Compacting oblivious agents on dynamic rings

Periodic configurations

Let us start with a general results, that holds also in case the ring is not dynamic.

Theorem 2 Given a ring $\mathcal{G}$, and a set of agents A initially placed on G₀ in a configuration that is periodic and not compacted, it is impossible to solve the CCP or the ColoredCCP problem, even in the Global Snapshot model.

Proof. In a periodic configuration, the ring can be partitioned into identical non-full segments. In case no edge is ever missing, the initial symmetry between the agents cannot be broken deterministically: in fact, agents occupying equivalent positions in different segments can only take the same action in each step; thus, the configuration can only stay periodic. Finally, by observing that any compacted configuration (with k < n) is not periodic, the theorem follows.

Therefore, in the following we will assume that the initial configuration is either asymmetric or contains a mirror symmetry.

Local snapshots

In this section we show that the compaction problem cannot be solved in the Local Snapshot model, even when the initial configuration is asymmetric. The visibility graph of a configuration C is defined as the graph G_vis = (A, E), where A is the set of agents and (a, b) ∈ E whenever agent b is within distance R from a.

Theorem 3 In the Local Snapshot model, starting from a configuration C such that C is asymmetric and has a connected visibility graph, there is no algorithm that solves CCP, avoiding collisions, even if the agents have unbounded memory.

Proof. The proof is based on the concept of local view of an agent: the local view is the part of the entire configuration that the agent can see. More formally, given an agent a at node v of G with visibility radius R, its local view is an ordered list of 2R elements; the element of the list in position j is of the form “distance $j$:(L:v_L, R:v_R)”, where v_L (resp., v_R) is either 0 or 1 depending on whether the j-th node to the left (resp., to the right) of a is empty or not (being the ring unoriented, left and right refer to the local notion that a has of left/right). For instance, let us consider the agent a₁ in Fig. 3 with visibility radius 3: its local view is [distance 1: (L:0, R:1), distance 2: (L:1, R:1), distance 3: (L:1, R:1)].

Let us now consider the configuration C depicted in Fig. 3, with R = 3, and let us first focus on agents a₃ and a₄: they both have the same local view [distance 1: (L:0, R:1), distance 2: (L:1, R:1), distance 3: (L:1, R:1)] (note that being the ring unoriented, they may have different notion of left/right and clockwise/counter-clockwise direction). Therefore, they either both move or they both stay still. In case they both move, there will be a collision. Thus, to avoid collision, they should not move. The same holds for the pair of agents a₅ and a₆, that also have the same view of a₃ and a₄, and for the pair of agents a₂, a₇. In contrast, a₁ and a₈ have a different view: [distance 1: (L:0, R:1), distance 2:(L:1, R:1), distance 3: (L:1, R:1)].

Let us now focus on agent a₁: before deciding to move to v₂, it has to be sure that a₂ does not decide to do the same. Therefore, a₁ tries to infer the local view of agent a₂. In particular:

• a₁ sees that a₂ has an empty node (i.e., v₂) and an occupied node (i.e., v₃) at distance 1;

• a₁ sees that a₂ has an occupied node at distance 2 (i.e., a₁), but cannot see whether the other node at distance 2 from a₂ (i.e., v₄) is occupied or not; and

• a₁ sees that a₂ has an occupied node at distance 3 (i.e., v₁), but cannot see whether the other node at distance 3 from a₂ (i.e., v₅) is occupied or not.

Therefore, a₁ can only infer a partial view of a₂ (i.e., [distance 1: (0, 1), distance 2: (1, ·), distance 3: (1, ·)]); and it cannot decide whether a₂ has a different view from its own view (notice that a₁ is not aware of the orientation of a₂). Hence, it cannot decide to safely move to v₂ without the risk of colliding with a₂.

The same argument also holds for a₂, a₇ and a₈; thus, agents a₁, a₂, …, a₈ cannot move. Therefore, none of the agents can move if they want to avoid collision. Hence it is not possible to reach a configuration in which agents form compact lines. The same argument can be extended for agents having any visibility radius R by using a sufficiently large ring.

We note that the previous theorem holds for any ring, even non dynamic ones.

The case of two agents

Finally, in this section we examine the very special case of having only two agents in the system. It is clear that in such a case only the monocromatic version of the problem makes sense: in fact, if the two agents have different colors, then the problem is solved by definition. Surprisingly, solving CCP with two agents, in arbitrary initial configurations, is impossible.

Theorem 4 Let us consider an arbitrary dynamic ring $\mathcal{G}$ and an arbitrary initial configuration with two agents. Then, it is impossible to solve CCP.

Proof. First, notice that if the two agents are antipodal, the configuration is periodic and, by Theorem 2, the theorem follows. Thus, let us assume that at the beginning the two agents are not antipodal.

Also, notice that with only two non-neighbors agents in the ring, the configuration has always an unique axis of symmetry, say ax. If ax, passes through two empty nodes, the problem cannot be solved: in fact, any movement of the agents would keep ax passing through two empty nodes (remember that a collision of agents with the same color is not admissible), the two agents can never become neighbors, and the theorem follows.

Hence, ax has to pass through at least one edge. Note that, the only possible strategy for the agents to form a full segment is to move towards one of the two edges crossed by ax, say e. Referring to the example depicted in Fig. 4, we distinguish the two possible cases:

1. ax passes through a node (Fig. 4A). As stressed before, to solve the problem the agents can only try to reach e. If, during this movements, one of the two agents cannot move because of a missing edge (Fig. 4B), either they stay still forever (and CCP cannot be solved), or one of them moves. In this second case, the configuration stays as a new axis of symmetry passing through an empty node and edge e′ that is antipodal with respect to e (Fig. 4C). In order to correctly achieve compaction, the agents now have to start converging towards e′. If, during these movements, one of the edge is missing, this argument can be iterated, hence a compacted configuration never achieved, and the theorem follows.

2. ax passes through two edges (Fig. 4D). Let e the edge elected by the agents (being the configuration aperiodic, agents can always elect e). If, during these movements, one of the two agents cannot move because of a missing edge (Fig. 4E), either they stay still forever (and compaction never achieved), or one of them keeps moving in the same direction: in this second case, the configuration has a new axis of symmetry passing now through two empty nodes (Fig. 4F). Now, by previous Case 1, CCP cannot be solved. Therefore, the other option they have is to switch direction, and start moving towards the edge e′ on the axis of symmetry, antipodal to e (Fig. 4G). In this case (Fig. 4H), we end up again in a scenario similar to the one in Fig. 4D. Hence, by iterating the argument, we can conclude that a compacted configuration is never achieved, and the theorem follows.

The asymmetric case

First, let us consider the case when the initial configuration is asymmetric. Let ${\mathcal{E}}_r$ be the empty segment of maximum size in the configuration at round r. If, at round r = 0, there is more than one empty segment of maximum size, we can deterministically elect one of these (since the initial configuration is asymmetric).

Let S₁ and S₂ be the maximal full segments of length at least 1 on the two sides of segment ${\mathcal{E}}_r$ (see Fig. 5A). In case |S₁| ≠ |S₂|, without loss of generality let |S₁| < |S₂|; we define the augmented S₁, denoted by S ⁺ ₁, as the block of nodes constituted by the nodes in S₁ (all non empty), plus the empty node v close to S₁ and not in ${\mathcal{E}}_r$, plus, if any, all agents between v and the next empty node (moving away from S₁, see Figs. 5B and 6A).

The algorithm for solving CCP tries to increase the length of the empty segment ${\mathcal{E}}_r$ in each step, while preserving the asymmetric configuration. This is done by moving either S₁ or S₂ or both. The details are reported in Algorithm 1.

Lemma 1 Starting from an asymmetric configuration, by executing Algorithm One Color Connected Formation , at any round r ≥ 0:

1. |ℰ_r| > |ℰ_r−1|, and

2. The configuration is either asymmetric or solves CCP.

Proof. Let S₁ and S₂ be as defined in Algorithm 1. We proceed by induction on the number of rounds. By the precondition, the starting configuration is asymmetric. Let us now assume that the inductive hypothesis is true for round r −1. Let us consider the possible cases of the algorithm starting from the configuration at the beginning of round r:

• 1. If the smallest distance between S₁ and S₂ is strictly greater than one, by construction, we have the following cases (refer to Fig. 5):

• If |S₁| = |S₂|, and neither S₁ or S₂ is blocked, they both move away from ${\mathcal{E}}_r$, and by induction both (i) and (ii) hold. What happens is that both distances d₁ and d₂ decrease by one. If one of the distances reaches value 0, then the respective full segment S_i increases, and the asymmetry is kept. If both distances go to 0, both segment increase, and the asymmetry is kept by induction hypothesis.

• If |S₁| = |S₂|, S₁ is blocked (the case when S₂ is blocked is symmetric), and d₁ = d₂, then S₂ moves away from ${\mathcal{E}}_r$, one of the distances became different from the other, thus introducing a new asymmetry. Moreover, apart from ${\mathcal{E}}_r$, no other empty segment increases its size. Therefore, both (i) and (ii) hold.

• If |S₁| = |S₂|, S₁ is blocked, and d₁ < d₂, then all agents not in S₁ move towards S₁ away from ${\mathcal{E}}_r$. Two cases may occur: if d₁ does not reach 0, then the asymmetry is kept (since d₂ does not change); otherwise, if d₁ becomes 0, we have that |S₁| ≠ |S₂|, i.e., the configuration stays asymmetric. Therefore, both (i) and (ii) hold.

• If |S₁| = |S₂|, S₁ is not blocked, and d₁ < d₂, then S₁ moves away from ${\mathcal{E}}_r$ . By using the same arguments of the previous case, it follows that both (i) and (ii) hold.

• If |S₁| < |S₂| then, $S_1^{+}$ and S₂ move away from ${\mathcal{E}}_r$. Thus, | ${\mathcal{E}}_r$| > | ${\mathcal{E}}_{r-1}$|. If before the movement d₁ >1 then after the movement the asymmetry is kept, since |S₁| ≠ |S₂|. Otherwise, d₁ = 1 before and after the movement, hence the asymmetry is preserved. Thus, both (i) and (ii) hold.

• 2. If the smallest distance between S₁ and S₂ is exactly one (see Fig. 7), let v the only empty node separating S₁ and S₂. Since by inductive hypothesis the configuration is asymmetric, we have |S₁| ≠ |S₂|. Also, let e₁ and e₂ be the edges between S₁ and v, and between S₂ and v, respectively. Since at most one between e₁ and e₂ can be missing, by the algorithm, one full segment is formed within one round, and the lemma follows.

In all cases, the lemma follows.

By previous lemma, since the size of ${\mathcal{E}}_r$ strictly increases at each round, we can state the following:

Theorem 5. If the initial configuration is asymmetric, the agents executing Algorithm ONE COLOR CONNECTED FORMATION , solve CCP within at most n rounds.

The case of mirror symmetry

Let us now consider the case where in the initial configuration C there exists an unique axis of symmetry. Note that if there are two or more axes of symmetry, then the configuration is periodic.

Theorem 6. Let the initial configuration be aperiodic with an unique axis of symmetry, and not compact. Then, if the axis of symmetry passes through two empty nodes, then CCP is not solvable.

Proof. Let us assume that the problem is solvable, and that, by contradiction, the axis of symmetry of the initial configuration passes through two empty nodes (see Fig. 8). If no edge is missing during the algorithm, the agents in both sides of the axis perform symmetric actions and the configuration stays with the same axis of symmetry. Since the agents avoid collision, no agent can move to the nodes on the axis; therefore, CCP cannot be solved in this case.

In Algorithm 2, we present a solution for CCP with more than 2 agents, when the initial configuration is aperiodic and the axis of symmetry either (a) passes through at least one edge and it does not pass through a non empty node, or (b) passes through at least one non empty node. By Algorithm 2, and by Theorem 5, we can state the following:

Theorem 7 If the initial configuration has an unique axis of symmetry, more than two agents, and the axis of symmetry either (a) passes through at least one edge and it does not pass through a non empty node, or (b) passes through at least one non empty node, then CCP is solvable.

Proof. We distinguish the two possible cases:

1. The axis of symmetry passes through at least one edge. Let a and b be the two agents that do not belong to the full segment S containing e. Note that a and b have to exist, otherwise the problem is solved. Also, if only one of them exists, then the configuration cannot be symmetric with the axis of symmetry passing through e, contradicting the assumption. These two agents move towards e using two distinct edges, let them be e_a and e_b. Now, two scenarios may occur. (i) Edges e_a and e_b are both alive: in this case the distance between a and b and S decreases; when this distance becomes 0, segment S increases. It is clear, that if this scenario always applies all agents eventually join S. (ii) Either e_a or e_b is missing: in this case only one among a and b moves leading to an asymmetric configuration. In this case, Algorithm 1 can be applied and, by Theorem 5, the theorem follows.

2. The axis of symmetry passes through at least one occupied node. Let us analyse the three possible cases of Algorithm 2(b).

   1. In this case a moves leading to an asymmetric configuration (note that the adversary cannot prevent a to move since it can block at most one edge). In this configuration, Algorithm 1 can be applied, and by Theorem 5 the theorem follows.

   2. The proof in this case is similar to the proof of the previous one.

   3. Let a be the agent and S⁻ and S⁺ be the two full maximal segments to the left and to the right of a. Both of them try to move away from a, and at most one of them can be blocked by an adversary. If none of them is blocked, then we reach a configuration in which both neighbour locations of a are empty, and thus the previous case applies. If only one can move, an asymmetric configuration is reached. In this configuration, Algorithm 1 can be applied and, by Theorem 5, the theorem follows.

Asymmetric initial configuration and h ≥ 3

The algorithm for this case builds segments around some specific points of the ring, called rally points. These points are identified during the execution of the algorithm, and to each color is assigned a specific rally point.

Definition 3. We say that agents are forming a compact line if they are forming a full segment of size h around the rally point of their color. We say that agents are forming an almost compact line if they are forming a full segment of size h − 1 around the rally point of their color; the only agent that is not part of the almost compact line is called a dangling agent.

Moreover, let FC denote the set of agents colored with first_color. We say that the current configuration is correctly placed if and only if both the following conditions hold on all the colors different from first_color:

• (i) There are at least c − 2 compact lines that do not overlap;

• (ii) There is at most one almost compact line.

The MULTI COLOR CONNECTED SEGMENT algorithm is split into three main steps, described in Algorithm 1, 4, and 6, respectively. Let us first describe the intuition behind each step.

• First Step (Algorithm 3). The main idea of the first step is to make an agent with color first_color move in such a way that all agents with color first_color become asymmetrically placed (this step is skipped if FC is already asymmetric). Once FC is asymmetric, the agents in FC do not move until the last phase of the algorithm: these agents are used as reference points to univocally identify both the rally points and a unique orientation of the ring.

• Second Step (Algorithm 4). In the second step, the algorithm proceeds by making each color but first_color to form a full segment around the respective rally point; that is, after this step the configuration becomes correctly placed.

• Third Step (Algorithm 6). Once the configuration is correctly placed, the only agents still to fix in order to solve the problem, are the agents in FC (that are still asymmetrically placed), and the (at most one) dangling agent (this agent has a color different from first_color). Note that, if there is no dangling agent, then there are c − 1 compact lines, and no almost compact line.

The idea here is to use the compact lines formed so far to establish a global chirality of the ring, and a rally point for FC. In particular, the already formed compact lines do not move, hence the computed chirality can be kept; the other agents (i.e., those in FC and the dangling agent) move following the same strategy used in the second step. The movements go on until either ColoredCCP is solved, or there are c − 1 compact lines and one almost compact line. In the latter case, the only dangling agent and the almost compact line (by construction, all these agents have the same color) move one towards each other until they form a compact line.

Since the initial configuration is asymmetric, we have the following:

Lemma 2 If in the initial configuration FC is not asymmetric, by executing Algorithm 3, within finite time agents in FC are placed asymmetrically on the ring.

Proof. The lemma follows by observing that there are at least two edges connecting agents in FC to nodes not occupied by any agent in FC, hence agent a can always be uniquely identified.

Once the agents in FC occupy asymmetric positions on the ring, it is possible to elect one of them as a leader, which provides a global orientation to the ring. Once a global orientation has been computed, the positions of agents in FC allow also to compute the rally points where all other agents will form their respective compact lines, as detailed in Algorithms 4. Let us denote these points by rp_i, 0 ≤ i ≤ c − 1. A color i is assigned to each rally point rp_i, 0 ≤ i ≤ c − 1, with color 0 = first_color assigned to FC.

Definition 4 Given a rally point rp_i, let us call the rally line of color i a maximally full segment of color i that is formed around rp_i. Extending Definition 3, we will call dangling any agent that is not part of a rally line.

ROUTINE RALLY POINTS CONNECTED FORMATION (Algorithm 5) makes all agents of color i gather around rp_i.

Lemma 3 Within finite time, by executing Routine RALLY POINTS CONNECTED FORMATION (Algorithm 5), the system reaches a configuration with c − 1 almost compact or compact rally lines.

Proof. If c − 1 rally lines are almost compact, the lemma trivially follows. Thus, let us assume that there exists at least one rally line, say rl_i, that has at least two dangling agents. By construction, only Pattern 1 of RALLY POINTS CONNECTED FORMATION can be executed. Let us consider only agents having color i. Let a be the closest agent in the counter-clockwise direction to rp_i that has not reached rl_i yet. We will show that, within finite time, the size of rl_i increases. Note that, as long as there is no missing edge between a and rp_i, a will always move towards its own rally line, even if other agents are blocked.

Therefore, if no edge on the path between a and rp_i is ever missing, within finite time the size of rl_i increases by one unit and the statement trivially follows. Otherwise, let a′ be the furthest agent from rl_i: according to Pattern 1, a′ switches direction and starts moving clockwise towards rl_i. As long as a is blocked by a missing edge on its path towards rp_i, a′ keeps approaching rl_i. If a′ becomes blocked before reaching rl_i, then a can perform at least one step (counter-clockwise) towards rl_i, thus decreasing its distance from rl_i. Thus, by iterating the above argument, within finite time either a or a′ will join rl_i.

In conclusion, within finite time, rl_i becomes almost compact, and the lemma follows.

Lemma 4 Let us assume that in the current configuration there exist m > 2 almost compact rally lines, and c − 1 − m compact lines. Within finite time, by executing Routine RALLY POINTS CONNECTED FORMATION in Algorithm 5, m decreases.

Proof. By construction, only Pattern 2 of RALLY POINTS CONNECTED FORMATION can be executed. Note that the agents that are already part of a rally line do not move any more. According to Pattern 2, each dangling agent moves towards its rally line, according to the counter-clockwise direction. First, let us assume that either m − 1 or m dangling agents are blocked by a missing edge (towards their way to their rally lines), and let a be the dangling agent that is closest to its rally line, according to the clockwise direction, and p be the counter-clockwise path that connects a with its own rally line. We distinguish the three possible cases:

• If m − 1 agents are blocked by a missing edge (on their counter-clockwise direction), and a is not one of them, and the missing edge is on p, then a moves clockwise towards its rally line. If a reaches its rally line, the lemma follows. Otherwise, when a becomes blocked (during its clockwise movements), the other m − 1 agents cannot be blocked anymore according to the counter-clockwise orientation, hence they can get closer (counter-clockwise) of at least one unit to their respective rally lines. Note that as long as one of the m − 1 agents does not reach its line, a will be the agent that is closer, clockwise, to its own rally line. By iterating this argument, within finite time m decreases, and the lemma follows.

• If m agents are blocked, then a is one of them: in this case, a moves of one step clockwise. Therefore, either a joins its line, and the lemma follows, or previous case applies.

• If m − 1 agents are blocked by a missing edge (on their counter-clockwise direction), and a is one of them, by construction there exists an agent, say b, that is not blocked, and that is moving counter-clockwise towards the m − 1 blocked agents. Within finite time, either b reaches its own rally line, or b reaches the the blocked edge, or the m − 1 agents become unblocked. In the first case, the lemma follows. In the second case, previous case applies. Otherwise, the m − 1 agents get closer to their rally lines. Thus, by iterating the argument, the lemma follows.

Now, let us assume that at most m − 2 dangling agents are blocked. One of the following holds: (1) one agent reaches its own rally line, thus m decreases and the lemma follows; (2) another agent will join the blocked ones, hence there will be either m − 1 or m blocked agents, and previous case applies.

Thus, by previous Lemmas 3 and 4, the following holds:

Lemma 5 Within finite time, by executing Algorithm 4, the configuration becomes correctly placed.

Finally, by executing Algorithm 6, agents are able to solve the problem. In particular, at the beginning of this step, there are at least c − 2 compact lines, at most one line with just one dangling agent, and the agents in FC that still needs to be compacted.

Lemma 6 If the current configuration is correctly placed, then, within finite time, by executing Algorithm 6 (Third Step), ColoredCCP is solved.

Proof. Let us call da the dangling agent of the almost compact line. By definition of Algorithm 6, as long as there is more than one dangling agent (i.e., da and the agents in FC), neither the agents in the compact lines nor those in the almost compact line move. Once the rally point of FC has been computed, the agents in FC and da move according to RALLY POINTS CONNECTED FORMATION, while all others stay still.

By previous Lemmas 3 and 4, within finite time the agents either reach a configuration where ColoredCCP is solved, and the lemma follows, or where there is only one almost compact line and c − 1 compact lines. In the latter case, the only dangling agent and all agents belonging to the almost compact line (note that all of them have the same color) start moving towards each other (according to the smallest distance). Since at most one edge can be missing, within finite time these two parts will meet. Moreover, since rally points computed in Algorithm 4 are distant (2h + 1) from each other, at most two compact lines can overlap. Hence, the lemma follows.

Combining all previous results from this section, we can conclude that:

Theorem 8 Starting from an asymmetric initial configuration, with c ≥ 3 and h ≥ 3, MULTI COLOR CONNECTED SEGMENT algorithm correctly solves the ColoredCCP problem.

Asymmetric initial configuration and h = 2

Now, let us focus on the case where there are c > 2 colors, but there are only two agents for each color (h = 2). In this case, the agents execute the MODIFIED MULTI COLOR CONNECTED SEGMENT algorithm, that follows the lines of MULTI COLOR CONNECTED SEGMENT algorithm, with a minor modification: the agents of the two first colors, say L₁ and L₂, act as a single group that has the same color. In other words, FC is the union of the agents having the first and the second color in the total ordering of the colors. This change ensures that there are at least three agents in FC, hence the three steps defined by the MULTI COLOR CONNECTED SEGMENT algorithm can still be executed.

Therefore, after the execution of the three steps, agents not in FC form compact lines, while the agents in FC form a segment where agents of two different colors might be interleaved. If the colors of the agents in FC are not interleaved, then ColoredCCP is solved.

Thus, let us assume that the colors of the agents in FC are interleaved: Configuration A in Fig. 11 is, up to symmetries, the only possible configuration. At this point, it is necessary to run a separation procedure that separates the agents of distinct colors, thus forming the remaining two compact lines.

As shown in Fig. 11, from Configuration A, it is possible to reach either Configuration B or Configuration C, by swapping the agents on either edge e₁ or edge e₃ (at least one of these edges must be available): in both configuration, c − 1 compact lines are formed. At this point, the last two agents (having the same color) have to be compacted: they move towards each other. Note that, since there are at least 2 compact lines of other colors, the configuration remains asymmetric during the movement of these two agents. Thus, since at most one edge at the time can be missing, these two agents will eventually become neighbors, thus solving ColoredCCP. Thus, we just showed the following:

Theorem 9 Starting from an asymmetric initial configuration, with c ≥ 3 and h = 2, the modified version of Modified Multi Color Connected Segment algorithm solves the ColoredCCP problem.

Initial configuration with a mirror symmetry and c > 2

We now consider the last remaining case for ColoredCCP with c > 2 colors: the initial configuration has a mirror symmetry.

Theorem 10 Starting from an initial configuration that has a mirror symmetry (hence, is not periodic), and not compact, the ColoredCCP problem for c > 2 is not solvable if either

1. The axis of symmetry passes through two empty nodes, or,

2. The axis of symmetry passes through one edge and one empty node, or,

3. The axis of symmetry passes through two edges and c > 3.

Proof. We prove each of the statements independently.

1. The proof follows directly from Theorem 6.

2. By hypothesis, the symmetry axis intersects the ring on a node v and an edge e. Therefore, the agents can form the compact lines either around v or around e. If the lines are formed around v, since the ring is not oriented, two agents with the same color would move to v, thus violating the no collision requirement of the problem. If the lines are formed around e, then there would be three intersecting compact lines of three different colors around e, thus violating the ColoredCCP specification.

3. Since the configuration has a mirror symmetry, the compact lines have to be centred around the symmetry axis. By construction, it is only possible to form two disjoint compact lines. Since there are more than three colors, by the pigeonhole principle, either three compact lines will intersect or there is a pair of intersecting compact lines, thus violating the specification of ColoredCCP.

Note that previous theorem holds for any ring, even non dynamic ones.

Algorithm 7 solves the two remaining cases: (a) the axis of symmetry passes through at least one occupied node; (b) there is an axis of symmetry passing through two edges, and c = 3. We can thus conclude that:

Theorem 11 If the initial configuration is aperiodic, it has a symmetry axis and either (a) the axis of symmetry passes through at least one occupied node, or (b) there is an axis of symmetry passing through two edges, and c = 3, then Algorithm 7 solves ColoredCCP.

Proof. Let us consider the two possible cases.

• (a) The axis of symmetry passes through at least one occupied node. First, note that by running Algorithm 7, within finite time the configuration becomes asymmetric. At this time, if h > 2 then algorithm MULTI COLOR CONNECTED SEGMENT (Asymmetric initial configuration and h ≥ 3) can be applied, and the proof follows by Theorem 8. If h = 2, the proof follows by Theorem 9.

• (b) There is an axis of symmetry passing through two edges, and c = 3. According to Algorithm 7, agents of different colours are compacted sequentially according to the total ordering of the colors. Let us consider the first color $\overline{c}$ such that the agents with colors $\overline{c}$ do not form a full segment. We will first show that, given the edge e elected for compacting the agents around it, with e as defined by Case (b) of Algorithm 7, the agents will indeed form within finite time a compact line centered around e.

Let a and b be two symmetric agents with color $\overline{c}$ that do not belong yet to the full segment of color $\overline{c}$ that contains e. If there is no such pair of agent, then segments for all colors are full, and the theorem follows. These two agents move towards e using two different edges, let them be e_a and e_b. Two possible scenarios may occur. (i) Either e_a or e_b is missing: in this case only one among a and b moves, leading to an asymmetric configuration. Now, Algorithm 1 can be applied and, by Theorem 5, the theorem follows. (ii) Edges e_a and e_b are both alive: in this case, after both a and b move, their distance to S decreases. Now, by iterating the argument, either the theorem follows by previous Case (i), or within finite time a and b will eventually join S. In the latter case, the number of symmetric pairs with color $\overline{c}$ not in the full segment being formed around e decreases. Hence, by induction on all pairs of symmetric agents with color $\overline{c}$, and on the number of colors, we can conlcude that within finite time the theorem follows.

Asymmetric initial configuration

Let us first consider the case of an asymmetric initial configuration.

Theorem 12. If c = 2 and the initial configuration is asymmetric, ColoredCCP is solvable.

Proof. We distinguish two cases:

• (a) If h = 2 (i.e., four agents in total, two for each color), by Theorem 5, within finite time Algorithm 1 lets the agents form a compact line (where agents of different colors might be interleaved). At this time, the agents can be separated within finite time by using the technique described in Asymmetric initial configuration and h = 2.

• (b) If h > 2, let c₁ and c₂ be the two colors. First note that, since the initial configuration is asymmetric, it is possible to establish a total order among all agents with color c₁ (resp., c₂). We again distinguish two possible cases. (i) If the agents of color c₁ (resp., c₂) are placed asymmetrically, they execute Algorithm 1; by Theorem 5, within finite time agents with color c₁ (resp., c₂) will form a compact line. (ii) Otherwise, the first agent with color c₁ (resp., c₂) that can move, makes a move that makes the set of all agents with color c₁ (resp., c₂) to become asymmetric, and previous Case (i) applies.

Hence, within finite time, the theorem follows.

Initial configuration with a mirror symmetry

Let us now consider the case of a symmetric initial configuration. By Theorem 6, it follows that if the initial configuration has a mirror symmetry, is aperiodic, and not compact and the axis of symmetry passes through two empty nodes, then ColoredCCP is not solvable. In the following, we will show that in all other cases the problem is solvable.

Theorem 13 If the initial configuration is aperiodic and has an unique axis of symmetry and either (a) the axis of symmetry passes through at least one occupied node, or (b) there is an axis of symmetry passing through one edge, and c = 2, then ColoredCCP is solvable.

Proof. We distinguish two cases.

(a) The axis of symmetry passes through at least one occupied node. In this case, by running Algorithm 7, within finite time an asymmetric configuration is reached. At this time, theorem follows by Theorem 12.

(b) The axis of symmetry passes through one edge. In this case, by running Algorithm 7, within finite time, two compact lines are formed, and the theorem follows; or the configuration becomes asymmetric, and the theorem follows by Theorem 12.