A* algorithm solver #4
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sswadkar
reviewed
Apr 28, 2025
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| val newNode = Node(currentNode, grid[newY][newX]) | ||
| newNode.g = currentNode.g + 1 | ||
| newNode.h = |
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manhattan distance might be a better heuristic to use here
sswadkar
reviewed
Apr 28, 2025
| ) { // Keep running until all the nodes are visited. In that case, no optimal path is | ||
| // found. | ||
| currentNode = | ||
| openList.minByOrNull { it.f } ?: openList.last() // Find the node with the lowest cost. |
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your openList should be a priority queue supported by a minheap -- will notice a performance difference for larger open lists
sswadkar
reviewed
Apr 28, 2025
| newNode.f = newNode.g + newNode.h | ||
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| // Discard node if it means going over the charge station. | ||
| if ((newNode.position.x - currentNode.position.x).absoluteValue.inMeters >= 10) { |
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old game so understandable but i think you want to construct your graph so that this isn't even a possibility. think about an actual graph, it can only follow edges that exist, an edge that goes over the charge station shouldn't exist
sswadkar
requested changes
Apr 28, 2025
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| import org.team4099.lib.geometry.Pose2d | ||
| import org.team4099.lib.units.base.inMeters | ||
| import kotlin.math.pow |
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does our value class not have a pow function?
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Uses a starting position and ending position (Pose2d objects) with a grid given (a 2D list of Pose2d objects) to find the most optimal path between the starting position and the ending position.