java,binary-search-tree,breadth-first-search

Your traversal function is correct. You may want to check this online tool https://www.cs.usfca.edu/~galles/visualization/BST.html It provides visualization of the insert, delete and find process as well. This is the resulting tree: ...

search,graph,depth-first-search,breadth-first-search

And how would you define "best" ? If you know that the goal node is at depth n from the root node (the node from which you begin the search), BFS - will ensure that the search won't iterate nodes with depth > n. That said, DFS might still "choose"...

algorithm,search,graph-theory,graph-algorithm,breadth-first-search

The Example Let G = (V,E) a graph with V = ℕ ∪ {-1, 0} and E = { {-1,t}, {t,0} | t ∈ ℕ } and let s = -1 and f = 0. There exist an infinite number of paths of length 2 from s to f, but...

php,tree,binary-tree,breadth-first-search

I wrote BFS search function myself. I used PHP's SplQueue global variable to store array of children. Private $queue = []; function for BFS public function traverseTree($rootNode, $dummyQueue) { if($rootNode->lft != 0) { $dummyQueue->enqueue($rootNode->lft); } if($rootNode->rgt != 0) { $dummyQueue->enqueue($rootNode->rgt); } if(!($dummyQueue->isEmpty())){ $nextId = $dummyQueue->dequeue(); $nextNode = //get next node...

c++,performance,graph,breadth-first-search,undirected-graph

An edge whose deletion disconnects two connected components is called a bridge and there are linear-time algorithms for finding all the bridges in a graph (usually based on depth-first search). The Wikipedia article lists one of them (due to Tarjan) as an example. This paper also gives a simple algorithm...

algorithm,binary-search-tree,graph-theory,breadth-first-search

For a 'proper' tree (see below), it's the same thing, at least by most definitions. Like Wikipedia, for example: Breadth-first See also: Breadth-first search Trees can also be traversed in level-order, ... ... a breadth-first (level-order) traversal ... Well, at least level-order traversal is the same as breadth-first traversal. There...

algorithm,time,breadth-first-search

Well, since it's a worst case your analysis is correct but not tight. The while loop runs as long as there are vertices in the queue. Only vertices whose color is WHITE are enqueued in which case their color becomes GRAY so they will never be enqueued again. This tells...

python,tree,artificial-intelligence,breadth-first-search

Hmm, BFS is not really designed to have multiple target endings. You couldn't really call it BFS after the monkeying around. If you're really interested in that you could always draw it out and see what happens if you follow the algorithm mentally. But you'd need to keep track of...

python,python-3.x,artificial-intelligence,breadth-first-search,water-jug-problem

I've made a smaller code example that reproduces your error: J1SIZE = 3 J2SIZE = 4 def fillJug(tempNode, jug): copyNode = tempNode copyNode = copyNode if jug == 1: copyNode[0] = J1SIZE elif jug == 2: copyNode[1] = J2SIZE return copyNode frontier = [[0, 0]] tempNode = frontier.pop(0) newNode =...

tree,binary-search-tree,graph-algorithm,depth-first-search,breadth-first-search

The problem you're trying to solve is called the maximum independent set problem applied to trees. As a hint, think about the following: For each complete subtree of the tree, think about the optimal solution for that subtree in two cases - the case where you include that node in...

time-complexity,shortest-path,breadth-first-search

Your analysis seems correct. Please refer to the following references. http://axon.cs.byu.edu/~martinez/classes/312/Slides/Paths.pdf https://courses.engr.illinois.edu/cs473/sp2011/lectures/03_class.pdf...

algorithm,graph,dijkstra,shortest-path,breadth-first-search

This is laborous case work: A < B and 0 and N-1 are joined by A -> trivial. B < A and 0 and N-1 are joined by B -> trivial. B < A and 0 and N-1 are joined by A -> Do BFS on graph with only K...

python,set,cluster-computing,depth-first-search,breadth-first-search

It seems from your example that for any X and Y, if X in graph[Y] then Y in graph[X}. If that's the case, then: def addNode(name, neighbors, graph, clusterlist): graph[name] = neighbors clusters = set() for n in neighbors: graph[n].add(name) for c in clusterlist: if n in c: c.add(name) clusters.add(frozenset(c))...

algorithm,matlab,matrix,breadth-first-search

You can use open and closed list like in the a star pathfinding algorithm. Check the all neighbours and if the neighbour is not a obstacle, put in the open list. Examine all neighbours and the neighbour which has a min cost, put it in the closed list. In the...

algorithm,graph,shortest-path,depth-first-search,breadth-first-search

If anyone is interested in how I solved this here is my solution: def dfs(self, v): if v.successor is None: # this is the root node v.cost = 0 v.visited = True for adj in v.adj: if adj.visited is False: # set the parent adj.successor = v adj.cost = v.cost...

java,algorithm,graph,breadth-first-search

I think your problem is how you build the edge list using the previous map. You store the last seen edge when queuing nodes, but this edge may not be lie on the shortest path. You check for dest when you pull it from the queue, but the edge stored...

algorithm,graph,tree,breadth-first-search

One standard way to get a BFS tree from a graph is to run BFS and, as you do so, keep a table mapping each node in the graph to its parent in the tree. You populate it as follows: the source node has no parent (it's the root, after...

c++,iterator,breadth-first-search

You are using a list in your struct containing copies of each child node. You need to be using pointers. Try using a current node pointer to hop over each child node: Node* currNode; std::list<node*> neighbors;...

algorithm,breadth-first-search,iterative-deepening

Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b : https://mhesham.wordpress.com/tag/depth-first-search/ Iterative Deepening DFS worst case # of nodes: N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. +...

You will need to distinguish three states that a node can be in: unprocessed (not yet seen) discovered (queued) processed (traversed, outputted) With the visited boolean array you might either tag nodes when they are discovered or when they have been traversed. While the term "visited" usually refers to the...

binary-search-tree,pseudocode,breadth-first-search

The code you posted is pseudo code, and is not intended to be valid C++. In C++, the assignment operator is =. In other languages such as Ada, BCPL, Cecil, Dylan, E, Eiffel, Maple, Mathematica, Modula-3, Pascal, Pliant, Sather, Simula, Smalltalk, SML, the assignment operator is :=. GNU make also...

java,algorithm,graph,breadth-first-search,adjacency-matrix

You hit the condition temp == end at the first iteration because your temp starts from 0, then it's 1 without inserting element to queue(because it's not adjacent to 0), then it's 2 without inserting, and here you are, under condition temp == end (both equal 2) you return Q...

c++,c,algorithm,depth-first-search,breadth-first-search

You can implement adjacency list with vectors like this, it is much easier than using pointers. Check the code, it is also much easier to understand how it works. #include <bits/stdc++.h> using namespace std; vector<int> edges[5]; bool visited[5]; void dfs(int x) { visited[x] = true; for(int i=0; i < edges[x].size();...

perl,dependencies,breadth-first-search

There's always the brute force method. I'll let someone else come up with something clever: use strict; use warnings; my @input = ("a:d,e,f", "b:c,d", "c:f", "d:e"); my %children; my %parents; for (@input) { my ($parent, @kids) = split /[:,]/; for (@kids) { $children{$parent}{$_}++; $children{$_} ||= {}; push @{$parents{$_}}, $parent; }...

graph,cycle,breadth-first-search,bfs,undirected-graph

Here below you will find the code to traverse a graph using BFS and find its cycles. #include <stdio.h> #include <queue> using namespace std; int nodes, edges, src; int graph[100][100], color[100], prev[100]; const int WHITE = 0; const int GRAY = 1; const int BLACK = 2; void print(int); int...