on turn $1$. On turn $1$, Arthur chooses to continue the game. On turn $2$, he runs runs a program enumerating $P$ while searching for either integer $n_1,n_2$ communicated to him by Merlin: if he finds $n_1$ in $P$ then he declares $0$, and if he finds $n_2$ in $P$ then he declares $1$, and one of the two will necessarily occur because of how Nimue played. The same strategy works, mutatis mutandis, if $Q$ is computably enumerable.
If ($P\cap Q = \varnothing$ and) $P$ or $Q$ is co-computably enumerable, Arthur+Nimue still have a winning strategy where this time Arthur enumerates the complement of whichever set is assumed to be co-c.e. until he can rule out one of the elements presented by Merlin.
If $z \in \mathbb{R}$ is any irrational number and, fixing a coding of rationals by integers, $P = \{r \in \mathbb{Q} : r < z\}$ and $Q = \{r \in \mathbb{Q} : r > z\}$, then Arthur+Nimue have a winning strategy: again, on turn $1$, Nimue chooses $\mathtt{P}$ if $b=0$ and $\mathtt{Q}$ if $b=1$ (and Arthur continues the game), and on turn $2$, Nimue chooses the opposite of what she chose on turn $1$. Arthur now wins by simply comparing the two rationals received from Merlin. Since $P$ and $Q$ have the same Turing degree as $z$, this example shows that Arthur+Nimue can have a winning strategy with $P,Q$ (complement to each other and) of arbitrarily Turing degree.
If $P \subseteq P'$ and $Q \subseteq Q'$ and Arthur+Nimue have a winning strategy for $(P',Q')$, then they obviously also have one for $(P,Q)$.
If $P$ is “sufficiently generic” and $Q = \mathbb{N}\setminus P$ is its complement, then Arthur+Nimue do not have a winning strategy. This is not completely obvious (or at least it wasn't to me…), so I give a proof of this proposition at the bottom for completeness of this question.
FURTHER REMARKS:
The case where $P$ and $Q$ are complement to each other seems most interesting. I don't know if this is a genuine simplification, however.
In all my examples where Arthur and Nimue win, Nimue's strategy is very simple: she ignores what Merlin does except for the initial $b$, and simply plays $\mathtt{P},\mathtt{Q}$ or $\mathtt{Q},\mathtt{P}$ according to the value of $b$ and Arthur uses a computable function that separates $P\times Q$ from $Q\times P$ inside their union (so the game is over in two turns). This leads to the specific question #1: are there $(P,Q)$ for which Arthur+Nimue have a winnning strategy but not one of this kind?
A trivial point that might be worth stating explicitly because it confused me: the game need not be determined in the usual sense. It is obvious that Merlin cannot have a winning strategy because Arthur could always guess the right $b$ by accident; but this does not imply that Arthur+Nimue have one.
Relation to the effective topos (and motivation for the question): for those who know what this is about, $(P,Q)$ defines a subobject $X \hookrightarrow \nabla 2$ in the effective topos or equivalently, its characteristic function $\chi\colon\nabla 2 \to \Omega$ (taking $0$ to $P$ and $1$ to $Q$ with the obvious abuse of language). If I didn't mess up my translation of the formalism, Arthur+Nimue have a winning strategy exactly when the smallest Lawvere-Tierney topology $j$ making $X \hookrightarrow \nabla 2$ be $j$-dense (viꝫ. which satisfies $\forall b:\nabla 2.\,(j(\chi(b)))$) coincides with double negation $(\neg\neg)$ topology. (The double negation topology is itself the smallest L-T topology such for which the subobject $2 \hookrightarrow \nabla 2$ defined by $\{0\}$ and $\{1\}$ is dense.)
For secondary motivation: the game is somewhat reminiscent of the indistinguishability game used to define the IND-CPA security level of in cryptography.
Proposition: There exists $P$ such that (in fact, for all “sufficiently generic”$P$) if $Q = \mathbb{N}\setminus P$ then Arthur+Nimue do not have a winning strategy.
Proof. Endow the powerset $\mathcal{P}(\mathbb{N})$, identified with $\{0,1\}^{\mathbb{N}}$, with the product topology. We wish to show that there is a countable intersection of dense open sets $G$ (so $G$ is dense by Baire's theorem) such that, if $P$ is taken in $G$, then Arthur+Nimue do not have a winning strategy.
Since there are countably many Arthur strategies $\tau$ (since they are supposed to be computable), it is enough to show that, for any given Arthur strategy $\tau$, there is a countable intersection of dense open sets $G_\tau$ such that if $P$ is taken in $G_\tau$ then Nimue does not have a winning strategy along with Arthur playing $\tau$.
So, fix an Arthur strategy $\tau$: it takes as input a finite sequence $s$ with values in $\mathbb{N}$ and (computably) returns an element of $\{0,1,\texttt{continue}\}$. Define the “game tree”$T$ (associated to $\tau$) to be the (countable) set of finite sequences $s$ of natural numbers such that $\tau$ returns $\texttt{continue}$ for every subsequence obtained by removing some of the final elements of $s$ (by convention, we agree that the empty sequence is in $T$ and $\tau$ returns $\texttt{continue}$ on it).
Note that we are now dealing with a perfect information game between Nimue and Merlin.
Let us now label each node $s\in T$ with either “$0$” or “$1$” or both or none, and we also define sets $K_{s,v}$, as follows. First, if $\tau$ returns $c\in\{0,1\}$ on $s$, then we label $s$ by $c$; and if $\tau$ fails to terminate on $s$, then we label it both $0$ and $1$ (this won't really matter). Next, recursively, if the set of $m \in \mathbb{N}$ such that $s^\frown m$ (the concatenation of $s$ and $m$) is labeled $c$, is infinite, then we also label $s$ itself by $c$, and we let $K_{s,c}$ be the (infinite) set of such $m$. Finally, if $s$ has remained unlabeled after all of this, then there must be infinitely many $m$ such that $s^\frown m$ is unlabeled (because at most finitely many are labeled $0$ and at most finitely many are labeled $1$), and in this case we let $K_{s,\bot}$ be the (infinite) set of such $m$.
Now note that if $K$ is an infinite subset of $\mathbb{N}$, then the set $W_K \subseteq \mathcal{P}(\mathbb{N})$ of the $E\subseteq\mathbb{N}$ such that both $K \cap E \neq \varnothing$ and $K \setminus E \neq \varnothing$ hold, is open and dense in $\mathcal{P}(\mathbb{N})$. So intersection $G_\tau$ of the $W_{K_{s,v}}$ for $s\in T$ and $v \in \{0,1,\bot\}$ such that $K_{s,v}$ has been defined (always an infinite set) in the previous paragraph, is a countable intersection of dense open sets.
I claim that if $P$ is in $G_\tau$ and $Q$ is its complement, then Merlin has a winning strategy against Arthur playing $\tau$ (whatever Nimue does). Indeed:
If the empty sequence $s$ is labeled $c\in\{0,1\}$ (if it is labeled both, pick one arbitrarily), then Merlin plays $b := 1-c$. Whether Nimue plays $\mathtt{P}$ or $\mathtt{Q}$, since $P$ and $Q$ both meet $K_{s,c}$, Merlin can play an $m \in K_{s,c}$, so that $s^\frown m$ is also labeled $c$. Continuing this way, two things can happen: either Arthur eventually declares $c$ and loses since $c\neq b$, or the play continues indefinitely (including when $\tau$ doesn't terminate), so Arthur also never wins. So in all cases, Merlin wins.
If the empty sequence $s$ is unlabeled, then Merlin plays an arbitrary $b$. Whether Nimue plays $\mathtt{P}$ or $\mathtt{Q}$, since $P$ and $Q$ both meet $K_{s,\bot}$, Merlin can play an $m \in K_{s,\bot}$, so that $s^\frown m$ is also unlabeled. Continuing this way, the play continues indefinitely, and Arthur also never wins. So again, Merlin wins.
So in all cases, if $P$ is in $G_\tau$ then Arthur playing $\tau$ cannot give Arthur+Nimue a winning strategy. So if $P$ is in $G := \bigcap_\tau G_\tau$, then Arthur+Nimue do not have a winning strategy. ∎
(The same reasoning shows that for all “sufficiently random”$P$ the same conclusion holds: just note that $W_K$ has full measure on top of being open dense.)
