递归蛮力迷宫求解器Java

Question

In an attempt to write a brute force maze solving C program, I've written this java program first to test an idea. I'm very new to C and intend to convert it after getting this right in java. As a result, I'm trying stick away from arraylists, fancy libraries, and such to make it easier to convert to C. The program needs to generate a single width path of shortest steps to solve a maze. I think my problem may be in fragmenting a path-storing array passed through each recursion. Thanks for looking at this. -Joe

maze:

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


Same maze solved by this program:
4 4 4 4 4 
4 4 4 4 4 
4 0 0 0 4 
3 0 3 3 4 
0 3 3 3 2 

number notation are explained in code

    public class javamaze {

static storage[] best_path;
static int best_count;
static storage[] path;

//the maze - 1 = start; 2 = finish; 3 = open path
static int maze[][] = {{1, 3, 3, 3, 3}, 
    {3, 3, 3, 3, 3},
    {0, 0, 0, 0, 3},
    {0, 0, 3, 3, 3},
    {3, 3, 3, 3, 2}};

public static void main(String[] args) {

    int count1;
    int count2;

    //declares variables used in the solve method
    best_count = 0;
    storage[] path = new storage[10000];
    best_path = new storage[10000];
    int path_count = 0;


    System.out.println("Here is the maze:");
    for(count1 = 0; count1 < 5; count1++) {
        for(count2 = 0; count2 < 5; count2++) {
            System.out.print(maze[count1][count2] + " ");   
        }                       
        System.out.println("");         
    }                       

    //solves the maze
    solve(findStart()/5, findStart()%5, path, path_count);  

    //assigns an int 4 path to the maze to visually represent the shortest path
    for(int count = 0; count <= best_path.length - 1; count++)
        if (best_path[count] != null)
            maze[best_path[count].getx()][best_path[count].gety()] = 4;

    System.out.print("Here is the solved maze\n");

    //prints the solved maze
    for(count1 = 0; count1 < 5; count1++) {
        for(count2 = 0; count2 < 5; count2++){
            System.out.print(maze[count1][count2] + " ");
        }
        System.out.print("\n");
    }
}

//finds maze start marked by int 1 - this works perfectly and isn't related to the problem
public static int findStart() {
    int count1, count2;
    for(count1 = 0; count1 < 5; count1++) {
        for(count2 = 0; count2 < 5; count2++) {
            if (maze[count1][count2] == 1)
                return (count1 * 5 + count2);
        }
    }
    return -1;
}

//saves path coordinate values into a new array
public static void save_storage(storage[] old_storage) {
    int count;
    for(count = 0; count < old_storage.length; count++) {
        best_path[count] = old_storage[count];
    }
}

//solves the maze
public static Boolean solve(int x, int y, storage[] path, int path_count) {

    //checks to see if grid squares are valid (3 = open path; 0 = wall
    if (x < 0 || x > 4) { //array grid is a 5 by 5
        //System.out.println("found row end returning false");
        return false;
    }
    if (y < 0 || y > 4) {
        //System.out.println("Found col end returning false");
        return false;
    }

    //when finding finish - records the number of moves in static int best_count
    if (maze[x][y] == 2) {
        if (best_count == 0 || best_count > path_count) {
            System.out.println("Found end with this many moves: " + path_count);
            best_count = path_count;
            save_storage(path); //copies path counting array into a new static array
        }
    }
    //returns false if it hits a wall
    if (maze[x][y] == 0)
        return false;

    //checks with previously crossed paths to prevent an unnecessary repeat in steps
    for(storage i: path) 
        if (i != null)
            if (i.getx() == x && i.gety() == y) 
                return false;

    //saves current recursive x, y (row, col) coordinates into a storage object which is then added to an array.
    //this array is supposed to fragment per each recursion which doesn't seem to - this may be the issue
    storage storespoints = new storage(x, y);
    path[path_count] = storespoints;

    //recurses up, down, right, left
    if (solve((x-1), y, path, path_count++) == true || solve((x+1), y, path, path_count++) == true ||
            solve(x, (y+1), path, path_count++) == true || solve(x, (y-1), path, path_count++) == true) {
        return true;
    }

    return false;
}
} 

//stores (x, y) aka row, col coordinate points
class storage {

private int x;
private int y;

public storage(int x, int y) {
    this.x = x;
    this.y = y;
}
public int getx() {
    return x;
}
public int gety() {
    return y;
}
public String toString() {
    return ("storage coordinate: " + x + ", " + y + "-------");
}

}

Answer 1

This wasn't originally intended to be an answer but it sort of evolved into one. Honestly, I think starting in Java and moving to C is a bad idea because the two languages are really nothing alike, and you won't be doing yourself any favors because you will run into serious issues porting it if you rely on any features java has that C doesn't (ie most of them)

That said, I'll sketch out some algorithmic C stuff.

Support Structures

typedef
struct Node
{
    int x, y;
    // x and y are array indices
}
Node;

typedef
struct Path
{
    int maxlen, head;
    Node * path;
    // maxlen is size of path, head is the index of the current node
    // path is the pointer to the node array
}
Path;

int    node_compare(Node * n1, Node * n2); // returns true if nodes are equal, else false

void   path_setup(Path * p, Node * n); // allocates Path.path and sets first node
void   path_embiggen(Path * p);        // use realloc to make path bigger in case it fills up
int    path_toosmall(Path * p);        // returns true if the path needs to be reallocated to add more nodes
Node * path_head(Path * p);            // returns the head node of the path
void   path_push(Path * p, Node * n);  // pushes a new head node onto the path
void   path_pop(Path * p);             // pops a node from path

You might to change your maze format into an adjacency list sort of thing. You could store each node as a mask detailing which nodes you can travel to from the node.

Maze Format

const int // these constants indicate which directions of travel are possible from a node
N = (1 << 0),       // travel NORTH from node is possible
S = (1 << 1),       // travel SOUTH from node is possible
E = (1 << 2),       // travel EAST  from node is possible
W = (1 << 3),       // travel WEST  from node is possible
NUM_DIRECTIONS = 4; // number of directions (might not be 4.  no reason it has to be)

const int
START  = (1 << 4),  // starting  node
FINISH = (1 << 5);  // finishing node

const int
MAZE_X = 4,         // maze dimensions
MAZE_Y = 4;

int maze[MAZE_X][MAZE_Y] = 
{
    {E,        S|E|W,    S|E|W,    S|W       },
    {S|FINISH, N|S,      N|START,  N|S       },
    {N|S,      N|E,      S|E|W,    N|S|W     },
    {N|E,      E|W,      N|W,      N         }
};

Node start  = {1, 2}; // position of start node
Node finish = {1, 0}; // position of end node

My maze is different from yours: the two formats don't quite map to each other 1:1. For example, your format allows finer movement, but mine allows one-way paths.

Note that your format explicitly positions walls. With my format, walls are conceptually located anywhere where a path is not possible. The maze I created has 3 horizontal walls and 5 vertical ones (and is also enclosed, ie there is a continuous wall surrounding the whole maze)

For your brute force traversal, I would use a depth first search. You can map flags to directions in a number of ways, like maybe the following. Since you are looping over each one anyway, access times are irrelevant so an array and not some sort of faster associative container will be sufficient.

Data Format to Offset Mappings

// map directions to array offsets
// format is [flag], [x offset], [y offset]
int mappings[][] =
{
    {N, -1,  0},
    {S,  1,  0},
    {E,  0,  1},
    {W,  0, -1}
}

Finally, your search. You could implement it iteratively or recursively. My example uses recursion.

Search Algorithm Pseudocode

int search_for_path(int ** maze, char ** visited, Path * path)
{
    Node * head = path_head(path);
    Node temp;
    int i;

    if (node_compare(head, &finish)) return 1; // found finish
    if (visited[head->x][head->y])   return 0; // don't traverse again, that's pointless

    visited[head->x][head->y] = 1;
    if (path_toosmall(path)) path_embiggen(path);

    for (i = 0; i < NUM_DIRECTIONS; ++i)
    {
        if (maze[head->x][head->y] & mappings[i][0]) // path in this direction
        {
            temp = {head->x + mappings[i][1], head->y + mappings[i][2]};
            path_push(path, &temp);
            if (search_for_path(maze, visited, path)) return 1; // something found end
            path_pop(path);
        }
    }
    return 0; // unable to find path from any unvisited neighbor
}

To call this function, you should set everything up like this:

Calling The Solver

// we already have the maze
// int maze[MAZE_X][MAZE_Y] = {...};

// make a visited list, set to all 0 (unvisited)
int visited[MAZE_X][MAZE_Y] = 
{
    {0,0,0,0},
    {0,0,0,0},
    {0,0,0,0},
    {0,0,0,0}
};

// setup the path
Path p;
path_setup(&p, &start);

if (search_for_path(maze, visited, &path))
{
    // succeeded, path contains the list of nodes containing coordinates from start to end
}
else
{
    // maze was impossible
}

It's worth noting that because I wrote this all in the edit box, I haven't tested any of it. It probably won't work on the first try and might take a little fiddling. For example, unless start and finish are declared globally, there will be a few issues. It would be better to pass the target node to the search function instead of using a global variable.