## 4 Numerical Examples
### Example 1: Seven-Wall Floor  
The first example has seven walls and eight joints. The shape of the corresponding flat is shown in Figure 7. Table 2 shows the joint coordinates and the start and end joints of walls. It is assumed that concrete is poured starting at Joint 1 on Wall 1 and progressing to Joint 2. The remaining part of the path is determined by the algorithm. As previously described, the movement of the nozzle between two joints when concrete is not poured is accomplished in two ways: one parallel to the coordinate axes x and y, the other by freely following the shortest distance between the two points, that is, by following the hypotenuse. The paths corresponding to these two cases are abbreviated in Figure 7 and onwards with the letters R and D. Here R means that nozzle movements are only parallel to x and y axes, D means that diagonal movements are permitted too.  
Figure 6: Number of path possibilities with respect to the number of walls.  
Figure 7: Example 1: seven-wall flat. The first R and D solutions are with T = 25.  
Six of the best solutions obtained for this problem are shown in Table 3, where three solutions are for rectangular movements of the nozzle (marked with the letter R) and three are for diagonal movements of the nozzle (marked with the letter D). It can be seen that the best R and D solutions for this problem are the same (indicated by arrows in Figure 7).  
In the runs for this example, the population number is 20, and the maximum number of cycles for each run is 5000. Allure of a typical run is shown in Figure 8. It can be seen that, for rectangular movements (i.e., for nozzle movements always parallel to the x and y axes), the shortest distance to travel is 25 units. The first solution shown in Table 3 indicates that after Wall 1, a move will occur without concreting. The pouring of concrete will continue with Walls 2 and 6 and then proceed with Wall 3 in a negative direction after a movement without concrete being poured is made. This is followed by the concreting of Walls 7, 4, and 5, the last two being in negative directions. The T value (i.e., the objective function to be minimized for this path) is the sum of the distances traveled without concrete being poured.  
\begin{table}
\begin{tabular}{l c c c c c} \hline \hline  & \multicolumn{2}{c}{**Vertex Coordinates**} & \multicolumn{3}{c}{**Connectivities**} \\ \hline
**Vertex** & **x-Coor** & **y-Coor** & **Wall** & **Start** & **End** \\ \hline
1 & 0 & 0 & 1 & 1 & 2 \\
2 & 20 & 0 & 2 & 3 & 4 \\
3 & 25 & 0 & 3 & 5 & 6 \\
4 & 35 & 0 & 4 & 7 & 8 \\
5 & 15 & 30 & 5 & 1 & 7 \\ \hline
6 & 35 & 30 & 6 & 4 & 6 \\
7 & 0 & 50 & 7 & 6 & 8 \\
8 & 35 & 50 & & & \\ \hline \hline \end{tabular}
\end{table}
Table 2: Joint coordinates and wall start/end points. Example 1 with seven walls.  
Figure 8: Typical run for Example 1 with seven walls: rectangular movements.  
\begin{table}
\begin{tabular}{l c c c c c c c c} \hline \hline \multirow{2}{*}{**Movement of the Nozzle**} & \multirow{2}{*}{**Path**} & \multicolumn{6}{c}{**Order**} & \multirow{2}{*}{**T**} \\ \cline{3-3} \cline{5-10}  & & **1** & **2** & **3** & **4** & **5** & **6** & **7** \\ \hline \multirow{3}{*}{R} & 1 & 1 & 2 & 6 & 3 & 7 & \(-4\) & \(-5\) & 25 \\  & 2 & 1 & 5 & 4 & \(-7\) & 3 & \(-6\) & \(-2\) & 40 \\  & 3 & 1 & \(-2\) & 6 & \(-3\) & 7 & \(-4\) & \(-5\) & 45 \\ \hline \multirow{3}{*}{D} & 1 & 1 & 2 & 6 & 3 & 7 & \(-4\) & \(-5\) & 25 \\  & 2 & 1 & 2 & 6 & \(-3\) & 7 & \(-4\) & \(-5\) & 25 \\  & 3 & 1 & 2 & 6 & 7 & \(-4\) & \(-5\) & 3 & 38.5410 \\ \hline \hline \end{tabular}
\end{table}
Table 3: Six solutions obtained for Example 1 with seven walls.  
In this problem, when diagonal movements are allowed, the best solution corresponds to that obtained for rectangular movements. In fact, the best solutions for both cases are exactly the same.  
For this problem, although the theoretical possibilities can be calculated as 46,080, as previously given, determining the best solution by intuition is possible when examining the flat graph.