Assuming the map is rectangular, you can loop over all border points, and start a flood fill to find the most distant point from the starting point: bestSolution = { start: (0,0), end: (0,0), distance: 0 }; for each point p on the border flood-fill all points in the map to find the most distant … Read more
I’ll try to keep this short and sweet. One approach that is often used to solve these problems in simulation is a Rapidly-Exploring Random Tree. To give at least a little credibility to my post, I’ll admit I studied these, and motion planning was my research lab’s area of expertise (probabilistic motion planning). The canonical … Read more
Just select the file tab and hit Ctrl+C to copy file basename into clipboard. Hit Alt+Enter to bring up the file properties with absolute path and copy it manually. Additionally you can use an Eclipse macro plugin that can do the steps in one go: http://sourceforge.net/projects/practicalmacro/files/
Wikipedia has an article on the topic: http://en.wikipedia.org/wiki/D* Also a D* Lite implementation in C is available from Sven Koenig’s page: http://idm-lab.org/code/dstarlite.tar However I find the impenetrable math much easier to read than the C source code 😉 Another implementation of D* Lite (in C++) is available here: http://code.google.com/p/dstarlite/
Well, I’ll add my comment here, since it is more optimal than the current highest-voted answer by @larsmans – but, I am convinced there must be something better (hence the bounty). If I multiply the heuristic with a constant, it’s no longer admissible The best I can come up with is (manhattenDistance/3)*6 + (manhattenDistance%3), where … Read more
You can use recursion to explore every “branch” that emerges when every possible letter containing the proper initial character is added to a running list: words = [‘giraffe’, ‘elephant’, ‘ant’, ‘tiger’, ‘racoon’, ‘cat’, ‘hedgehog’, ‘mouse’] def get_results(_start, _current, _seen): if all(c in _seen for c in words if c[0] == _start[-1]): yield _current else: for … Read more
The short answer is yes, there are situations in which A* is not the best algorithm to solve a problem. However, there are a number of ways to assess what constitutes the best algorithm for finding a solution. If you are considering best in terms of performance of multiple searches from a single source to … Read more
I present a greedy approach and it’s maybe close to the optimal (but I couldn’t find approximation factor). Idea is simple, we should block the cells which are in critical places of the Maze. These places can help to measure the connectivity of maze. We can consider the vertex connectivity and we find minimum vertex … Read more
It says A* is faster than using dijkstra and uses best-first-search to speed things up. A* is basically an informed variation of Dijkstra. A* is considered a “best first search” because it greedily chooses which vertex to explore next, according to the value of f(v) [f(v) = h(v) + g(v)] – where h is the … Read more
Actually, I’d say your approach is a pretty awesome solution, with almost zero-run time cost compared to any sort of pathfinding. If you need it to generalise to arbitrary maps, you could use any pathfinding algorithm – breadth-first search is simple to implement, for example – and use that to calculate which directions to encode … Read more