breadth-first-search

I am making a program that solves a puzzle game, and it finds all the possible moves on a board and puts all the possible resulting boards in an object. Then it finds all the possible moves for the resulting boards, and so on. The object will look something like this: { "board": { "starts": [[0,0],[0,3]], "blocks": [[3,0],[3,3]], "ends": [[2,4]] }, "possibleMoves": [ { "board": { "starts": [[0,0],[2,3]], "blocks": [[3,0],[3,3]], "ends": [[2,4]] }, "possibleMoves":[ { "board": {}, "possibleMoves": [{}] } ] }, { "board": { "starts": [[0,3]], "blocks": [[3,0],[3,3]], "ends": [[2,4]] },

How do you use a Bidirectional BFS to find the shortest path? Let's say there is a 6x6 grid. The start point is in (0,5) and the end point is in (4,1). What is the shortest path using bidirectional bfs? There are no path costs. And it is undirected. amit How does Bi-directional BFS work? Simultaneously run two BFS's from both source and target vertices, terminating once a vertex common to both runs is discovered. This vertex will be halfway between the source and the target. Why is it better than BFS? Bi-directional BFS will yield much better results than simple BFS in most cases. Assume the

The site http://web.eecs.utk.edu/~huangj/CS302S04/notes/graph-searching.html describes that when an adjacency list is used then, DFS and BFS have complexity O(V+E), and if an adjacency matrix is used, the complexity is O(V 2 ). Why is this? In both cases, the runtime depends on how long it takes to iterate across the outgoing edges of a given node. With an adjacency list, the runtime is directly proportional to the number of outgoing edges. Since each node is visited once, the cost is the number of nodes plus the number of edges, which is O(m + n). With am adjacency matrix, the time required

Why are the running times of BFS and DFS O(V+E), especially when there is a node that has a directed edge to a node that can be reached from the vertex, like in this example in the following site http://www.personal.kent.edu/~rmuhamma/Algorithms/MyAlgorithms/GraphAlgor/depthSearch.htm E is the set of all edges in the graph, as G={V,E}. So, |E| is count of all edges in the graph. This alone should be enough to convince you that the overall complexity can't be |V| times |E|, since we are not iterating over all the edges in the graph for each vertex. In the adjacency list representation, for each

Wikipedia about Depth First Search: Depth-first search (DFS) is an algorithm for traversing or searching a tree, tree structure, or graph. One starts at the root (selecting some node as the root in the graph case) and explores as far as possible along each branch before backtracking. So what is Breadth First Search? "an algorithm that choose a starting node, checks all nodes backtracks , chooses the shortest path, chose neighbour nodes backtracks , chose the shortest path, finally finds the optimal path because of traversing each path due to continuous backtracking. Regex find 's pruning --

Time complexity to go over each adjacent edges of a vertex is say O(N) , where N is number of adjacent edges. So for V number of vertices time complexity becomes O(V*N) = O(E) , where E is the total number of edges in the graph. Since removing and adding a vertex from/to Queue is O(1) , why it is added to the overall time complexity of BFS as O(V+E) . Considering the following Graph we see how the time complexity is O(|V|+|E|) but not O(V*E). Adjacency List V E v0:{v1,v2} v1:{v3} v2:{v3} v3:{} Operating How BFS Works Step by Step Step1: Adjacency lists: V E v0: {v1,v2} mark, enqueue v0 v1: {v3

I am bit confused after reading examples at DFS where output is printed in different fashion for both DFS approaches though both are said to be DFS. In fact DFS recursion approach print the output just in same fashion as BFS . Then what's the difference ? Is recursion approach given here not an example of DFS ? Using Stack // Iterative DFS using stack public void dfsUsingStack(Node node) { Stack<Node> stack=new Stack<Node>(); stack.add(node); node.visited=true; while (!stack.isEmpty()) { Node element=stack.pop(); System.out.print(element.data + " "); List<Node> neighbours=element.getNeighbours