Today, we will cover the first algorithm and sometime in the future I will also get into the second one that is more advanced. I will implement yet another Graph algorithm and this time we are talking about the Shortest Path Problem that can be solved mainly through Dijkstra and Bellman-Ford. So instead of setting pacman as the target vertice we set the vertices immediatetly reachable by pacman as target, the result will be that the available ghosts will try to cover up themajor escape routes of pacman and catch him.Hello its me drifter1 again! Today we get into Java again. Eventually the distance will reduce and you will be able to catch pacman.Īnother heuristic that can be used it to find all the immediate edges reachable from pacman and try to cover as many of these vertices as possible by ghosts. I would suggest finding the shortest path from each ghost to pacman on every movement and move the ghost in that direction.

There can be many optimizations done to this algorithm to speed up the things like taking into account where the pacman was in its previous position and in which direction it moved so that you can get some heiristics in the algorithm. The first path reached is the shortest path. Then on the next vertice you do the same and keep on doing until to you get to the target. The algorithm finds the shortest path from source vertice to target vertice by moving one step down each immediate reachable edge. Each edge has a wait (in your case all the edges have same weight). Just to summarize: Visualize the maze as a graph with vertices and edges. A very good implementation is Dijkstra's algorithm. I think go for the shortest path algorithm at every move made by pacman. And many others! But first you need to frame your problem in terms of a graph, and then toy with how you'd go from node A (pacman) to node B (ghost). By constructing a graph (use an adjency matrix or a list of nodes), you can use the A* algorithm to find the shortest path. The path between two intersections is an edge, and the length of that path is the weight of that edge.Įnter A*. It would be easy to translate any game-board setup into a graph. However, to move toward the bottom only takes 1 hop. Now, to go from a node in the middle to a node at the top, you need 3 hops. To go from the center to the bottom-right requires 2 'hops' rather than one. Something more closer to reality might be this: *-*-*

If Pacman is in the center, it can move to any other node very easily. Each node is connected to its neighbors by an edge, and each edge has equal 'weight' (moving to the node on 'above' is no slower than moving to the node 'below'). In the case above, consider each pixel to be a node on a graph. (I'm assuming knowledge of graphs, some graph theory, adj.

If you just have a grid of pixels - an 'big field' on which pacman and ghost can move about freely - then the shortest path is easy - a straight line between the ghost and the pacman.īut 'shortest path' invariably means we're trying to solve a graph-theory problem.

Path finder algorithm java