# 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 find an exit to a virtual maze when starting at a pre-defined location. 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 starting position is dropped down somewhere into the middle of the maze, a fair distance from any exit. Look at Figure 2 to get an idea of where we are going in this section.

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. We can only pass through the open squares of the maze. If we bump into a wall, we must try a different direction. We will require a systematic procedure to find our way out of the maze. Here is the procedure:

• From our starting position we will first try going North one square and then recursively try our procedure from there.

• If we are not successful by trying a Northern path as the first step then we will take a step to the South and recursively repeat our procedure.

• If South does not work then we will try a step to the West as our first step and recursively apply our procedure.

• If North, South, and West have not been successful then apply the procedure recursively from a position one step to our East.

• If none of these directions works then there is no way to get out of the maze and we fail.

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. We have run into a wall. Since the square is occupied by a wall no further exploration can take place.

2. We have 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. In this instance, we will stick to a text-only representation (ASCII).

• __init__ Initializes basic variables to default values, and calls readMazeFile

• readMazeFile Reads the text of the maze text file, and calls findStartPosition.

• findStartPosition Finds the row and column of the starting position.

• isOnEdge Checks to see if the current position is on the edge, and therefore an exit.

• print Prints the text of the maze to the screen.

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

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 def searchFrom(maze, startRow, startColumn): # Check for base cases (Steps 1, 2, and 3): # 1. We have run into an obstacle, return false if maze[startRow][startColumn] == MAZE_OBSTACLE: return False # 2. We have found a square that has already been explored if maze[startRow][startColumn] == MAZE_TRIED: return False # 3. Success, an outside edge not occupied by an obstacle if maze.isOnEdge(startRow, startColumn): maze[startRow][startColumn] = MAZE_PATH return True # 4. Indicate that the currently visited space has been tried. # Refer to step two. maze[startRow][startColumn] = MAZE_TRIED # 5. Otherwise, check each cardinal direction (North, south, east, and west). # We are checking one space in each direction, thus the plus or minus one below. found = searchFrom(maze, startRow - 1, startColumn) or \ searchFrom(maze, startRow + 1, startColumn) or \ searchFrom(maze, startRow, startColumn - 1) or \ searchFrom(maze, startRow, startColumn + 1) # 6. Mark the location as either part of the path or a dead end, # depending on whether or not an exit has been found. if found: maze[startRow][startColumn] = MAZE_PATH else: maze[startRow][startColumn] = MAZE_DEAD_END return found 

As you look through the algorithm you will see that the first thing the code does (steps 1 and 2) is determine if the space should be visited. This is done by checking if the spot is an obstacle (MAZE_OBSTACLE), or has already been visited (MAZE_TRIED). The algorithm then determines if it has found an exit (step 3). If none of these cases are true, it continues 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 searchFrom method uses this internal representation to traverse throughout the maze.

Figure 3: An Example Maze Data File

Data file: maze1.txt
++++++++++++++++++++++
+   +   ++ ++     +   ‏‏‎ ‎
+ +   +       +++ + ++
+ + +  ++  ++++   + ++
+++ ++++++    +++ +  +
+          ++  ++    +
+++++ ++++++   +++++ +
+     +   +++++++  + +
+ +++++++      S +   +
+                + +++
++++++++++++++++++ +++


Finally, the isOnEdge method uses our current position to test for an exit condition. An exit condition occurs whenever we have navigated to the edge of the maze, either row zero or column zero, or the far right column or the bottom row.

Listing 4

MAZE_OBSTACLE = '+'
MAZE_START = 'S'
MAZE_PATH = 'O'
MAZE_TRIED = '.'

class Maze:
def __init__(self, mazeFileName):
# Initialize all of our default variables.
self.mazeList = []
self.totalRows = 0
self.totalColumns = 0

self.startRow = 0
self.startColumn = 0

# And read the maze file.

# The maze list is a list of strings.
# Components of the maze are indicated by specific characters.
# These characters are listed at the top of the file.

# The line below says the following:
# For every line of text in our maze text file, add every single character to a list.
# The final result is a list of lists, where each element is a single character.
self.mazeList = [[char for char in line] for line in open(mazeFileName).read().split("\n")]

# The total number of rows is the total number of strings in the list.
self.totalRows = len(self.mazeList)

# The total number of columns is the length of a single line.
# We can assume all lines of text for the maze are the same length.
self.totalColumns = len(self.mazeList[0])

# Lastly, find the start position.
self.findStartPosition()

def findStartPosition(self):
# Iterate through every individual character in the maze list.
# If we come across the MAZE_START character ('S'),
# we save the row and column of where it was found, and stop looking.

# enumerate(...) is very much like using a typical list,
# except it gives you two pieces of information instead of one.
# It assumes the format of (index_of_item, item).
for (row, text) in enumerate(self.mazeList):
for(column, component) in enumerate(text):
if component == MAZE_START:
self.startRow = row
self.startColumn = column
return

def isOnEdge(self, row, column):
return (row == 0 or
row == self.totalRows - 1 or
column == 0 or
column == self.totalColumns - 1)

# This allows us to use the Maze class like a list, e.g, maze[index]
def __getitem__(self, index):
return self.mazeList[index]


The complete program is shown in ActiveCode 1. This program uses the data file maze1.txt shown above. Feel free to also attempt to use maze2.txt from up above. Note that it is a much more simple example file in that the exit is very close to the starting position.

Data file: maze2.txt
++++++++++++++++++++++
+   +   ++ ++        +
+     ++++++++++      ‏‏‎ ‎‎‏‏‎
+ +    ++  ++++ +++ ++
+ +   + + ++    +++  +
+          ++  ++  + +
+++++ + +      ++  + +
+++++ +++  + +  ++   +
+          + + 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: You can also use CTurtle to visualize the traversal throughout the maze.

7 5

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