On the Subtle Nature of a Simple Logic of the Hide and Seek Game

Abstract

We discuss a simple logic to describe one of our favourite games from childhood, hide and seek, and show how a simple addition of an equality constant to describe the winning condition of the seeker makes our logic undecidable. There are certain decidable fragments of first-order logic which behave in a similar fashion and we add a new modal variant to that class of logics. We also discuss the relative expressive power of the proposed logic in comparison to the standard modal counterparts.

Appendix: Proof of Theorem 2

Proof

Let $T = \{T^1,\cdots ,T^n\}$ be a finite set of tile types. For each $T^i\in T$ we use $u(T^i), d(T^i), l(T^i), r(T^i)$ to represent the colors of its up, down, left and right edges, respectively. We are going to define a formula $\varphi _{\mathsf {tile}}$ such that:

$$\begin{aligned} \varphi _{\mathsf {tile}}\,\,\text {is satisfiable iff }T\text { tiles }\mathbb {N}\times \mathbb {N}. \end{aligned}$$

To do so, we will use three relations in models $(W,R^s,R^r,R^u)$ in the proof to follow. In line with this, syntactically we have six operators $[\mathsf {left}]^{\star }$ and $[\mathsf {right}]^{\star }$ for $\star \in \{s,r,u\}$. Intuitively, all the relations describe the transitions of the left evaluation point and the right evaluation point of a graph model: in what follows, we are going to construct a spy point over relation $R^s$, and the relations $R^u$ and $R^r$ represent moving up and to the right, respectively, from the corresponding tile to the other.

These three relations are useful to present the underlying intuitions of the formulas that will be constructed. Also, they are helpful in making these formulas short and readable, facilitating a better understanding of the same. Crucially, this does not change the computational behavior of the original $\mathsf {LHS}$: the three relations can be reduced to one relation as that of our standard models. For instance, given that we have three relations and the evaluation gap, we can mimic the three relations with a singular relation and $3\times 2$ fresh propositional letters encoding, e.g., $[\mathsf {left}]^{\star }$ and $\langle \mathsf {right} \rangle ^{\star }$ as $[\mathsf {left}](p^{\star }_E\rightarrow \cdots )$ and $\langle \mathsf {right} \rangle (p^{\star }_A\wedge \cdots )$, respectively. Definitely, to preserve the structure of those relations and truth of formulas, we need to be careful when defining the new relation and the valuation function. However, due to page-limit constraints, we forego those details here. Now, we proceed to present the details of $\varphi _{\mathsf {tile}}$, whose components will be divided into four groups. Let us begin with the first one.

Group 1: Infinite many states induced by $R^s$ and their ‘scope’

$$\begin{aligned} (U1)&\qquad I\wedge [\mathsf {left}]^s \lnot I \\ (U2)&\qquad \langle \mathsf {left} \rangle ^s[\mathsf {left}]^s \bot \\ (U3)&\qquad [\mathsf {right}]^s\langle \mathsf {left} \rangle ^s(\lnot I \wedge \langle \mathsf {left} \rangle ^s\top \wedge [\mathsf {left}]^s I)\\ (U4)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s([\mathsf {left}]^s\bot \wedge [\mathsf {right}]^s\bot \rightarrow I) \\ (U5)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s (\langle \mathsf {left} \rangle ^s\langle \mathsf {right} \rangle ^s I \rightarrow I) \end{aligned}$$

Notice that formulas (U1)–(U3) are just the $R^s$-version of the formulas in Theorem 1 that are used to create infinite models. Immediately, there exists an infinite sequence of states as follows:

$$\begin{aligned} w_0,w_1,w_2,\cdots \end{aligned}$$

such that $R^s(w_{i+1})=\{w_i\}$ and $R^s(w_0)=\emptyset$. Also, for the current evaluation pair (e.g., (s, s)), we have $\{w_i\mid i\in \mathbb {N}\}\subseteq R^s(s)$.

Now let us spell out what (U4) and (U5) express. Essentially, both the formulas establish a ‘border’ for the scope of nodes that are (directly or indirectly) reachable from s via relation $R^s$. Specifically, the formula (U4) shows that $R^s(s)$ contains only a dead end which is exactly $w_0$ listed above, and moreover, the formula (U5) indicates that for any $w_i,w_j\in R^s(s)$, if they can reach the same state, then we have $w_i=w_j$. See Fig. 4 for two counterexamples without the properties of (U4) or (U5). From the two formulas, we know that $R^s(s)=\{w_i\mid i\in \mathbb {N}\}$.

Fig. 4.

Two impossible cases of the $R^s$-structure $R^s(s)$: Case 1 cannot satisfy (U4), while Case 2 cannot satisfy (U5).

Intuitively, we will use these $w_i's$ to represent tiles. To make this precise, beyond the simple linear order of $R^s$ among those states, we still need to structure them with $R^r$ and $R^u$ in a subtler way. Our next group of formulas concerns some basic features of the two relations:

Group 2: Basic features of $R^u$ and $R^r$

$$\begin{aligned} (U6)&\qquad [\mathsf {left}]^s[\mathsf {left}]^\dagger \langle \mathsf {right} \rangle ^sI&\quad \dagger \in \{u,r\}\\ (U7)&\qquad [\mathsf {left}]^s (\langle \mathsf {left} \rangle ^u\top \wedge \langle \mathsf {left} \rangle ^r\top ) \\ (U8)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s (I\rightarrow [\mathsf {left}]^u\lnot I\wedge [\mathsf {left}]^r\lnot I)\\ (U9)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s(I\rightarrow [\mathsf {left}]^\dagger \lnot \langle \mathsf {left} \rangle ^\dagger I)&\quad \dagger \in \{u,r\} \end{aligned}$$

Before listing more formulas, let us briefly comment on these properties.

For all $i\in \mathbb {N}$, the formulas of (U6) essentially give $R^r(w_i)$ and $R^u(w_i)$ a ‘scope’. Specifically, they guarantee that $R^r(w_i),R^u(w_i)\subseteq \{w_0,w_1,\cdots \}$. Therefore, when considering the two relations, we only need to consider those $w_i's$, and there do not exist other states that are involved.

The formula (U7) states that every $w_i$ has successors via $R^u$ and $R^r$, i.e., $R^u(w_i)\not =\emptyset$ and $R^r(w_i)\not =\emptyset$. Intuitively, this expresses that every tile has at least one tile above it and at least one tile to its right.

Also, the formula (U8) indicates that for all $i\in \mathbb {N}$, we do not have $R^rw_iw_i$ or $R^uw_iw_i$. Moreover, formulas in (U9) show that both the relations $R^r$ and $R^u$ are asymmetric.

Except those basic features captured by formulas of Group 2, what might be more important is our next group of formulas, which structure the states in a grid with $R^r$ and $R^u$:

Group 3: Grid formed by $R^u$ and $R^r$

$$\begin{aligned} (U10)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s (\langle \mathsf {left} \rangle ^\dagger I\rightarrow [\mathsf {left}]^\dagger I)&\quad \dagger \in \{u,r\}\\ (U11)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s (I \rightarrow [\mathsf {left}]^u[\mathsf {right}]^r\lnot I) \\ (U12)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s (I \rightarrow [\mathsf {left}]^u[\mathsf {left}]^r\lnot I \wedge [\mathsf {left}]^r[\mathsf {left}]^u\lnot I) \\ (U13)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s (I \rightarrow [\mathsf {left}]^u[\mathsf {left}]^r[\mathsf {left}]^u\lnot I) \\ (U14)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s( I\rightarrow [\mathsf {left}]^u[\mathsf {right}]^r\langle \mathsf {left} \rangle ^r\langle \mathsf {right} \rangle ^u I ) \end{aligned}$$

Whereas (U7) tells us that all the $w_i's$ have $R^r$- and $R^u$-successors, formulas in (U10) state that every $w_i$ has at most one $R^r$-successor and at most one $R^u$-successor. Thus, both (U7) and (U10) ensure that the transitions between those $w_i's$ via $R^u$ and $R^r$ are essentially functions: precisely, for all $i\in \mathbb {N}$ and $\dagger \in \{u,r\}$, $R^{\dagger }(w_i)$ is a singleton.

Moreover, formula (U11) suggests that the $R^r$-successor and the $R^u$-successor of a tile are different: for all $i\in \mathbb {N}$, $R^r(w_i)\cap R^u(w_i)=\emptyset$. That is, a tile cannot be above as well as to the right of another tile.

Additionally, (U12) shows that no tile can be both above/below and to the right/left of another tile, and (U13) disallows cycles following successive steps of the $R^u$, $R^r$ and $R^u$ relations, in this order. Formula (U14) states the property of ‘confluence’: for all tiles $w_i,w_j,w_k$, if $R^uw_iw_j$ and $R^rw_iw_k$ hold, then there exists another tile $w_n$ such that $R^rw_jw_n$ and $R^uw_kw_n$ hold. Now, the tiles are arranged in a grid.

Now, it remains to set a genuine tiling, which can be achieved by our fourth group of formulas. Very roughly, in usual cases this work is often routine when we have an infinite grid-like model (cf. [9]). Let us present the details here:

Group 4: Tiling the model

$$\begin{aligned} (U15)&\qquad [\mathsf {left}]^s(\bigvee \limits _{1\le i\le n} t^i_E \wedge \bigwedge \limits _{1\le i<j\le n} \lnot (t^i_E\wedge t^{j}_E) ) \\ (U16)&\qquad [\mathsf {right}]^s(\bigvee \limits _{1\le i\le n} t^i_A \wedge \bigwedge \limits _{1\le i<j\le n} \lnot (t^i_A\wedge t^{j}_A)) \\ (U17)&\qquad [\mathsf {left}]^s[\mathsf {right}]^s( I \rightarrow \bigvee \limits _{1\le i\le n} (t^i_E\wedge t^i_A) ) \\ (U18)&\qquad [\mathsf {left}]^s (\bigwedge \limits _{1\le i\le n} (t^i_E\rightarrow \langle \mathsf {left} \rangle ^u\bigvee \limits _{1\le j\le n,u(T_i)=d(T_j)} t^j_E)) \\ (U19)&\qquad [\mathsf {left}]^s (\bigwedge \limits _{1\le i\le n} (t^i_E\rightarrow \langle \mathsf {left} \rangle ^r\bigvee \limits _{1\le j\le n,r(T_i)=l(T_j)} t^j_E )) \end{aligned}$$

Formulas (U15)–(U16) indicate that a node can be occupied ‘two’ tiles $t^i_E$ and $t^j_A$. As one node can only be occupied by exactly one tile, the statement here may look a bit strange. However, we would like to argue that essentially there exists no problem, see our discussion on formula (U17) below.

By formula (U17), for every fixed i, when both $t^i_E$ and $t^j_A$ hold at a node, then we have $i=j$, i.e., they are of the same type $T^i$. In this sense, we can say that ‘E’ and ‘A’ are just ‘position-labels’ to refer to the evaluation nodes in the current graph model, and a node in the model is essentially occupied by exactly one tile. Moreover, for the same reason, although for each $T^i$, we have different propositional atoms $t^i_A$ and $t^i_E$, all types of tiles we use are exactly those given by the original T, but not any extra ones.

Finally, the ideas of formulas (U18) and (U19) are routine: the former one states that colors match going up, while the latter expresses that they match going right.

Now, let $\varphi _{\mathsf {tile}}$ be the conjunctions of all formulas listed in the four groups. Based on our analyses above, any model satisfying $\varphi _{\mathsf {tile}}$ is a tiling of $\mathbb {N}\times \mathbb {N}$.

On the other hand, we still need to show the other direction. Now suppose that a function $f:\mathbb {N}\times \mathbb {N}\rightarrow T$ is a tiling of $\mathbb {N}\times \mathbb {N}$. Define a model $\mathbf {M}_t=(W,R^s,R^u,R^r,\mathsf {V})$ in the following:

• $W:=\{s\}\cup (\mathbb {N}\times \mathbb {N})$

• $R^s$ consists of the following:

  • For all $x\in \mathbb {N}\times \mathbb {N}$, $\langle s,x \rangle \in R^s$

  • For all $\langle n,0 \rangle \in \mathbb {N}\times \mathbb {N}$ with $1\le n$, $\langle \langle n+1,0 \rangle ,\langle 0,n \rangle \rangle \in R^s$

  • For all other $\langle n,m+1 \rangle \in \mathbb {N}\times \mathbb {N}$, $\langle \langle n,m+1 \rangle ,\langle n+1,m \rangle \rangle \in R^s$

• $R^u:=\{\langle \langle n,m \rangle ,\langle n,m+1 \rangle \rangle \mid n,m\in \mathbb {N}\}$

• $R^r:=\{\langle \langle n,m \rangle ,\langle n+1,m \rangle \rangle \mid n,m\in \mathbb {N}\}$

• $\mathsf {V}(t^i_E)=\mathsf {V}(t^i_A)=\{\langle n,m \rangle \in \mathbb {N}\times \mathbb {N}\mid f(\langle n,m \rangle )=T^i\}$, for all $i\in \{1,\cdots ,n\}$

• $\mathsf {V}(p_E)=\mathsf {V}(q_A)=\emptyset$, for all other $p_E,q_A\in \mathsf {P_E}\cup \mathsf {P_A}$.

Figure 5 presents a crucial fragment of the model. By construction, it is not hard to check that $\mathbf {M}_t,s,s\vDash \varphi _{\mathsf {tile}}$. This completes the proof.    $\square$

Fig. 5.

The restriction of the structure of $\mathbf {M}_t$ to $\mathbb {N}\times \mathbb {N}$, where the resulting $R^s,R^u,R^r$ are represented by dotted-, dashed- and solid-arrows respectively. To obtain the whole structure, we just need to add the state s and draw a dotted-arrow from s to each of the members of $\mathbb {N}\times \mathbb {N}$.