Recursion: Solving a Maze

The Problem

A robot is asked to navigate a maze. It is placed at a certain position (the starting position) in the maze and is asked to try to reach another position (the goal position). Positions in the maze will either be open or blocked with an obstacle. Positions are identified by (x,y) coordinates.

At any given moment, the robot can only move 1 step in one of 4 directions. Valid moves are:

• Go North: (x,y) -> (x,y-1)
• Go East: (x,y) -> (x+1,y)
• Go South: (x,y) -> (x,y+1)
• Go West: (x,y) -> (x-1,y)

Note that positions are specified in zero-based coordinates (i.e., 0...size-1, where size is the size of the maze in the corresponding dimension).

The robot can only move to positions without obstacles and must stay within the maze.

The robot should search for a path from the starting position to the goal position (a solution path) until it finds one or until it exhausts all possibilities. In addition, it should mark the path it finds (if any) in the maze.

Representation

To make this problem more concrete, let's consider a maze represented by a matrix of characters. An example 6x6 input maze is:

 '' - where the robot can move (open positions) '' - obstacles (blocked positions) '' - start position (here, x=0, y=0) '' - goal (here, x=5, y=4)

Aside: Remember that we are using x and y coordinates (that start at 0) for maze positions. A y coordinate therefore corresponds to a row in the matrix and an x coordinate corresponds to a column.

A path in the maze can be marked by the '' symbol...

 A path refers to either a partial path, marked while the robot is still searching: (i.e., one that may or may not lead to a solution). Or, a solution path: which leads from start to goal.

Algorithm

We'll solve the problem of finding and marking a solution path using recursion.

Remember that a recursive algorithm has at least 2 parts:

• Base case(s) that determine when to stop.

• Recursive part(s) that call the same algorithm (i.e., itself) to assist in solving the problem.

Recursive parts

Because our algorithm must be recursive, we need to view the problem in terms of similar subproblems. In this case, that means we need to "find a path" in terms of "finding paths."

Let's start by assuming there is already some algorithm that finds a path from some point in a maze to the goal, call it `FIND-PATH(x, y)`.

Also suppose that we got from the start to position x=1, y=2 in the maze (by some method):

 What we now want to know is whether there is a path from x=1, y=2 to the goal. If there is a path to the goal from x=1, y=2, then there is a path from the start to the goal (since we already got to x=1, y=2).

To find a path from position x=1, y=2 to the goal, we can just ask `FIND-PATH` to try to find a path from the , , , and of x=1, y=2:

• `FIND-PATH(insertCoord(1, 2, 0))`
• `FIND-PATH(insertCoord(1, 2, 1))`
• `FIND-PATH(insertCoord(1, 2, 2))`
• `FIND-PATH(insertCoord(1, 2, 3))`

Generalizing this, we can call `FIND-PATH` recursively to move from any location in the maze to adjacent locations. In this way, we move through the maze.

Base cases

It's not enough to know how to use `FIND-PATH` recursively to advance through the maze. We also need to determine when `FIND-PATH` must stop.

One such base case is to stop when it reaches the goal.

The other base cases have to do with moving to invalid positions. For example, we have mentioned how to search of the current position, but disregarded whether that position is legal. In order words, we must ask:

• Is the position in the maze (...or did we just go outside its bounds)?
• Is the position open (...or is it blocked with an obstacle)?

Now, to our base cases and recursive parts, we must add some steps to mark positions we are trying, and to unmark positions that we tried, but from which we failed to reach the goal:

``` FIND-PATH(x, y) ```

``` if (x,y outside maze) return false if (x,y is goal) return true if (x,y not open) return false mark x,y as part of solution path if (FIND-PATH(insertDir(0) of x,y) == true) return true if (FIND-PATH(insertDir(1) of x,y) == true) return true if (FIND-PATH(insertDir(2) of x,y) == true) return true if (FIND-PATH(insertDir(3) of x,y) == true) return true unmark x,y as part of solution path return false ```
``` ``` All these steps together complete a basic algorithm that finds and marks a path to the goal (if any exists) and tells us whether a path was found or not (i.e., returns true or false). This is just one such algorithm--other variations are possible.

Note: `FIND-PATH` will be called at least once for each position in the maze that is tried as part of a path.

Also, after going to another position (e.g., ):

``` if (FIND-PATH(insertDir(0) of x,y)¹ == true) return true² ```
if a path to the goal was found, it is important that the algorithm stops. I.e., if going of x,y finds a path (i.e., returns true¹), then from the current position (i.e., current call of `FIND-PATH`) there is no need to check , or . Instead, `FIND-PATH` just need return true² to the previous call.

Path marking will be done with the '' symbol and unmarking with the '' symbol.

Using Algorithm

To use `FIND-PATH` to find and mark a path from the start to the goal with our given representation of mazes, we just need to:

1. Locate the start position (call it startx, starty).
2. Call `FIND-PATH(startx, starty)`.
3. Re-mark* the start position with ''.

*In the algorithm, the start position (marked '') needs to be considered an open position and must be marked as part of the path for `FIND-PATH` to work correctly. That is why we re-mark it at the end.

Backtracking

An important capability that the recursive parts of the algorithm will give us is the ability to backtrack.

 For example, suppose the algorithm just marked position x=2, y=3 in this maze. I.e, it is in the call to ```FIND-PATH(x=2, y=3)```. After marking...

 First, it will try to find a path to the goal from the position of x=2, y=3, calling `FIND-PATH(insertCoord(2, 3, 0))`. Since the position is not open, the call `FIND-PATH(insertCoord(2, 3, 0))` will return false, and then it will go back (backtrack) to `FIND-PATH(x=2, y=3)` and resume at the step just after it went .

 Next, it will go of x=2, y=3, calling `FIND-PATH(insertCoord(2, 3, 1))`. This position is not open, so it will backtrack to `FIND-PATH(x=2, y=3)` and resume at the step just after it went .

 Next, it will go of x=2, y=3, calling call `FIND-PATH(insertCoord(2, 3, 2))`. This position is not open, so it will backtrack to `FIND-PATH(x=2, y=3)` and resume at the step just after it went .

 Finally, it will go of x=2, y=3, calling `FIND-PATH(insertCoord(2, 3, 3))`. This position is not open, so it will backtrack to `FIND-PATH(x=2, y=3)` and resume at the step just after it went . Since is the last direction to search from x=2, y=3, it will unmark x=2, y=3, and backtrack to the previous call, `FIND-PATH(x=1, y=3)`.

To better illustrate this backtracking, let's look at an example run in which a dead end is reached at some point in the search:

 Maze: y x
Symbols: '' = open, '' = blocked, '' = start, '' = goal, '' = path
 Order: 1.North 1.East 1.South 1.West 2.North 2.East 2.South 2.West 3.North 3.East 3.South 3.West 4.North 4.East 4.South 4.West

Run Speed: Slow Medium Fast

Follow: Algorithm Recursive Calls

 Current: Step description: Goal: Search: x= y=

Feel free to follow the search algorithm using other input mazes and parameters.

Questions

In order to demonstrate an understanding of this problem and solution, you should be able to answer the following questions:

• What happens if instead of searching in the order , , , , `FIND-PATH` searchs , , , ?

• In all cases when `FIND-PATH` returns false, does that mean there is no path from the start to the goal?

• Can parts of the maze be searched by `FIND-PATH` more than once? If so, how can this be rectified? If not, how does the algorithm prevent that?

• What can you say about the kinds of paths that `FIND-PATH` finds?

BU CAS CS - Recursion: Solving a Maze