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Understanding the Jump Point Search algorithm

Started by Markk Jun 20, 2012 at 8:04 AM 2 replies 15.3k views
Original Post
Markk
Markk
Hi! I'm trying to implement the Jump Point Search algorithm into Java, but having a hard time understanding one part of it.The algorithm is explained in this paper, and I'm using the source of this JPS implementation as an example.

When using my example implementation with to find the path around a wall, I get this:
p1gdl.jpg

The green and red squares represent the start and end "tiles", gray is wall, blue and lightgreen are jump points (I guess the lightgreens are in the open list, blue in the closed). Now, I get why the tiles at the green arrows came out as jump points, since they all got one forced neighbour, but can someone explain to me why the tiles at the red arrows are jump points?

Thanks!
bloodisblue
bloodisblue
It seems like the algorithm used is this g(y) = g(x) + dist(x; y) so it would always attempt to go diagonally first and than level off because that would shorten the distance to the target earlier in the program. This is why the first thing the algorithm would do is go diagonally until it reaches either the end of the wall or the goal node and than continue going horizontal.
pithlit
pithlit
Hi Markk,

Consider what would happen if the blue nodes were not jump points: you would continue to recurse diagonally away from the current node and miss a potential turn of the optimal path. To avoid this problem JPS only recurses diagonally if it determines there are no jump points in a vertical or horizontal direction. If it finds such a node, the recursion stops and the current node returned as a jump point.
kruncher
kruncher
Hello Markk

[attachment=12145:jump-node-explained.png]

The above image was derived from images from the whitepaper: http://grastien.net/...s/hg-aaai11.pdf

Please see lines 8-11 on Algorithm 2 (function jump) for the logic behind this. I hope that this helps you to understand why the successor node is forced in this particular scenario.

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