递归函数导致堆栈溢出

Question

I am having a problem with my function made to find a path in a maze of 1s and 0s, return true if it is on that path or has found the exit, and return false if the maze is unsolvable. I am getting a stack overflow error any time I try checking for the "- 1s" of my variables but my base cases should be preventing that. Is there a way to use less stack space with recursion? Here is my code

bool Pathfinder::check(string& maze, stack<string>& path, int x, int y, int z) 
{int checking = 0;
    if ((x == 4) && (y == 4) && (z == 4))
    {
        path.push(this->createCoords(x, y, z));
        return true;
    }
    else
    {
        if ((x + 1) < 1 || (x + 1) > columns)
        {
            return false;
        }
        if ((y + 1) < 1 || (y + 1) > rows)
        {
            return false;
        }
        if ((z + 1) < 1 || (z + 1) > floors)
        {
            return false;
        }
        if ((x < 0) || (y < 0) || (z < 0))
        {
            return false;
        }
        if (this->getValue(maze, x, y, z) == 1)
        {
            this->setValue(maze, x, y, z, 2);
        }
        else
        {
            return false;
        }
    }


    if (this->check(maze, path, x + 1, y, z) ||
        this->check(maze, path, x, y + 1, z) ||
        this->check(maze, path, x, y, z + 1))
    {
        checking++;
    }
    if (this->check(maze, path, x - 1, y, z) && checking == 1) //Overflow error comes from here
    {
        checking++;
    }
    if (this->check(maze, path, x, y - 1, z) && checking == 2)
    {
        checking++;
    }
    if (this->check(maze, path, x, y, z - 1) && checking == 3)
    {
        path.push(this->createCoords(x, y, z));

        return true;
    }

    return false;
}

Answer 1

You cannot use "less stack space" for the simple reason that when the amount of required stack space is infinite, anything less than infinite is still infinite. That's what infinity means. The shown algorithm is logically flawed, and clearly results in infinite recursion. Let's label some of these recursive calls as follows:

if (this->check(maze, path, x + 1, y, z) ||   // A
    this->check(maze, path, x, y + 1, z) ||   // B
    this->check(maze, path, x, y, z + 1))     // C
{
    checking++;
}

if (this->check(maze, path, x - 1, y, z) && checking == 1) // D
{
    checking++;
}

The initial value of checking is 0. We can draw the following conclusions based on the above code.

1) Recursive call A always takes place.

2) there's some set of coordinates for which either A, or B, or C recursive cals returns true

3) And when condition 2) holds true, recursive call D always takes place.

Therefore, whenever condition "2)" is true, this guarantees infinite recursion. Recursive call A which then makes recursive call D is logically infinite recursion.

This is because prior to the recursive call that results in condition 2) being true, recursive call A passes x+1 . And inside the recursive call condition "2)" evaluates to true, which results in recursive call D.

But this results in two consecutive nested recursive calls, first passing x+1 , then x+1-1 on the second one, with the same y and z values. The second recursive call then passes the same values for x , y , and z as its grandparent call, and there's nothing here that will stop this infinite recursion.

The shown algorithm is fundamentally broken. You will need to figure out why it's broken, and what should be the correct algorithm. This cannot be answered based on the limited information in the question, the only thing that can be trivially determined is the reason, and the explanation why infinite recursion occurs; it's rather obvious.