EXPLORATION WITH RETURN OF HIGHLY DYNAMIC NETWORKS

In this paper, we study the necessary and sufficient time to explore with return constantly connected dynamic networks modelled by a dynamic graphs. Exploration with return consists, for an agent operating in a dynamic graph, of visiting all the vertices of the graph and returning to the starting vertex. We show that for constantly connected dynamic graphs based on a ring of size n , 3n-4 time units are necessary and sufficient to explore it. Assuming that the agent knows the dynamics of the graph.


ISSN: 2320-5407
Int. J. Adv. Res. 9 (10), 315-319 316 More precisely, the authors study the exploration time for a single agent which knows the dynamics of the graph for the next steps in its -hop neighborhood, for given parameters and .
The problem of perpetual exploration is considered in [2,7]. In [2], the authors consider that all agents are active at each time step and show that to solve the problem, one agent is sufficient in the rings of size two, two agents are sufficient in the rings of size three, and three agents are sufficient for all other rings. The authors define a ring of size two as a two-node path if the graph is simple, or as two nodes linked by two bidirectional edges otherwise. In [7] the authors consider time varying graphs whose topology is arbitrary and unknown to the agents and investigates the number of agents that are necessary and sufficient to explore such graphs. In addition to the problem of exploration, the problem of dispersion of a team of agents [1], gathering [5] and patrolling by a team of agents [3] are studied, considering constantly connected dynamic graphs based on the ring.

Our results.
We consider the problem of exploration with return of constantly connected dynamic graphs the case where the underlying graph is a ring of vertices. We show that to explore that dynamic graphs, 3 − 4 time units are necessary and sufficient, where is the size of the ring.

Definitions and model
This section provides precise definitions of the concepts and models informally mentioned above. We also give some previous results from the literature on the problem studied. The proofs of the theorems mentioned in this section are given in [14].

Definition 1. (Dynamic graph)
A dynamic graph is a pair = , ℰ , where is a set of static vertices, and ℰ is a function which maps every integer ≥ 1 to a set ℰ( ) of undirected edges on .

Definition 2. (Underlying graph)
Given a dynamic graph = , ℰ , the static graph = ( , ∪ ℰ ) is called the underlying graph of . Conversely, the dynamic graph is said to be based on the static graph .

Definition 3. (Constant connectivity))
A dynamic graph is said to be constantly connected if, for any integer i, the static graph = ( , ℰ ) is connected. We give in the following definitions the different types of exploration of graphs by a mobile agent that exists, namely perpetual exploration, periodic exploration, exploration with stop and exploration with return.

Definition 5. (Periodic exploration)
In a periodic exploration, the agent must visit all the vertices of the graph periodically. In this type of exploration too, the agent does not have to stop after visiting all the vertices of the graph.

Definition 6. (Exploration with stop) In this type of exploration, the agent must visit each vertex of the graph at least once. It must then stop once it has finished visiting all the vertices (not necessarily immediately after the last unknown vertex has been visited). Definition 7. (Exploration with return)
The exploration with return is an exploration with stop with an additional constraint: the mobile entity (agent) must stop at its starting position.
In this paper, we study the time complexity of the exploration with return of constantly connected dynamic graphs based on a ring of size . A mobile entity, called agent, operates on these dynamic graphs. The agent can traverse at most one edge per time unit. We say that an agent explores with return the dynamic graph if and only if he visits all its vertices and returns to his starting vertex. We also assume that the agent knows the dynamics of the graph, that is to say, the times of appearance and disappearance of the edges of the dynamic graph. In this article, we will use the following results from the literature.

Theorem 1. (From [12])
For any integer ≥ 3 and for any constantly connected dynamic graph based on a ring with vertices, there exists an agent (algorithm), Explore-ring, exploring this dynamic graph in time at most 2 − 3 (the agent knows the dynamics of the graph). 317

Sketch of proof.
Consider virtual agents placed on the vertices (one agent on each vertex). Make all agents move in the clockwise direction for − 1 time units from time . Since at most one edge is removed at a time, it holds that, at each time, at most one such virtual agent is blocked at this time without having been blocked before. Thus, one of the n virtual agents is never blocked during the − 1time units, and the starting vertex of this agent is the vertex ( ) we are looking for. This can be done in at most − 1time units.

⋄ Theorem 2. [15]
For any constantly connected dynamic graph on vertices, at most − 1time units are sufficient for an agent to go from any vertex to any other vertex in the graph, when the agent knows the dynamics of the graph.

Sketch of proof.
Let be some arbitrary vertex of the dynamic graph. For any integer ≥ 0, let be the set of vertices reachable from in at most time units. We have that ⊊ + 1until contains all the vertices. Indeed, before all vertices are reachable, there exists a vertex not in which is neighbor of a vertex in , because the dynamic graph is constantly connected. ⋄

Upper bound
This section gives an upperbound on the exploration time of constantlyconnecteddynamic graphs based on a ring of size . The algorithmexecuted bythe agent namedExplore-ring-with-return to explorewith return anyconstantlyconnecteddynamic graph based on a ring of verticesisvery simple,and we show in the next section thatitis optimal. Let'sdescribeit.
From the start vertex, the agent explores the ring using the algorithmExplore-ringwhosecomplexityisgiven by Theorem 1. Once the exploration isfinished,hereturns to hisstarting vertex. Theorem 3. An agent executing the algorithm Explore-ring-with-return requires at most 3 − 4 time units to explore with return any constantly connected dynamic graph based on a ring of vertices.

Proof.
First of all, note that the algorithm is composed of two procedures: a) Exploring-the-ring b) Returning-to-the-starting-vertex The complexity of the algorithm is given by the sum of the complexities of this two procedures. The complexity of the procedure Exploring-the-ring is 2 − 3 time units (exploration time of a ring with st vertices cf. Theorem1).
And the complexity of the procedure Returning-to-the-starting-vertex is st − 1 time units cf. Theorem2. The sum of these two complexities gives the claimed bound. Which concludes the proof of the theorem. □

Lower bound
This section gives the lower bound on the exploration time of constantly connected dynamic graphs based on a ring of size . We prove that the simple algorithm described in the section 2 is optimal. For that, we show that for any agent (algorithm), there exists a constantly connected dynamic graph based on a ring of size such that the agent must pay at least 3 − 4 time units to explore it with return. We have the following Theorem. Consider any agent (algorithm). We will now prove that the time the agent uses to explore with return is at least 3 − 4 time units. To explore the dynamic graph with return, the agent must visit all vertices of in particular the vertices 1 , 2 and return at 0 . We consider the following cases. 1. CASE1. The agent explore 2 before 1 .
To visit 2 without going through 1 , the agent must go around the ring and pay at least − 2time units. After visiting 2 , to visit 1 , the agent has two choices. Either he goes around the cycle and pay − 1 other time units or wait on 2 , − 1time units, the time that the edge 1 , 2 appears. In both cases, the agent pays at least 2 − 3time units to visit 1 for the first time. After visiting 1 , the agent must return to 0 to complete his exploration with return. To do this, two ways are available to him. Fist, he can return to 0 without going through 2 after visiting 1 .
To do this, he must wait on 1 until the edge 0 , 1 reappears. This after 3 − 4 time units. The other way is to return to 0 by going around the ring through 2 . In this case, he pays − 1other time units to return to 0 . In both cases, the agent pays at least 3 − 4time units to explore with return .

CASE2
The agent explore 1 before 2 . So the agent will arrive on 0 and finish his exploration with return after paying at least 3 − 4time units which gives the claimed bound. This concludes the proof of the theorem.

Conclusion:-
In this paper, we studied the time complexity for exploring with return constantly connected dynamic graphs based on a ring, under the assumption that the agent knows the dynamics of the graph. We gave a simple exploration algorithm for dynamic graphs that we called Explore-ring-with-return, and we have shown that it is optimal. For exploring with return constantly connected dynamic graphs based on a ring of vertices 3 − 4time units are necessary and sufficient. This study opens several perspectives. An interesting question to investigate would be if Tinterval connectivity (for ≥ 1) allows to save a significant factor in the exploration time. A further perspective is to consider the exploration problem of dynamic graphs using more than one agent, assuming standard models of communication between the agents. The objective would be to study whether dynamic graph exploration can be performed more efficiently by using more than one agent.