Question

I’m working with a TreeSetthat is meant to store pathfind locations used during the execution of a A* algorithm.

Basically until there are “open” elements (still to be exhaustively visited) the neighbours of every open element are taken into consideration and added to a SortedSetthat keeps them ordered by their cost and heuristic cost. This means that I have a class like:

public class PathTileInfo implements Comparable<PathTileInfo>
{
  int cost;
  int hCost;
  final int x, y;

  @Override
  public int compareTo(PathTileInfo t2) {
    int c = cost + hCost;
    int c2 = t2.cost + t2.hCost;
    int costComp = c < c2 ? -1 : (c > c2 ? 1: 0);

    return costComp != 0 ? costComp : (x < t2.x || y < t2.y ? -1 : (x > t2.x || y > t2.y ? 1 : 0));
  }

  @Override
  public boolean equals(Object o2) {
    if (o2 instanceof PathTileInfo) {
      PathTileInfo i = (PathTileInfo)o2;
      return i.cost + i.hCost == cost + hCost && x == i.x && y == i.y;
    }

    return false;
  }
}

In this way first the total cost is considered, then, since a total ordering is needed (consistency with equals) a ordering according to the x,y coordinate is taken into account.

This should work but simply it doesn’t, if I iterate over the TreeSet during the algorithm execution like in

for (PathTileInfo t : openSet)
  System.out.print("("+t.x+","+t.y+","+(t.cost+t.hCost)+") ");

I get results in which the right ordering is not kept, eg:

(7,7,6) (7,6,7) (6,8,6) (6,6,7) (5,8,7) (5,7,7) (6,7,6) (6,6,7) (6,5,7) (5,7,7) (5,5,8) (4,7,7) (4,6,8) (4,5,8)

is there something subtle I am missing?
Thanks!

Answer 1

Please ensure NOT to change ‘Comparable’ value of element after adding to the TreeSet. If you change the value after adding to the TreeSet, TreeSet cannot maintain the new order.

I think your PathTileInfo is correct, although equals() override is not necessary in this case.