Solving a maze in 2d array

Question

I had referred to many articles and questions that answered how to solve a maze effectively but here I want to confirm what's going wrong in my code. Consider the maze:

2 1 0 0 3
0 1 0 1 1
0 1 0 0 1
0 1 1 0 0
0 0 0 0 0

where the 1's represent the walls and 0's represent the path.(source is 2 and destination is 3). I have to output whether there is a path or not.

int y=0;
        while(y==0)
        {
            robo1(n,m,maze);//this function adds 2 to any '0'/'3' in (i,j+1),(i+1,j),(i-1,j),(i,j-1) (if exists),where (i,j) is 2
            robo2(n,m,k2,maze);//this function adds 3 to any '0'/'2' in (i,j+1),(i+1,j),(i-1,j),(i,j-1) (if exists), where (i,j) is 3
            if(find5(n,m,maze)==1)//this function returns 1 if there is '5' in the maze
                y++;
            if(find0(n,m,maze)==0)//this function returns 0 if there are no '0' in the maze
                break;
        }
        if(find0(n,m,maze)==0 && y==0)
            printf("-1\n");//no path
        else
            printf("1\n");//there is a path

My idea is that if after any number of loops a five is found in the maze, then it would mean there is a path. But while implementing this function in code I get wrong answers and sometimes run-time errors. Is there any flaw in the above logic?

Answer 1

The general idea should almost work, but of course everything is in the details.

One case in which your approach will not work even if implemented correctly is however this:

  2 1 0 0 0
  1 1 0 1 1 
  0 0 0 1 3

ie if both 2 and 3 are "closed" by walls but there are 0s in the room. Your loop will never end because despite having 0s around neither of the two robo function will change anything.

A simple solution is returning 0/1 from robos if they actually changed at least a value in the matrix and quitting when this doesn't happen.

Note that this is not a very efficient way of solving a maze (your code will keep checking the same cells over and over many times).