A-star-Pathfinding | A-star pathfinding with pygame visualization

it means map the id to a Node and put (collect) it into a set

this::getNode translates to: from this class, use getNode on the id which is just syntactic sugar for .map(id -> getNode(id)).collect(Collectors.toSet())

A* can work with any number of dimensions; it's a graph traversal algorithm, and no matter how many dimensions your problem space has, connecting one position to another still produces a graph.

You have two problems with generating new nodes, however.

• You included (0, 0, 0) in the list, so no change. You keep putting the current position back into the queue for consideration. That’s just busy work, because the current position is already in the closed list.

• You never subtract from any of your coordinates, you only add. So your x, y and z values can only ever go up. If reaching your goal requires a path around an obstacle, then you have a problem here because all your version can ever do is move in one direction along any given axis.

In a 3 by 3 by 3 3D matrix, with the current position in the middle, there are 3 times 3 times 3 minus 1 == 26 positions you can reach with a single step. Your code reaches just 7 of those, plus one that stays put.

If you extract your tuples in the for new_position in [...] list into a separate variable and add some newlines, and re-arrange them a bit to group them by how many 1s you have in the tuple, you get the following definition. I renamed this to deltas, because it's not a new position, it's the change relative to the old position, or delta. I re-arranged your tuples to make it easier to group them logically:

The point is that each node is visited at most one time when running Dijkstra or A*.
That is because each time a node is being visited (after we pop it from the queue), we 'mark' this node as being already visited. In the implementation you gave, the marking is made by adding the node to closed_list:

I believe you have a typo here: