How to do dijkstra's algorithm on a grid matrix

Question

So I have a gird that can be any given size (i.e. matrix or 2d array). Each element contains a value and simply I need to find the shortest path. However, the problem I am having is trying to represent this grid as a graph or adj matrix or what ever you are meant to do. For example this is my code:

public int shortestPath (int[][] matrix, int sr, int sc, int dr, int dc)
{
    int p = matrix.length * matrix[0].length;
    int[] distances = new int[p];
    boolean[] visited = new boolean[p]; //I think all are set to false by default

    for (int i = 0; i < p; i++)
    {
        distances[i] = Integer.MAX_VALUE;
    }

    PriorityQueue<Node> pq = new Priority<>(); // a Node holds the int row, int col and the distance from source cell. It also orders by distances, so therefore a comparable was created to change natural order.
    pq.add(new Node(sr,sc,0);
    distances[(sr * matrix[0].length) + sc] = 0;
    visited[(sr * matrix[0].length) + sc] = true;

    while(!pq.isEmpty())
    {
        Node n = pq.poll();
        int row = n.getRow();
        int col = n.getColumn();
        int dist = n.getDistance();

     //Now here with a normal graph I would search neighbours via a for loop and find in the adj list where an element = 1 and is not visited. This is where I am stuck
    }
}

So obviously with a grid, I will need to find neighbours of left/right/up/down (not doing diagonals). Thus, I need to take into account boundaries. How can a create an Adj matrix or what is the correct way to start searching neighbours for a grid like this?

I am having bad luck with this today because most examples show in graph form to adj matrix and not from grid form to adj matrix.

Dijkstra has nothing to do with how the graph is actually represented. It seems that you did not fully understand the algorithm yet, I suggest studying it more. All you have to do is to relax all outgoing edges per step, it is completely up to you how to get hands on those. Take a look at its pseudo-code rather than actual implementations and watch some animations, maybe then you will understand the algorithm better. — Zabuzard, May 01 '20 at 11:06
@Zabuzard I am new so I will make mistakes. But how do I search for the neighbours? Normally its done one neighbour at a time (BFS) on an Adj matrix. But a grid is hard to perform a loop to find neighborus because I know that i + 1, i -1, j + 1 an j - 1 are possible neighbours. Also I know it compares distance of stored path with the distance of potential new path. — Willdomybest18, May 01 '20 at 11:09
https://stackoverflow.com/questions/61444628/does-every-matrix-correspond-to-a-graph-conceptually/61448978#61448978 — Matt Timmermans, May 01 '20 at 11:54
You know the coordinate of the node. So, as you shared, you also know all the coordinates of all neighbors. Now simply check with `if` whether they are still in-bounds. If so, proceed. If not, skip that non-existing-neighbor. — Zabuzard, May 01 '20 at 14:26

Photon · Answer 1 · 2020-05-02T07:55:19.577

3

There's a trick for grid graphs:

lets say {x,y} denotes difference between 2 neighbour cells
You know neighbours will be in {0,-1}, {0,1}, {1,0} or {-1,0} (assuming no diagonal moves), or those cells will be out of bounds
Save arrays of those differences:

int differenceX[] = {0,0,1,-1};
int differenceY[] = {-1,1,0,0};

Now you can use for loop for neighbours:

    for(int i=0; i<4; i++)
    {
        int neighRow = row + differenceY[i];
        int neighCol = col + differenceX[i];
        if(min(neighRow, neighCol) >= 0 && neighRow < matrix.length && neighCol < matrix[0].length){
            //process node
        }
    }

edited May 02 '20 at 07:55

answered May 01 '20 at 11:55

Photon

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The important key aspect which OP was not aware of is `if(notOutOfBound(...))`. It would help OP to show him exactly that. I.e. a coordinate-bounds-check. – Zabuzard May 01 '20 at 14:27
@Photon - Thanks that makes sense now. I would just like to ask, if the gird is not symmetrical i.e. k * k where k is an integer. How do you check boundaries efficiently there? For example a 3 * 4 grid. – Willdomybest18 May 02 '20 at 02:00
@Willdomybest18 updated if statement for general matrix case – Photon May 02 '20 at 07:55

How to do dijkstra's algorithm on a grid matrix

1 Answers1