Optimal Number of Pursuers in Differential Games on the 1-Skeleton of an Orthoplex

Abstract

We study a differential game of many pursuers and one evader. All the players move only along the one-skeleton graph of an orthoplex of dimension $d+1$. It is assumed that the maximal speeds of the pursuers are less than the speed of the evader. By definition, the pursuit is completed if the position of a pursuer coincides with the position of the evader. Evasion is said to be possible in the game if the movements of players are started from some initial positions and the position of the evader never coincides with the position of any pursuer. We found the optimal number of pursuers in the game. The symmetry of the orthoplex plays an important role in the construction of the players’ strategies.

1. Introduction

There are many papers devoted to pursuit and evasion differential games in ${\mathbb{R}}^n$ (see, for example, [1,2,3,4,5,6,7]).

An essential part of differential games is the differential games of many players. In the case of geometric constraints, interesting results were obtained by [8,9,10,11,12,13,14,15,16,17,18,19]. The paper [20] was devoted to a survey of such differential games.

In the work [21], a self-triggered pursuit strategy was proposed. It was assumed that the state information was available to the pursuer and evader. In a differential game of two evaders and one faster pursuer considered in [22], the plane was divided into two half-planes, the play and goal regions. The pursuer tries to protect the goal region from the evaders, and the evaders try to reach this region. A strategy for the pursuer was constructed based on the Apollonius circle.

Differential games of many players with integral constraints on the control functions of the players are also of increasing interest. For example, the works [23,24,25,26,27,28] dealt with the evasion differential games of many pursuers.

The papers [29,30] were devoted to differential games with state constraints.

There are some differential games in ${\mathbb{R}}^n$ where, for any behavior of the evader, the pursuit can be completed from some initial position. At the same time, the evader, by choosing his/her control, may delay the capture time as long as he/she wishes. However, this does not happen for the pursuit and evasion differential games on finite graphs. For finite graphs, it is possible either that there is a number $\theta$ such that, for any initial state of the players, the game ends by time $\theta$ or evasion is possible in the game forever.

The following two types of games on graphs should be mentioned. In the first type of games on graphs, players move from one vertex of the graph to an adjacent vertex by jumping [31,32,33,34,35,36,37]. In the second type of games, players move along the edges of the graph [38,39,40,41,42,43]. Both types of games can be formulated in a minimax form, and each of them is a model of the search problem of a moving object [43,44,45].

In the paper [39], pursuit and evasion differential games of n pursuers and one evader were studied on the one-skeletons of the regular polyhedrons in ${\mathbb{R}}^3$. All players can move with the speed not exceeding one. It was shown that the optimal number of pursuers for the tetrahedron, cube, and octahedron is two and, for the icosahedron and dodecahedron, is three. A similar differential game was studied in [40] on the one-skeletons of the d-dimensional regular simplex, orthoplex, and cube, and it was proven that the optimal number of pursuers for these polyhedrons are two, two, and $[d/2]+1$, respectively. Later on, it was shown in the work [41] that the optimal number of pursuers for the regular twenty-four-gone and one-hundred-twenty-gone in ${\mathbb{R}}^4$ is equal to three.

The purpose of the present paper is to study a pursuit and evasion game on the edge graph $K_{d+1}$ of the orthoplex ${\Sigma }^{2(d+1)}$ in the Euclidean space ${\mathbb{R}}^{d+1}$. A $(d+1)$-dimensional orthoplex ${\Sigma }^{2(d+1)}$ is a convex hull of $2(d+1)$ points:

$$e_{2i-1}=(0,\dots ,0,1,0,\dots ,0),e_{2i}=(0,\dots ,0,-1,0,\dots ,0)\in {\mathbb{R}}^{d+1},i=1,2,\dots ,(d+1),$$

where 1 in $e_{2i-1}$ and $-1$ in $e_{2i}$ is in the i-th place. Its edges of length $\sqrt{2}$ form a finite graph $K_{d+1}$ with $2(d+1)$ vertices.

Let n pursuing points $x_k$, $k=1,2,\dots ,n$, whose velocities do not exceed in absolute value ${\rho }_k$, ${\rho }_k>0$, $k=1,2,\dots ,n$, respectively, and one evading point E, whose velocity does not exceed $\sigma$, $\sigma >0$, move along the graph $K_{d+1}$. These data define the differential game. The main difference of the present work from [40] is that we study differential games of slow pursuers. In [40], the construction of the strategies of pursuers was based on the fact that a pursuer can move symmetrically toward the evader with respect to some hyperplane, but in the case of slow pursuers, this is impossible. The construction of the strategies of pursuers in the present paper is based on the fact that each pursuer with speed $\ge 1/2$ can guard two vertices of an edge of the orthoplex, and each pursuer with speed $<1/2$ can guard only one vertex of an edge of the orthoplex. In constructing the evader’s strategy, this fact plays a key role as well.

Usually, in pursuit differential games, the pursuer has an advantage over the evader. For example, the control set of the pursuer may contain that of the evader, and vice versa, in evasion differential games, it is natural that the evader has an advantage over the pursuer. Note that the fact that the players cannot leave a given finite graph is itself an advantage for the pursuer in differential games on finite graphs. Therefore, we can consider a pursuit differential game with slow pursuers as well. Furthermore, we can obtain a pair of winning strategies of players in pursuit and evasion problems [2,6]. In the present paper, we use the $\pi$-strategy (see, for example, [46]) in constructing the pursuit strategies. Note that if a pursuer applies the $\pi$-strategy on some time interval $[t_0,t_1]$, then the straight line passing through the states of the pursuer and evader at any time $t\in [t_0,t_1]$ remains parallel with the straight line passing through the states of the pursuer and evader at time $t_0$.

2. Statement of the Problem

We considered a differential game of n pursuers $x_1$, $x_2$, …, $x_n$, $n\ge 2$, and one evader y, whose dynamics are given by the following equations:

$$\begin{array}{cc}{\dot{x}}_i=u_i,\hfill & x_i\left(0\right)=x_{i0},i=1,...,n,\hfill \\ \dot{y}=v,\hfill & y\left(0\right)=y_0,\hfill \end{array}$$

(1)

where $x_{i0},y_0\in K_{d+1}$, $x_{i0}\ne y_0$, $i=1,\dots ,n$; $u_i$ is the control parameter of the i-th pursuer; v is the control parameter of the evader. All the players move along the edges of orthoplex $K_{d+1}$. The maximal speeds of the pursuers $x_1$, $x_2$,…, $x_n$ are ${\rho }_1$, ${\rho }_2$, …, ${\rho }_n$, respectively, and that of the evader y is $\sigma$, i.e., $|u_i|\le {\rho }_i$, $i=1,\dots ,n$, $\left|v\right|\le \sigma$.

A function $u_i(·)$, $u_i:[0,\infty )\to B\left({\rho }_i\right)$ is called an admissible control of the i-th pursuer, $i\in \{1,\dots ,n\}$, if for the solution $x_i(·)$ of the equation:

$${\dot{x}}_i=u_i,x_i\left(0\right)=x_{i0},$$

we have $x_i\left(t\right)\in K_{d+1}$, $t\ge 0$.

A function $v(·)$, $v:[0,\infty )\to B\left(\sigma \right)$ is called the admissible control of the evader, if for the solution $y(·)$ of the equation:

$$\dot{y}=v,y\left(0\right)=y_0,$$

we have $y\left(t\right)\in K_{d+1}$, $t\ge 0$.

We considered pursuit and evasion games. In the pursuit differential game, pursuers apply some strategies, and the evader uses an arbitrary admissible control. Let us define the strategies of the pursuers.

Functions $(t,x_1,\dots ,x_n,y,v)\to U_i(t,x_1,...,x_n,y,v)$, $i=1,2,\dots ,n$, are called the strategies of the pursuers $x_i$, $i=1,2,...,n$, if the initial value problem (1) has a unique solution $x_1\left(t\right),\dots ,x_n\left(t\right)$, $y\left(t\right)\in K_{d+1}$, $t\ge 0$, for $u_i=U_i(t,x_1,\dots ,x_n,y,v)$, $i=1,2,\dots ,n$, and for any admissible control $v=v\left(t\right)$ of the evader.

If, for some number $T>0$, there exist strategies of pursuers such that $x_i\left(\tau \right)=y\left(\tau \right)$ at some $0<\tau \le T$ and $i\in \{1,\dots ,n\}$, then the pursuit is said to be completed. The pursuers are interested in completing the pursuit as earlier as possible.

A function $(t,x_1,\dots ,x_n,y)\to V(t,x_1,\dots ,x_n,y)$ is called a strategy of the evader y if the initial value problem (1) has a unique solution $x_1\left(t\right),\dots ,x_n\left(t\right)$, $y\left(t\right)\in K_{d+1}$, $t\ge 0$, for $v=V(t,x_1,\dots ,x_n,y)$ and for any admissible controls of the pursuers $u_i=u_i\left(t\right)$, $i=1,2,\dots ,n$.

If, for some initial states of the players $x_{10},\dots ,x_{n0},y_0\in K_{d+1}$, there exists a strategy of the evader such that $x_i\left(t\right)\ne y\left(t\right)$ for all $t\ge 0$, and $i=1,\dots ,n$, then we say that evasion is possible in the game in $K_{d+1}$. The evader is interested in maintaining the inequality $x_i\left(t\right)\ne y\left(t\right)$ as long as possible. Since for some initial states, the evader may be trapped by the pursuers and the pursuit can be completed by the pursuers easily, therefore this definition contains the phrase “for some initial states of players $x_{10},\dots ,x_{n0},y_0\in K_{d+1}$”.

The number $N=N\left(K_{d+1}\right)$ is called the optimal number of pursuers for the game on cocube $K_{d+1}$ if, for any initial states of the players, the pursuit can be completed in the game with N pursuers and evasion is possible in the game with $N-1$ pursuers.

The problem is to find the optimal number of pursuers N in the game, to construct strategies for the pursuers in the pursuit game, and the evasion strategy.

3. Main Result

The one-skeleton $K_{d+1}$ of the orthoplex ${\Sigma }^{2(d+1)}$ can be obtained as follows. We call the symmetry vertices $e_{2i-1}$ and $e_{2i}$ of ${\Sigma }^{2(d+1)}$ antipodal. For each $i=1,2,\dots ,2(d+1)$, we connect $e_i$ with all vertices with segments, which are not antipodal to $e_i$, and we obtain $K_{d+1}$.

If ${\rho }_{i_0}>\sigma$ for some $i_0\in \{1,2,\dots ,n\}$, then, clearly, only one pursuer $x_{i_0}$ can capture the evader. If ${\rho }_{i_0}=\sigma$ and ${\rho }_{j_0}=\sigma$, for some $i_0,j_0\in \{1,2,\dots ,n\}$ and $i_0\ne j_0$, then it is shown that only two pursuers $x_{i_0}$ and $x_{j_0}$ can capture the evader [40]. Furthermore, it can be shown that if ${\rho }_{i_0}=\sigma$ and $n\ge 2$, then the pursuit can be completed. Therefore, in the present paper, we considered the case where ${\rho }_1$, ${\rho }_2$, …, ${\rho }_n$ less than $\sigma$, that is $0<{\rho }_i<\sigma$.

We denote the vectors corresponding to the points $x_1$, $x_2$, …, $x_n$, and E by $x_1$, $x_2$, …, $x_n$, and y, respectively. Let:

$$1/2\le {\rho }_i<1,i=1,2,...,k;0<{\rho }_i<1/2,i=k+1,...,n.$$

If the inequality $1/2\le {\rho }_i<1$ is not satisfied for all $i=1,2,\dots ,n$, then we consider $k=0$. Clearly, $k\le n$. For the pursuit differential game, the following statement is true.

Theorem 1.

If either (i) $n=k=d+1$ or (ii) $n\ne k$ and $n+k\ge 2d$, then the pursuit can be completed in the differential game on the orthoplex $K_{d+1}$.

Proof. 

Case 1. Let $n=k=d+1$, that is the maximum speeds of all pursuers greater than or equal to 1/2 and the number of pursuers is $d+1$. We temporarily remove from the orthoplex $K_{d+1}$ the two symmetric with respect to the origin vertices $S=e_{2i-1}=(0,\dots ,0,1,0,\dots ,0)$, $S_1=e_{2i}=(0,\dots ,0,-1,0,\dots ,0)$ with $i=d+1$ and all the edges with the endpoint at S or $S_1$. As a result, we obtain an orthoplex $K_d$, which has $2d$ vertices.

Let $A_1{B}_1$, $A_2{B}_2$, …, $A_d{B}_d$ be the edges of the orthoplex $K_d$, any two of which have no common vertex (Figure 1), where $A_1$, $B_1$, $A_2$, $B_2$, …, $A_d$, $B_d$ are all the vertices of the orthoplex $K_d$. For example, if we denote the points corresponding to the vectors $e_1$, $e_3$, …, $e_{2d-1}$, $e_2$, $e_4$, …, $e_{2d}$, respectively, by $A_1$, $B_1$, $A_2$, $B_2$, …, $A_d$, $B_d$, then we obtain such edges, where $A_i$ is not antipodal to $B_i$, $i=1,2,\dots ,d$, and so, the vectors $e_p$ and $e_q$ corresponding to $A_i$ and $B_i$ are orthogonal.

We show that any two edges of the orthoplex $K_d$ are either parallel, or orthogonal, or form an angle equal to $\pi /3$. To this end, it is sufficient to find the angle between any two vectors $e_i-e_j$ and $e_k-e_l$, where $e_i\ne e_j$, $e_i\ne -e_j$, $e_k\ne e_l$, $e_k\ne -e_l$. Note that any two distinct vectors $e_p$ and $e_q$ are either symmetric with respect to the origin or orthogonal. We have:

$$cos\alpha =\frac{(e_i-e_j)(e_k-e_l)}{|e_i-e_j\left|\right|e_k-e_l|}=\frac{e_i{e}_k-e_i{e}_l-e_j{e}_k+e_j{e}_l}{2}.$$

A. Let $e_i=-e_k$, $e_j=-e_l$. Then, $e_i{e}_k=-1$, $e_j{e}_l=-1$, $e_i{e}_l=0$, $e_j{e}_k=0$, and so, $cos\alpha =-1$. Consequently, $\alpha =\pi$, and so, the vector $(e_i-e_j)$ is parallel with the vector $(e_k-e_l)$.

B. Let $e_i=-e_k$, $e_j\ne -e_l$. Then, $e_i{e}_k=-1$, $e_j{e}_l=0$, $e_i{e}_l=0$, $e_j{e}_k=0$, and so, $cos\alpha =-1/2$. Consequently, $\alpha =2\pi /3$. This means the angle between the edges with vertices $e_i$, $e_j$ and $e_k$, $e_l$ forms an angle equal to $\pi /3$.

C. Let $e_i\ne -e_k$, $e_j=-e_l$. Then, $e_i{e}_k=0$, $e_j{e}_l=-1$, $e_i{e}_l=0$, $e_j{e}_k=0$, and so, $cos\alpha =-1/2$. Consequently, $\alpha =2\pi /3$. This means the angle between the edges with vertices $e_i$, $e_j$ and $e_k$, $e_l$ forms an angle equal to $\pi /3$.

D. Let $e_i\ne -e_k$, $e_j\ne -e_l$. Then, $e_i{e}_k=0$, $e_j{e}_l=0$, $e_i{e}_l=0$, $e_j{e}_k=0$, and so, $cos\alpha =0$. Consequently, $\alpha =\pi /2$. This means the angle between the edges with vertices $e_i$, $e_j$ and $e_k$, $e_l$ forms an angle equal to $\pi /2$.

Thus, the vector $(e_i-e_j)$ is parallel with the vector $(e_k-e_l)$ only in Case 1. Hence, for each edge of $K_d$, there is at most one edge parallel with this edge.

Next, we construct strategies for the pursuers $x_1$, $x_2$,…, $x_d$ as follows. First, we let the pursuers $x_1$, $x_2$, …, $x_d$ come to the vertices $A_1$, $A_2$, …, $A_d$, respectively. The pursuer reaches his/her vertex and waits until the other pursuers reach their vertices. Let all the pursuers reach the vertices $A_1$, $A_2$, …, $A_d$ at some time $t_1$. Without any loss of generality, we assumed that if the edge $A_{j_0}{B}_{j_0}$ is parallel with $A_{i_0}{B}_{i_0}$, $i_0\ne j_0$, then the vector $\overrightarrow{A_{j_0}{B}_{j_0}}$ is codirected toward $\overrightarrow{A_{i_0}{B}_{i_0}}$.

Start from the time $t_1$ each pursuer $x_i$ walks along the edge $A_i{B}_i$ with the speed ${\rho }_i^{\prime }=min\{{\rho }_i,{\rho }_j\}$, where ${\rho }_j$ is the maximal speed of the pursuer $x_j$ moving along the edge $A_j{B}_j$ parallel with $A_i{B}_i$ (if there is such an edge), until $x_i$ captures the projection $y_i$ of the evader y on the edge $A_i{B}_i$, $i=1,2,\dots ,d$. Thus, the pursuers $x_i$ and $x_j$ that move in parallel edges $A_i{B}_i$ and $A_j{B}_j$ move with the same speed and direction until they capture the projections of the evader on those edges, respectively.

As $x_i\left(t\right)=y_i\left(t\right)$ at some time $t=t_{2i}$, we suggest the following strategy to the pursuer $x_i$:

$$u_i\left(t\right)=(m_i,v\left(t\right))m_i,t_{2i}<t\le t_2,$$

(2)

where $m_i$ is the unit vector of $\overrightarrow{A_{i_0}{B}_{i_0}}$, that is $m_i=\frac{1}{\sqrt{2}}\overrightarrow{A_{i_0}{B}_{i_0}}$; $t_2$ is the time when $x_i\left(t\right)=y_i\left(t\right)$ for all $i=1,\dots ,d$. Then, clearly, (2) ensures that $x_i\left(t\right)=y_i\left(t\right)$, $t_{2i}<t\le t_2$.

The strategy of pursuer (2) is admissible since (i) if the evader moves along an edge $A_i{B}_i$ orthogonal to $m_i$, then $u_i\left(t\right)=0$ since $(m_i,v\left(t\right))=0$; (ii) if the edge where the evader moves on forms an angle $\pi /3$ with the edge $A_i{B}_i$, then:

$$|u_i\left(t\right)|=|(m_i,v\left(t\right))m_i|\le |m_i||v\left(t\right)|cos\frac{\pi }{3}\le \frac{1}{2}$$

(3)

since $|m_i|=1$ and $\left|v\right(t\left)\right|\le 1$.

Note that if the evader is on an edge $A_j{B}_j$ parallel with $A_i{B}_i$ on the time interval $t_{2i}\le t\le t_2$, then the evader is captured by a pursuer at the time $t_{2i}$ since one of the points $y_i\left(t_{2i}\right)$ coincides with the real evader and the pursuers $x_i$ and $x_j$ move on the interval $[t_1,t_{2i}]$ on the parallel edges with the same speed and direction. Therefore, the evader cannot be on the edge parallel with $A_i{B}_i$ on the time interval $t_{2i}\le t\le t_2$. Thus, (2) is admissible. Using the strategy (2), the pursuer $x_i$ controls the edge $A_i{B}_i$, meaning that the evader is not on $A_i{B}_i$, and if he/she reaches one of the points $A_i$, $B_i$, he/she is captured by the pursuer $x_i$.

If the evader is on an edge of $K_d$ at the time $t_2$, then he/she is trapped by two pursuers and the pursuit can be completed by these pursuers. To this end, these pursuers just move towards the evader to catch him/her. Let now the evader be not in $K_d$ at the time $t_2$. Then, we let the pursuers $x_i$, $i=1,\dots ,d$, further use strategies (2) for $t\ge t_2$. Then, clearly, $x_i\left(t\right)=y_i\left(t\right)$, $t\ge t_2$.

If we remove $K_d$ from $K_{d+1}$, then we obtain two trees, one of which contains the vertex S and the other of which contains $S_1$. Therefore, the pursuer $x_{d+1}$ moving towards the evader will catch him/her or force the evader to reach $K_d$, in which case the evader will be caught by a pursuer in $K_d$.

Case 2. Let $n\ne k$ and $n+k\ge 2d$. It follows from $n\ne k$ that $n>k$. It suffices to consider the case $n+k=2d$, that is if $n+k=d$, then n pursuers can complete the game. Let $A_1{B}_1$, $A_2{B}_2$, …, $A_k{B}_k$ be the edges of the orthoplex $K_d$, any two of which have no common vertex (Figure 2), where $A_1$, $B_1$, $A_2$, $B_2$, …, $A_k$, $B_k$ are some distinct vertices of the orthoplex $K_d$.

First, we bring the pursuers $x_1$, $x_2$, …, $x_k$ to the vertices $A_1$, $A_2$, …, $A_k$. Then, as in Case 1, these pursuers move towards $B_1$, $B_2$, …, $B_k$, respectively, to catch the projections on the edges $A_1{B}_1$, $A_2{B}_2$, …, $A_k{B}_k$. As the pursuer $x_i$ catches the projection $y_i$ of the evader y on $A_i{B}_i$, this pursuer further moves on the projection, keeping it. In this way, pursuers control the edges $A_1{B}_1$, $A_2{B}_2$, …, $A_k{B}_k$ (Figure 2). In particular, these pursuers can control $2k$ vertices $A_1$, $B_1$, $A_2$, $B_2$, …, $A_k$, $B_k$ of $K_d$.

The rest $n-k$ of the pursuers $x_{k+1}$, …, $x_n$ go to the $2d-2k$ vertices $A_{k+1}$, $B_{k+1}$, …, $A_d$, $B_d$. This is possible since $n-k=2d-2k$. Thus, all the pursuers can now control all the vertices of $K_d$. $K_{d+1}\backslash K_d$ is a union of two disjoint trees and one of them contains S and another contains $S_1$. The pursuer $x_{k+1}$ moving towards the evader either catches the evader or forces him/her to reach $K_d$. In the latter case, the evader will be captured by some of the pursuers $x_1$, …, $x_k$, $x_{k+2}$, …, $x_d$. □

Theorem 2.

If either (i) $n=k<d+1$ or (ii) $n\ne k$ and $n+k<2d$, then the evader E can avoid the pursuers $x_1$, $x_2$, …, $x_n$ in the game on orthoplex $K_{d+1}$ from some initial states.

Proof. 

Case 1. Let $n=k<d+1$ and the initial state of the evader be at the vertex S of $K_{d+1}$. Hence, $n=k\le d$. Let the evader stay at the vertex S until the time $t_0$ when $\underset{i=1,\dots ,n}{min}|x_i\left(t_0\right)-S|\le 1/3$ for the first time. Furthermore, it is possible that $t_0=0$. For the definiteness, assume that $|x_1\left(t_0\right)-S|\le 1/3$.

Pursuer $x_1$ then cannot control more than one vertex of the orthoplex $K_d$ in $\sqrt{2}$ units of time. Each of the rest $(n-1)$ of the pursuers $x_2$, …, $x_n$ can control at most two vertices of $K_d$, and so, together, they can control at most $2(n-1)\le 2(d-1)$ vertices of $K_d$ in $\sqrt{2}$ units of time.

Thus, all the pursuers $x_1$, …, $x_n$ can control $2(d-1)+1=2d-1$ vertices of orthoplex $K_d$ in $\sqrt{2}$ units of time. Hence, the pursuers cannot control one of the vertices of $K_d$ because $K_d$ has $2d$ vertices, and so, the evader can reach that vertex of $K_d$ in $\sqrt{2}$ units of time not being caught. The evader repeats this procedure over and over and can walk not being caught for an infinite period of time.

Case 2. Let $n\ne k$ and $n+k<2d$ and the evader be at the vertex S of $K_{d+1}$ at the initial time. Clearly, k pursuers $x_1$, $x_2$, …, $x_k$ can control $2k$ vertices of $K_d$ in $\sqrt{2}$ units of time. We now think about the rest $2d-2k$ of the vertices of $K_d$.

Each of the rest $n-k$ of the pursuers can control at most one vertex of $K_d$ in $\sqrt{2}$ units of time, and so, all these pursuers together can control at most $(n-k)$ vertices. Hence, all the pursuers can control at most $2k+(n-k)=n+k$ vertices of $K_d$ in $\sqrt{2}$ units of time. Since $n+k<2d$, therefore, the evader can come to a vertex of $K_d$ not being caught. The evader repeats this procedure over and over and can walk not being caught for an infinite period of time. The proof of the theorem is complete. □

4. Discussion and Conclusions

In the present paper, we studied pursuit and evasion differential games of many slow pursuers and one evader on the edge graph of an orthoplex. If either (i) $n=k=d+1$ or (ii) $n\ne k$ and $n+k\ge 2d$, then we proved that the pursuit can be completed in the differential game on the orthoplex $K_{d+1}$. If this condition is not satisfied, that is either (i) $n=k<d+1$ or (ii) $n\ne k$ and $n+k<2d$, then we proved that evasion is possible. In the case of pursuit game, we constructed explicit strategies for the pursuers, and in the case of the evasion game, we constructed an evasion strategy for the evader. Thus, we solved the pursuit and evasion game problems on the orthoplex completely.

A differential game on the one-skeletons of the d-dimensional orthoplex studied in [40] considered the case where the dynamical possibilities of the pursuers and evader being equal. It was shown in [40] that the optimal number of pursuers is equal to two. This is not difficult to prove using the same idea of that paper that the two pursuers, one of them having the same maximal speed as the evader and another pursuer having a maximal speed less than that of the evader, can complete the game as well. If the maximal speeds of all pursuers are less than that of evader, then the problem of the optimal number of pursuers is open, and in the present paper, we found a formula for this number.

Based on Theorems 1 and 2, we can conclude that the optimal number of pursuers in the game on the one-skeleton $K_{d+1}$ is $N\left(K_{d+1}\right)=d+1$ if $n=k$, and $N\left(K_{d+1}\right)=2d-k$ if $n>k$ (recall that $n\ge k$).

As an open problem, we suggest solving such differential game problems for the n-cube, that is in the n-dimensional cube in ${\mathbb{R}}^n$.