ProceduralGridMover with Procedural tiles?

General Grid graph
#21

You also need to know the origin point for the chunk. How is that stored?

#22

Each chunk has 2 variables called chunkPosX and chunkPosY. They aren’t in world coordinates though so you would have to multiply them by the chunk size.

#23

So you would have to do something like

Chunk chunk = GameManager.worldController.world.GetChunkAt((int)nodePositions[dataIndex].x, (int)nodePositions[dataIndex].z);
int gridX = bounds.min.x + dataX;
int gridZ = bounds.min.z + dataZ;

int chunkCellX = gridX - chunk.chunkPosX*25;
int chunkCellZ = gridZ - chunk.chunkPosY*25;
int chunkCellIndex = chunkCellX*25 + chunkCellZ;
Tile t = chunk.chunkTiles[chunkCellIndex];

This assumes that chunkPosX=0 and chunkPosY=0 implies that the chunk starts at the world origin. And that the node size is 1 world unit.
Might be easier for you to write a world position to tile function? Instead of just world position to chunk.

#24

I managed to get it working! In a different way though, I realized since I have the positions of your grid nodes in world position I could just use my GetTileAt() method which simply takes a world position and gives a Tile from that position, going into chunks and all that.

Which made it as simple as this:

Tile t = GameManager.worldController.world.GetTileAt((int)nodePositions[dataIndex].x, (int)nodePositions[dataIndex].y);

nodeWalkable[dataIndex] = t.terrainType.isWalkable && t.buildingType.isNull;

Though am having a bit of a smaller issue, the graph is off by 0.5 or something, which you can see here: https://gyazo.com/9468751fde5b9e83aca484650334ac1c

#25

Nice!

You can probably want to add (0.5, 0.5) to the graph center in the grid graph settings. That way it will be aligned with your underlaying data.

#26

That fixed it! Thank you so much! This will work very well :smile:

1 Like
#27

How would I recalculate a single node?

#28

Hi

You can do something like

AstarPath.active.UpdateGraphs(new Bounds((Vector3)node.position, new Vector3(1, 1, 1));

That will recalculate the graph in a small box around a given position.

Note that updating the graph has some overhead, so it is much faster to recalculate a large group of nodes at the same time compared to updating each node individually.

#29

Thank you! It works, though yeah I do notice some overhead.