Optimising Robotic Pool-Cleaning with a Genetic Algorithm

Abstract

Demand for Genetic Algorithms (GA) in research and market applications has been increasing considerably. This can be explained through big data and the necessity of interpreting them in an automatic, efficient and intelligent way. In the case of intelligent systems for unmanned vehicles there are two well-defined subsystems: autonomous and robotic navigation. Despite the present day’s good development of the latter, taking the best decisions is an essential robot’s attribute that can only be acquired with the former. This work presents a fully programmed GA for a robot to walk around a planar graph G with the highest efficiency. Each robot’s action is one among five kinds of genes, and fourteen of them build a chromosome, namely a sequence of actions for the robot to walk all around G. The efficiency of a chromosome is given by the number of visited vertices and the amount of saved energy, which are both computed by a fitness function. Our GA returns near-optimal chromosomes for the robot to clean the whole reflection pool with the least energy consumption. Our Coverage Path Planning differs from others in the literature because they consider obtaining a near-optimal sequence of vertices for the robot to follow in that order. Moreover, for such a sequence they do not allow vertex repetitions, whereas in our developed algorithm the robot can pass more than once in a same vertex, with the objective of guaranteeing a lower energy consumption. This objective already makes our best chromosomes avoid repeating vertices, as we have observed in our experiments.

Appendix

Here we prove some important results previously cited in the text.

1.1 Preliminaries

In Section 5.2 we presented the equalisation procedure that results in G = (V, E). Each closed polygonal v₁ → v₂ →⋯ → v_k → v₁ =: ℙ ⊂ E such that the interior of ℙ has at most one element (v, e) ∈ V × E is called a cell of G. Without going into details, the equalisation will always produce G with the following properties $\mathcal {P}_{1}$ −$\mathcal {P}_{3}$:

$\mathcal {P}_{1}$::

4 ≤ k ≤ 6.

$\mathcal {P}_{2}$::

∀v ∈ V |deg(v) = 1, the unique $w\in V | \overline {vw}=e\in E\Rightarrow deg(w)= 4$.

$\mathcal {P}_{3}$::

if v, w and e are as in $\mathcal {P}_{2}$, then ∃{w₁,…w₄}⊂ V | the polygonal w → w₁ →…w₄ → w has only v and e in its interior.

In the next subsection we study the special chromosome chr.

1.2 Visiting All Vertices

At the end of Section 5.3 we introduced two important exceptions $\mathcal {E}_{1}$ and $\mathcal {E}_{2}$, which together with Table 2 will guarantee that chr makes the robot visit all vertices in V. The proof is by contradiction. If there is v ∈ V that the robot never accesses, then take w ∈ V such that $e=\overline {vw}\in E$. Hence w is also never accessed, because $\mathcal {E}_{2}$ guarantees that trajectories are rewound. Since G is connected then we apply the same argument successive times until concluding that the robot never accesses any element of V, which is a contradiction. Therefore, the robot will indeed visit all vertices with chr.

But this chromosome can be quite expensive. In Fig. 12 we show G = (V, E) with |V | = 76, and the robot follows chr by starting at 1.

Fig. 12

Example of |S| > 2.27 nv

The robot surrounds the outer contour and reaches vertex 52 exactly at the 52^nd step. Then it goes to the boundary of the shaded region, an obstacle that the robot finally surrounds completely at the 80^th step. Namely, the robot repeated 80 − 76 = 4 vertices until now. Vertex number 56 is the last one the robot visits without repetition, because the 57^th step is again vertex 55. The robot has not cleaned many vertices yet, so it will rewind to vertex 47 at the 112^th step (the next one is indicated as 113). The robot will finish cleaning the whole pool at the 173^rd step, which is precisely the vertex between numbers 1 and 2. Hence it takes a number of steps that is 173/76 ≅ 2.2763 times nv, where nv = 76.

1.3 Optimal Trajectories

Now we prove that with an optimal trajectory the robot visits all vertices with |S| < 1.25 nv. Fig. 13 depicts all topological kinds of cells that appear in G = (V, E) due to the equalisation procedure explained in Section 5.2. One of them is presented as a double-cell, two indicate (v, e) contained in their interior. Black and white bullets fit together, and one begins with either 4 or 6 vertices to re-construct G by starting from one of its cells.

Fig. 13

Kinds of cells and the way they fit together to form G = (V, E)

In Fig. 13 the top left cell needs 2 steps to be cleaned. If we do not count the double-cell, then cleaning the others will take 4 to 5 steps each. Hence, in the worst case we need a number m of steps that is 5 times the number N of cells to clean all vertices.

If we re-construct G by starting from one cell then 4(N − 1) vertices will be added to the initial 6 in the worst case, so that nv = 4N + 2. Hence

$$ m \le 5N=\frac{5}{4}\cdot(4N + 2)-\frac{5}{2}<1.25 {\tt nv}. $$

(5)

However, this optimal trajectory can lead to a high battery consumption. This is because each cell with its interior (v, e) is completely cleaned before the robot goes to the next one. According to our scores the robot spends at most (2 + 4 + 3)J = 9J to go, clean and return from a spot. Hence we set our battery to start with at least 9 ⋅ 1.5 ⋅nv J = 13.5 nv J, where the factor 1.5 is explained in Section 5.4. But this starting value will be enough providingdev > -13.5 nv, where dev is defined in Section 5.7.