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copied!<p>One thing to try would be converting from isometric coordinates to a square grid coordinate set for all calculations. </p> <p>Say that 0,0 stays the root of the map. 0,1 stays the same, 1,2 becomes 0,2; 1,3 becomes 0,3; 2,3 becomes 1,4; 3,3 becomes 2,5; 0,2 becomes -1,1; etc. This puts you back into a square grid such that the coordinates and heuristics should work again.</p> <p>Y coordinate becomes Y+sourceX offset (3,3 is at x=2; so becomes 2,5); finding sourceX mathmatically is proving itself more difficult.</p> <p>[Stream of consciousness; ignore] Isometric coordinates at Y=0 are accurate for source X. So, to calculate source X you need to 'move left/up Y times' which should net a change of Y/2; rounded down, in the X coordinate.... roughly suggesting that the square coordinates would be:</p> <pre><code>sourceX = X - Y/2 sourceY = Y + sourceX </code></pre> <p>Where sourceX and sourceY are the coordinates in a normal square grid; and Y/2 is integer arithmetic/rounded down.</p> <p>So, in theory, this becomes:</p> <pre><code>double DistanceToEnd(Point at, Point end) { Point squareStart = squarify(at); Point squareEnd = squarify(end); int dx=squareStart.X-squareEnd.X; int dy=squareStart.Y-squareEnd.Y; return Math.Sqrt(dx*dx+dy*dy); } Point squarify(Point p1) { return new Point(p1.X-p1.Y/2, p1.Y+(p1.X-p1.Y/2)); } </code></pre> <p><strong>Update</strong> based on new bits of question:</p> <p>Assuming that you are trying to get distance(3,2; 3,3) < (distance(2,3; 3,3) = distance(3,1; 3,3)); the following should work: (translated from C#; typos not guaranteed to be non present)</p> <pre><code>inline int Pathfinder::calculateDistanceEstimate(const CellCoord& coord) const { int cx=coord.x - coord.y/2; int cy=coord.y + cx; int gx=goal->position.x - goal->position.y/2; int gy=goal->position.y + gx; int diagonal = std::min(abs(cx-gx), abs(cy-gy)); int straight = (abs(cx-gx) + abs(cy-gy)); return 14 * diagonal + 10 * (straight - 2 * diagonal); } </code></pre> <p><strong>EDIT:</strong> Fixed horrible typo.... again.</p>

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