5.11. Exploring a Maze

In this section we will look at a problem that has relevance to the expanding world of robotics: How do you find your way out of a maze? If you have a Roomba vacuum cleaner for your dorm room (don’t all college students?) you will wish that you could reprogram it using what you have learned in this section. The problem we want to solve is to help our turtle find its way out of a virtual maze. The maze problem has roots as deep as the Greek myth about Theseus who was sent into a maze to kill the minotaur. Theseus used a ball of thread to help him find his way back out again once he had finished off the beast. In our problem we will assume that our turtle is dropped down somewhere into the middle of the maze and must find its way out. Look at Figure 2 to get an idea of where we are going in this section.


Figure 2: The Finished Maze Search Program

To make it easier for us we will assume that our maze is divided up into “squares.” Each square of the maze is either open or occupied by a section of wall. The turtle can only pass through the open squares of the maze. If the turtle bumps into a wall it must try a different direction. The turtle will require a systematic procedure to find its way out of the maze. Here is the procedure:

Now, that sounds pretty easy, but there are a couple of details to talk about first. Suppose we take our first recursive step by going North. By following our procedure our next step would also be to the North. But if the North is blocked by a wall we must look at the next step of the procedure and try going to the South. Unfortunately that step to the south brings us right back to our original starting place. If we apply the recursive procedure from there we will just go back one step to the North and be in an infinite loop. So, we must have a strategy to remember where we have been. In this case we will assume that we have a bag of bread crumbs we can drop along our way. If we take a step in a certain direction and find that there is a bread crumb already on that square, we know that we should immediately back up and try the next direction in our procedure. As we will see when we look at the code for this algorithm, backing up is as simple as returning from a recursive function call.

As we do for all recursive algorithms let us review the base cases. Some of them you may already have guessed based on the description in the previous paragraph. In this algorithm, there are four base cases to consider:

  1. The turtle has run into a wall. Since the square is occupied by a wall no further exploration can take place.
  2. The turtle has found a square that has already been explored. We do not want to continue exploring from this position or we will get into a loop.
  3. We have found an outside edge, not occupied by a wall. In other words we have found an exit from the maze.
  4. We have explored a square unsuccessfully in all four directions.

For our program to work we will need to have a way to represent the maze. To make this even more interesting we are going to use the turtle module to draw and explore our maze so we can watch this algorithm in action. The maze object will provide the following methods for us to use in writing our search algorithm:

The Maze class also overloads the index operator [] so that our algorithm can easily access the status of any particular square.

Let’s examine the code for the search function which we call searchFrom. The code is shown in Listing 3. Notice that this function takes three parameters: a maze object, the starting row, and the starting column. This is important because as a recursive function the search logically starts again with each recursive call.

Listing 3

def searchFrom(maze, startRow, startColumn):
    maze.updatePosition(startRow, startColumn)
   #  Check for base cases:
   #  1. We have run into an obstacle, return false
   if maze[startRow][startColumn] == OBSTACLE :
        return False
    #  2. We have found a square that has already been explored
    if maze[startRow][startColumn] == TRIED:
        return False
    # 3. Success, an outside edge not occupied by an obstacle
    if maze.isExit(startRow,startColumn):
        maze.updatePosition(startRow, startColumn, PART_OF_PATH)
        return True
    maze.updatePosition(startRow, startColumn, TRIED)

    # Otherwise, use logical short circuiting to try each
    # direction in turn (if needed)
    found = searchFrom(maze, startRow-1, startColumn) or \
            searchFrom(maze, startRow+1, startColumn) or \
            searchFrom(maze, startRow, startColumn-1) or \
            searchFrom(maze, startRow, startColumn+1)
    if found:
        maze.updatePosition(startRow, startColumn, PART_OF_PATH)
        maze.updatePosition(startRow, startColumn, DEAD_END)
    return found

As you look through the algorithm you will see that the first thing the code does (line 2) is call updatePosition. This is simply to help you visualize the algorithm so that you can watch exactly how the turtle explores its way through the maze. Next the algorithm checks for the first three of the four base cases: Has the turtle run into a wall (line 5)? Has the turtle circled back to a square already explored (line 8)? Has the turtle found an exit (line 11)? If none of these conditions is true then we continue the search recursively.

You will notice that in the recursive step there are four recursive calls to searchFrom. It is hard to predict how many of these recursive calls will be used since they are all connected by or statements. If the first call to searchFrom returns True then none of the last three calls would be needed. You can interpret this as meaning that a step to (row-1,column) (or North if you want to think geographically) is on the path leading out of the maze. If there is not a good path leading out of the maze to the North then the next recursive call is tried, this one to the South. If South fails then try West, and finally East. If all four recursive calls return false then we have found a dead end. You should download or type in the whole program and experiment with it by changing the order of these calls.

The code for the Maze class is shown in Listing 4, Listing 5, and Listing 6. The __init__ method takes the name of a file as its only parameter. This file is a text file that represents a maze by using “+” characters for walls, spaces for open squares, and the letter “S” to indicate the starting position. Figure 3 is an example of a maze data file. The internal representation of the maze is a list of lists. Each row of the mazelist instance variable is also a list. This secondary list contains one character per square using the characters described above. For the data file in Figure 3 the internal representation looks like the following:

[ ['+','+','+','+',...,'+','+','+','+','+','+','+'],
  ['+',' ',' ',' ',...,' ',' ',' ','+',' ',' ',' '],
  ['+',' ','+',' ',...,'+','+',' ','+',' ','+','+'],
  ['+',' ','+',' ',...,' ',' ',' ','+',' ','+','+'],
  ['+','+','+',' ',...,'+','+',' ','+',' ',' ','+'],
  ['+',' ',' ',' ',...,'+','+',' ',' ',' ',' ','+'],
  ['+','+','+','+',...,'+','+','+','+','+',' ','+'],
  ['+',' ',' ',' ',...,'+','+',' ',' ','+',' ','+'],
  ['+',' ','+','+',...,' ',' ','+',' ',' ',' ','+'],
  ['+',' ',' ',' ',...,' ',' ','+',' ','+','+','+'],
  ['+','+','+','+',...,'+','+','+',' ','+','+','+']]

The drawMaze method uses this internal representation to draw the initial view of the maze on the screen.

Figure 3: An Example Maze Data File

+   +   ++ ++     +
+ +   +       +++ + ++
+ + +  ++  ++++   + ++
+++ ++++++    +++ +  +
+          ++  ++    +
+++++ ++++++   +++++ +
+     +   +++++++  + +
+ +++++++      S +   +
+                + +++
++++++++++++++++++ +++

The updatePosition method, as shown in Listing 5 uses the same internal representation to see if the turtle has run into a wall. It also updates the internal representation with a “.” or “-” to indicate that the turtle has visited a particular square or if the square is part of a dead end. In addition, the updatePosition method uses two helper methods, moveTurtle and dropBreadCrumb, to update the view on the screen.

Finally, the isExit method uses the current position of the turtle to test for an exit condition. An exit condition is whenever the turtle has navigated to the edge of the maze, either row zero or column zero, or the far right column or the bottom row.

Listing 4

class Maze:
    def __init__(self,mazeFileName):
        rowsInMaze = 0
        columnsInMaze = 0
        self.mazelist = []
        mazeFile = open(mazeFileName,'r')
        rowsInMaze = 0
        for line in mazeFile:
            rowList = []
            col = 0
            for ch in line[:-1]:
                if ch == 'S':
                    self.startRow = rowsInMaze
                    self.startCol = col
                col = col + 1
            rowsInMaze = rowsInMaze + 1
            columnsInMaze = len(rowList)

        self.rowsInMaze = rowsInMaze
        self.columnsInMaze = columnsInMaze
        self.xTranslate = -columnsInMaze/2
        self.yTranslate = rowsInMaze/2
        self.t = Turtle(shape='turtle')

Listing 5

def drawMaze(self):
    for y in range(self.rowsInMaze):
        for x in range(self.columnsInMaze):
            if self.mazelist[y][x] == OBSTACLE:

def drawCenteredBox(self,x,y,color):
    for i in range(4):

def moveTurtle(self,x,y):

def dropBreadcrumb(self,color):

def updatePosition(self,row,col,val=None):
    if val:
        self.mazelist[row][col] = val

    if val == PART_OF_PATH:
        color = 'green'
    elif val == OBSTACLE:
        color = 'red'
    elif val == TRIED:
        color = 'black'
    elif val == DEAD_END:
        color = 'red'
        color = None

    if color:

Listing 6

def isExit(self,row,col):
     return (row == 0 or
             row == self.rowsInMaze-1 or
             col == 0 or
             col == self.columnsInMaze-1 )

def __getitem__(self,idx):
     return self.mazelist[idx]

The complete program is shown in ActiveCode 1. This program uses the data file maze2.txt shown below. Note that it is a much more simple example file in that the exit is very close to the starting position of the turtle.

+   +   ++ ++        +
      +     ++++++++++
+ +    ++  ++++ +++ ++
+ +   + + ++    +++  +
+          ++  ++  + +
+++++ + +      ++  + +
+++++ +++  + +  ++   +
+          + + S+ +  +
+++++ +  + + +     + +

Self Check

Now that you’re familiar with this simple maze exploring algorithm, use what you’ve learned about file handling, classes, and IO to implement this in C++! To visualize the exploration, print out the characters using cout to create an ASCII representation of your cave. For example, your program should be able to read and operate from a file formatted as follows:

7 5

Next Section - 5.12. Dynamic Programming