## Following the walls

This time, the maze is a bit more complicated. Random won't be enough, we ough to be smart!

The good news is that this maze is simpler that it seems at the first glance: every wall are connected to each other. To get out of this kind of maze, the buggle only have to follow a wall (the one on its left or the one on its right, it doesn't matter). While keeping a paw on the wall, the buggle must move forward until it finds the maze exit and this biscuit it loves so much.

This works here because there is no island of isolated walls, so our buggle cannot loop around for ever without encountering its baggle.

### Traguardo dell'esercizio

The goal of this exercise is to write an algorithm allowing the buggle to get out of this maze.

As said earlier, it does not matter whether you decide to follow the left wall or the right one. Simply, the demo follows the left one, so you should do the same in your solution to ease the comparison of your solution and the demo.

Write a method `keepHandOnSideWall()` which lets the buggle move one step forward while keeping the paw on the wall of the selected side. You must ensure that the buggle always keep the paw on the wall, but also that it won't crash into a wall. You can check the tip for more info on this, but only do so if you're stuck. Try to do it without the tip first.

Then, write the whole algorithm to traverse the maze step by step (using `keepHandOnSideWall()`) until it finds the biscuit and the exit. Don't forget to pick the baggle up once you've found it.

When your buggle has a wall on the left, there is three situations to consider, depending on the surrounding walls. The following table depicts each initial situation, and where you should let your buggle end after one step.

| Case 1 | Case 2 | Case 3
Initial situation |  |  |
Where is the next step |  |  |

If you do a `right()` in any case at the end of your function, it's possible to write it in 3 lines with a `while` loop.