Consider, however, the observers in the remaining three rooms. Each of these rooms has five walls. If the solution line starts in one of these rooms, its observer will see the line leave (through one wall), re-enter and leave again (two more walls) and enter and leave a second time (the last two walls). If the solution line starts somewhere else, the observer will see the solution line enter and leave (two walls), enter and leave a second time (two more walls) and finally enter through the fifth wall and end (all five walls have been crossed, so the line cannot get back out of the room again). So, we see that for the rooms with five walls, the solution line must either start inside the room or it must end inside the room. There is no other possibility. In our arguments, we have said nothing about exactly which walls the solution line crosses, the order in which it crosses them or where the line goes when it is outside a particular room. Therefore, these arguments apply to all solutions that obey the rules. Again, for the rooms with five walls, the solution line must either start or end inside the room.
But, we have three rooms with five walls. The solution line has one start and one end, so it can pass through all five walls of two of these rooms. However, having run out of ends, the line can not pass through all of the walls of the third five-walled room. Therefore, the solution line cannot be drawn to obey the rules.