Floyd Warshall complexity

Question

Someone can give to me the time complexity of this procedure inside the for iteration? This piece of code is the "reconstruction path" part of FloydWarshall algorithm. prev[n][n] is the matrix of the nodes that are between source node and destination in the shortest path. printAllSP runs n^2 times in the iteration, what i can't really figure out is the number of recursive calls that it does every time, only understood that depends by i value (max n), j value(max n) and the number of k (???) nodes that interject in the shortest path between i and j, by my evaluations including the for cycles this part runs in On^4 but the total algorithm complexity is On^3.Enlight me pls!

 void PrintAllSP(int i, int j, int prev[n][n]){
if (i == j)
    print i
else  {
    PrintAllSP(i, prev[i][j], prev);
    print j
}
}

//PSEUDOPART OF MAIN
for i = 1 TO n
    for j = 1 TO n
        PrintAllSP(i, j, prev);
    }
}

score 2 · Accepted Answer · answered Jul 23 '14 at 20:40

There are n nodes total, which means O(n^2) shortest paths are printed. Each shortest path can only have up to n nodes in it. Therefore, only O(n^3) nodes can be printed. Since each base recursive step results in one node being printed, this means only O(n^3) base recursive steps total.

Floyd Warshall complexity

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