## Discrete two-countour system with one-directional motion

Yashina M.V.1,2, Tatashev A.G.1,2, Fomina M.J.1
${}^{1}\leavevmode\nobreak\ $Moscow Automobile and Road Construction State
Technical University (MADI)
${}^{2}\leavevmode\nobreak\ $Moscow Technical University of Communications and
Informatics (MTUCI)
<EMAIL_ADDRESS>
Abstract. This paper studies a dynamical system, which contains two contours.
There is a cluster on each contour. The cluster contains particles, located in
adjacent cells. The clusters move under prescribed rules. The delays of
clusters are due to that the clusters cannot pass through the node
simultaneously. The dynamical system belongs to the class of contour networks
introduced by A.P. Buslaev.

## 1\. Introduction

In versions of Nagel–Schreckenberg model, [1], the particles move on a closed
or infinite one-dimensional lattice under prescribed rule. In of these
versions, at any discrete moment, each particle moves onto one cell in the
direction of movement if the cell ahead is vacant. This rule is elementary
cellular automaton CA 184 in the classification of S. Wolfram, [2]. In
[3[–[7], analytical results are obtained for one-contour models of this type.

A two-dimensional traffic model with a toroidal supporter (BML traffic model)
has been introduced in [7]. Conditions of self-organization (all particles
move without delays after some moment) and collapse (no particle moves after
some moment) for BML model have been obtained in [8]–[10].

In [12], the concept of a traffic model with cluster movement has been
introduced. In the discrete version of the cluster model, each contour
contains a given quantity of cells. There are clusters of particles on each
contour. Particles of each cluster occupy neighboring cells. All particles of
each cluster move simultaneously in accordance with the rule of the cellular
automaton 240. Clusters can be delayed at nodes. In the continuous version of
the model a cluster is a segment moving on the contour with constant velocity
in a given direction.

The concept of a contour network has been introduced by A.P. Buslaev, [13]. In
papers A.P. Buslaev, basic approaches for the study of complex networks has
been developed. In accordance with these approaches, models are dynamical
systems. The supporters of these systems are contours systems with network
structures. Particles (clusters) move on contours in accordance with some
rules. There are common nodes of contours called nodes. Delays in movement at
nodes are due to that two contours cannot move through the same node
simultaneously. Analytical results have been obtained for systems with two
contours and for systems with regular periodic structures and any number of
contours.

Analytical results for contour networks were obtained in [13]–[23]. In
partucular, in [22], [23] a two-contour system with two nodes is studied under
the assumtion that the length of both clusters is the same. Two versions of
the system were considered. In the system with co-directional motion one
cluster moves counter-clockwise and the other cluster moves clockwise. In the
system with one-directional motion both the clusters move counter-clockwise.

In this paper we study a discrete two-contour system with two nodes and
clusters of different lengths. We consider the case of one-dimensional motion.

## 2\. System description

Suppose that the system contains two closed contours —-contour 1 and contour
2, Fig. 1. On contour $i$ there are the same number $n$ of cells and cluster
containing $l_{i}<n$ particles, which are located in $l_{i}$ neighboring cells
and move simultaneously at discrete times $t=0,1,2,\dots$, $i=1,2.$ The cells
have numbers $0,1,\dots,n-1$ increasing in the direction of cluster motion
module $n.$ There are two common points of the contours, called nodes. The
node 1 is located between the cells $0$ and $1$ on the contour 1 and between
the cells $d$, $d+1$ $(d\leq n/2)$ on the contour 2. The node 2 is located
between cells $0$ and $1$ on contour 2 and between cells $d$, $d+1$ on contour
1. If there is no delay, all particles of the cluster move one cell in the
direction of motion at each discrete moment. The delays are caused by the
impossibility of simultaneous passage of particles of different contours
through the node. We suppose the cluster of contour $i$ (cluster $i$) is at
node $i$ at time $t$, if at this moment the front particle of this cluster is
in cell 0, and cluster $i$ is at node $j\neq i$, if the front particle of the
cluster is in cell $d$, $i=1,2$. We suppose a cluster of contour $i$ (cluster
$i$) occupies node $i$ at time $t$, if at this moment there are particles in
cells 0, 1 of this contour and occupies node $j\neq i)$ if at the moment there
are particles in cells $d$, $d+1$ of contour $i$, $i=1,2$. If at time $t$
cluster $i$ is at one of the nodes, and another cluster occupies this node,
then there is a delay in the movement of cluster $i$ and in the moment $t+1$
the particles of this cluster are in the same cells as at the moment $t$. If
the clusters are located at the same time at the node $i$ (a competition
occurs), then the cluster $j\neq i$ passes through the node first. In
accordance with this competition resolution rule, the cluster passing to the
arc of not less length wins the competition.

Figure 1: A two-contour system

We suppose that $d\leq\frac{1}{2},$ $l_{1}\leq l_{2},$ which, due to symmetry,
does not limit the generality.

The state of the system at time $t$ is a vector
$A(t)=(\alpha_{1}(t),\alpha_{2}(t)),$ where $\alpha_{i}(t)$ — the number of
the cell in which the front particle of the cluster $i$, $i=1,2$ is located.
The state of the system is acceptable, if, when the system is in this state,
the clusters do not occupy the same node at the same time. Initial state, i.e.
the state at the initial moment of time $t=0,$ is specified. This state must
be valid. It is clear that with the given rules of motion, the system can only
go from valid state to valid one.

## 3\. Limit cycles. Average cluster speed. States of free motion and collapse

Since the number of states is finite and the behavior of the system is
deterministic, starting from a certain moment its states are repeated after
$T$ cycles. We have a limit cycle with period $T.$

Let $A_{i}$ — the number of transitions of the cluster $i$ on the limit cycle
with the period $T.$ Then the value $v_{i}=A_{i}/T$ is called the average
speed of the cluster $i,$ $i=1,2.$

The average cluster velocities generally depend on the values of $n,$ $l_{1},$
$l_{2},$ $d_{1},$ $d_{2}$ and the initial state of the system.

If at any moment $t\geq t_{0}$ all particles move, then we say that from the
moment $t_{0}$ the system is in a state of free motion. We will call self-
organization the property of a system to enter a state of free movement from
any initial state. If the system is in a state of free motion, then
$v_{1}=v_{2}=1.$

If starting from the moment $t_{0}$ no particle moves, then we say that from
the moment $t_{0}$ the system is in the state of collapse. If the system is in
the state of collapse, then $v_{1}=v_{2}=0.$

## 3\. Some lemmas

Let us prove some lemmas. Lemma 1. If the inequality $l_{1}+l_{2}>n$ is
satisfied the system cannot be in a state of free motion. Proof. If at the
limit cycle the system is in a state of free motion, then for a cycle period
equal to $n,$ each node is crossed by $l_{1}$ of particles of cluster 1 and
$l_{2}$ of particles of cluster 2. But this is impossible when the inequality
$l_{1}+l_{2}>n$ is satisfied, since the same node cannot simultaneously cross
more than one particle. Lemma 1 is proved. Lemma 2. For any limit cycle, the
average velocities of both clusters are not equal to 0, or the system is in a
state of collapse. Proof. If one cluster does not move on a spectral cycle,
then it is at one of the nodes, which is occupied by another cluster
throughout the cycle. But this is impossible, since, under the assumptions of
the lemma, the second cluster moves and must completely traverse the contour
at least once during the limit cycle. This contradiction proves the lemma.

Lemma 3. If at the limit cycle the system is not in a state of free motion or
in a state of collapse, then the system falls on this limit cycle in one of
the states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1}+d-n,0),$ $(0,l_{2}+d-n),$
$(l_{1},d),$ $(d,l_{2})$ Proof.It follows from Lemma 2 and terms of Lemma 3
that on the considered limit cycle, the average velocities of both clusters
are not equal, and at least one cluster is delayed. If there is a cluster 1
delay at node 1, then at the end of this delay the system is in the state
$(0,l_{2}+d)$ (addition module $n$). If there is a cluster 1 delay at node 2,
then this delay ends when the system is in state $(d,l_{2}).$ If there is a
delay of cluster 2 at node 1, then at the end of this delay the system is in
state $(l_{1}+d,0)$ (addition over module $n).$ If there is a delay of the
cluster 2 at the node 2, then this delay ends when the system is in the state
$(l_{l},d).$ It follows that the system falls on the limit cycle in at least
one of terms of the lemma 3 state. Lemma 3 is proved. Lemma 4. collapse state
exists if and only if terms $l_{1}>d$, $l_{2}>d$ are met. Proof. If terms of
Lemma 4 are met, the collapse state is $(d,d).$

If the system is in a state of collapse, then each cluster is at one node and
occupies another, but this is impossible under the terms of Lemma 4, since if
this terms is not met, at least one of the clusters cannot simultaneously be
at one node and occupy another node. Lemma 4 is proved.

## 4.System behavior options

### 4.1. Self-organization on terms $l_{2}\leq d,$ $l_{1}+l_{2}\leq n-2d$

We prove a theorem that gives a sufficient terms for self-organization of the
system.

Theorem 1. If terms

$l_{2}\leq d,$ $l_{1}+l_{2}\leq n-2d,$

are met, then the system is in a state of free movement at any initial state
from a certain point in time. Proof. Under the assumptions of Theorem 1, the
system, in accordance with Lemma 4, cannot be in a state of collapse. If on
some limit cycle the system is also not in a state of free motion, then in
accordance with Lemma 3 on any limit cycle the system is in at least one of
the states $(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ Let us
prove that from any of this state the system enters the state of free motion
in some time.

Let at the moment of time $t_{0}$ the system is in the state

$A(t_{0})=(l_{1}+d,0).$

Then, under the assumptions of the theorem, we have the following sequence of
states:

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+n)=(l_{1}+d,n).$

Thus, in $n$ steps, the system returns to the state $(l_{1}+d,0),$ i.e. it is
in a state of free motion.

Due to symmetry, the state $(0,l_{2}+d)$ is also a state of free motion.

Suppose that

$A(t_{0})=(l_{1},d).$

So on

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+n-d)=(n-d+l_{1},0),\
A(t_{0}+n)=(l_{1},d).$

Thus, the state $(l_{1},d_{1}),$, and by virtue of symmetry, the state
$(d_{1},l_{2}),$ are states of free motion. Theorem 1 is proved. Terms of
Theorem 1 are met, for example, for $n=10,$ $l_{1}=1,$ $l_{2}=2,$ $d=3.$ vskip
3pt In this case, we have a sequence of transitions

$(4,0)\to(5,1)\to(6,2)\to(7,3)\to(8,4)\to(9,5)\to$
$\to(0,6)\to(1,7)\to(2,8)\to(3,9)\to(4,0),$
$(0,5)\to(1,6)\to(2,7)\to(3,8)\to(4,9)\to(5,0)\to$
$\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(0,5),$
$(1,3)\to(2,4)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to$
$\to(7,9)\to(8,0)\to(9,1)\to(0,2)\to(1,3),$
$(3,2)\to(4,3)\to(5,4)\to(6,5)\to(7,6)\to(8,7)\to$
$\to(9,8)\to(0,9)\to(1,0)\to(2,1)\to(3,2),$

### 4.2.System behavior on terms $l_{2}\leq d,$ $l_{2}\leq n-2d,$
$l_{1}+l_{2}>n-2d,$

Under terms of the following theorem, the average cluster velocity depends on
the initial state. Theorem 2. If terms

$l_{2}\leq d,\ l_{2}\leq n-2d,\ l_{1}+l_{2}>n-2d,$

are met then, depending on the initial state of the system, its enters the
state of free motion or the average velocity of the clusters is

$v=\frac{n}{l_{1}+l_{2}+2d}.$ $None$

Proof. Under the assumptions of Theorem 2, the system, as follows from Lemma
4, cannot be in a state of collapse. If on some limit cycle the system is also
not in a state of free motion, then, in accordance with Lemma 3, on any limit
cycle the system is in at least one of the states $(l_{1}+d,0),$
$(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ Let us prove that from any of this
state the system enters in some time either the state of free motion, or a
limit cycle is realized with the average speed of each cluster, calculated by
formula (1).

Suppose that at the moment $t_{0}$ the system is in the state

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(t_{0}+l_{2}+d)=(0,l_{2}+d),$
$A(t_{0}+n)=(n-l_{2}-d,0),\ A(t_{0}+l_{1}+l_{2}+2d)=(l_{1}+d,0).$

Thus, for a time interval of $l_{1}+l_{2}+2d$, all clusters complete a
complete revolution, i.e. $n$ movements of each cluster are performed. Thus,
on the limit cycle containing the state $l_{1}+d,$ $l_{2}+d,$ the average
cluster speed is calculated by the formula (1).

If at the time $t_{0}$

$A(t_{0})=(l_{1},d),$

then we have

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+n-l_{1})=(0,d-l_{1}),\
A(t_{0}+n)=(l_{1},d).$

Therefore, the state $(l_{1},d_{2})$ is a state of free movement. Similarly,
the $(d,l_{2})$ state is also a free movement state. Thus, each of the four
initial states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2})$ belongs
to the limit cycle on which the system either is in a state of free motion, or
the average velocity of the clusters is determined by formula (1). Theorem 2
is proved. The conditions of Theorem 2 are satistified, for example, for
$n=12,$ $l_{1}=2,$ $l_{2}=4,$ $d=4.$

In this case, we have a sequence of transitions

$(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to(11,5)\to(0,6)\to(0,7)\to$
$\to(0,8)\to(1,9)\to(2,10)\to(3,11)\to(4,0)\to(5,0)\to(6,0),$
$(2,4)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to(7,9)\to(8,10)\to$
$\to(9,11)\to(10,0)\to(11,1)\to(0,2)\to(1,3)\to(2,4),$
$(4,4)\to(5,5)\to(6,6)\to(7,7)\to(8,8)\to(9,9)\to(10,10)\to$
$(11,11)\to(0,0)\to(1,1)\to(2,2)\to(3,3)\to(4,4).$

### 4.3. System self-organization on terms $l_{2}\leq d,$ $l_{1}\leq n-2d,$
$l_{2}>n-2d$

Under terms of the following theorem the system enters the state of self-
organization.

Theorem 3. If terms

$l_{2}\leq d,\ l_{1}\leq n-2d,\ l_{2}>n-2d,$

are met then the system for any initial state is in a state of free motion
from some moment in time Proof. Under the assumptions of Theorem 4, the
system, in accordance with Lemma 4, cannot be in a state of collapse. If on
some limit cycle the system is also not in a state of free motion, then, in
accordance with Lemma 3, on any limit cycle the system is in at least one of
the states $(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ Let us
prove that from any of this state the system enters the state of free motion
in some time.

Let at the moment of time $t_{0}$ the system is in the state

$A(t_{0})=(l_{1}+d,0).$

Then, under the assumptions of the theorem, we have the following sequence of
states:

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),\ A(t_{0}+n)=(n-l_{2}-d,0),$
$A(t_{0}+l_{2}+2d)=(d,l_{2}+2d-n),\ A(t_{0}+n+l_{2})=(d,l_{2}),$
$A(t_{0}+n+d)=(2d-l_{2},d),\ A(t_{0}+2n+l_{2}-d)=(0,n-d+l_{2}),$
$A(t_{0}+2n+l_{2})=(d,l_{2}).$

So,

$A(t_{0}+n+l_{2})=A(t_{0}+2n+l_{2}),$

i.e. the system enters a state of free motion.

Thus, the state $(d,l_{2})$ is a state of free motion and from the states
$(l_{1}+d,0),$ $(0,l_{2}+d)$ the system enters the state of free motion.

Let’s suppose that

$A(t_{0})=(l_{1},d).$

So on

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+n-d)=(n+l_{1}-d,0),$
$A(t_{0}+n-l_{1})=(0,d-l_{1}),\ A(t_{0}+n)=(l_{1},d).$

Therefore, the state $(l_{1},d)$ is a state of free motion. Theorem 3 is
proved.

Terms of Theorem 3 is satisfied, for example, for $n=12,$ $l_{1}=1,$
$l_{2}=4,$ $d=5.$ In this case, we have a sequence of transitions

$(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to(11,5)\to$
$\to(0,6)\to(0,7)\to(0,8)\to(0,9)\to(1,10)\to$
$\to(2,11)\to(3,0)\to(4,1)\to(5,2)\to(5,3)\to(5,4)\to$
$\to(6,5)\to(7,6)\to(8,7)\to(9,8)\to(10,9)\to$
$\to(11,10)\to(0,11)\to(1,0)\to(2,1)\to(3,2)\to$ $\to(4,3)\to(4,5),$
$(1,5)\to(2,6)\to(3,7)\to(4,8)\to(5,9)\to(6,10)\to$
$\to(7,11)\to(8,0)\to(9,1)\to(10,2)\to(11,3)\to$ $(0,4)\to(1,5).$

### 4.4. System self-organization on terms $l_{2}\leq d,$ $l_{1}>n-2d$

Under terms of the following theorem the system enters the state of self-
organization.

Theorem 4. If terms

$l_{2}\leq d,$ $l_{1}>n-2d,$

are met then the system for any initial state is in a state of free motion
from some moment in time. Proof. Under the assumptions of Theorem 4, the
system, in accordance with Lemma 4, cannot be in a state of collapse. If on
some limit cycle the system is also not in a state of free motion, then, in
accordance with Lemma 3, on any limit cycle the system is in at least one of
the states $(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ Let us
prove that from any of this state the system enters the state of free motion
in some time.

Let at the moment of time $t_{0}$ the system is in the state

$A(t_{0})=(l_{1}+d,0).$

Then, under the assumptions of the theorem, we have the following sequence of
states:

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(t_{0}+d)=(l_{1}+2d-n,d),$
$A(t_{0}+n-d)=(l_{1},d),\ A(t_{0}+n-l_{1})=(d,2d-l_{1}),$
$A(t_{0}+2n-2d)=(n+l_{1}-d,0),\ A(t_{0}+2n-l_{1}-d)=(0,d-l_{1}),$
$A(t_{0}+2n-d)=(l_{1},d).$

Thus, the state $(d,l_{1})$ is the state of free motion, and from the state
$(l_{1}+d,0)$ the system enters the state of free motion.

Let’s suppose that

$A(t_{0})=(0,l_{2}+d).$

So on

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+n-d)=(n-d+l_{1},0)$
$A(t_{0}+n-l_{1})=(0,d-l_{1}),\ A(t_{0}+l_{2}+d)=(0,l_{2}+d),$
$A(t_{0}+n)=(l_{1},d).$

Hence, the state $(l_{1},d)$ is the state of free motion Similarly, the state
of free motion is the state $(d,l_{2}).$ Theorem 4 is proved.

Terms of Theorem 4 are satisfied, for example, for $n=12,$ $l_{1}=3,$
$l_{2}=4,$ $d=5.$ In this case, we have a sequence of transitions:

$(8,0)\to(9,1)\to(10,2)\to(11,3)\to(0,4)\to(1,5)\to$
$\to(2,5)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to$
$\to(7,9)\to(8,10)\to(9,11)\to(10,0)\to(11,1)\to(0,2)\to$
$\to(1,3)\to(2,4)\to(3,5)\to(4,6)\to(5,7),$
$(0,9)\to(1,10)\to(2,11)\to(3,0)\to(4,1)\to(5,2)\to$
$\to(5,3)\to(5,4)\to(6,5)\to(7,6)\to(8,7)\to$
$\to(9,8)\to(10,9)\to(11,10)\to(0,11)\to(1,0)\to(2,1)\to$
$\to(3,2)\to(4,3)\to(5,4).$

### 4.5. System behavior on terms $l_{1}\leq d,$ $l_{2}>d,$ $l_{1}+l_{2}\leq
2d$

Under the conditions of the following theorem, the average cluster velocity
depends on the initial state.

Theorem 5. If terms

$l_{1}\leq d,\ l_{2}>d,\ l_{1}+l_{2}\leq 2d,$

are met then the system for any initial state is in a state of free motion
from some moment in time. Proof. In accordance with Lemma 4, under the
assumptions of Theorem 4, the system cannot be in the state of collapse. If,
on a limit cysle, the system is not also in a state of free motion, then, in
accordance with Lemma 3, then any limit cycle contains at least one of the
states $(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ Let us
prove that, from any of these states, the system results in a state of free
motion.

Assume that

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+n-d)=(l_{1},n-d),$
$A(t_{0}+n)=(l_{1}+d,0).$

Thus the state $(l_{1}+d,0)$ if a state of free motion.

We can prove analogously that the state $(0,l_{2}+d)$ is also a state of free
motion.

Suppose

$A(t_{0})=(d,l_{2}).$

We have

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n-d)=(0,n+l_{2}-d),$
$A(t_{0}+n)=(d,l_{2}).$

Thus the state $(d,l_{2})$ is a stare of free movement. Theorem 5 has been
proved

The condition of Theorem 5 is fulfilled, for example, in the case $n=12,$
$l_{1}=1,$ $l_{2}=3,$ $d=2.$ We have

$(3,0)\to(4,1)\to(5,2)\to(6,3)\to(7,4)\to(8,5)\to$
$\to(9,6)\to(10,7)\to(11,8)\to(0,9)\to$ $\to(1,10)\to(2,11)\to(3,0),$
$(0,5)\to(1,6)\to(2,7)\to(3,8)\to(4,9),$
$(5,10)\to(6,11)\to(7,0)\to(8,1)\to(9,2)\to(10,3)\to$ $\to(11,4)\to(0,5),$
$(1,2)\to(2,3)\to(3,4)\to(4,5)\to(5,6)\to(6,7)\to$
$\to(7,8)\to(8,9)\to(9,10)\to(10,11)\to$ $\to(11,0)\to(0,1)\to(1,2),$
$(2,3)\to(3,4)\to(4,5)\to(5,6)\to(6,7)\to(7,8)\to$
$\to(8,9)\to(9,10)\to(10,11)\to(11,0)\to(0,1)\to$ $\to(1,2)\to(2,3),$

### 4.6. System behavior on terms $l_{1}\leq d,$ $d<l_{2}\leq 2d,$
$l_{1}+l_{2}\leq 2d$

Under the assumption of the following theorem, the average speed of clusters
depends on the initial state.

Theorem 6. Suppose

$l_{1}\leq d,\ d<l_{2}<2d,$ $l_{1}+l_{2}\leq 2d,$

then $v=1$ or

$v_{1}=v_{2}=\frac{n}{n-d+l_{1}+l_{2}}.$ $None$

Proof. It follows from Lemma 4 that the system cannot be in the state of
collapse. If, on a limit cycle, the system is not in a state of free motion,
then, in accordance with Lemma 3, on any limit cycle, the system is in at
least of one states $(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$
$(d,l_{2}).$ We shall prove that for any of these states the following is
true. Either the system results in a state of free motion or the average speed
satisfies (2).

Suppose

$A(t_{0})=(l_{1}+d,0).$

We have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+n)=(l_{1}+d,0).$

Thus the state $(l_{1}+d,0)$ is a state of free motion.

It is proved analogously that the state $(0,l_{2}+d)$ is a state of free
motion.

Assume that

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+l_{2}-d)=(d,l_{2}),$
$A(t_{0}+n-d)=(n+d-l_{2},0),\ A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d)$
$A(t_{0}+n)=(2d-l_{2},d)\ A(t_{0}+n-2d+l_{1}+l_{2})=(l_{1},d).$

If

$A(t_{0})=(d,l_{2}),$

then

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n-d)=(0,l_{2}-d),$
$A(t_{0}+n+d-l_{2})=(2d-l_{2},d)\ A(t_{0}+n+l_{1}-d)=(l_{1},d).$

Then, from the state $(d,l_{2}),$ the system results in the state $(l_{1},d),$
which belongs to a limit cycle with the average speed, calculated with (2).
Theorem 6 has been proved.

The condition of Theorem 6 is fulfilled if $n=14,$ $l_{1}=2,$ $l_{2}=5,$
$d=3.$ In this case, we have

$(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(12,7)\to(13,8)\to(0,9)\to(1,10)\to$
$\to(2,11)\to(3,12)\to(4,13)\to(5,0),$
$(0,8)\to(1,9)\to(2,10)\to(3,11)\to(4,12)\to(5,13)\to$
$\to(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to$
$\to(11,5)\to(12,6)\to(13,7)\to(0,8),$
$(2,3)\to(3,4)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to$
$\to(7,9)\to(8,10)\to(9,11)\to(10,12)\to(11,13)\to$
$\to(12,0)\to(13,1)\to(0,2)\to(1,3)\to(2,3),$
$(3,5)\to(4,6)\to(5,7)\to(6,8)\to(7,9)\to(8,10)\to$
$\to(9,11)\to(10,12)\to(11,13)\to(12,0)\to(13,1)\to(0,2)$ $\to(1,3)\to(2,3).$

### 4.7. Self-organization in the case $l_{1}\leq d,$ $l_{2}\geq 2d,$
${}_{1}+l_{2}\leq 2d$

The follows theorem gives a condition of self-organization.

Theorem 7. If

$l_{1}\leq d,\ l_{2}\geq 2d,\ l_{1}+l_{2}\leq 2d,$

then $v=1.$ vskip 3pt Proof. We shall prove that from any of states
$(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2})$ the system results
in a state of free motion.

Suppose

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+n)=(l_{1}+d,0).$

Therefore the state $(l_{1}+d,0)$ is a state of free motion.

It is proved analogously that the state $(0,l_{2}+d)$ is also a state of free
motion.

If

$A(t_{0})=(l_{1},d),$

then

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+l_{2}-d)=(d,l_{2}),$
$A(t_{0}+n-d)=(n+d-l_{2},0),\ A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d)$
$A(t_{0}+n)=(n+2d-l_{2},d),\ A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d),$
$A(t_{0}+n+l_{2})=(0,l_{2}+d).$

Hence, from the state $(l_{1},d),$ the system results in the state
$(0,l_{2}+d),$ which is a state of free movement.

If

$A(t_{0})=(d,l_{2}),$

then

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n-d)=(0,l_{2}-d),$
$A(t_{0}+n+d)=(0,l_{2}+d).$

Therefore, from the state $(d,l_{2}),$ the system results in the state
$(0,l_{2}+d),$ which is a state of free movement.

Thus, from the state $(d,l_{2}),$ the state results in the state in the state
$(0,l_{2}+d),$ which is a state of free motion. The conditions of Theorem 7
are satistified, for example, for $n=12,$ $l_{1}=1,$ $l_{2}=5,$ $d=2.$ We have

$(3,0)\to(4,1)\to(5,2)\to(6,3)\to(7,4)\to(8,5)\to$
$\to(9,6)\to(10,7)\to(11,8)\to(0,9)\to(1,10)\to$ $\to(2,11)\to(3,0),$
$(0,7)\to(1,8)\to(2,9)\to(3,10)\to(4,11)\to(5,0)\to$
$\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$ $\to(11,6)\to(0,7),$
$(1,2)\to(2,3)\to(2,4)\to(2,5)\to(3,6)\to(4,7)\to$
$\to(5,8)\to(6,9)\to(7,10)\to(8,11)\to(9,0)\to$
$\to(10,1)\to(11,2)\to(0,3)\to(1,4)\to(0,5)\to$ $\to(0,6)\to(0,7),$

### 4.8. System behavior on terms $l_{1}\leq d,$ $d<l_{2}<2d,$ $l_{2}\leq
n-2d,$ $l_{1}+l_{2}>n-2d,$

Under the conditions of the following theorem the system cannot result in a
state of free motion or collapse, and there are two possible states. Theorem
8. If

$l_{1}\leq d,\ d<l_{2}<2d,\ l_{2}\leq n-2d,$ $l_{1}+l_{2}>n-2d,$ $None$

then

$v=v_{1}=v_{2}=\frac{n}{l_{1}+l_{2}+2d}$ $None$

or

$v=\frac{n}{l_{1}+l_{2}+n-2d}.$ $None$

Proof. It follows from Theorem 8 that the system cannot be in a state of
collapse. Then there exist moments such that the leading particle of the
cluster 2 is in the cell 0.

Suppose

$A(t_{0})=(\alpha,0),$

and

$0\leq\alpha<d.$

We have

$A(d-\alpha)=(d,d-\alpha).$

Therefore, at time $t=d-\alpha,$ a delay of the cluster 1 begins.

If

$d\leq\alpha<l_{1}+d,$

then, at time $t=0,$ the cluster 2 does not move.

If

$l_{1}+d<\alpha\leq n-d,$ $None$

then

$A(n-\alpha)=(0,n-\alpha)$

and, taking into account that (3), (6), we get that, at time $t=n-\alpha,$ the
cluster 1 does not move.

If

$\alpha>n-d,$

then

$A(d)=(d,d-\alpha)$

and, therefore, at time $t=d,$ the cluster 1 does not move.

From Lemma 3, it follows that there exists a moment sush that the system is in
one of the states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$

Suppose

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),\ A(t_{0}+l_{2}+2d)=(d,l_{2}+2d),$
$A(t_{0}+n)=(n-l_{2}-d,0),\ A(t_{0}+l_{1}+l_{2}+2d)=(l_{1}+d,0).$

Thus the states $(l_{1}+d,0),$ $(0,l_{2}+d)$ belong to a limit cycle with the
average velocity belongs to a limit cycle with the average velocity satisfied
(4).

Assume that

$A(t_{0})=(l_{1},d).$

We have

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+l_{2}-d)=(d,l_{2}),$
$A(t_{0}+n-d)=(n+d-l_{2},0),\ A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d),$
$A(t_{0}+n)=(2d-l_{2},d),\ A(t_{0}+n+l_{1}+l_{2}-2d)=(l_{1},d).$

Thus the states $(l_{1},d),$ $(d,l_{2})$ belong to a limit cycle with the
average speed satisfied (5).

Conditions of Theorem 8 are satisfied if $n=18,$ $l_{1}=4,$ $l_{2}=7,$ $d=4.$
We have

$(8,0)\to(9,1)\to(10,2)\to(11,3)\to(12,4)\to(13,5)\to$
$\to(14,6)\to(15,7)\to(16,8)\to(17,9)\to(0,10)\to$
$\to(0,11)\to(1,12)\to(2,13)\to(3,14)\to(4,15)\to$
$\to(5,16)\to(6,17)\to(7,0)\to(8,0),$
$(4,4)\to(4,6)\to(4,7)\to(5,8)\to(6,9)\to(7,10)\to$
$\to(8,11)\to(9,12)\to(10,13)\to(11,14)\to(12,15)\to$
$\to(13,16)\to(14,17)\to(15,0)\to(16,1)\to(17,2)\to$
$\to(0,3)\to(1,4)\to(2,4)\to(3,4)\to(4,4).$

### 4.9.System behavior on terms $l_{1}\leq d,$ $2d<l_{2}\leq n-2d,$
$l_{1}+l_{2}>n-2d,$

Under the conditions of Theorem 9, the average velocity does not depend on the
initial state. Theorem 9. If

$l_{1}\leq d,\ 2d\leq l_{2}\leq 2n-d,\ l_{1}+l_{2}>n-2d,$

then there is a single limit cycle with the average velocity that satisfies
(4).

Proof. From Lemma 4 it follows that, under the assumptions of Theorem 9, the
system does not result in a state of collapse. Therefore there exist moments
such that the leading particle of the cluster 2 is in the cell 0. Assume that

$A(t_{0})=(\alpha,0).$

If

$0\leq\alpha<d,$

then

$A(d-\alpha)=(d,d-\alpha)$

and, therefore, at time $t=d-\alpha,$ a delay of the cluster 1 begins.

If

$d\leq\alpha<l_{1}+d,$

then, at time $t=0,$ then the cluster 2 does not move.

If (6) is true, then

$A(n-\alpha)=(0,n-\alpha).$

From this and (3), (6), it follows that, at time $t=n-\alpha,$ then the
cluster 1 does not move.

If

$\alpha>n-d,$

then

$A(d)=(d,d-\alpha)$

and, therefore, at time $t=d,$ the cluster 1 does not move. From this and
Lemma 3, it follows that there are exist a moment such that the system is in
one of the states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$

If

$A(t_{0})=(l_{1}+d,0),$

then

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),\ A(t_{0}+l_{2}+2d)=(d,l_{2}+2d),$
$A(t_{0}+n)=(n-l_{2}-d,0),\ (t_{0}+l_{1}+l_{2}+2d)=(l_{1}+d,0).$

Hence the states $(l_{1}+d,0),$ $(0,l_{2}+d)$ belong to a limit cycle with the
average velocity calculated with the formula (4).

If

$A(t_{0})=(l_{1},d),$

then

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+l_{2}-d)=(d,l_{2}),$
$A(t_{0}+n-d)=(n+d-l_{2},0),\ A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d)$
$A(t_{0}+n)=(n+2d-l_{2},d),$

(addition and subtraction modulo $n),$

$A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d),\ A(t_{0}+n+l_{2})=(0,l_{2}+d).$

Thus, from any of states $(l_{1},d),$ $(d,l_{2}),$ the system results in the
state $(0,l_{2}+d),$ which belongs to a limits cycle with the average velocity
that satisfies (4).

Conditions of Theorem 9 are satisfied if $n=14,$ $l_{1}=2,$ $l_{2}=7,$ $d=3.$
In this case,

$(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(12,7)\to(13,8)\to(0,9)\to(0,10)\to$
$\to(1,11)\to(2,12)\to(3,13)\to(4,0)\to(5,0),$
$(2,3)\to(3,4)\to(3,5)\to(3,6)\to$
$\to(3,7)\to(4,8)\to(5,9)\to(6,10)\to(7,11)\to(8,12)\to$
$\to(9,13)\to(10,0)\to(11,1)\to(12,2)\to(13,3)\to$
$\to(0,4)\to(0,5)\to(0,6)\to(0,7)\to(0,8)\to$ $\to(0,9)\to(0,10).$

### 4.10. System behavior on terms $l_{1}\leq d,$ $d<l_{2}<2d,$ $l_{1}\leq
n-2d,$ $n-2d<l_{2}\leq n-d,$ $l_{1}+l_{2}\leq 2d$

Under the conditions of the following theorem, the system results in a state
of free motion from any initial states.

Theorem 10. If

$l_{1}\leq d,\ l_{1}\leq n-2d,\ n-2d<l_{2}\leq n-d,\ l_{1}+l_{2}\leq 2d,$

then the system results in a state of free motion from any initial state.
Proof. From Lemma 4 it follows that, under the assumptions of Theorem 9, the
system does not result in a state of collapse. Therefore, in accordance with
Lemma 3, on any limit cycle, the system is in one of the states
$(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$

If

$A(t_{0})=(l_{1},d),$

then

$A(t_{0}+d)=(d,2d-l_{1}),\ A(t_{0}+n-d)=(n+l_{1}-d,0),$
$A(t_{0}+n-l_{1})=(0,d-l_{1}),\ A(t_{0}+n)=(l_{1},d).$

Hence the state $(l_{1},d)$ is a state of free motion.

If

$A(t_{0})=(d,l_{2}),$

then

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n-d)=(0,l_{2}-d),$
$A(t_{0}+n+d-l_{2})=(2d-l_{2},d),\ A(t_{0}+n)=(d,l_{2}).$

Therefore the state $(d,l_{2})$ is also a state of free motion.

Suppose

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n+l_{1}-d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),\ A(t_{0}+l_{2}+2d)=(d,l_{2}+2d-n),$
$A(n+l_{2})=(d,l_{2}).$

Thus, from the states $(l_{1}+d,0),$ $(0,l_{2}+d),$ the system results in the
state $(l_{2},d),$ which is a state of free motion. Theorem 10 has been
proved. The condition of Theorem is satisfied if $n=12,$ $l_{1}=1,$ $l_{2}=6,$
$d=4.$ In this, we have the following sequences of states

$(1,4)\to(2,5)\to(3,6)\to(4,7)\to(5,8)\to(6,9)\to$
$\to(7,10)\to(8,11)\to(9,0)\to(10,1)\to(11,2)\to$ $\to(0,3)\to(1,4),$
$(4,6)\to(5,7)\to(6,8)\to(7,9)\to(8,10)\to(9,11)\to$
$\to(10,0)\to(11,1)\to(0,2)\to(1,3)\to(2,4)\to$ $\to(3,5)\to(4,6),$
$(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(0,7)\to(0,8)\to(0,9)\to(0,10)\to$
$\to(1,11)\to(2,0)\to(3,1)\to(4,2)\to(4,3)\to$ $\to(4,4)\to(4,5)\to(4,6).$

### 4.11. System behavior on terms $l_{1}\leq d,$ $l_{1}\leq n-2d,$
$d<l_{2}<2d,$ $n-2d<l_{2}\leq n-d,$ $l_{1}+l_{2}>2d$

Under the assumptions of Theorem 11, there are two possible values of the
average velocity. Theorem 11. If

$l_{1}\leq d,\ l_{1}\leq n-2d,$ $d<l_{2}<2d,\ n-2d<l_{2}\leq n-d,\
l_{1}+l_{2}>2d,$

then there exists a single limit cycle, and the average velocity calculatted
with the formula (5).

Proof. It follows from Theorem 11 that the system cannot be in a state of
collapse. Then there exist moments such that the leading particle of the
cluster 2 is in the cell 0.

Assume that at time $t_{0}$ the system is in the state

$(\alpha,0).$

If

$0\leq\alpha<d,$

then

$A(d-\alpha)=(d,d-\alpha).$

Therefore, at time $t=d-\alpha,$ a delay of the cluster 1 begins.

If

$d\leq\alpha<l_{1}+d,$

then, at time $t=0,$ the cluster 2 does not move.

If

$l_{1}+d<\alpha\leq n-d,$ $None$

then

$A(n-\alpha)=(0,n-\alpha).$

Ffom this and (3), (7), we get that, at time $t=n-\alpha,$ the cluster 1 does
not move.

If

$\alpha>n-d,$

then

$A(d)=(d,d-\alpha)$

and, therefore, at time $t=d,$ the cluster 1 does not move.

From this and Lemma 3, it follows that there exists a moment such that the
system is in one of states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$
$(d,l_{2}).$

If

$A(t_{0})=(l_{1},d),$

then

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),$ $A(t_{0}+l_{2}-d)=(d,l_{2}),$
$A(t_{0}+n-d)=(n+d-l_{2},0),$ $A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d),$
$A(t_{0}+n)=(2d-l_{2},d),$ $A(t_{0}+n+l_{1}+l_{2}-2d)=(l_{1},d).$

Therefore the states $(l_{1},d),$ $(d,l_{2})$ belong to a limit cycle with the
average velocity calculated with the formula (5).

If

$A(t_{0})=(l_{1}+d,0),$

then

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n)=(n-l_{2}-d,0),\ A(t_{0}+n+2d-l_{2})=(d,2d-l_{2}).$
$A(t_{0}+l_{2})=(d,l_{2}).$

Therefore, from the state $(l_{1}+d,0),$ the system results in a state
belonging to a limit cycle with the average velocity calculated with the
formula (5).

If

$A(t_{0})=(0,l_{2}+d),$

then

$A(t_{0}+n-l_{2}-d)=(n-l_{2}-d,0),\ A(t_{0}+d)=(d,l_{2}+2d-n),$
$A(t_{0}+n-d)=(d,l_{2}).$

Thus, from the state $(0,l_{2}+d),$ the system results in a state belonging to
a limit cycle with the average velocity that satisfies (5). Theorem 11 has
been proved.

The condition of Theorem 11 is satisfied if $n=12,$ $l_{1}=3,$ $l_{2}=6,$
$d=4.$ In this case,

$(3,4)\to(4,5)\to(4,6)\to(5,7)\to(6,8)\to(7,9)\to$
$\to(8,10)\to(9,11)\to(10,0\to(11,1)\to(0,2)\to$ $\to(1,3)\to(2,4)\to(3,4),$
$\to(5,16)\to(6,17)\to(7,0)\to(8,0),$
$(7,0)\to(8,1)\to(9,2)\to(10,3)\to(11,4)\to(0,5)\to$
$\to(0,6)\to(0,7)\to(0,8)\to(0,9)\to(0,10)\to$
$\to(1,11)\to(2,0)\to(4,2)\to(4,3)\to(4,4)\to$ $\to(4,5)\to(4,6),$
$(0,10)\to(1,11)\to(2,0)\to(3,1)\to(4,2)\to(4,3)\to$
$\to(4,4)\to(4,5)\to(4,6).$

### 4.12. System behavior on terms $l_{1}\leq d_{2},$ $l\leq n-2d$,
$l_{2}>n-d,$ $d<l_{2}<2d$

Under the conditions of the following theorem, the system results in a state
of free motion from any initial state. Theorem 12. Suppose

$l_{1}\leq d,\ l_{1}\leq n-2d,\ l_{2}>n-d,$ $d<l_{2}<2d,\ l_{1}+l_{2}\leq 2d.$

Then the system results in the state oof free motion from any initial state.

Proof. From Lemma 4 it follows that, under the assumptions of Theorem 9, the
system does not result in a state of collapse. Therefore, in accordance with
Lemma 3, on any limit cycle, the system is in one of the states
$(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ We shall prove
that the system results in a state of free motion from any initial state.

If

$A(t_{0})=(l_{1},d),$

then

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+n-d)=(n-d+l_{1},0),$
$A(t_{0}+n-l_{1})=(0,d-l_{1}),\ A(t_{0}+n-l_{1})=(l_{1},d).$

Whence the state $(l_{1},d)$ is a state of free motion.

Suppose

$A(t_{0})=(d,l_{2}).$

Then

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n-d)=(0,l_{2}-d),$
$A(t_{0}+n+d-l_{2})=(2d-l_{2},d),\ A(t_{0}+n)=(d,l_{2}).$

Therefore the state $(l,d_{2})$ is also the state of free motion.

Assume that

$A(t_{0})=(0,l_{2}+d-n).$

Then

$A(t_{0}+n-l_{2}-d)=(n-l_{2}-d,0),\ A(t_{0}+d)=(d,l_{2}+2d-n),$
$A(t_{0}+n-d)=(d,l_{2}).$

Therefore, from the state $(0,l_{2}+d),$ the system results in the state
$(d,l_{2}),$ which is a state of free motion.

If

$A(t_{0})=(l_{1}+d,0),$

then

$A(t_{0}+n-l_{2})=(n-l_{2},d),\ A(t_{0}+d)=(d,l_{2}+2d-n),$
$A(t_{0}+n-d)=(d,l_{2}).$

Thus, from the state $(l_{1}+d,0),$ the system results in a state
$(0,l_{2}+d),$ from which the system results in the state $(d,l_{2}),$ and the
state $(d,l_{2})$ is a state of free movement. Theorem 12 has been proved.

The condition of Theorem 12 is satisfied for $n=12,$ $l_{1}=1,$ $l_{2}=9,$
$d=5.$ In this case,

$(1,5)\to(2,6)\to(3,7)\to(4,8)\to(5,9)\to(6,10)\to$
$\to(7,11)\to(8,0)\to(9,1)\to(10,2)\to(11,3)\to$ $\to(0,4)\to(1,5),$
$(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to(11,5)\to$
$\to(0,6)\to(0,7)\to(0,8)\to(0,9)\to(0,10)\to$
$(0,11)\to(0,0)\to(0,1)\to(0,2)\to(1,3)\to(2,4)\to$
$\to(3,5)\to(4,6)\to(5,7)\to(5,8)\to(5,9)\to$
$\to(6,10)\to(7,11)\to(8,0)\to(9,1)\to(10,2)\to$
$\to(11,3)\to(0,4)\to(1,5)\to(2,6)\to(3,7)\to$ $\to(4,8)\to(5,9).$

### 4.13. System behavior on terms $l_{1}\leq d,$ $l_{1}\leq n-2d,$
$d<l_{2}<2d,$ $l_{2}>n-d,$ $2d<l_{1}+l_{2}\leq n$

Under the conditions of the following theorem, the system results in a state
of free motion from any initial state. Theorem 13. Assume that

$l_{1}\leq d,\ l_{1}\leq n-2d,\ l_{2}>n-d,$ $d<l_{2}<2d,\ l_{2}>n-d,\
l_{1}+l_{2}>2d.$

Then the system results in the state of free motion from any iinitial state.
Proof. From Lemma 4 it follows that, under the assumptions of Theorem 9, the
system does not result in a state of collapse. Therefore, in accordance with
Lemma 3, on any limit cycle, the system is in one of the states
$(l_{1}+d_{1},0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ We shall prove
that the system results in a state of free motion from any initial state.

Suppose

$A(t_{0})=(l_{1},d).$

We have

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+n+l_{2}-d)=(d,l_{2}),\
A(t_{0}+n-d)=(n-l_{2}+d,0),$ $A(t_{0}+n+l_{2}-2d)=(0,l_{2}-d),\
A(t_{0}+n+l_{2}-d)=(d,l_{2}).$

Hence the state $(d,l_{2})$ is a state of free motion, and, from the state
$(l_{1},d),$ the system results in the state of free motion.

$A(t_{0})=(l_{1}+d,0).$

Then we have the following sequence of states

$A(t_{0}+n+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\
A(t_{0}+n+l_{2}+d)=(0,l_{2}+d-n),$ $A(t_{0}+n+d)=(n-l_{2},d),\
A(t_{0}+l_{2}+2d)=(d,l_{2}+2d-n),\ A(l_{2}+d+n)=(0,l_{2}+d-n).$

Thus the state $(0,l_{2}+d-n)$ if a state of free motion, and, from the state
$(l_{1}+d,d),$ the system results in a state of free motion.

If

$A(t_{0})=(0,l_{2}+d-n),$

then

$A(t_{0}+n-l_{2}-d)=(n-l_{2}-d,0),\ A(t_{0}+d)=(d,l_{2}+2d-n),\
A(t_{0}+n-d)=(d,l_{2}).$

Thus, from the state $(0,l_{2}+d),$ the system results in the state
$(d,l_{2}),$ which is a state of free motion. The condition of Theorem 13 is
satisfied if $n=20,$ $l_{1}=3,$ $l_{2}=14,$ $d=8.$ We have

$(11,0)\to(12,1)\to(13,2)\to(14,3)\to(15,4)\to(16,5)\to$
$\to(17,6)\to(18,7)\to(19,8)\to(0,9)\to(0,10)\to$
$\to(0,11)\to(0,12)\to(0,13)\to(0,14)\to(0,15)\to(0,16)\to$
$\to(0,17)\to(0,18)\to(0,19)\to(0,0)\to(0,1)\to$
$(0,2)\to(1,3)\to(2,4)\to(3,5)\to(4,6)\to(5,7)\to$
$\to(6,8)\to(7,9)\to(8,10)\to(9,11)\to(10,12)\to$
$\to(11,13)\to(12,14)\to(13,15)\to(14,16)\to(15,17)\to$
$\to(16,18)\to(17,19)\to(18,0)\to(19,1)\to(0,2),$
$(3,8)\to(4,9)\to(5,10)\to(6,11)\to(7,12)\to(8,13)\to$
$\to(8,14)\to(9,15)\to(10,16)\to(11,17)\to(12,18)\to$
$\to(13,19)\to(14,0)\to(15,1)\to(16,2)\to(17,3)\to(18,4)\to$
$\to(19,5)\to(0,6)\to(1,7)\to(2,8)\to(3,9)\to$
$(4,10)\to(5,11)\to(6,12)\to(7,13)\to(8,14).$

### 4.14. System behavior on terms $l_{1}\leq d,$ $l_{1}\leq n-2d,$
$l_{2}>n-d,$ $l_{1}+l_{2}>n,$

Under the condition of the following theorem, there is a single limit cycle,
and the value of the cluster 1 velocity is equal to the value of the cluster 2
multiplied by 2. Theorem 14. If

$l_{1}\leq d,\ l_{1}\leq n-2d,\ l_{2}>n-d,\ l_{1}+l_{2}>n,$

then the velocities of the clusters are equal to

$v_{1}=\frac{n}{2(l_{1}+l_{2})},$ $None$ $v_{2}=\frac{n}{l_{1}+l_{2}}.$ $None$

Proof. In accordance with Lemmas 1 and 4, the system cannot be in a state of
collapse or free motion. In accordance with Lemma 3, on a limit cycle results
in one of the states $(l_{1}+d_{1},0),$ $(0,l_{2}+d-n),$ $(l_{1},d),$
$(d,l_{2}).$ We shall prove that average velocities of clusters are calculated
with (8), (9).

If

$A(t_{0})=(l_{1}+d,0),$

then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n)=(0,0),\ A(t_{0}+l_{2}+d)=(0,l_{2}+d-n),$
$A(t_{0}+n+d)=(n-l_{2},d),\ A(t_{0}+l_{1}+l_{2}+d)=(l_{1},d),$
$A(t_{0}+l_{2}+2d)=(d,2d-l_{1}),\ A(t_{0}+l_{1}+2l_{2})=(d,l_{2}),$
$A(t_{0}+n+l_{1}+l_{2})=(n+d-l_{2},0),\ A(t_{0}+2(l_{1}+l_{2}))=(l_{1}+d,0).$

Thus there exists a single limit cycle. The period of the cycle equals
$2(l_{1}+l_{2}).$ On the cycle, there are two turns of the cluster 1 and a
turn of the cluater 2. From this, Theorem 14 follows. For example, the
condition of Theorem 14 is satisfied if $n=12,$ $l_{1}=2,$ $l_{2}=11,$ $d=3.$
In this case, we have the following sequence of transitions

$(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(0,7)\to(0,8)\to(0,9)\to(0,10)\to$
$\to(0,11)\to(0,0)\to(0,1)\to(0,2)\to(1,3)\to$
$\to(2,3)\to(3,4)\to(3,5)\to(3,6)\to(3,7)\to(3,8)\to$
$\to(3,9)\to(3,10)\to(3,11)\to(4,0)\to(5,0).$

### 4.15. System behavior on terms $l_{1}\leq d,$ $n-2d<l_{1}\leq n-d,$
$l_{2}>n-d,$ $l_{1}+l_{2}>n$

If the condition of the following theorem there exists a single limit cycle
with the average velocity less than 1. Theorem 15. If

$l_{1}\leq d,\ n-2d<l_{1}<n-d,\ l_{2}>n-d,$ $l_{1}+l_{2}>n,$

then, from any initial state, the system results in a state of a single limit
cycle with the average velocities of clusters equal to

$v_{1}=v_{2}=\frac{n}{n-2d+l_{1}+l_{2}}.$ $None$

Proof. Under the assumptions of Theorem 15, in accordance with Lemma 1, the
system cannot be in a state of free motion and, in accordance with Lemma 1,
the system cannot be in a state of

If

$A(t_{0})=(l_{1}+d,0),$

then

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(t_{0}+d)=(l_{1}+2d-n,d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+l_{2})=(d,l_{2}),$
$A(t_{0}+n)=(n+d-l_{2},0),\ (t_{0}+n+l_{2}-d)=(0,l_{2}-d),$
$A(n+d)=(2d-l_{2},d),\ A(n+l_{1}+l_{2}-d)=(l_{1},d),$
$A(n+l_{2})=(d,2d-l_{1}),\ A(n+l_{1}+2l_{2}-2d)=(d,l_{2}).$

Therefore the states $(l_{1},d),$ $(d,l_{2})$ belong to a limit cycle such
that the average velocity is calculated with the formula (9).

If

$A(t_{0})=(0,l_{2}+d-n),$

then

$A(t_{0}+d)=(d,l_{2}+2d-n),\ A(t_{0}+n-d)=(d,l_{2}).$

Thus, from the state $(0,l_{2}+d-n),$ the system results in the state, which
belongs to a limit cycle with the average velocity satisfied (9). Theorem 15
has been proved.

### 4.16. System behavior on terms $l_{1}\leq d,$ $l_{1}\leq n-2d,$
$n-2d<l_{2}\leq n-d,$ $l_{2}\geq 2d,$ $l_{1}+l_{2}\leq n$

If the conditions of the following theorem holds, then there exists a single
limit cycle such that the average velocity of the cluster 1 is equal to 1/2,
and the velocityaverage velocity of the cluster 2 with speed 1. Theorem 16. If

$l_{1}\leq d,\ l_{1}\leq n-2d,\ l_{2}>n-d,\ l_{2}\geq 2d,\ l_{1}+l_{2}\leq n,$

then, from any initial state, the system results in a state such that this
state belongs to a limit cycle with the average velicities of clusters

$v_{1}=\frac{1}{2},\ v_{2}=1.$ $None$

Proof. Under the assumptions of Lemma 1, the system cannot be in a state of
free movement, and, in accordance with Lemma 4, the system cannot be in a
stete of collapse.

If

$A(t_{0})=(l_{1}+d,0),$

In this case,

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(t_{0}+n)=(0,0),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d-n),\ A(t_{0}+n+d)=(n-l_{2},d),$
$A(t_{0}+l_{2}+2d)=(d,l_{2}+2d-n),\ A(t_{0}+n+l_{2})=(d,l_{2}),$
$A(t_{0}+2n)=(n+d-l_{2},0),\ A(2n+d)=(n+2d-l_{2},d),$
$A(2n+l_{2}-d)=(0,l_{2}-d),\ A(2n+l_{2}+d)=(0,l_{2}+d-n).$

Whence the states $(l_{1}+d,0),$ $(0,l_{2}+2d-n),$ $(d,l_{2})$ belong to a
limit cycle with the average velocity satisfied (10).

Soppuse

$A(t_{0})=(l_{1},d)$

Then we have

$A(t_{0}+d-l_{1})=(d,2d-l_{1}),\ A(t_{0}+l_{2})=(d,l_{2}).$

Thus, from the state $(l_{1},d),$ the system results in the state $(d,l_{2}),$
which belongs to a limit cycle with the average veelocities of clusters,
claculated with the formula (10). Theorem 16 has been proved.

For example, the condition of Theorem 16 is satisfied for $n=12,$ $l_{1}=1,$
$l_{2}=10,$ $d=3.$

We have

$(4,0)\to(5,1)\to(6,2)\to(7,3)\to(8,4)\to(9,5)\to$
$\to(10,6)\to(11,7)\to(0,8)\to(0,9)\to(0,10)\to(0,11)\to$
$\to(0,0)\to(0,1)\to(1,2)\to(2,3)\to(3,4)\to(3,5)\to$
$\to(3,6)\to(3,7)\to(3,8)\to(3,9)\to(3,10)\to(4,11)\to$
$\to(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(0,7)\to(0,8\to(0,9)\to(0,10)\to(0,11)\to$ $\to(0,0)\to(0,1),$
$(1,3)\to(2,4)\to(3,5)\to(4,6)\to(5,7)\to(5,8)\to$ $\to(5,9)\to(5.10.$

### 4.17. System behavior on terms $d<l_{1}<2d,$ $d<l_{2}<2d,$
$l_{1}+l_{2}\leq n-2d$

If th condition of the following theorem holds, the average velocities of both
clusters are equal to 0 or 1. Theorem 17. Suppose

$d<l_{1}<2d,\ d<l_{2}<2d,$ $l_{1}+l_{2}\leq n-d_{1}-d_{2},$

then, depending on the initial state, the system results in the state of free
motiion or collapse. Proof. If, on a limit cycle, the system is not in the
state neither free movement or collapse, then, in accordance with Lemma 3, on
any limit cycle, the system results in at least on of the states
$(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ We shall prove that,
from this state, the system results in the state of free motion or collapse.

If

$A(t_{0})=(l_{1}+d,0),$

then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+n)=(l_{1}+d,0).$

Thus the state $(l_{1}+d,0)$ is a state of free motion.

Assume that

$A(t_{0})=(l_{1},d).$

Then we have the following sequences of states

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+n-l_{1}+d)=(d,2d-l_{1}),$
$A(t_{0}+n)=(d,d).$

Therefore, from the state $(l_{1},d),$ the system results in the state
$(d,d),$ which is a state of collapse.

Assume that

$A(t_{0})=(d,l_{2}).$

Then the following sequence of states is realized

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n-l_{2}+d)=(2d-l_{2},d),$
$A(t_{0}+n)=(d,d).$

Thus, from the state $(l_{1},d),$ the state results in the state $(d,d),$
i.e., the system results in the state of collapse. Theorem 17 has been proved.
Theorem 17 has been proved. The condition of Theorem 17 is satisfied, for
example, in the case $n=12,$ $l_{1}=3,$ $l_{2}=3,$ $d=2.$

We have the following sequence of transitions

$(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(0,7)\to(1,8)\to(2,9)\to(3,10)\to$ $\to(4,11)\to(5,0),$
$(3,2)\to(4,3)\to(5,4)\to(6,5)\to(7,6)\to(8,7)\to$
$\to(9,8)\to(10,9)\to(11,10)\to(0,11)\to(1,0)\to(2,1)\to$ $\to(3,2).$

### 4.18.System behavior on terms $d<l_{1}<2d,$ $l_{2}\geq 2d,$
$l_{1}+l_{2}\leq n-2d$

Under the condition of the following theorem, depending on the initial state,
the average velocity of clusters is equal to 0 or 1. Theorem 18. If the
conditions

$d<l_{1}\leq d,\ l_{2}\geq 2d,\ l_{1}+l_{2}\leq n-d_{1}-d_{2}$

holds, then, depending on the initial state, the system results in a state of
free movement or collapse. Proof. If, on a limit cycle, the system is not in a
state of free motion or collapse, then, in accordance with Lemma 3, on this
limit cycle, the system results in at least one of states $(l_{1}+d,0),$
$(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ We shall prove that, from any of
these state, the system results in a state of free motion or collapse.

Suppose

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(t_{0}+n)=(l_{1}+d,0).$

Whence the state $(l_{1}+d,0)$ is a state of free motion.

Assume that

$A(t_{0})=(l_{1},d).$

We have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+n-l_{1}+d)=(d,2d-l_{1}),$
$A(t_{0}+n)=(d,d).$

Whence, from the state $(l_{1},d),$ the systems results in a state of free
$(d,d),$ which is a state of free motion.

If

$A(t_{0})=(d,l_{2}),$

then we have

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),\ A(t_{0}+n)=(0,l_{2}-d),$
$A(t_{0}+n+2d)=(0,l_{2}+d).$

Hence, from the state $(d,l_{2}),$ the system results in the state
$(0,l_{2}+d),$ which is the state of free motion.

The state $(d,d)$ is a state of collapse. Theorem 18 has been proved. The
condition of Theorem 18 holds if, for example, $n=12,$ $l_{1}=3,$ $l_{2}=4,$
$d=2.$

In this case, we have

$(5,0)\to(6,1)\to(7,2)\to(8,3)\to(9,4)\to(10,5)\to$
$\to(11,6)\to(0,7)\to(1,8)\to(2,9)\to(3,10)\to$ $\to(4,11)\to(5,0),$
$(0,6)\to(1,7)\to(2,8)\to(3,9)\to(4,10)\to(5,11)\to$
$\to(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to(11,5)\to$ $\to(0,6),$
$(3,2)\to(4,3)\to(5,4)\to(6,5)\to(7,6)\to(8,7)\to$
$\to(9,8)\to(10,9)\to(11,.10)\to(0,11)\to(1,0)\to$ $\to(2,1)\to(2,2).$
$(2,4)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to(7,9)\to$
$\to(8,10)\to(9,11)\to(10,0)\to(11,1)\to(0,2)\to$
$\to(0,3)\to(0,4)\to(0,5)\to(0,6).$

### 4.19. System behavior on terms $l_{1}\geq 2d,$ $l_{2}\geq 2d,$
$l_{1}+l_{2}\leq n-2d$

Under the assumptions of the following theorem, depending on the initial
state, the average of clusters is equal to 0 or 1. Theorem 19. If

$l_{2}\geq 2d,$ $l_{1}+l_{2}\leq n-2d$

then, depending on the initial state, the system results in the state of free
motion or collapse. Proof. If, on a limit cycle, the system is not in a state
of free motion or collapse, then, in accordance with Lemma 3, this system
results in one of states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$
$(d,l_{2}).$ We shall prove that, from any of these states, the system results
in a state of free motion or collapse.

Assume that

$A(t_{0})=(l_{1}+d,0),$

Then we have the following sequence of transitions

$A(t_{0}+d)=(l_{1}+2d,d),$ $A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+n-l_{1})=(d,n-l_{1}),$ $A(t_{0}+n)=(l_{1}+d,0).$

Therefore the state $(l_{1}+d,0)$ is a state of free motion.

Suppose

$A(t_{0})=(0,l_{2}+d).$

Then we have

$A(t_{0}+d)=(0,l_{2}+2d),$ $A(t_{0}+n-l_{2}-d)=(n-l_{2}-d,0),$
$A(t_{0}+n-l_{2})=(n-l_{2},d),$ $A(t_{0}+n)=(l_{2}+d,0).$

Hence the state $(0,l_{2}+d)$ is a state of free motion.

Suppose

$A(t_{0})=(l_{1},d).$

Then we have the following sequence of states

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),$ $A(t_{0}+n)=(l_{1}-d,0),$
$A(t_{0}+n+2d)=(l_{1}+d,0).$

Therefore, from the state $(l_{1},d),$ the system results in the state
$(l_{1}+d,d),$ which is a state of free movement.

Suppose

$A(t_{0})=(d,l_{2}).$

Then we have

$A(t_{0}+n-l_{2})=(n-l_{2}+d,0),$ $A(t_{0}+n)=(0,l_{2}-d),$
$A(t_{0}+n+2d)=(0,l_{2}+d).$

Thus, from the state $(d,l_{2}),$ the system results in the state
$(0,l_{2}+d),$ which is a state of free movement. Theorem 19 has been proved.

The condition of Theorem 19 holds on, for example, when $n=12,$ $l_{1}=6,$
$l_{2}=7,$ $d=2.$

We have the following sequence of states

$(8,0)\to(9,1)\to(10,2)\to(11,3)\to(0,4)\to(1,5)\to$
$\to(2,6)\to(3,7)\to(4,8)\to(5,9)\to(6,10)\to$ $\to(7,11)\to(5,0),$
$(0,6)\to(1,7)\to(2,8)\to(3,9)\to(4,10)\to(5,11)\to$
$\to(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to(11,5)\to$ $\to(0,6),$
$(3,2)\to(4,3)\to(5,4)\to(6,5)\to(7,6)\to(8,7)\to$
$\to(9,8)\to(10,9)\to(11,.10)\to(0,11)\to(1,0)\to$ $\to(2,1)\to(2,2).$
$(2,4)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to(7,9)\to$
$\to(8,10)\to(9,11)\to(10,0)\to(11,1)\to(0,2)\to$
$\to(0,3)\to(0,4)\to(0,5)\to(0,6).$

### 4.20. System behavior on terms $l_{1}\geq 2d,$ $l_{2}\leq n-2d,$
$n-2d<l_{1}+l_{2}\leq n,$

Under the assuptions of the following theorem, depending on the initial state,
the velocity of both clusters is equal to 0 or $v=\frac{n}{l_{1}+l+2+2d}.$
Theorem 20. If the conditions

$l_{1}\geq 2d,\ l_{2}\leq n-2d,$ $n-2d<l_{1}+l_{2}\leq n,$

holds, then, depending on the initial state, either the system results in the
state of collapse or a limit cycle with the average velocity calculted with
(1) is realized. Proof. In accordance with Lemma 1, if the condition of
Theorem 20, then, from any initial state, the system does not results in a
sraye of free movement. If, on a limit cycle, the system is not in a state of
free motion or collapse, then, in accordance with Lemma 3, this system results
in one of states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ We
shall prove that, from any of these states, the either the system results in a
state of free motion or the average velocity of clusters satisfies (1).

If

$A(t_{0})=(l_{1}+d,0),$

then we have

$A(t_{0}+d)=(l_{1}+2d,d),\ A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),\ A(t_{0}+l_{1}+l_{2}+2d)=(d_{1}+l,0).$

Therefore, the states $(l_{1}+d_{1},0)$ and $(0,l_{2}+d_{2})$ belong to a
limit cycle with the average velocity calculated with the formula (1).

If

$A(t_{0})=(l_{1},d),$

then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+l_{2}+d)=(0,l_{2}+d).$

Therefore, from the state $(l_{1},d_{2}),$ the system results in a state
$(0,l_{2}+d),$ and, from this state, the system results in a state pf a limit
cycle with the average velocity calculated with the formula (1). From the
state $(d,l_{2}),$ the system results in a state $(l_{1}+d,0),$ from which the
system also results in a state of a limit cycle calculated with the with the
average velocity formula (1).

The state $(d,d)$ is a state of collapse. Theorem 20 has been proved. The
condition of Theorem 20 holds, for example, in the case $n=12,$ $l_{1}=4,$
$l_{2}=5,$ $d=2.$

We have the following sequences of transitions

$(6,0)\to(7,1)\to(8,2)\to(9,3)\to(10,4)\to(11,5)\to$
$\to(0,6)\to(0,7)\to(1,8)\to(2,9)\to(3,10)\to$ $\to(4,11)\to(5,0)\to(6,0),$
$(4,2)\to(5,3)\to(6,4)\to(7,5)\to(8,6)\to(9,7)\to$
$\to(10,8)\to(11,9)\to(0,10)\to(1,11)\to(2,0)\to(11,5)\to$ $\to(0,6)$
$(3,2)\to(4,3)\to(5,4)\to(6,5)\to(7,6)\to(8,7)\to$
$\to(9,8)\to(10,9)\to(11.10)\to(0,11)\to(1,0)\to$ $\to(2,1)\to(3,2).$
$(2,4)\to(3,5)\to(4,6)\to(5,7)\to(6,8)\to(7,9)\to$
$\to(8,10)\to(9,11)\to(10.0)\to(11,1)\to(0,2)\to$ $\to(1,3)\to(2,4).$

### 4.21. Behavior of the systems in the case $l_{1}\geq 2d,$ $l_{2}\leq
n-2d,$ $l_{1}+l_{2}>n,$

If the condition of the following theorem holds, then, depending on the
initial state, the average velocity equals 0 or $v=\frac{n}{l_{1}+l+2+2d}.$
Theorem 21. If the conditions

$l_{1}\geq 2d,\ l_{2}\leq n-2d,\ l_{1}+l_{2}>n,$

holds, then, depending on the initial state, either the system results in the
state of collapse or a limit cycle with the average velocity calculted with
(1) is realized. Proof. In accordance with Lemma 1, if the condition of
Theorem 20, then, from any initial state, the system does not results in a
sraye of free movement. If, on a limit cycle, the system is not in a state of
free motion or collapse, then, in accordance with Lemma 3, this system results
in one of states $(l_{1}+d,0),$ $(0,l_{2}+d),$ $(l_{1},d),$ $(d,l_{2}).$ We
shall prove that, from any of these states, the either the system results in a
state of free motion or the average velocity of clusters satisfies (1).

Suppose

$A(t_{0})=(l_{1}+d,0).$

Then we have

$A(t_{0}+d)=(l_{1}+2d,0),$ $A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),$ $A(t_{0}+l_{2}+2d)=(d,l_{2}+2d),$
$A(t_{0}+n)=(n-l_{2}-2d,0),$ $A(t_{0}+l_{1}+l_{2}+2d)=(l_{1}+d,0).$

Therefore the states $(l_{1}+d,0)$ and $(0,l_{2}+d)$ belong to a limit cycle
with the average velocity that is calculated, which is calculated with (1).

Suppose

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),$ $A(t_{0}+l_{2}+d)=(0,l_{2}+d).$

Thus, from the state $(l_{1},d),$ the system results in the state
$(0,l_{2}+d),$ from which the system results in a state of a limit cycle with
the average velocity calculted with (1). From the state $(d,l_{2})$ the system
results in the state $(l_{1}+d,0),$ from which the system results in a state
of a limit cycle with the average velocity calculted with (1).

The state $(d,d)$ is a state of collapse. Theorem 21 has been proved.

The condition of Theorem 21 holds for $n=12,$ $l_{1}=6,$ $l_{2}=7,$ $d=2.$

In this case, the following sequences of the transitions are realized

$(8,0)\to(9,1)\to(10,2)\to(11,3)\to(0,4)\to(0,5)\to$
$\to(0,6)\to(0,7)\to(0,8)\to(0,9)\to(1,10)\to$
$\to(2,11)\to(3,0)\to(4,0)\to(5,0)\to(6,0)\to(7,0\to(8,0),$
$(6,2)\to(7,3)\to(8,4)\to(9,5)\to(10,6)\to(11,7)\to$ $\to(0,8)\to(0,9),$
$(2,7)\to(3,8)\to(4,9)\to(5,10)\to(6,11)\to(7,0)\to$ $\to(8,0).$

### 4.22. Resulting in a state of collapse in the case $l_{2}\geq 2d,$
$l_{1}\leq n-2d,$ $n-2d<l_{2}\leq n-d$

Under the assumptions of the following theorem, the system results in a state
of collapse from any initial state. Theorem 22. Suppose

$l_{1}\geq 2d,\ l_{1}\leq n-2d,\ n-2d<l_{2}\leq n-d.$

Then the system results in a state of collapse from any initial state. Proof.
In accordance with Lemma 1, the system cannot be in a state of free movement.

If, on a limit cycle, the system is not in neither a state of free motion nor
collapse, then, in accordance with Lemma 3, on the limit cycle, the system
results in at least one of the states $(l_{1}+d,0),$ $(0,l_{2}+d),$
$(l_{1},d),$ $(d,l_{2}).$ We shall prove that the system results in a state of
collapse from any initial state.

Under the assumption that

$A(t_{0})=(l_{1}+d,0),$

we have

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(l_{2}+d)=(0,l_{2}+d),$
$A(t_{0}+l_{2}+d)=(0,l_{2}+d),\ A(t_{0}+n)=(n-l_{2}-d,0),$
$A(t_{0}+l_{2}+2d)=(d,l_{2}+2d-n),\ A(t_{0}+n+d)=(d,d).$

Thus, from the states $(l_{1}+d_{1},0),$ $(0,l_{2}+d_{2}),$ the state results
in a state of collapse.

Suppose

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+l_{2})=(0,l_{2}+d).$

Therefore, from the state $(l_{1},d),$ the system results in the state
$(0,l_{2}+d),$ and, from this state, the system results in a state of
collapse.

Assume that

$A(t_{0})=(d,l_{2}).$

We have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+l_{2})=(0,l_{2}+d).$

Thus, from the state $(d,l_{2}),$ the system results in the state
$(l_{1}+d,0),$ and, from this state, the system results in a state of
collapse. Theorem 22 has been proved.

The condition of Theorem 22 holds, if, for example, $n=12,$ $l_{1}=5,$
$l_{2}=9,$ $d=2.$

In this case,

$(7,0)\to(8,1)\to(9,2)\to(10,3)\to(11,4)\to(0,5)\to$
$\to(0,6)\to(0,7)\to(0,8)\to(0,9)\to(0,10)\to$
$\to(1,11)\to(2,0)\to(3,1)\to(3,2)\to(3,3),$
$(6,2)\to(7,3)\to(8,4)\to(9,5)\to(10,6)\to(11,7)\to$ $\to(0,8)\to(0,9),$
$(2,7)\to(3,8)\to(4,9)\to(5,10)\to(6,11)\to(7,0)\to$ $\to(8,0).$

### 4.23. Resulting in a state of collapse in the case $l_{1}\geq 2d,$
$n-2d<l_{1}\leq n-2d,$ $n-2d<l_{2}\leq n-d$

If the condition of the following theorem holds, then the system results in a
state of collapse from any initial state. Theorem 23. If the condition

$n-2d<l_{1}\leq l_{2}\leq n-d,$

holds, then the system results in a state of collapse from any initial state.
Proof. In accordance with Lemma 1 the system cannot be in a state of free
movement.

If, on a limit cycle, the system is not in neither a state of free motion nor
collapse, then, in accordance with Lemma 3, on the limit cycle, the system
results in at least one of the states $(l_{1}+d,0),$ $(0,l_{2}+d),$
$(l_{1},d),$ $(d,l_{2}).$ We shall prove that the system results in a state of
collapse from any initial state.

Assume that

$A(t_{0})=(l_{1}+d,0).$

In this case,

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(l_{2}+d)=(l_{1}+2d-n,d),\
A(t_{0}+n-l_{1})=(d,d).$

Whence, from the state $(l_{1}+d,0),$ the system results in a state of
collapse.

It is proved analogously that the system results in a state of collapse from
the state $(0,l_{2}+d).$

If

$A(t_{0})=(l_{1},d),$

then the following sequence of states is realized

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(l_{2}+d)=(l_{1}+2d-n,d),\
A(t_{0}+n-l_{1})=(d,d).$

Therefore, from the state $(l_{1}+d,0),$ the system results in a state of
collapse.

Suppose

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+l_{2})=(0,l_{2}+d).$

Whence, from the state $(l_{1},d),$ the system results in the state
$(0,l_{2}+d),$ from which the system results in a state of collapse.

From the state $(d,l_{2}),$ the system results in the state $(l_{1}+d,0),$
from which the system results in a state of collapse. Theorem 23 has been
proved.

The condition of Theorem 23 holds, for example, under the assumptions that
$n=14,$ $l_{1}=9,$ $l_{2}=10,$ $d=3.$

In this case, we have the following sequences of states

$(12,0)\to(13,1)\to(0,2)\to(1,3)\to(2,3)\to(3,3),$
$(0,13)\to(1,0)\to(2,1)\to(3,2)\to(3,3),$
$\to(1,11)\to(2,0)\to(3,1)\to(3,2)\to(3,3),$
$(9,3)\to(10,4)\to(11,5)\to(0,6)\to(0,9)\to(0,10)\to$
$\to(0,11)\to(0,12)\to(0,13),$
$(3,10)\to(4,11)\to(5,12)\to(6,13)\to(7,0)\to(8,0)\to$ $\to(9,0)\
to(10,0)\to(11,0)\to(12\to 0)\to(13,0).$

### 4.24. Resulting in a state of collapse in the case $n-2d<l_{1}\leq n-d,$
$l_{2}>n-d$

If the condition of the following theorem holds, then the system results in a
state of collapse from any initial state. Theorem 24. If the condition

$n-2d<l_{1}\leq n-d,\ l_{2}>n-d,$

holds, then the system results in a state of collapse from any initial state.
Proof. In accordance with Lemma 1 the system cannot be in a state of free
movement.

If, on a limit cycle, the system is not in neither a state of free motion nor
collapse, then, in accordance with Lemma 3, on the limit cycle, the system
results in at least one of the states $(l_{1}+d,0),$ $(0,l_{2}+d),$
$(l_{1},d),$ $(d,l_{2}).$ We shall prove that the system results in a state of
collapse from any initial state.

Suppose

$A(t_{0})=(l_{1}+d,0).$

In this case,

$A(t_{0}+n-l_{1}-d)=(0,n-l_{1}-d),\ A(t_{0}+n-d)=(l_{1}-d,d),\
A(t_{0}+n-l_{1})=(d,d).$

Whence, from the state $(l_{1}+d,0),$ the system results in a state of
collapse.

Assume that

$A(t_{0})=(0,l_{2}+d-n).$

We have

$A(t_{0}+n-l_{2})=(n-l_{2},d),\ A(t_{0}+d)=(d,d).$

Hence, from the state $(0,l_{2}+d-n),$ the system results in a state of
collapse.

Suppose

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(l_{2}+d)=(l_{1}+2d-n,d),$
$A(t_{0}+n-l_{1})=(d,d).$

Hence, from the state $(l_{1}+d,0),$ the system results in a statte of
collapse.

If

$A(t_{0})=(0,l_{2}+d-n),$

then we have

$A(t_{0}+n-l_{2})=(n-l_{2},d),\ A(t_{0}+l_{2})=(d,d).$

Hence, from the state $(0,l_{2}+d-n),$ the state results in a state of
collapse.

Assume that

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+n-l_{1})=(0,n-l_{1}+d),$
$A(t_{0}+l_{2})=(0,l_{2}+d-n),$

Hence, from the state $(l_{1},d),$ the system results in the state
$(0,l_{2}+d-n),$ from which the system results in a state of collapse. Theorem
24 has been proved.

The condition of Theorem 24 holds if, for example, $n=12,$ $l_{1}=9,$
$l_{2}=11,$ $d=2.$

The following sequences of states are realized

$(11,0)\to(0,1)\to(1,2)\to(2,2),$ $(0,1)\to(1,2)\to(2,2),$
$(1,11)\to(2,0)\to(3,1)\to(3,2)\to(3,3),$
$(9,2)\to(10,3)\to(11,4)\to(0,5)\to(0,6)\to(0,7)\to$
$\to(0,8)\to(0,9)\to(0,10)\to(0,11)\to(0,0)\to(0,1),$
$(2,11)\to(3,0)\to(4,0)\to(5,0)\to(6,0)\to(7,0)\to$
$\to(8,0)\to(9,0)\to(10,0)\to(11\to 0).$

### 4.25. Resulting in a state of collapse under the assumptions that
$l_{1}>n-d,$ $l_{2}>n-d$

If the condition of the following theorem holds, then the system results from
a state of collapse from any initial sttate. Theoremc 25. If

$l_{1}>n-d,\ l_{2}>n-d,$

then the system results in a state of collapse from any iniitial state. Proof.
If the condition of Lemma 1 holds, then the system cannot be in a state of
free motion..

If, on a limit cycle, the system is not in either a state of free motion or
collapse, then, in accordance of Lemma 3, on the limit cycle, the system
results in at least one of states $(l_{1}+d-n,0),$ $(0,l_{2}+d-n),$
$(l_{1},d),$ $(d,l_{2}).$ We shall result in a state of collapse from any
initial state. Assume that

$A(t_{0})=(l_{1}+d-n,0).$

In this case,

$A(t_{0}+n-l_{1})=(d,n-l_{1}),\ A(l_{2}+d)=(d,d).$

Whence, from the state $(l_{1}+d,0),$ the system results in a state of
collapse..

It is proved analogously that the system results in the state of collapse from
the state $(0,l_{2}+d-n).$

Suppose

$A(t_{0})=(l_{1},d)$

In this case, we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}-d),\ A(l_{2}+l_{2})=(0,l_{2}+d-n).$

Hence, from the state $(0,l_{2}+d-n),$ the system results in the state
$(0,l_{2}+d-n),$ from which the system results in a state of collapse.

Suppose

$A(t_{0})=(l_{1},d).$

Then we have

$A(t_{0}+n-l_{1})=(0,n-l_{1}+d),\ A(t_{0}+l_{2})=(0,l_{2}+d-n).$

Thus, from the state $(l_{1},d),$ the state results in the state
$(0,l_{2}+d-n),$ from which the system results in a state of collapse.

It is proved analogously that, from the state $(d,l_{2}),$ the state results
in a state $1+d-n,0),$ from which the system results in a state of collapse.
Theorem 25 has been proved.

The condition of Theorem 25 holds, for example, for $n=10,$ $l_{1}=8,$
$l_{2}=9,$ $d=3.$

In this case, we have the following sequences of states

$(1,0)\to(2,1)\to(3,2)\to(3,3),$ $(0,2)\to(1,3)\to(2,3)\to(3,3),$
$(8,3)\to(9,4)\to(0,5)\to(0,6)\to(0,7)\to(0,8)\to$
$\to(0,9)\to(0,0)\to(0,1)\to(0,2),$
$(3,9)\to(4,0)\to(5,0)\to(6,0)\to(7,0)\to(8,0)\to$ $\to(9,0)\
\to(0,0)\to(1,0).$

## 5\. Conclusion

We consider different scenarios of behavior of two-contours system with
different lengths of clusters. The number of these scenarios is sufficiently
greater than in the case of clusters of the same length. In the later case,
there 5 scenarios, [22]. In the former case, the scenarios are the following,
Fig. 2–5.

Figure 2: Scenarios of system behavior for different $0<l_{1}\leq l_{2}<n$ and
$d<\frac{n}{4}$

Figure 3: Scenarios of system behavior for different $0<l_{1}\leq l_{2}<n$ and
$\frac{n}{4}<d<\frac{n}{3}$

Figure 4: Scenarios of system behavior for different $0<l_{1}\leq l_{2}<n$ and
$d=\frac{n}{3}$

Figure 5: Scenarios of system behavior for different $0<l_{1}\leq l_{2}<n$ and
$\frac{n}{3}<d<\frac{n}{2}$

1) The system results in a state of free motion from any initial state. 2)
Depending on the initial state, $v=1$ or

$v=\frac{n}{l_{1}+l_{2}+2d}.$

3) The average speed of each cluster is equal to

$v=\frac{n}{l_{1}+l_{2}+2d}$

or

$v=\frac{n}{l_{1}+l_{2}+n-2d}.$

4) The average speed of clusters is equal to

$v=\frac{n}{l_{1}+l_{2}+2d}.$

5) The average speed of clusters is equal to

$v=\frac{n}{l_{1}+l_{2}+n-2d}.$

6) The average speed of the cluster 1 is equal to

$v_{1}=\frac{n}{2(l_{1}+l_{2})},$

and the average speed of the cluster 2 is equal to

$v_{2}=\frac{2}{l_{1}+l_{2}}.$

7) The average speed of the cluster 1 is equal to

$v_{1}=\frac{1}{2},$

and the cluster 2 is equal to

$v_{2}=1.$

8) Depending on the initial state, the system results in the state of free
motion or collapse. 9) The average speed equals 0 or

$v=\frac{n}{l_{1}+l_{2}+2d}.$

10) The system results in a state of collapse from any initial state.

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