# EXPTIME Hardness of an \(n\) by \(n\) Custodian Capture Game
## 3 Gates
### Switches
: If [PERSON] will switch ON, he must move \((\mathcal{O}_{b},\bullet_{b})=(E^{\prime},Y^{\prime})\rightarrow(F^{\prime},Y ^{\prime})\) for the following reason in any case of \(Y^{\prime}=L^{\prime}\) or \(Y^{\prime}\
eq L^{\prime}\) (i.e., \(Y^{\prime}\in S2U^{\prime}-L^{\prime}\)). Any horizontal move \(\mathcal{O}_{b}:(E^{\prime},Y^{\prime})\rightarrow(X1^{\prime},Y^{\prime}), X1^{\prime}\in S1M^{\prime}-\{L^{\prime}\}\) forces WHITE to lose w4\({}^{\prime}\) by \(Y^{\prime}=L^{\prime}\Rightarrow d_{\bullet}(L^{\prime},w4^{\prime})=3\leq 4=d_{ \circ}(X1^{\prime},w4^{\prime})\), or incurring BLACK's winning repetition by \(Y^{\prime}\
eq L^{\prime}\Rightarrow(\mathcal{O}_{b},\bullet_{b})=(E^{\prime },Y^{\prime})\rightarrow(X1^{\prime},Y^{\prime})\rightarrow(X1^{\prime},L^{ \prime})\rightarrow(E^{\prime},L^{\prime})\rightarrow(E^{\prime},Y^{\prime})\) where the same analysis forces WHITE's comeback \((X1^{\prime},L^{\prime})\rightarrow(E^{\prime},L^{\prime})\). The \(\mathcal{O}_{b}\) could never win T1\({}^{\prime}\) against \(\bullet_{b}\)'s protection by \(d_{\bullet}(Y^{\prime},J^{\prime})\leq 3=d_{\circ}(E^{\prime},T1^{\prime})\). \(\overline{\mathbb{R}.4}\) forbids \(\mathcal{O}_{b}\) to stay in F71\({}^{\prime}\) since \(\bullet_{b}\) can wait for it in N1\({}^{\prime}\) on the assumption \(|\)N1\({}^{\prime}\)L\({}^{\prime}\)\(|=0\) and \(|\)F71\({}^{\prime}\)\(|\leq|\)N1\({}^{\prime}\)\(|\). Eventually, \(\mathcal{O}_{b}\) must go to F\({}^{\prime}\) immediately to not delay the game. Similarly, BLACK must respond as follows: \(Y^{\prime}\
eq L^{\prime}\Rightarrow(\mathcal{O}_{b},\bullet_{b})=(F^{\prime },Y^{\prime})\rightarrow(F^{\prime},L^{\prime})\) for surviving T2\({}^{\prime}\) by \(d_{\circ}(F^{\prime},T2^{\prime})=2<3=d_{\circ}(Y^{\prime},T2^{\prime\prime}\text{ s protector})<4=d_{\circ}(Y^{\prime},w4^{\prime})\). The Y\({}^{\prime}=L^{\prime}\) case forces either \(\bullet_{b}\)'s vertical move (by the same analysis) or \(\bullet_{f}\)'s horizontal one since \(\bullet_{f}\
ot\in\ell(S2,U)\Rightarrow d_{\circ}(F^{\prime},T2^{\prime})=2<1+2= d_{\circ}(\bullet_{f},\ell(S2,U))+d_{\bullet}(L^{\prime},T2^{\prime\prime}\text{ s protector})\). Lemma 13 makes \((\mathcal{O}_{b},\bullet_{b})=(F^{\prime},Y^{\prime})\rightarrow(V^{\prime},F^{ \prime})\) happening in the second turn. The same argument induces \((\mathcal{O}_{f},\bullet_{f})=(E,L)\stackrel{{*}}{{\rightarrow}}( V,F)\) if WHITE will switch OFF. The free men may get into Figure 105 from the left and reach to \((\mathcal{O}_{f},\bullet_{f})=(\text{A},\text{H})\) for changing its state. Notice that the left side of Figure 105 connects to the outside area via Figure 101's right side structure as \((\text{A},\text{H})=(\text{C},\text{K})\). **Lemma 21**. : _Figure 105 forces \((\mathcal{O}_{f},\bullet_{f})=(A,H)\rightarrow(B,I)\rightarrow(C,IK-\{I\})\) or \((A,H)\rightarrow(B,I)\rightarrow(T1,J)\rightarrow(C,IK-\{I\})\)._  
Proof. : Similar to Lemma 14's proof. In the first turn, \(\mathcal{O}_{f}\) must proceed \((\text{A},\text{H})\rightarrow(\text{B},\text{H})\) to survive w1 by \(\mathcal{O}_{f}\
ot\in\ell(\text{B},\text{C})\Rightarrow d_{\bullet}(\text{H},\text{w1})=2\leq d_{\circ}(\mathcal{O},\text{T1})\), then \(\bullet_{f}:(\text{B},\text{H})\rightarrow(\text{B},\text{I})\) to survive T1 by \(\bullet_{f}\in\ell(\text{I},\text{H})-\{\text{I}\}\Rightarrow d_{\circ}(\text{B },\text{T1})=1\leq d_{\bullet}(\bullet_{f},\text{w1})\leq 2=d_{\bullet}(\bullet_{f},J)\). In the  
Figure 105: W-switch OFF. second turn, WHITE can choose either \((B,I)\rightarrow(T1,I)\rightarrow(T1,J)\rightarrow(C,J)\rightarrow(C,IK-\{I\})\) or \((B,I)\rightarrow(C,I)\rightarrow(C,IK-\{I\})\); \(\overline{R4}\) forbids \(O_{f}\) to stay in BT1 since \(\bullet_{f}\) can wait for it in IK on the assumption \(|BT1|\leq|IK|\).