The one-cop-moves game on planar graphs

Abstract

Cops and Robbers is a vertex-pursuit game played on graphs. In this game, a set of cops and a robber occupy the vertices of the graph and move alternately along the graph’s edges with perfect information about each other’s positions. If a cop eventually occupies the same vertex as the robber, then the cops win; the robber wins if she can indefinitely evade capture. Aigner and Fromme established that in every connected planar graph, three cops are sufficient to capture a single robber. In this paper, we consider a recently studied variant of the cops and robbers game, alternately called the one-active-cop game, one-cop-moves game or the lazy cops and robbers game, where at most one cop can move during any round. We show that Aigner and Fromme’s result does not generalize to this game variant by constructing a connected planar graph on which a robber can indefinitely evade three cops in the one-cop-moves game.

Appendices

Detailed analysis of Algorithm 4 for Case (A)

Case (1). There is at least one corner $v^{\prime }$ of $L_{U,49}$ such that $d_{\mathcal {D}}(v^{\prime },u^{\prime }) \ge 99$ for all $u^{\prime } \in \{u_2,u_3\}$ and $d_{\mathcal {D}}(v^{\prime },u_1) \ge 98$. First, assume that $d_{\mathcal {D}}(v_1,u^{\prime }) \ge 99$ for all $u^{\prime } \in \{u_2,u_3\}$; since $u_1 \in \{p_1,p_2,p_3\}$ (Fig. 9), it holds that $d_{\mathcal {D}}(v_1,u_1) \ge 99$. $\gamma$ first moves to $v_1$ in 98 rounds. If, at the end of the 98-th round, both $\lambda _2$ and $\lambda _3$ are at least 101 edges away from $q_1$, then $\gamma$ can safely reach $q_1$ in another 100 rounds and Lemma 5.7 may then be applied to $U_1 \cup U_2 \cup U_6$.

Suppose, on the other hand, that either $\lambda _2$ or $\lambda _3$ is at most 100 edges away from $q_1$ at the end of the 98-th round. Without loss of generality, assume that at the end of the 98-th round, $\lambda _2$ is in $U_1$ and is at most 100 edges away from $q_1$. This implies that $\lambda _3$ could have moved at most 1 step between the 1-st round and the 98-th round, and so $\lambda _3$ is at most 1 edge closer to $o_2$ than $\gamma$ is at the end of the 98-th round. $\gamma$ now starts moving towards $o_2$. If $\lambda _3$ skips more than 1 turn as $\gamma$ is approaching $o_2$, then $\gamma$ can safely reach $o_2$; else, $\gamma$ continues moving towards $o_2$ until she reaches vertex $t^{\prime \prime }$ on $L_{U_2,23}$, as shown in Fig. 10. $\gamma$ then moves along the path $t^{\prime \prime } \rightsquigarrow t^{\prime \prime \prime }$ highlighted in Fig. 10. Note that the length of the path $t^{\prime \prime } \rightsquigarrow t^{\prime \prime \prime }$ is $2 \cdot 23 + 2 = 48$, while $\lambda _2$ is at least $98 - 2 \cdot 23 = 52$ edges away from $t^{\prime \prime \prime }$ and $\lambda _3$ is at least $4 \cdot 23 - 1 = 91$ edges away from $t^{\prime \prime \prime }$ when $\gamma$ is at $t^{\prime \prime }$. It follows that $\gamma$ can safely reach $t^{\prime \prime \prime }$. Furthermore, suppose that between the round when $\gamma$ is at $t^{\prime \prime }$ and the round when $\gamma$ is at $t^{\prime \prime \prime }$, $\lambda _2$ moves i steps and $\lambda _3$ moves j steps. Since $i+j \le 2\cdot 23 + 2 = 48$, at least one of the following holds: (i) $i \le 24$; (ii) $j \le 24$. If (i) holds, then, since $\lambda _2$ is at least $97 - i \ge 73$ edges away from $q_1$ (and $\lambda _3$ is even further away from $q_1$) while $\gamma$ is $98 - 2 \cdot 23 = 52$ edges away from $q_1$ when $\gamma$ is at $t^{\prime \prime \prime }$, $\gamma$ can safely move to $q_1$; Lemma 5.7 may then be applied to $U_1 \cup U_2 \cup U_6$. If (ii) holds, then $\gamma$ moves along the path $t^{\prime \prime \prime } \rightsquigarrow t^{\prime \prime \prime \prime } \rightsquigarrow t_{12}$ highlighted in Fig. 10. Note that the length of the path $t^{\prime \prime \prime } \rightsquigarrow t^{\prime \prime \prime \prime } \rightsquigarrow t_{12}$ is 100, whereas $\lambda _3$ is at least $97 + 2 \cdot 23 - j \ge 119$ edges away from $t_{12}$ (and $\lambda _2$ is even further away from $t_{12}$) when $\gamma$ is at $t^{\prime \prime \prime }$. Consequently, $\gamma$ can safely move to $t_{12}$ in another 100 rounds after reaching $t^{\prime \prime \prime }$. Upon reaching $t_{12}$, either $\gamma$ may safely move to $o_7$ in another 98 rounds, or (if $\lambda _3$ uses up at least $99 - j \ge 75$ turns to move towards $U_7$ as $\gamma$ starts moving from $t^{\prime \prime \prime }$ to $t^{\prime \prime \prime \prime }$) Lemma 5.7 may be applied to $U_2 \cup U_6 \cup U_7$.

Fig. 9

Initial positions of $\lambda _1$ and $\gamma$

Fig. 10

The escape path of $\gamma$ in Case (A.1)

$\mathbf{Case}\, \mathbf{(1')}.$ For some $v^{\prime \prime } \in \{v_2,v_3,v_4,v_5\}$, $d_{\mathcal {D}}(v^{\prime \prime },u^{\prime }) \ge 99$ for all $u^{\prime } \in \{u_2,$ $u_3\}$ and $d_{\mathcal {D}}(v^{\prime \prime },u_1) \ge 99$. Notice that $\gamma$’s strategy in Case (1) still applies (with an appropriate transformation of vertices; for example, if $v^{\prime \prime } = v_2$, then we apply the mapping $v_2 \rightarrow v_1, v_1 \rightarrow v_5, v_5 \rightarrow v_4, v_4 \rightarrow v_3, v_3 \rightarrow v_2$, and extend this mapping so as to obtain an automorphism of $\mathcal {D}$).

Case (2). For every corner $v^{\prime }$ of $L_{U,49}$, there is some $u^{\prime } \in \{u_2,u_3\}$ such that $d_{\mathcal {D}}(u^{\prime },v^{\prime }) \le 98$ or $d_{\mathcal {D}}(u_1,v^{\prime }) \le 97$ (or both inequalities hold). Without loss of generality, assume that $d_{\mathcal {D}}(u_1,v_4) = 97$, $d_{\mathcal {D}}(u_2,v_1) \le 98, d_{\mathcal {D}}(u_2,v_5) \le 98, d_{\mathcal {D}}(u_3,v_2)$ $\le 98$ and $d_{\mathcal {D}}(u_3,v_3) \le 98$. Recall that q is the vertex of $L_{U,49}$ that is one edge away from $m_5$ and p is the vertex of $L_{U,49}$ that is one edge away from $m_4$ (as shown in Fig. 4). Note that by Lemma 5.1, the condition imposed on the positions of $\lambda _2$ and $\lambda _3$, and the fact that neither $\lambda _2$ nor $\lambda _3$ is at o, it holds that $d_{\mathcal {D}}(u_2,B_9) \ge 99$ and $d_{\mathcal {D}}(u_3,B_{10}) \ge 99$.

Case (2.1). $d_{\mathcal {D}}(p,u_3) \ge 99$. As was observed earlier, $d_{\mathcal {D}}(u_2,B_9) \ge 99$. We show that $d_{\mathcal {D}}(u_3,B_9) \ge 50$. Take any $x \in V(B_9)$. If x lies between $v_4$ and $m_4$ inclusive, then $d_{\mathcal {D}}(u_3,x) \ge d_{\mathcal {D}}(x,v_2) - d_{\mathcal {D}}(u_3,v_2) \ge 150 - 98 = 52$. If x lies between p and $v_3$ inclusive, then $d_{\mathcal {D}}(u_3,x) \ge d_{\mathcal {D}}(p,u_3) - d_{\mathcal {D}}(p,x) \ge 99 - 49 = 50$. Since $d_{\mathcal {D}}(u_2,B_9) + d_{\mathcal {D}}(u_3,B_9) \ge 149$, Lemma 5.5 shows that $\gamma$ can reach the centre of a pentagonal face.

Case (2.2). $d_{\mathcal {D}}(q,u_2) \ge 99$. One can establish in a way similar to that used in Case (2.1) the inequality $d_{\mathcal {D}}(u_2,B_{10}) \ge 48$, so that $d_{\mathcal {D}}(u_2,B_{10}) + d_{\mathcal {D}}(u_3,B_{10}) \ge 147$. An application of Lemma 5.5 then gives the required result.

Case (2.3). $d_{\mathcal {D}}(p,u_3) \le 98$ and $d_{\mathcal {D}}(q,u_2) \le 98$. Then, by Lemma 5.1 and the fact that $\{u_2,u_3\} \cap \{o\} = \emptyset$, $d_{\mathcal {D}}(u_2,m_2) \ge 99$ and $d_{\mathcal {D}}(u_3,m_2) \ge 99$. For any $x \in V(B_7)$, $d_{\mathcal {D}}(u_2,x)$ $\ge d_{\mathcal {D}}(x,q) - d_{\mathcal {D}}(q,u_2) \ge 151 - 98 = 53$ and $d_{\mathcal {D}}(u_3,x) \ge d_{\mathcal {D}}(x,p) - d_{\mathcal {D}}(u_3,p) \ge 149 - 98 = 51$. Thus $d_{\mathcal {D}}(u_2,B_7) + d_{\mathcal {D}}(u_3,B_7) \ge 104$, and so one may conclude from Lemma 5.5 that $\gamma$ can move to $m_2$ and safely reach the centre of a pentagonal face.

Detailed analysis of Algorithm 5 for Case (B)

Case (1). There is at least one corner $v_i \in \{v_1,v_2,v_3,v_4,v_5\}$ of $L_{U,49}$ such that $d_{\mathcal {D}}(v_i,u^{\prime }) \ge 99$ for all $u^{\prime } \in \{u_1,u_2,u_3\}$. We consider the case $i = 1$; the proofs for the cases $i = 2,3,4,5$ are similar. Define $F := U_{10} \cup U_6 \cup U_1 \cup U_2 \cup U_7$.

Case (1.1). Both $u_2$ and $u_3$ are in F.

Case (1.1.1). $d_{\mathcal {D}}(u_2,U_1 \cup U_2) + d_{\mathcal {D}}(u_3,U_1 \cup U_2) \ge 100$. First, suppose that at least one of $u_2$ and $u_3$ belongs to $V(U_1) \cup V(U_2)$. Without loss of generality, assume that $u_2 \in V(U_1) \cup V(U_2)$. Then $d_{\mathcal {D}}(u_3,U_1 \cup U_2) \ge 100$.

Suppose $u_2 \in V(B_1)$. $\gamma$ first moves to $v_3$ in 98 rounds. If $\lambda _1$ does not reach $U_4$ by the end of the 98-th round, then $\gamma$ can safely reach $o_4$ in another 98 rounds. If $\lambda _1$ does reach $U_4$ by the end of the 98-th round, using up at least 97 turns in the process, then after $\gamma$ reaches $v_3$, Lemma 5.7 may be applied to $U \cup U_3 \cup U_4$.

Suppose $u_2 \notin V(B_1)$. $\gamma$ first moves to $v_1$ in 98 rounds. If $\lambda _2$ does not move to $B_1$ as $\gamma$ is moving to $v_1$, then $B_1$ does not contain any cop at the end of the 98-th round and at least one of $U_1$ and $U_2$, say $U_i$, does not contain any cop at the end of the 98-th round; thus $\gamma$ can safely reach $o_i$ in another 98 rounds. If $\lambda _2$ does move to $B_1$ as $\gamma$ is moving to $v_1$, using up at least 1 turn in the process, then both $\lambda _3$ and $\lambda _1$ are at least 2 edges away from $U_1 \cup U_2$ at the end of the 98-th round, and so Lemma 5.6 may be applied to $U_1 \cup U_2$.

Second, suppose that neither $u_2$ nor $u_3$ belongs to $V(U_1) \cup V(U_2)$. $\gamma$ then moves to $v_1$ in 98 rounds. Since $d_{\mathcal {D}}(u_i,U_1 \cup U_2) + d_{\mathcal {D}}(u_j,U_1 \cup U_2) \ge 100$ for any distinct $i,j \in \{1,2,3\}$, at least two of the cops are more than 1 edge away from $U_1 \cup U_2$ at the end of the 98-th round. Hence after $\gamma$ reaches $v_1$, Lemma 5.6 may be applied to $U_1 \cup U_2$.

Case (1.1.2). $d_{\mathcal {D}}(u_2,U_1 \cup U_2) + d_{\mathcal {D}}(u_3, U_1 \cup U_2) \le 99$.

Case (1.1.2.1). At least one of $u_2$ and $u_3$ is more than 100 edges away from $U_3 \cup U_4 \cup U_8$. Without loss of generality, assume that $u_2$ is at least 100 edges away from $U_3 \cup U_4 \cup U_8$. $\gamma$ first moves to $v_3$ in 98 rounds. If $\lambda _1$ does not reach $U_4$ by the end of the 98-th round, then $\gamma$ can safely reach $o_4$ in another 98 rounds. Suppose $\lambda _1$ does reach $U_4$ by the end of the 98-th round, using up at least 97 turns in the process. Then, since $u_3$ is at most 99 edges away from $U_1 \cup U_2$, $\lambda _3$ must be at least 100 edges away from $q_3$ at the end of the 98-th round. $\gamma$ now continues moving towards $o_4$ until she reaches $L_{U_4,48}$ as shown in Fig. 11; she then uses 98 turns to move from $p^{\prime \prime }$ to $p^{\prime \prime \prime }$. Suppose that as $\gamma$ is moving from $p^{\prime \prime }$ to $p^{\prime \prime \prime }$, $\lambda _3$ moves at most 97 steps. Then $\gamma$ can safely move from $p^{\prime \prime \prime }$ to $q_3$ in another 2 rounds; after $\gamma$ reaches $q_3$, Lemma 5.7 may be applied to $U_3 \cup U_4 \cup U_8$. Suppose that as $\gamma$ is moving from $p^{\prime \prime }$ to $p^{\prime \prime \prime }$, $\lambda _3$ moves exactly 98 steps. Then $\gamma$ moves from $p^{\prime \prime \prime }$ to $p^{\prime \prime \prime \prime }$ and then to $t_9$ along the path highlighted in Fig. 11. After reaching $t_9$, $\gamma$ may then safely move to $o_9$ in another 98 rounds.

Case (1.1.2.2). Both $u_2$ and $u_3$ are at most 100 edges away from $U_3 \cup U_4 \cup U_8$. If $d_{\mathcal {D}}(u_2,B_1) \ge 99$, then $\gamma$ moves to $v_1$ and then to $o_1$ in 196 rounds.

Suppose $d_{\mathcal {D}}(u_2,B_1) \le 98$. Since $d_{\mathcal {D}}(u_2,U_3 \cup U_4 \cup U_8) \le 100$, $d_{\mathcal {D}}(u_2,B_1) = 98$. If $d_{\mathcal {D}}(u_3,U_1 \cup U_2) \ge 2$, $\gamma$ moves to $v_1$; after reaching $v_1$, either $\gamma$ can safely move to one of $o_1$ and $o_2$, or Lemma 5.6 may be applied to $U_1 \cup U_2$. If $d_{\mathcal {D}}(u_3,U_1 \cup U_2) \le 1$, $\gamma$ moves to $v_3$; then, arguing as in Case (1.1.2.1), either $\gamma$ may safely reach $o_4$ in another 98 rounds, or $\gamma$ may safely reach $q_3$ and then apply the strategy in Lemma 5.7.

Fig. 11

The escape path of $\gamma$ in Case (B.1.1.2.1)

Case (1.2). Neither $u_2$ nor $u_3$ is in F. Note that $d_{\mathcal {D}}(u_2,q_1) \ge 197$ and $d_{\mathcal {D}}(u_3,q_1) \ge 197$. $\gamma$ first moves to $v_1$ in 98 rounds. If at least one of $u_2$ and $u_3$ is not in $U_3 \cup U_5$, Lemma 5.6 may be applied to $U_1 \cup U_2$ after $\gamma$ reaches $v_1$. We will therefore assume that $u_2$ is in $U_5$ and $u_3$ is in $U_3$ (the remaining cases can be dealt with in a very similar way).

If both $\lambda _2$ and $\lambda _3$ are at least 101 edges away from $q_1$ at the end of the 98-th round, then $\gamma$ can safely reach $q_1$ in another 100 rounds. After $\gamma$ reaches $q_1$, either $\gamma$ may safely move to $o_6$ in another 98 rounds or Lemma 5.7 may be applied to $U_1 \cup U_2 \cup U_6$.

Suppose, on the other hand, that at least one of $\lambda _2$ and $\lambda _3$ is at most 100 edges away from $q_1$ at the end of the 98-th round. Without loss of generality, assume that $\lambda _2$ is at most 100 edges away from $q_1$ at the end of the 98-th round. If $\lambda _2$ is at most 99 edges away from $q_1$ at the end of the 98-th round, then neither $\lambda _1$ nor $\lambda _3$ could have moved between the 1-st and the 98-th round, and therefore $\lambda _2$ can safely reach $o_2$ in another 98 rounds. Suppose that $\lambda _2$ is exactly 100 edges away from $q_1$ and $\lambda _3$ is in $U_2$ at the end of the 98-th round. $\gamma$ then starts moving towards $o_2$ until she reaches vertex $x^{\prime }$ on $L_{U_2,23}$ as shown in Fig. 12. She then moves from $x^{\prime }$ to $x^{\prime \prime }$ in 48 rounds (see Fig. 12). If, after $\gamma$ reaches $x^{\prime \prime }$, $\lambda _2$ is still at least 53 edges away from $q_1$ (meaning that $\lambda _2$ did not move during 1 round as $\gamma$ went from $x^{\prime }$ to $x^{\prime \prime }$), $\gamma$ can safely reach $q_1$ in another 52 rounds and Lemma 5.7 may then be applied to $U_1 \cup U_2 \cup U_6$. Suppose that after $\gamma$ reaches $x^{\prime \prime }$, $\lambda _2$ is 52 edges away from $q_1$. $\gamma$ then continues moving along the path highlighted in Fig. 12 until she reaches $x^{\prime \prime \prime }$.

Again, if $\lambda _3$ skips at least one turn as $\gamma$ is moving from $x^{\prime \prime }$ to $x^{\prime \prime \prime }$, then $\gamma$ can safely move to $x^{\prime \prime \prime \prime }$ and then move to $t_{12}$; she may then apply the strategy in Lemma 5.7 to $U_2 \cup U_6 \cup U_7$.

On the other hand, if $\lambda _3$ does not skip any turn as $\gamma$ is moving from $x^{\prime \prime }$ to $x^{\prime \prime \prime }$, then $\gamma$ continues moving along the path highlighted in Fig. 12 until she reaches $m_7$. If, just after $\gamma$ reaches $m_7$, $\lambda _2$ is still at least 1 edge away from $U_6$, then $\gamma$ can safely reach $o_6$. If $\lambda _2$ is in $U_6$ just after $\gamma$ reaches $m_7$, then $\lambda _3$ must still be at least 52 edges away from $t_{12}$ when $\gamma$ is at $m_7$. $\gamma$ may thus safely move from $m_7$ to $t_{12}$ in 50 rounds, and then apply the strategy in Lemma 5.7 to $U_2 \cup U_6 \cup U_7$.

Fig. 12

The escape path of $\gamma$ in Case (B.1.2)

Case (1.3). Exactly one of $u_2$ and $u_3$ is in F. We will assume that $u_2$ is in $U_1$ and $u_3$ is in $U_3$ (the other cases are trivial or similar).

Case (1.3.1). $d_{\mathcal {D}}(u_3,F) \ge 99$. First, suppose $u_2 \in V(B_1)$. Then $\gamma$ can safely move to $v_5$ in 98 rounds. If, at the end of the 98-th round, $\lambda _1$ is not in $U_5$, then $\gamma$ can safely reach $o_5$ in another 98 rounds. Suppose $\lambda _1$ does reach $U_5$ by the end of the 98-th round, using up at least 97 turns in the process. $\gamma$ can then safely reach $q_5$ by moving along $B_5$. After $\gamma$ reaches $q_5$, Lemma 5.6 may be applied to $U_1 \cup U_5 \cup U_{10}$.

Second, suppose $u_2 \notin V(B_1)$. $\gamma$ first moves to $v_1$ in 98 rounds. Suppose $\lambda _2$ does not move as $\gamma$ is moving to $v_1$. Then, since $d_{\mathcal {D}}(u_2,U_1 \cup U_2) \ge 99$ and $d_{\mathcal {D}}(u_3,F) \ge 99$, no cop occupies a vertex belonging to $V(B_1)$ at the end of the 98-th round; furthermore, both $\lambda _2$ and $\lambda _3$ are at least 1 edge away from $U_1 \cup U_2$ at the end of the 98-th round. Hence, after reaching $v_1$, $\gamma$ can safely reach either $o_1$ or $o_2$.

Now suppose $\lambda _2$ uses up at least 1 turn as $\gamma$ is moving to $v_1$. Then both $\lambda _2$ and $\lambda _3$ must be at least 2 edges away from $U_1 \cup U_2$ at the end of the 98-th round. $\gamma$ may now apply the strategy in Lemma 5.6 to $U_1 \cup U_2$.

Case (1.3.2). $d_{\mathcal {D}}(u_3,F) \le 98$.

Case (1.3.2.1). $d_{\mathcal {D}}(u_3,v_3) \ge 12$. $\gamma$ begins moving towards $m_4$. We further distinguish two cases.

Case (1.3.2.1.1). $\lambda _1$ moves at least 47 steps as $\gamma$ is moving towards $m_4$. Suppose that as $\gamma$ is approaching $m_4$, $\lambda _1$ moves z steps for some $z \ge 47$. $\gamma$ then continues moving until she reaches $m_4$ in 98 rounds. Note that $\lambda _2$ and $\lambda _3$ can move a total of at most 51 steps between the turn $\gamma$ moves away from o and the turn after $\gamma$ reaches $m_4$. So $\lambda _3$ is at most 39 edges closer to $o_4$ than $\gamma$ is after $\gamma$ reaches $m_4$. We may assume that at least one of $\lambda _1,\lambda _3$ reaches $U_4$ just after $\gamma$ reaches $m_4$ (otherwise, $\gamma$ can safely reach $o_4$ in another 98 rounds).

Case (1.3.2.1.1.1). $\lambda _1$ reaches $U_4$ before $\lambda _3$. Note that $\lambda _3$ is still at least 11 edges away from $U_4$ just after $\gamma$ reaches $m_4$. $\gamma$ starts moving towards $o_4$ until she reaches $L_{U_4,4}$; $\gamma$ then moves along the path highlighted in Fig. 13. An argument very similar to those used in earlier cases shows that either $\gamma$ can move to $o_8$ without being caught after reaching $r^{\prime \prime \prime }$, or $\gamma$ can continue moving until she safely reaches $t_9$, at which point Lemma 5.7 may be applied to $U_4 \cup U_8 \cup U_9$.

Case (1.3.2.1.1.2). $\lambda _3$ reaches $U_4$ before $\lambda _1$. Suppose that $\lambda _3$ is $\ell$ edges closer to $o_4$ than $\gamma$ is during the turn after $\gamma$ reaches $m_4$. Note that $\ell \le 39$. $\gamma$ starts by moving towards $o_4$. Suppose that as $\gamma$ is approaching $o_4$, $\lambda _3$ skips j turns. If $j > \ell$ then $\gamma$ can safely reach $o_4$. So assume that $j \le \ell$. $\gamma$ continues moving towards $o_4$ until she reaches $L_{U_4,4+\ell -j}$. She then moves along the path highlighted in Fig. 14. One can verify that after reaching $q_3$, either $\gamma$ can safely reach $o_8$ in another 98 rounds, or Lemma 5.7 may be applied to $U_3 \cup U_4 \cup U_8$.

Fig. 13

An escape path of $\gamma$ in Case (B.1.3.2.1.1.1)

Fig. 14

An escape path of $\gamma$ in Case (B.1.3.2.1.1.2)

Case (1.3.2.1.2). $\lambda _1$ moves at most 46 steps as $\gamma$ is moving towards $m_4$. Suppose that $\lambda _1$ moves $\ell$ steps towards $v_4$, where $\ell \le 46$. $\gamma$ first moves to $L_{U,\ell +3}$; she then moves along the side path of $L_{U,\ell +3}$ parallel to $B_9$ until she reaches the corner of $L_{U,\ell +3}$ that is $92-2\ell$ edges away from $v_4$. $\gamma$ then moves to $v_4$ in $92-2\ell$ rounds. Since $d_{\mathcal {D}}(u_3,v_3) \ge 12$, $\gamma$ can safely reach at least one of $\{o_4,o_5,q_4\}$ after reaching $v_4$. Note that if $\gamma$ moves to $q_4$ using the preceding strategy, then she requires a total of $202+\ell$ rounds (starting at the round when she moves away from o). On the other hand, the cops need at least 196 rounds to reach $U_9$, $\lambda _3$ needs at least 12 rounds to reach $U_4$, and $\lambda _1$ needs at least 96 rounds to reach a neighbour of $U_4 \cup U_5 \cup U_9$. Thus if $\gamma$ safely reaches $q_4$ in another 100 rounds, then Lemma 5.7 may be applied to $U_4 \cup U_5 \cup U_9$.

Case (1.3.2.2). $d_{\mathcal {D}}(u_3,v_3) \le 11$.

Case (1.3.2.2.1) $d_{\mathcal {D}}(u_2,q_1) \ge 110$ and $d_{\mathcal {D}}(u_2,U_6) \ge 12$. $\gamma$ first moves to $v_1$ in 98 rounds. If $\lambda _3$ is not in $U_2$ at the end of the 98-th round, then $\gamma$ can safely reach $o_2$ in another 98 rounds. Suppose $\lambda _3$ does reach $U_2$ by the end the 98-th round, using up at least 89 turns in the process. After reaching $v_1$, $\gamma$ continues moving along $B_1$ until she reaches $q_1$ in another 100 rounds. Since $d_{\mathcal {D}}(u_2,q_1) \ge 110$ and $d_{\mathcal {D}}(u_3,q_1) \ge 199$, $\gamma$ can safely reach $q_1$. Furthermore, since $d_{\mathcal {D}}(u_2,U_6) \ge 12$, $\lambda _1$ needs at least 98 rounds to reach a neighbour of $U_1 \cup U_2 \cup U_6$ and $\lambda _3$ uses up at least 89 turns to reach $U_2$, either $\gamma$ may safely reach $o_6$ after reaching $q_1$ or Lemma 5.7 may be applied to $U_1 \cup U_2 \cup U_6$.

Case (1.3.2.2.2). $d_{\mathcal {D}}(u_2,q_1) \le 109$ or $d_{\mathcal {D}}(u_2,U_6) \le 11$. Then $d_{\mathcal {D}}(u_2,v_5) \ge 12$. $\gamma$ may thus apply a strategy similar to that in Case (1.2.2.1), first moving towards $m_5$ and then either safely reaching $o_5$ or moving to one of $\{z_2,q_4\}$ and subsequently applying the strategy in Lemma 5.7 to either $U_5 \cup U_9 \cup U_{10}$ or $U_4 \cup U_5 \cup U_9$.

Case (2). There does not exist a corner $v \in \{v_1,v_2,v_3,v_4,v_5\}$ of $L_{U,49}$ such that $d_{\mathcal {D}}(u_1,v) \ge 98$, $d_{\mathcal {D}}(u_2,v) \ge 99$ and $d_{\mathcal {D}}(u_2,v) \ge 99$. Without loss of generality, assume that $u_2$ is in $U_1$ while $u_3$ is in $U_3$. Since $d_{\mathcal {D}}(u_1,v_4) \le 98$, we have $d_{\mathcal {D}}(u_1,v_4) = 97$.

Case (2.1). $d_{\mathcal {D}}(u_3,v_3) \le 11$. The following two cases are distinguished.

Case (2.1.1) $d_{\mathcal {D}}(u_2,v_5) \ge 12$. $\gamma$ begins moving towards $m_5$. We further distinguish two cases.

Case (2.1.1.1). $\lambda _1$ moves at least 47 steps as $\gamma$ is moving towards $m_5$. Suppose that as $\gamma$ is approaching $m_5$, $\lambda _1$ moves z steps for some $z \ge 47$. $\gamma$ then continues moving until she reaches $m_5$ in 98 rounds. Note that $\lambda _2$ and $\lambda _3$ can move a total of at most 51 steps between the turn $\gamma$ moves away from o and the turn after $\gamma$ reaches $m_5$. So $\lambda _2$ is at most 39 vertices closer to $o_5$ than $\gamma$ is after $\gamma$ reaches $m_5$. We may assume that at least one of $\lambda _1,\lambda _2$ reaches $U_5$ just after $\gamma$ reaches $m_5$ (otherwise, $\gamma$ can safely reach $o_5$ in another 98 rounds).

Case (2.1.1.1.1). $\lambda _1$ reaches $U_5$ before $\lambda _2$. Note that $\lambda _2$ is still at least 11 edges away from $U_5$ just after $\gamma$ reaches $m_5$. $\gamma$ starts by moving towards $o_5$ until she reaches $L_{U_5,4}$; $\gamma$ then moves along the path highlighted in Fig. 15. An argument very similar to those used in earlier cases shows that either $\gamma$ can move to $o_{10}$ without being caught after reaching $t_{19}$, or $\gamma$ can continue moving until she safely reaches $z_2$, at which point Lemma 5.7 may be applied to $U_5 \cup U_9 \cup U_{10}$.

Case (2.1.1.1.2). $\lambda _2$ reaches $U_5$ before $\lambda _1$. Suppose that $\lambda _2$ is $\ell$ vertices closer to $o_5$ than $\gamma$ is during the turn after $\gamma$ reaches $m_5$. Note that $\ell \le 39$. $\gamma$ starts by moving towards $o_5$. Suppose that as $\gamma$ is approaching $o_5$, $\lambda _2$ skips j turns. If $j > \ell$ then $\gamma$ can safely reach $o_5$. So assume that $j \le \ell$. $\gamma$ continues moving towards $o_5$ until she reaches $L_{U_5,4+\ell -j}$. She then moves along the path highlighted in Fig. 16. One can verify that after reaching $q_4$, either $\gamma$ can safely reach $o_9$ in another 98 rounds, or Lemma 5.7 may be applied to $U_4 \cup U_5 \cup U_9$.

Case (2.1.1.2). $\lambda _1$ moves at most 46 steps as $\gamma$ is moving towards $m_5$. Suppose that $\lambda _1$ moves $\ell$ steps towards $v_4$, where $\ell \le 46$. $\gamma$ first moves to $L_{U,\ell +3}$; she then moves along the side path of $L_{U,\ell +3}$ parallel $B_{10}$ until she reaches the corner of $L_{U,\ell +3}$ that is $92-2\ell$ edges away from $v_4$. $\gamma$ then moves to $v_4$ in $92-2\ell$ rounds; note that $\lambda _3$ cannot catch $\gamma$ just after $\gamma$ reaches $v_4$ because he is at least 4 edges away from $U_4$. Since $d_{\mathcal {D}}(u_2,v_5) \ge 12$, $\gamma$ can safely reach either $o_5$ or $q_4$ after reaching $v_4$. Note that if $\gamma$ moves to $q_4$ using the preceding strategy, then she requires a total of $202+\ell$ rounds (starting at the round when she moves away from o). On the other hand, the cops need at least 196 rounds to reach $U_9$, $\lambda _2$ needs at least 12 rounds to reach $U_5$, and $\lambda _1$ needs at least 96 rounds to reach a neighbour of $U_4 \cup U_5 \cup U_9$. Thus if $\gamma$ can safely reach $q_4$ in another 100 rounds, then Lemma 5.7 may be applied to $U_4 \cup U_5 \cup U_9$.

Fig. 15

An escape path of $\gamma$ in Case (B.2.1.1.1.1)

Fig. 16

An escape path of $\gamma$ in Case (B.2.1.1.1.2)

Case (2.1.2). $d_{\mathcal {D}}(u_2,v_5) \le 11$. Both $d_{\mathcal {D}}(u_2,v_5) \le 11$ and $d_{\mathcal {D}}(u_3,v_3) \le 11$ hold. $\gamma$ starts moving towards $m_2$. Note that at most one of $\lambda _2$ and $\lambda _3$ can reach $U_2$ before or just after $\gamma$ reaches $m_2$. We may assume that either $\lambda _2$ or $\lambda _3$ reaches $U_2$ before or just after $\gamma$ reaches $m_2$.

Suppose that $\lambda _3$ reaches $U_2$ before $\lambda _2$. Suppose $\lambda _3$ is $\ell$ vertices closer to $o_2$ than $\gamma$ is just after $\gamma$ reaches $m_2$. Note that $\ell \le 9$. $\gamma$ starts moving towards $o_2$. Suppose $\lambda _3$ skips j turns as $\gamma$ is approaching $o_2$. If $j > \ell$, then $\gamma$ can safely reach $o_2$. Assume now that $j \le \ell$. $\gamma$ moves towards $o_2$ until she reaches $L_{U_2,4+\ell -j}$, continuing along the path highlighted in Fig. 17 until she reaches $q_1$. One may directly verify (in a way that is similar to earlier cases) that either $\gamma$ can safely reach $o_6$, or Lemma 5.7 may be applied to $U_1 \cup U_2 \cup U_6$. The case that $\lambda _2$ reaches $U_2$ before $\lambda _3$ may be handled similarly; in this case $\gamma$ should move from $t_{22}$ to $q_2$ instead.

Fig. 17

An escape path of $\gamma$ in Case (B.2.1.2)

Case (2.2). $d_{\mathcal {D}}(u_3,v_3) \ge 12$. Observe that this case is almost symmetrical to Case (2.1.1) and a parallel argument may be applied. More precisely, note that if one maps the set of corner vertices of U to itself as follows: $v_4 \rightarrow v_4, v_5 \rightarrow v_3, v_3 \rightarrow v_5, v_1 \rightarrow v_2, v_2 \rightarrow v_1$, and extend this mapping so as to obtain an automorphism $\sigma$ of $\mathcal {D}$, then $\gamma$ may apply a strategy similar to that in Case (2.1.1) for $\sigma (\mathcal {D})$ (with the appropriate transformed vertices).

Detailed analysis of Algorithm 6 for Case (C)

Without loss of generality, assume that $\lambda _2$ is currently not in U while both $\lambda _1$ and $\lambda _3$ are currently in U. As the proof techniques in the present case are so similar to those in Cases (A) and (B), we will omit many proof details and refer to strategies for $\gamma$ in previous cases.

Case (1). There is at least one corner $v_i$ of $L_{U,49}$ such that $d_{\mathcal {D}}(v_i,u^{\prime }) \ge 99$ for all $u^{\prime } \in \{u_2,u_3\}$ and $d_{\mathcal {D}}(v_i,u_1) \ge 98$. We first assume that $i = 1$. As in Case (B), define $F := U_{10} \cup U_6 \cup U_1 \cup U_2 \cup U_7$.

Case (1.1). $u_2$ is in F. First, suppose that $d_{\mathcal {D}}(u_2,v_5) \le 98$. (This implies that $u_2$ is in $U_1$.) If $d_{\mathcal {D}}(u_3,B_{10}) \le 98$, then $\gamma$ moves to $v_2$ in 98 rounds; Lemma 5.6 may then be applied to $U_2 \cup U_3$. Now suppose that $d_{\mathcal {D}}(u_3,B_{10}) \ge 99$. If $d_{\mathcal {D}}(u_2,v_5) \ge 12$, then $\gamma$ may apply a winning strategy similar to that in Case (B.2.1.1). Now suppose that $d_{\mathcal {D}}(u_2,v_5) \le 11$. If $d_{\mathcal {D}}(u_3,B_7) \le 50$, then $\gamma$ may apply the winning strategy in Lemma 5.5, first moving to p in 98 rounds. Now suppose $d_{\mathcal {D}}(u_3,B_7) \ge 51$. $\gamma$ first moves to $v_1$ in 98 rounds. Note that $d_{\mathcal {D}}(u_2,q_1) \ge 185$. If $\lambda _3$ is not in $U_2$ just after the round when $\gamma$ reaches $v_1$, then $\gamma$ can safely reach $o_2$ in another 98 rounds. If $\lambda _3$ is in $U_2$ just after the round when $\gamma$ reaches $v_1$, then $\lambda _2$ could have moved at most 47 steps between the 1-st and the 98-th round. Thus $\lambda _2$ is at least 138 edges away from $q_1$ just after the 98-th round. $\gamma$ can now safely move to $q_1$ in 100 rounds, and then to $o_6$ in another 98 vertices.

Second, suppose that $d_{\mathcal {D}}(u_2,v_5) \ge 99$. We distinguish the following cases.

Case (1.1.1). $d_{\mathcal {D}}(u_3,v_5) \le 101$, $d_{\mathcal {D}}(u_3,v_4) \le 101$ and for all $i \in \{1,2,3\}$, $d_{\mathcal {D}}(u_3,v_i) \ge 100$. First, suppose that $d_{\mathcal {D}}(u_2,v_2) \le 98$. If $d_{\mathcal {D}}(u_3,U_4) + d_{\mathcal {D}}(u_2,U_3) \ge 5$, then $\gamma$ first moves to $v_3$ in 98 rounds. If neither $\lambda _1$ nor $\lambda _3$ reaches $U_4$ at the end of the 98-th round, then $\gamma$ can move to $o_4$ without being caught in another 98 rounds. If either $\lambda _1$ or $\lambda _3$ reaches $U_4$ at the end of the 98-th round (or if both $\lambda _1$ and $\lambda _3$ reach $U_4$ at the end of the 98-th round), then $\gamma$ can safely reach $q_3$ in another 100 rounds. After reaching $q_3$, either $\gamma$ can safely reach $o_8$ in another 98 rounds, or Lemma 5.7 may be applied to $U_3 \cup U_4 \cup U_8$.

If $d_{\mathcal {D}}(u_3,U_4) + d_{\mathcal {D}}(u_2,U_3) \le 4$, then $\gamma$ first moves to $v_1$ in 98 rounds. If $\lambda _3$ does not reach $U_1$ at the end of the 98-th round, then $\gamma$ can safely reach $o_1$ in another 98 rounds. If $\lambda _3$ reaches $U_1$ at the end of the 98-th round, then, since $d_{\mathcal {D}}(u_2,q_1) + d_{\mathcal {D}}(u_3,U_1) \ge 100 - d_{\mathcal {D}}(u_3,U_4) + 196 - d_{\mathcal {D}}(u_2,U_3) \ge 288$, $\gamma$ can move safely towards $q_1$ in another 100 rounds, and then safely reach $o_6$ using an additional 98 rounds.

Second, suppose $d_{\mathcal {D}}(u_2,v_2) \ge 99$. Suppose $u_2 \in V(U_7)$. If $u_2 \ne q_2$, then $\gamma$ first moves to $v_2$ in 98 rounds. If $\lambda _2$ does not reach $B_2$ by the end of the 98-th round, then $\gamma$ can safely reach either $o_2$ or $o_3$. If $\lambda _2$ does reach $B_2$ by the end of the 98-th round, then both $\lambda _1$ and $\lambda _3$ are still at least 2 edges away from $U_2 \cup U_3$ at the end of the 98-th round, and therefore Lemma 5.6 may be applied to $U_2 \cup U_3$. Suppose $u_2 = q_2$. If $d_{\mathcal {D}}(u_3,U_4) \ge 4$, then $\gamma$ first moves to $v_3$ in 98 rounds. If neither $\lambda _1$ nor $\lambda _3$ is in $U_4$ at the end of the 98-th round, then $\gamma$ can safely reach $o_4$ in another rounds. If either $\lambda _1$ or $\lambda _3$ is in $U_4$ at the end of the 98-th round, then $\gamma$ continues moving along $B_3$ until she reaches $q_3$ using another 100 rounds. Then either $\gamma$ can safely move to $o_8$ in another 98 rounds, or Lemma 5.7 may be applied to $U_3 \cup U_4 \cup U_8$. If $d_{\mathcal {D}}(u_3,U_4) \le 3$, then $\gamma$ first moves to $v_1$ in 98 rounds. After reaching $v_1$, $\gamma$ can either safely reach $o_1$ using another 98 rounds, or $\gamma$ can move to $q_1$ in another 100 rounds and then apply the strategy in Lemma 5.7 to $U_1 \cup U_2 \cup U_6$.

Now suppose $u_2 \notin V(U_7)$. If $d_{\mathcal {D}}(u_3,U_4) + d_{\mathcal {D}}(u_2,U_3) \ge 3$, then $\gamma$ first moves to $v_3$ in 98 rounds. If $U_3$ (resp. $U_4$) does not contain any cop at the end of the 98-th round, then $\gamma$ safely moves to $o_3$ (resp. $o_4$) using another 98 rounds. Suppose each of $U_3$ and $U_4$ contains a cop at the end of the 98-th round. Since $d_{\mathcal {D}}(u_i,U_4) + d_{\mathcal {D}}(u_2,U_3) \ge 3$ whenever $i \in \{1,3\}$, it follows that at the end of the 98-th round, $\gamma$ can safely move along $B_3$ to $q_3$ using another 100 rounds. After reaching $q_3$, $\gamma$ can either safely reach $o_8$ using another 98 rounds or apply the strategy in Lemma 5.7 to $U_3 \cup U_4 \cup U_8$. If $d_{\mathcal {D}}(u_3,U_4) + d_{\mathcal {D}}(u_2,U_3) \le 2$, then $\gamma$ first moves to $v_1$ in 98 rounds. If no cop is in $U_1$ at the end of the 98-th round, then $\gamma$ can safely reach $o_1$ in another 98 rounds; otherwise, $\gamma$ can move along $B_1$ and reach $q_1$ without being caught; she can then move safely to $o_6$ in another 98 rounds.

Case (1.1.2). $d_{\mathcal {D}}(u_3,v_3) \le 101$, $d_{\mathcal {D}}(u_3,v_4) \le 101$ and for all $i \in \{1,2,5\}$, $d_{\mathcal {D}}(u_3,v_i) \ge 100$. Note that $u_3$ and $u_1$ are each at least 99 edges away from $U_1 \cup U_2$. First, suppose that $u_2 \ne q_1$. $\gamma$ moves to $v_1$ in 98 rounds. If $\lambda _2$ does not move between the 1-st and the 98-th round, then some $U_i \in \{U_1,U_2\}$ does not contain any cop at the end of the 98-th round. $\gamma$ may then safely reach $o_i$ in another 98 rounds. If $\lambda _2$ moves at least one step between the 1-st and the 98-th round, then both $\lambda _3$ and $\lambda _1$ are each at least 2 edges away from $U_1 \cup U_2$ at the end of the 98-th round. One may then apply Lemma 5.6 to $U_1 \cup U_2$.

Second, suppose that $u_2 = q_1$. If $d_{\mathcal {D}}(u_3,B_{10}) \ge 50$, then $\gamma$ moves to $v_5$ in 98 rounds. At the end of the 98-th round, $\gamma$ can either safely reach $o_5$ in another 98 rounds, or move towards $q_5$ and then to $o_{10}$ in another 198 rounds. If $d_{\mathcal {D}}(u_3,B_8) \ge 50$, then $\gamma$ moves to $v_2$ in 98 rounds. An argument similar to that in the preceding case (that is, when $d_{\mathcal {D}}(u_3,B_{10}) \ge 50$) shows that $\gamma$ can either safely reach $o_3$ in another 98 rounds or safely reach $o_7$ in another 198 rounds.

Case (1.1.3). $d_{\mathcal {D}}(u_3,v_2) \le 101$, $d_{\mathcal {D}}(u_3,v_3) \le 101$ and for all $i \in \{1,4,5\}$, $d_{\mathcal {D}}(u_3,v_i) \ge 100$. $\gamma$ moves to $v_5$ in 98 rounds. An argument very similar to that in Case (1.1.2) shows that $\gamma$ can either safely reach $o_5$ in another 98 rounds, or safely reach $o_{10}$ in another 198 rounds.

Case (1.1.4). $d_{\mathcal {D}}(u_3,v_1) \le 101$, $d_{\mathcal {D}}(u_3,v_2) \le 101$ and for all $i \in \{3,4,5\}$, $d_{\mathcal {D}}(u_3,v_i) \ge 100$. $\gamma$ pursues the same winning strategy as that in Case (1.1.3).

Case (1.1.5). $d_{\mathcal {D}}(u_3,v_1) \le 101$, $d_{\mathcal {D}}(u_3,v_5) \le 101$ and for all $i \in \{2,3,4\}$, $d_{\mathcal {D}}(u_3,v_i) \ge 100$. First, suppose that $d_{\mathcal {D}}(u_2,v_2) \le 98$. $\gamma$ moves to $v_3$ in 98 rounds. If neither $\lambda _1$ nor $\lambda _3$ is in $U_4$ at the end of the 98-th round, then $\gamma$ may safely reach $o_4$ in another 98 rounds. If either $\lambda _1$ or $\lambda _3$ is in $U_4$ at the end of the 98-th round, then $\lambda _2$ must still be at least 195 edges away from $q_3$ at the end of the 98-th round. Thus $\gamma$ may safely move to $q_3$ in 100 rounds, and then to $o_8$ in another 98 rounds.

Second, suppose that $d_{\mathcal {D}}(u_2,v_2) \ge 99$. If $u_2 = q_2$, then $\gamma$ employs the winning strategy in the preceding case (that is, the case when $d_{\mathcal {D}}(u_2,v_2) \le 98$). If $u_2 \ne q_2$, then $\gamma$ moves to $v_2$ in 98 rounds. If $\lambda _2$ moves at least one step between the 1-st and the 98-th round, then $\lambda _1$ and $\lambda _3$ are each at least 2 edges away from $U_2 \cup U_3$ at the end of the 98-th round (note that since $d_{\mathcal {D}}(u_3,v_1) \ge 99$ by the case assumption and $v_1$ is the vertex of $U_2$ that is closest to $u_3$, $d_{\mathcal {D}}(u_3,U_2) \ge 99$) and therefore Lemma 5.6 may be applied to $U_2 \cup U_3$. If $\lambda _2$ does not move between the 1-st and the 98-th round, then there is some $U_i \in \{U_2,U_3\}$ such that $U_i$ does not contain any cop at the end of the 98-th round. Thus $\gamma$ may safely reach $o_i$ in another 98 rounds.

Case (1.2). $u_2$ is not in F. First, suppose that $d_{\mathcal {D}}(u_2,U_1\cup U_2) \ge 3$ or $d_{\mathcal {D}}(u_3,U_1 \cup U_2) \ge 3$. $\gamma$ first moves to $v_1$ in 98 rounds. If some $U_i \in \{U_1,U_2\}$ does not contain any cop at the end of the 98-th round, then $\gamma$ moves to $o_i$ in another 98 rounds. If both $U_1$ and $U_2$ contain at least one cop at the end of the 98-th round, then $\gamma$ continues moving towards $q_1$. Since each cop requires at least 196 rounds (from his starting position) to reach $q_1$ but at least 2 cops need more than 2 rounds to reach $U_1 \cup U_2$ (and no cop can reach $v_1$ in 98 rounds), $\gamma$ can safely get from $v_1$ to $q_1$ in 100 rounds, and then move from $q_1$ to $o_6$ in another 98 rounds.

Second, suppose that $d_{\mathcal {D}}(u_2,U_1 \cup U_2) \le 2$ and $d_{\mathcal {D}}(u_3,U_1 \cup U_2) \le 2$. $\gamma$ then moves to p in 98 rounds. One may then apply Lemma 5.5 to obtain a winning strategy for $\gamma$.

Case (2). For each corner $v^{\prime }$ of $L_{U,49}$, it holds that $d_{\mathcal {D}}(v^{\prime },u^{\prime }) \le 98$ for some $u^{\prime } \in \{u_2,u_3\}$ or $d_{\mathcal {D}}(v^{\prime },u_1) \le 97$ (or both inequalities hold).

Case (2.1). $d_{\mathcal {D}}(u_2,v_2) \le 98,d_{\mathcal {D}}(u_2,v_3) \le 98, d_{\mathcal {D}}(u_3,v_1) \le 98,d_{\mathcal {D}}(u_3,v_5) \le 98$ and $d_{\mathcal {D}}(u_1,v_4)$ $= 97$. Suppose that $d_{\mathcal {D}}(u_2,v_3) \ge 12$. $\gamma$ may then apply the winning strategy in Case (B.2.2). Now suppose that $d_{\mathcal {D}}(u_2,v_3) \le 11$. Then $d_{\mathcal {D}}(u_2,U_2) \ge 89$ and $d_{\mathcal {D}}(u_2,m_2) \ge 139$. Consider the following case distinction: (i) $d_{\mathcal {D}}(u_3,m_2) \le 98$ and (ii) $d_{\mathcal {D}}(u_3,m_2) \ge 99$.

(i) Notice that in this case, $d_{\mathcal {D}}(u_3,m_5) \ge 99$ and $d_{\mathcal {D}}(u_3,v_5) \ge 49$. $\gamma$ may then apply a winning strategy similar to that in Case (B.2.1.1).

(ii) It follows from the inequalities $d_{\mathcal {D}}(u_3,m_2) \ge 99$ and $d_{\mathcal {D}}(u_3,v_5)\le 98,d_{\mathcal {D}}(u_3,v_1) \le 98$ that $d_{\mathcal {D}}(u_3,v_1) \ge 49$. $\gamma$ may then apply the winning strategy in Lemma 5.5, first moving to $m_2$ in 98 rounds (note that the winning strategy applies in this case even though $\lambda _2$ is not in U at the start).

Case (2.2). $d_{\mathcal {D}}(u_2,v_1),d_{\mathcal {D}}(u_2,v_5) \le 98, d_{\mathcal {D}}(u_3,v_2),d_{\mathcal {D}}(u_3,v_3) \le 98$ and $d_{\mathcal {D}}(u_1,v_4)$ $= 97$. As in Case (B.2.1), we first suppose that $d_{\mathcal {D}}(u_2,v_5) \ge 12$. $\gamma$ may then employ the winning strategy in Case (B.2.1.1). Now suppose that $d_{\mathcal {D}}(u_2,v_5) \le 11$. Then $d_{\mathcal {D}}(u_2,m_2) \ge 139$ and $d_{\mathcal {D}}(u_2,v_1) \ge 89$.

(i) $d_{\mathcal {D}}(u_3,m_2) \ge 99$. Then $d_{\mathcal {D}}(u_3,v_2) \ge 49$. $\gamma$ moves to $m_2$ in 98 rounds, employing the winning strategy in Lemma 5.5.

(ii) $d_{\mathcal {D}}(u_3,m_2) \le 98$. Then $d_{\mathcal {D}}(u_3,m_4) \ge 99$. $\gamma$ moves to $m_4$ in 98 rounds, employing the win ning strategy in Case (B.2.2).