How to use the A* path finding algorithm on a grid less 2D plane?

Question

How can I implement the A* algorithm on a gridless 2D plane with no nodes or cells? I need the object to maneuver around a relatively high number of static and moving obstacles in the way of the goal. My current implementation is to create eight points around the object and treat them as the centers of imaginary adjacent squares that might be a potential position for the object. Then I calculate the heuristic function for each and select the best. The distances between the starting point and the movement point, and between the movement point and the goal I calculate the normal way with the Pythagorean theorem. The problem is that this way the object often ignores all obstacle and even more often gets stuck moving back and forth between two positions. I realize how silly mu question might seem, but any help is appreciated.

Answer 1

Create an imaginary grid at whatever resolution is suitable for your problem: As coarse grained as possible for good performance but fine-grained enough to find (desirable) gaps between obstacles. Your grid might relate to a quadtree with your obstacle objects as well.

Execute A* over the grid. The grid may even be pre-populated with useful information like proximity to static obstacles. Once you have a path along the grid squares, post-process that path into a sequence of waypoints wherever there's an inflection in the path. Then travel along the lines between the waypoints.

By the way, you do not need the actual distance (c.f. your mention of Pythagorean theorem): A* works fine with an estimate of the distance. Manhattan distance is a popular choice: |dx| + |dy|. If your grid game allows diagonal movement (or the grid is "fake"), simply max(|dx|, |dy|) is probably sufficient.

Answer 2

Uh. The first thing that come to my mind is, that at each point you need to calculate the gradient or vector to find out the direction to go in the next step. Then you move by a small epsilon and redo.

This basically creates a grid for you, you could vary the cell size by choosing a small epsilon. By doing this instead of using a fixed grid you should be able to calculate even with small degrees in each step -- smaller then 45° from your 8-point example.

Theoretically you might be able to solve the formulas symbolically (eps against 0), which could lead to on optimal solution... just a thought.