Path finding Exercise 12-1 Visibility graph: Nodes are corner - - PowerPoint PPT Presentation

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Sep 16, 2022 122 likes •769 views

Path finding Exercise 12-1 Visibility graph: Nodes are corner points of all obstacles, start and target An edge between a pair of nodes (a,b) exists if and only if a and b can see another, i.e. if the line [a,b] is not interrupted by

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Path finding

Exercise 12-1

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target a) Visibility graph:

Nodes are corner points of all obstacles, start and target
An edge between a pair of nodes (a,b) exists if and only if a

and b can see another, i.e. if the line [a,b] is not interrupted by an obstacle. Edges of polygons are edges in the visibility graph, too.

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target a) Visibility graph:

Edge costs correspond to the Euclidean distance
Shortest path search in the visibility graph (e.g. with Dijkstra)

delivers shortest path between start and target

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target b) For extended objects which are invariant to point reflection:

Pick centroid of the object and expand all obstacles by the corresponding

Minkowski-sum

Like that the path finding of the extended object can be reduced to the path finding
f a point: the shortest path of a circle through the obstacles is equivalent to the

shortest path of the center of the circle through the expanded obstacles.

Problem: Circles have infinite many angles which means that the visibility graph is

not computable precisely.

Solution: Approximate the circle with a polygon which is invariant to point reflection

(e.g. hexagon, octagon)

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target b) Not computable, but obvious solution:

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target Annotation: The triangle is not invariant to point reflection- If approximated with a circle, there would be no path to the target. c)

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target c) Approach from robotics (motion planning) which may be too expensive for MMOs:

Choose reference point in the polygon, e.g. a corner.
Calculate Minkowski difference of the polygon (relating to the reference

point) to every polygon (=Minkowski sum of the object point reflected by the reference point)

Like that the polygon itself „shrinks“ to a point, obstacles „grow“ (see

also robotic work space-> configuration space (C-Space, e.g. all positions of a roboter in a room), C-obstacles, free space)

Move point to the target through free space.

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target c) Now the shortest path to the target is again the shortest path in the visibility graph.

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target c) Now the shortest path to the target is again the shortest path in the visibility graph.

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target c) Now the shortest path to the target is again the shortest path in the visibility graph.

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target c) Path finding of such a triangle is thus not solvable with the rather coarse approximation of a circle. But it is possible using the Minkowski difference, which though depends

n the shape and can be computationally intensive for a high variety of

shapes in a game

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target c) Addition: if another corner of the triangle is chosen as reference point, it looks like this:

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target c) Addition: if another corner of the triangle is chosen as reference point, it looks like this:

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target c) Addition: if another corner of the triangle is chosen as reference point, it looks like this:

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target c) Addition: Comparison: The orange path was the shortest path found previously. As we can see, it is the same, so the shortest path does not depend on the reference point chosen.

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Tutorial Exercise

Alpha Beta Pruning

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4 7 2 4 6 5 9 2 6 1 4 2 1 2 7 3 max min min max

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max min min max

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7 β=7 max min min max

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4 7 max min min max

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4 7 max min min max

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4 7 α=4 max min min max

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4 7 2 α=4 max min min max

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4 7 2 β=2 α=4 max min min max

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4 7 2 β=2 α=4 β< α max min min max

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4 4

4 7 2 β=2 β< α max min min max

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4 4

4 7 2 β=4 max min min max

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4 4

4 7 2 6 β=4 max min min max

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4 4

4 7 2 6 β=4 β=6 max min min max

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4 4

4 7 2 6 5 β=4 β=6 max min min max

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4 4 5

4 7 2 6 5 β=4 α=5 max min min max

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4 4 5

4 7 2 6 5 β=4 α=5 β<α max min min max

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4 4 4 5

4 7 2 6 5 α=4 max min min max

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4 4 4 5

4 7 2 6 5 6 α=4 β=6 max min min max

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4 4 4 5 1

4 7 2 6 5 6 1 α=4 max min min max

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4 4 4 5 1

4 7 2 6 5 6 1 α=4 α=1 max min min max

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4 4 4 5 1

4 7 2 6 5 6 1 4 α=4 α=1 β=4 max min min max

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4 2 4 4 5 1 2

4 7 2 6 5 6 1 4 2 α=4 max min min max

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4 2 4 4 5 1 2

4 7 2 6 5 6 1 4 2 α=4 β=2 max min min max

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4 2 4 4 5 1 2

4 7 2 6 5 6 1 4 2 α=4 β=2 β<α max min min max

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4 4 2 4 4 5 1 2

4 7 2 6 5 6 1 4 2 max min min max

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Search strategy: Post-order (right partial tree first)

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4 7 2 4 6 5 9 2 6 1 4 2 1 2 7 3 max min min max

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3 β=3 max min min max

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7 3 α=3 max min min max

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2 7 3 α=3 β=2 max min min max

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2 7 3 α=3 β=2 β<α max min min max

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3 3

2 7 3 β=3 max min min max

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3 3

2 2 7 3 β=3 β=2 max min min max

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3 2 3

1 4 2 2 7 3 β=3 α=2 β=1 max min min max

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3 2 3

1 4 2 2 7 3 β=3 α=2 β=1 β<α max min min max

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3 2 2 3

4 2 2 7 3 β=3 1 max min min max

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3 2 2 2 3

4 2 2 7 3 α=2 1 max min min max

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3 2 2 2 3

2 4 2 2 7 3 α=2 β=2 1 max min min max

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3 2 2 2 2 3

9 2 4 2 2 7 3 α=2 α=2 1 max min min max

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3 2 2 2 2 3

5 9 2 4 2 2 7 3 α=2 α=2 β=5 1 max min min max

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3 2 2 5 2 2 3

6 5 9 2 4 2 2 7 3 α=2 α=2 1 max min min max

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5 3 2 2 5 2 2 3

6 5 9 2 4 2 2 7 3 α=2 1 max min min max

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5 3 2 2 5 2 2 3

6 5 9 2 4 2 2 7 3 α=2 1 β=5 max min min max

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5 3 2 2 5 2 2 3

4 6 5 9 2 4 2 2 7 3 α=2 1 β=5 β=4 max min min max

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5 3 2 2 2 5 2 2 3

2 4 6 5 9 2 4 2 2 7 3 α=2 1 β=5 α=2 max min min max

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5 3 2 2 2 5 2 2 3

4 2 4 6 5 9 2 4 2 2 7 3 α=2 1 β=5 α=2 β=4 max min min max

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4 5 4 3 2 4 2 4 2 5 2 2 3

4 7 2 4 6 5 9 2 4 2 2 7 3 1 max min min max