Question

I’m doing my first algorithm (A* Pathfinding) and part of it involves checking all nodes adjacent to a different node. Is there a quick and easy way to do this or must it be done manually for each node?

Edit:

By adjacent I mean this:

Each X is adjacent to the parent node, O

XXX  
XOX  
XXX

Answer 1

There’s a nice double-for-loop you can use:

for (int i = x - 1; i <= x + 1; i++) {
    for (int j = y - 1; j <= y + 1; j++) {
        /* Skip the point itself! */
        if (i == x && j == y) continue;

        /* Process the location here */
    }
}

This can also be generalized to only consider points adjacent by cardinal directions (i.e. directly up/down/left/right). To do that, you use a modification of the above for loop where you visit all eight neighbors, but then skip points that either

• Are identically where you are (both i == x and j == y), or
• Have neither x nor y in common with the start point (both i != x and j != y)

Interestingly, the above two tests can be rolled into one line: ((i == x) == (j == y)). This tests whether both values are the same (you’re at the same place you started) or both values are different (you’re on a diagonal):

for (int i = x - 1; i <= x + 1; i++) {
    for (int j = y - 1; j <= y + 1; j++) {
        if ((i == x) == (j == y)) continue;

        /* Process the location here */
    }
}

Of course, in both cases you should ensure that you’re within the bounds of the world, but since I don’t know how those are specified I’ll leave it as an exercise to the reader. 🙂

Hope this helps!