breadth-first-search

问题 I know there's a similar question in stack overflow, where one person has asked, why time complexity of BFS/DFS is not simply O(V). The appropriate answer given was that E can be as large as V^2 in case of complete graph, and hence it is valid to include E in time complexity. But, if V cannot be greater than E+1. So, in that case not having V in the time complexity, should work? 回答1: If it is given that E = kV + c , for some real constants k and c then, O(E + V) = O(kV + c + V) = O(V) = O(E)

问题 I know there's a similar question in stack overflow, where one person has asked, why time complexity of BFS/DFS is not simply O(V). The appropriate answer given was that E can be as large as V^2 in case of complete graph, and hence it is valid to include E in time complexity. But, if V cannot be greater than E+1. So, in that case not having V in the time complexity, should work? 回答1: If it is given that E = kV + c , for some real constants k and c then, O(E + V) = O(kV + c + V) = O(V) = O(E)

问题 I recently started using Haskell and it will probably be for a short while. Just being asked to use it to better understand functional programming for a class I am taking at Uni. Now I have a slight problem I am currently facing with what I am trying to do. I want to build it breadth-first but I think I got my conditions messed up or my conditions are also just wrong. So essentially if I give it [“A1-Gate”, “North-Region”, “South-Region”, “Convention Center”, “Rectorate”, “Academic Building1”

问题 I recently started using Haskell and it will probably be for a short while. Just being asked to use it to better understand functional programming for a class I am taking at Uni. Now I have a slight problem I am currently facing with what I am trying to do. I want to build it breadth-first but I think I got my conditions messed up or my conditions are also just wrong. So essentially if I give it [“A1-Gate”, “North-Region”, “South-Region”, “Convention Center”, “Rectorate”, “Academic Building1”

问题 My task was to find if a path exist from source to destination in a given graph(adjacency matrix). The source is 1 and the destination is 2 and the path can travel only through number 3 in the matrix. I have used BFS to solve this problem but I am getting SIGABRT error. If anyone could please help me, I am still a beginner in coding. I have attached the code here. The question is as follows Given a N X N matrix (M) filled with 1, 0, 2, 3. The task is to find whether there is a path possible

问题 I am currently using networkx library for Python with BFS and DFS. I need to get a tree and then explore it to get a path from a start node to an end node. For the BFS part I am using bfs_successors and it returns an iterator of successors in breadth-first-search from source. For the DFS part I am using: dfs_successors and it returns a dictionary of successors in depth-first-search from source. I need to get a list of nodes from source to end from both the algorithms. Each node is (x, y) and

问题 Say I have the following Haskell tree type, where "State" is a simple wrapper: data Tree a = Branch (State a) [Tree a] | Leaf (State a) deriving (Eq, Show) I also have a function "expand :: Tree a -> Tree a" which takes a leaf node, and expands it into a branch, or takes a branch and returns it unaltered. This tree type represents an N-ary search-tree. Searching depth-first is a waste, as the search-space is obviously infinite, as I can easily keep on expanding the search-space with the use

问题 Studying the Breadth-first search algorithm, I encountered the following pseudo-code: 1 Breadth-First-Search(G, v): 2 3 for each node n in G: 4 n.distance = INFINITY 5 n.parent = NIL 6 7 create empty queue Q 8 9 v.distance = 0 10 Q.enqueue(v) 11 12 while Q is not empty: 13 14 u = Q.dequeue() 15 16 for each node n that is adjacent to u: 17 if n.distance == INFINITY: 18 n.distance = u.distance + 1 19 n.parent = u 20 Q.enqueue(n) My question is regarding line 19 (n.parent = u): How could n's

问题 I wrote some function to search the list of possible paths from a starting node to an ending node. The function list_of_paths correctly returns all the possible paths from a starting point to an ending point but the same path inside the list is repeated even if this has already been found. For example calling the function: list_of_paths 2 7 (List.rev (bfs g1 2)) (node_succ g1) 2 returns: [[2; 3; 6; 7]; [2; 3; 6; 7]; [2; 3; 4; 6; 7]; [2; 3; 6; 7]; [2; 1; 5; 6; 7]; [2; 3; 6; 7]; [2; 1; 5; 6; 7]

问题 I'm trying to build up a program comparing number of strokes of algorithms BFS, DFS, A* (which has two heuristics) on a game of fifteen. My counter count the same result for the two arrays of counters of BFS and DFS and both A*. Yet, I'm using actually using four different arrays from a main (class Project) and I'm assigning four different variables for those strokes. The part of the code that isn't right is, to my mind, a while loop which explores the son of a vertice as far as possible (for