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I am given a $N$ x $M$ sized grid and $K$ start points $S = (s_1, s_2, .. s_k)$ where each $s_k = (x_k,y_k)$ representing the position on the grid. I am also given a single endpoint $(x_{end}, y_{end})$. Some cells of the grid contain obstacles which are undesirable to pass through.

Given that we have a routine getPaths(x_k, y_k) to compute all paths from each $(x_k,y_k)$ to $(x_{end}, y_{end})$, I need to choose a path for each $(x_k, y_k)$ to $(x_{end}, y_{end})$ such that the total number of obstacles encountered along the union of paths is minimised.

I was thinking if the following greedy algorithm would give an optimal answer. The psuedo steps are as follows:

U = {}
allPaths = flatten([getPaths(x_i, y_i) for all i = 1, ... k])
obstacleSet = set()
while U does not contain all k paths
    // best path that when merged with obstacle Set returns the least number of obstacles
    [bestPath, obstacleSet] = findBestPath(allPaths, obstacleSet)

// filter from allPaths paths which contain the source of `bestPath`
allPaths = filter(allPaths, bestPath.source)

// add best path to U
U.add(bestPath)



end

Would greedy algorithm be able to give an optimal solution in this scenario ?

calveeen
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  • https://cs.stackexchange.com/q/59964/755 – D.W. Oct 27 '23 at 06:18
  • No. This greedy algorithm does not give the optimal answer. Try to create a counter example with three start vertices. For each vertex there might exists an exclusive path with $2$ obstacles each. Total obstacles will be $6$. However, there might exits a path with $3$ obstacles that is shared by all the three vertices. Then, the total obstacles will be $3$. – Inuyasha Yagami Nov 02 '23 at 08:56

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