# Constant-Time Complete Visibility for Robots with Lights: The Asynchronous Case +
## 6 Stage 1-Interior Depletion
### Stage 1.1-Making Corners of \(\mathbf{P}\) the Corners of \(\mathbf{P}^{\prime}\)
#### 6.2.1 Stage 1.2.1: Moving Four Interior Robots in \(\mathbf{P^{\prime}}\) to \(L_{i}\)
Furthermore, no more than four robots will move to \(L_{i}\) because the interior robot in \(CP(c_{i})\) that is closest to \(L_{i}\) after four robots are in \(L_{i}\) sees all the robots in \(L_{i}\) and, hence, simply waits. We now show that these four robots on \(L_{i}\) are correctly positioned on \(EL_{i}\). Note that \(c_{i}\) sees all four robots on \(L_{i}\). Robot \(c_{i}\) simply waits until four robots are on \(L_{i}\). After four robots are on \(L_{i}\), \(c_{i}\) can wait for two of them to take color eligible. At this time, the two robots \(t_{1},t_{4}\) are on a line in the triangular area divided by line \(L_{i}\) towards \(c_{i}\) and the line \(\overleftrightarrow{t_{1}t_{4}}\) is parallel to \(L_{i}\) (Figure 11b). As \(c_{i}\) now computes \(EA(c_{i})\) and moves to the point \(x_{i}\) on one side \(\overleftrightarrow{c_{i}t_{i-1}}\) (Figure 11c), the line parallel to \(\overleftrightarrow{t_{1}t_{4}}\) going through \(x_{i}\) must pass through \(EA(c_{i})\). Now, since the four robots are arranged on two lines \(\overleftrightarrow{t_{1}t_{4}}\) and \(L_{i}\) with two robots on each of them, they provide the bearing to move \(t_{1},t_{4}\) to \(EL_{i}\). After that, the color eligible of \(t_{1},t_{4}\) will provide the bearing for \(t_{2},t_{3}\) to move to \(EL_{i}\). The colors \(\{\texttt{corner3},\texttt{corner5},\texttt{corner}\}\) of the corners are immediate. We now show that Stage 1.2.1 finishes in \(\mathcal{O}(1)\) epochs. The four robots can move to \(L_{i}\) in, at most, 8 epochs, i.e., a robot takes, at most, 2 epochs. In the first epoch, a robot moves to \(L_{i}\). In the next epoch, it changes its color to transit. After that, in one epoch, corner \(c_{i}\) changes its color to corner21. The robots in \(CP(c_{i})\) then color themselves internal in one epoch. Each corner \(c_{i}\) then colors itself corner22 or corner5 in one epoch. Therefore, Stage 1.2.1.1 finishes in, at most, 11 epochs. Similarly, since we are moving four robots on \(L_{i}\) to \(EL_{i}\), Stage 1.2.1.2 also needs \(\mathcal{O}(1)\) epochs. We now show that the execution of Stage 1.2.1 is collision-free. Note that the robots moving to two different corners do not collide since they are closer to those corners than others. The robots moving to \(L_{i}\) in Stage 1.2.1.1 do not collide because the lines joining them with the corner \(c_{i}\) intersect only at \(c_{i}\) (and they do not reach \(c_{i}\)). For Stage 1.2.1.2, it is clear from the construction (see Figure 11) that the robots never share the same positions, and their paths do not cross. Finally, since all the interior robots of \(\mathbf{P}\) see all the corners of \(\mathbf{P}^{\prime}\), the interior robots do not start Stage 1.2.1 since they see at least a robot with color corner1 until all corners of \(\mathbf{P}^{\prime}\) are colored corner2, corner3, or corner. **Lemma 16**. : _Let \(r_{i}\) be a robot with color internal in the interior of \(\mathbf{P}^{\prime}\). When Stage 1.2.1 finishes, \(r_{i}\) sees \(c_{i}\) and all four eligible colored robots in the eligible line \(EL_{i}\)._  
Proof. : We have from Lemma 1 that, in the beginning of Stage 1.2, all interior robots of \(\mathbf{P}^{\prime}\) can see \(c_{i}\) as \(c_{i}\) is positioned at a point inside \(EA(c_{i})\). Since all the interior robots of \(\mathbf{P}\) can see \(c_{i}\), a unique line joins \(c_{i}\) with each interior robot \(r_{j}\) of \(\mathbf{P}^{\prime}\). Therefore, it is easy to see that two lines \(\overrightarrow{c_{i}r_{j}}\) and \(\overrightarrow{c_{i}r_{k}}\) joining \(c_{i}\) with two interior robots \(r_{j}\), \(r_{k}\) of \(\mathbf{P}^{\prime}\) intersect only at \(c_{i}\). In Stage 1.2.1.1, when four interior robots \(t_{1},\ldots,t_{4}\) of \(\mathbf{P}^{\prime}\) move to \(L_{i}\), they move in their lines \(\overrightarrow{c_{i}t_{j}}\), \(1\leq j\leq 4\), to reach \(L_{i}\); hence, they do not block any other internal robot in \(\mathbf{P}^{\prime}\) to see \(c_{i}\) (and vice versa).