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  • BFS traversal of directed graph from a given node

    - by p1
    Hi, My understanding of basic BFS traversal for a graph is: BFS { Start from any node . Add it to que. Add it to visited array While(que is not empty) { remove head from queue. Print node; add all unvisited direct subchilds to que; mark them as visited } } However, if we have to traverse a DIRECTED graph from a given node and not all nodes are accessible from the given node [directly or indirectly] how do we use BFS for the same. Can you please explain in this graph as well: a= b = d = e = d a= c = d Here if the starting node is b , we never print a and c. Am I missing something in the algorithm. P.S: I used "HashMap adj = new HashMap();" to create the adjacencey list to store graph Any pointers are greatly appreciated. Thanks.

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  • Explain BFS and DFS in terms of backtracking

    - by HH
    Wikipedia about DFS 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 is BFS? "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 continuos backtracking. Regex, find's pruning -- backtracking? The term backtracking confuseses due to its variety of use. UNIX find's pruning an SO-user explained with backtracking. Regex Buddy uses the term "catastrophic backtracking" if you do not limit the scope of your Regexes. It seems to be too wide umbrella-term. So: how do you define "Backtracking" GRAPH-theoretically? what is "backtracking" in BFS and DFS?

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  • Finding all the shortest paths between two nodes in unweighted directed graphs using BFS algorithm

    - by andra-isan
    Hi All, I am working on a problem that I need to find all the shortest path between two nodes in a given directed unweighted graph. I have used BFS algorithm to do the job, but unfortunately I can only print one shortest path not all of them, for example if they are 4 paths having lenght 3, my algorithm only prints the first one but I would like it to print all the four shortest paths. I was wondering in the following code, how should I change it so that all the shortest paths between two nodes could be printed out? class graphNode{ public: int id; string name; bool status; double weight;}; map<int, map<int,graphNode>* > graph; int Graph::BFS(graphNode &v, graphNode &w){ queue <int> q; map <int, int> map1; // this is to check if the node has been visited or not. std::string str= ""; map<int,int> inQ; // just to check that we do not insert the same iterm twice in the queue map <int, map<int, graphNode>* >::iterator pos; pos = graph.find(v.id); if(pos == graph.end()) { cout << v.id << " does not exists in the graph " <<endl; return 1; } int parents[graph.size()+1]; // this vector keeps track of the parents for the node parents[v.id] = -1; // there is a direct path between these two words, simply print that path as the shortest path if (findDirectEdge(v.id,w.id) == 1 ){ cout << " Shortest Path: " << v.id << " -> " << w.id << endl; return 1; } //if else{ int gn; map <int, map<int, graphNode>* >::iterator pos; q.push(v.id); inQ.insert(make_pair(v.id, v.id)); while (!q.empty()){ gn = q.front(); q.pop(); map<int, int>::iterator it; cout << " Popping: " << gn <<endl; map1.insert(make_pair(gn,gn)); //backtracing to print all the nodes if gn is the same as our target node such as w.id if (gn == w.id){ int current = w.id; cout << current << " - > "; while (current!=v.id){ current = parents[current]; cout << current << " -> "; } cout <<endl; } if ((pos = graph.find(gn)) == graph.end()) { cout << " pos is empty " <<endl; continue; } map<int, graphNode>* pn = pos->second; map<int, graphNode>::iterator p = pn->begin(); while(p != pn->end()) { map<int, int>::iterator it; //map1 keeps track of the visited nodes it = map1.find(p->first); graphNode gn1= p->second; if (it== map1.end()) { map<int, int>::iterator it1; //if the node already exits in the inQ, we do not insert it twice it1 = inQ.find(p->first); if (it1== inQ.end()){ parents[p->first] = gn; cout << " inserting " << p->first << " into the queue " <<endl; q.push(p->first); // add it to the queue } //if } //if p++; } //while } //while } I do appreciate all your great help Thanks, Andra

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  • BFS Shortest Path: Edge weight either 1 or 2

    - by Hackster
    I am trying to implement a shortest path algorithm using BFS. That is I am trying to find the shortest path from a specified vertex to every other vertex. However, its a special case where all edge weights are either 1 or 2. I know it could be done with Dijkstra's algorithm but I must use Breadth First Search. So far I have a working version of BFS that searches first for a vertex connected with an edge of weight 1. If it cannot find it, then returns a vertex connected with an edge of weight 2. After thinking about it, this is not the correct way to find the shortest path. The problem is I cannot think of any reasoning why BFS would work with weights 1 or 2, as opposed to any weight. Here is the code: public void addEdge(int start, int end, int weight) { adjMat[start][end] = 1; adjMat[end][start] = 1; edge_weight[start][end] = weight; edge_weight[end][start] = weight; } // ------------------------------------------------------------- public void bfs() // breadth-first search { // begin at vertex 0 vertexList[0].wasVisited = true; // mark it displayVertex(0); // display it theQueue.insert(0); // insert at tail int v2; while( !theQueue.isEmpty() ) // until queue empty, { int v1 = theQueue.remove(); // remove vertex at head // until it has no unvisited neighbors while( (v2=getAdjUnvisitedVertex(v1)) != -1 ){// get one, vertexList[v2].wasVisited = true; // mark it displayVertex(v2); // display it theQueue.insert(v2); // insert it } } // end while(queue not empty) // queue is empty, so we're done for(int j=0; j<nVerts; j++) // reset flags vertexList[j].wasVisited = false; } // end bfs() // ------------------------------------------------------------- // returns an unvisited vertex adj to v -- ****WITH WEIGHT 1**** public int getAdjUnvisitedVertex(int v) { for (int j = 0; j < nVerts; j++) if (adjMat[v][j] == 1 && vertexList[j].wasVisited == false && edge_weight[v][j] == 1){ //System.out.println("Vertex found with 1:"+ vertexList[j].label); return j; } for (int k = 0; k < nVerts; k++) if (adjMat[v][k] == 1 && vertexList[k].wasVisited == false && edge_weight[v][k] == 2){ //System.out.println("Vertex found with 2:"+vertexList[k].label); return k; } return -1; } // end getAdjUnvisitedVertex() // ------------------------------------------------------------- } //////////////////////////////////////////////////////////////// public class BFS{ public static void main(String[] args) { Graph theGraph = new Graph(); theGraph.addVertex('A'); // 0 (start for bfs) theGraph.addVertex('B'); // 1 theGraph.addVertex('C'); // 2 theGraph.addEdge(0, 1,2); // AB theGraph.addEdge(1, 2,1); // BC theGraph.addEdge(2, 0,1); // AD System.out.print("Visits: "); theGraph.bfs(); // breadth-first search System.out.println(); } // end main() } The problem then is, that I don't know why BFS can work for the shortest path problem with edges of weight 1 or 2 as opposed to any edges of any weight. Any help is appreciated. Thanks!

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  • Python networkx DFS or BFS missing?

    - by sadawd
    Dear Everyone I am interested in finding a path (not necessarily shortest) in a short amount of time. Dijsktra and AStar in networkx is taking too long. Why is there no DFS or BFS in networkx? I plan to write my own DFS and BFS search (I am leaning more towards BFS because my graph is pretty deep). Is there anything that I can use in networkx's lib to speed me up? Thx

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  • Difference between Breadth First Search, and Iterative deepening

    - by theraven
    I understand BFS, and DFS, but for the life of me cannot figure out the difference between iterative deepening and BFS. Apparently Iterative deepening has the same memory usage as DFS, but I am unable to see how this is possible, as it just keeps expanding like BFS. If anyone can clarify that would be awesome. tree to work on if required: A / \ B C / / \ D E F

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  • BFS algorithm problem

    - by Gorkamorka
    The problem is as follows: A wanderer begins on the grid coordinates (x,y) and wants to reach the coordinates (0,0). From every gridpoint, the wanderer can go 8 steps north OR 3 steps south OR 5 steps east OR 6 steps west (8N/3S/5E/6W). How can I find the shortest route from (X,Y) to (0,0) using breadth-first search? Clarifications: Unlimited grid Negative coordinates are allowed A queue (linked list or array) must be used No obstacles present

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  • Efficiently finding the shortest path in large graphs

    - by Björn Lindqvist
    I'm looking to find a way to in real-time find the shortest path between nodes in a huge graph. It has hundreds of thousands of vertices and millions of edges. I know this question has been asked before and I guess the answer is to use a breadth-first search, but I'm more interested in to know what software you can use to implement it. For example, it would be totally perfect if it already exist a library (with python bindings!) for performing bfs in undirected graphs.

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  • Gave a talk at SoCal Code Camp at USC today titled “Linq to Objects A-Z”

    - by dotneteer
    I gave a talk at SoCal Code Camp on Linq to Objects. With careful categorization of Linq functions, I was able to cover the entire set of Linq functions in only 35 minutes. I was able to spend the rest time on demos. In my first demo, I show I was able to write a top 20 URL type of query using 4 lines of library code and 9 line of Linq code without tools like Log Parser. I also demonstrated that I only need to change 2 lines of code from querying a single log file to a whole directory of log files. It would be as simple to run the query against multiple servers in parallel. In my second demo, I discussed how to turn into graph depth-first-search (DFS) and breath-first-search (BFS) in the a Linq queryable problem. The class LingToGraph contains the only DFS and BFS code I ever have to write; the rest could be done the the lambda passed to the DFS or BFS calls. In future blogs, I will provide more details explanation of code. Links: Link to Powerpoint slides. Link to demos.

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  • Breadth first search all paths

    - by Amndeep7
    First of all, thank you for looking at this question. For a school assignment we're supposed to create a BFS algorithm and use it to do various things. One of these things is that we're supposed to find all of the paths between the root and the goal nodes of a graph. I have no idea how to do this as I can't find a way to keep track of all of the alternate routes without also including copies/cycles. Here is my BFS code: def makePath(predecessors, last): return makePath(predecessors, predecessors[last]) + [last] if last else [] def BFS1b(node, goal): Q = [node] predecessor = {node:None} while Q: current = Q.pop(0) if current[0] == goal: return makePath(predecessor, goal) for subnode in graph[current[0]][2:]: if subnode[0] not in predecessor: predecessor[subnode[0]] = current[0] Q.append(subnode[0]) A conceptual push in the right direction would be greatly appreciated. tl;dr How do I use BFS to find all of the paths between two nodes?

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  • Algorithm to map an area [on hold]

    - by user37843
    I want to create a crawler that starts in a room and from that room to move North,East,West and South until there aren't any new rooms to visit. I don't want to have duplicates and the output format per line to be something like this: current room, neighbour 1, neighbour 2 ... and in the end to apply BFS algorithm to find the shortest path between 2 rooms. Can anyone offer me some suggestion what to use? Thanks

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  • PCLinuxOS 2010 Release Available for Download

    <b>PCLinuxOS:</b> "PCLinuxOS 2010 Edition is now available for download. Features: Kernel 2.6.32.11-bfs kernel for maximum desktop performance. Full KDE 4.4.2 Desktop. Nvidia and ATI fglrx driver support. Multimedia playback support for many popular formats."

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  • Hopcroft–Karp algorithm in Python

    - by Simon
    I am trying to implement the Hopcroft Karp algorithm in Python using networkx as graph representation. Currently I am as far as this: #Algorithms for bipartite graphs import networkx as nx import collections class HopcroftKarp(object): INFINITY = -1 def __init__(self, G): self.G = G def match(self): self.N1, self.N2 = self.partition() self.pair = {} self.dist = {} self.q = collections.deque() #init for v in self.G: self.pair[v] = None self.dist[v] = HopcroftKarp.INFINITY matching = 0 while self.bfs(): for v in self.N1: if self.pair[v] and self.dfs(v): matching = matching + 1 return matching def dfs(self, v): if v != None: for u in self.G.neighbors_iter(v): if self.dist[ self.pair[u] ] == self.dist[v] + 1 and self.dfs(self.pair[u]): self.pair[u] = v self.pair[v] = u return True self.dist[v] = HopcroftKarp.INFINITY return False return True def bfs(self): for v in self.N1: if self.pair[v] == None: self.dist[v] = 0 self.q.append(v) else: self.dist[v] = HopcroftKarp.INFINITY self.dist[None] = HopcroftKarp.INFINITY while len(self.q) > 0: v = self.q.pop() if v != None: for u in self.G.neighbors_iter(v): if self.dist[ self.pair[u] ] == HopcroftKarp.INFINITY: self.dist[ self.pair[u] ] = self.dist[v] + 1 self.q.append(self.pair[u]) return self.dist[None] != HopcroftKarp.INFINITY def partition(self): return nx.bipartite_sets(self.G) The algorithm is taken from http://en.wikipedia.org/wiki/Hopcroft%E2%80%93Karp_algorithm However it does not work. I use the following test code G = nx.Graph([ (1,"a"), (1,"c"), (2,"a"), (2,"b"), (3,"a"), (3,"c"), (4,"d"), (4,"e"),(4,"f"),(4,"g"), (5,"b"), (5,"c"), (6,"c"), (6,"d") ]) matching = HopcroftKarp(G).match() print matching Unfortunately this does not work, I end up in an endless loop :(. Can someone spot the error, I am out of ideas and I must admit that I have not yet fully understand the algorithm, so it is mostly an implementation of the pseudo code on wikipedia

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  • What are some good, simple examples for queues?

    - by Michael Ekstrand
    I'm teaching CS2 (Java and data structures), and am having some difficulty coming up with good examples to use when teaching queues. The two major applications I use them for are multithreaded message passing (but MT programming is out of scope for the course), and BFS-style algorithms (and I won't be covering graphs until later in the term). I also want to avoid contrived examples. Most things that I think of, if I were actually going to solve them in a single-threaded fashion I would just use a list rather than a queue. I tend to only use queues when processing and discovery are interleaved (e.g. search), or in other special cases like length-limited buffers (e.g. maintaining last N items). To the extent practical, I am trying to teach my students good ways to actually do things in real programs, not just toys to show off a feature. Any suggestions of good, simple algorithms or applications of queues that I can use as examples but that require a minimum of other prior knowledge?

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  • shortest directed odd cycle

    - by gleb-pendler
    6.1.4 Describe an algorithm based on breadth-first search for finding a shortest odd cycle in a graph. 6.3.5 Describe an algorithm based on directed breadth-first search for finding a shortest directed odd cycle in a digraph. what is most importent is that it must be a directed graph not necessary bfs but must be the shortest directed odd cycle!!! Question was taken from "Graph Theory" by J.A. Bondy and U.S.R. Murty thanks in advance!!!

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  • finding shortest valid path in a colored-edge graphs

    - by user1067083
    Given a directed graph G, with edges colored either green or purple, and a vertex S in G, I must find an algorithm that finds the shortest path from s to each vertex in G so the path includes at most two purple edges (and green as much as needed). I thought of BFS on G after removing all the purple edges, and for every vertex that the shortest path is still infinity, do something to try to find it, but I'm kinda stuck, and it takes alot of the running time as well... Any other suggestions? Thanks in advance

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  • count of paths from A[a,b] to A[c,d] without duplicating?

    - by Sorush Rabiee
    I write a sokoban solver for fun and practice, it uses a simple algorithm (something like BFS). now i want to estimate its running time ( O and omega). but i need to know how to calculate count of paths from a vertex to another in a network. each path from a to b is a sequence of edges with no circuit. for example this is a correct path: http://www.imgplace.com/viewimg143/4789/501k.png but this is not: http://www.imgplace.com/viewimg143/6140/202.png

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  • System.InvalidOperationException in Output Window

    - by user318068
    I constantly get the following message in my output/debug windows. The app doesn't crash but I was wondering what the deal with it is: A first chance exception of type 'System.InvalidOperationException' occurred in System.dll my code :sol.cs using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication1 { class Sol { public LinkedList<int> tower1 = new LinkedList<int>(); public LinkedList<int> tower2 = new LinkedList<int>(); public LinkedList<int> tower3 = new LinkedList<int>(); public static LinkedList<string> BFS = new LinkedList<string>(); public static LinkedList<string> DFS = new LinkedList<string>(); public static LinkedList<string> IDS = new LinkedList<string>(); public int depth; public LinkedList<Sol> neighbors; public Sol() { } public Sol(LinkedList<int> tower1, LinkedList<int> tower2, LinkedList<int> tower3) { this.tower1 = tower1; this.tower2 = tower2; this.tower3 = tower3; neighbors = new LinkedList<Sol>(); } public virtual void getneighbors() { Sol temp = this.copy(); Sol neighbor1 = this.copy(); Sol neighbor2 = this.copy(); Sol neighbor3 = this.copy(); Sol neighbor4 = this.copy(); Sol neighbor5 = this.copy(); Sol neighbor6 = this.copy(); if (temp.tower1.Count != 0) { if (neighbor1.tower2.Count != 0) { if (neighbor1.tower1.First.Value < neighbor1.tower2.First.Value) { neighbor1.tower2.AddFirst(neighbor1.tower1.First); neighbor1.tower1.RemoveFirst(); neighbors.AddLast(neighbor1); } } else { neighbor1.tower2.AddFirst(neighbor1.tower1.First); neighbor1.tower1.RemoveFirst(); neighbors.AddLast(neighbor1); } if (neighbor2.tower3.Count != 0) { if (neighbor2.tower1.First.Value < neighbor2.tower3.First.Value) { neighbor2.tower3.AddFirst(neighbor2.tower1.First); neighbor2.tower1.RemoveFirst(); neighbors.AddLast(neighbor2); } } else { neighbor2.tower3.AddFirst(neighbor2.tower1.First); neighbor2.tower1.RemoveFirst(); neighbors.AddLast(neighbor2); } } //------------- if (temp.tower2.Count != 0) { if (neighbor3.tower1.Count != 0) { if (neighbor3.tower2.First.Value < neighbor3.tower1.First.Value) { neighbor3.tower1.AddFirst(neighbor3.tower2.First); neighbor3.tower2.RemoveFirst(); neighbors.AddLast(neighbor3); } } else { neighbor3.tower1.AddFirst(neighbor3.tower2.First); neighbor3.tower2.RemoveFirst(); neighbors.AddLast(neighbor3); } if (neighbor4.tower3.Count != 0) { if (neighbor4.tower2.First.Value < neighbor4.tower3.First.Value) { neighbor4.tower3.AddFirst(neighbor4.tower2.First); neighbor4.tower2.RemoveFirst(); neighbors.AddLast(neighbor4); } } else { neighbor4.tower3.AddFirst(neighbor4.tower2.First); neighbor4.tower2.RemoveFirst(); neighbors.AddLast(neighbor4); } } //------------------------ if (temp.tower3.Count() != 0) { if (neighbor5.tower1.Count() != 0) { if (neighbor5.tower3.ElementAtOrDefault(0) < neighbor5.tower1.ElementAtOrDefault(0)) { neighbor5.tower1.AddFirst(neighbor5.tower3.First); neighbor5.tower3.RemoveFirst(); neighbors.AddLast(neighbor5); } } else { neighbor5.tower1.AddFirst(neighbor5.tower3.First); neighbor5.tower3.RemoveFirst(); neighbors.AddLast(neighbor5); } if (neighbor6.tower2.Count() != 0) { if (neighbor6.tower3.ElementAtOrDefault(0) < neighbor6.tower2.ElementAtOrDefault(0)) { neighbor6.tower2.AddFirst(neighbor6.tower3.First); neighbor6.tower3.RemoveFirst(); neighbors.AddLast(neighbor6); } } else { neighbor6.tower2.AddFirst(neighbor6.tower3.First); neighbor6.tower3.RemoveFirst(); neighbors.AddLast(neighbor6); } } } public override string ToString() { string str; str = "tower1" + tower1.ToString() + " tower2" + tower2.ToString() + " tower3" + tower3.ToString(); return str; } public Sol copy() { Sol So; LinkedList<int> l1 = new LinkedList<int>(); LinkedList<int> l2 = new LinkedList<int>(); LinkedList<int> l3 = new LinkedList<int>(); for (int i = 0; i <= this.tower1.Count() - 1; i++) { l1.AddLast(tower1.ElementAt(i)); } for (int i = 0; i <= this.tower2.Count - 1; i++) { l2.AddLast(tower2.ElementAt(i)); } for (int i = 0; i <= this.tower3.Count - 1; i++) { l3.AddLast(tower3.ElementAt(i)); } So = new Sol(l1, l2, l3); return So; } public bool Equals(Sol sol) { if (this.tower1.Equals(sol.tower1) & this.tower2.Equals(sol.tower2) & this.tower3.Equals(sol.tower3)) return true; return false; } public virtual bool containedin(Stack<Sol> vec) { bool found = false; for (int i = 0; i <= vec.Count - 1; i++) { if (vec.ElementAt(i).tower1.Equals(this.tower1) && vec.ElementAt(i).tower2.Equals(this.tower2) && vec.ElementAt(i).tower3.Equals(this.tower3)) { found = true; break; } } return found; } public virtual bool breadthFirst(Sol start, Sol goal) { Stack<Sol> nextStack = new Stack<Sol>(); Stack<Sol> traversed = new Stack<Sol>(); bool found = false; start.depth = 0; nextStack.Push(start); while (nextStack.Count != 0) { Sol sol = nextStack.Pop(); BFS.AddFirst("poped State:" + sol.ToString() + "level " + sol.depth); traversed.Push(sol); if (sol.Equals(goal)) { found = true; BFS.AddFirst("Goal:" + sol.ToString()); break; } else { sol.getneighbors(); foreach (Sol neighbor in sol.neighbors) { if (!neighbor.containedin(traversed) && !neighbor.containedin(nextStack)) { neighbor.depth = (sol.depth + 1); nextStack.Push(neighbor); } } } } return found; } public virtual bool depthFirst(Sol start, Sol goal) { Stack<Sol> nextStack = new Stack<Sol>(); Stack<Sol> traversed = new Stack<Sol>(); bool found = false; start.depth = 0; nextStack.Push(start); while (nextStack.Count != 0) { //Dequeue next State for comparison //And add it 2 list of traversed States Sol sol = nextStack.Pop(); DFS.AddFirst("poped State:" + sol.ToString() + "level " + sol.depth); traversed.Push(sol); if (sol.Equals(goal)) { found = true; DFS.AddFirst("Goal:" + sol.ToString()); break; } else { sol.getneighbors(); foreach (Sol neighbor in sol.neighbors) { if (!neighbor.containedin(traversed) && !neighbor.containedin(nextStack)) { neighbor.depth = sol.depth + 1; nextStack.Push(neighbor); } } } } return found; } public virtual bool iterativedeepening(Sol start, Sol goal) { bool found = false; for (int level = 0; ; level++) { Stack<Sol> nextStack = new Stack<Sol>(); Stack<Sol> traversed = new Stack<Sol>(); start.depth = 0; nextStack.Push(start); while (nextStack.Count != 0) { Sol sol = nextStack.Pop(); IDS.AddFirst("poped State:" + sol.ToString() + "Level" + sol.depth); traversed.Push(sol); if (sol.Equals(goal)) { found = true; IDS.AddFirst("Goal:" + sol.ToString()); break; } else if (sol.depth < level) { sol.getneighbors(); foreach (Sol neighbor in sol.neighbors) { if (!neighbor.containedin(traversed) && !neighbor.containedin(nextStack)) { neighbor.depth = sol.depth + 1; nextStack.Push(neighbor); } //end if } //end for each } //end else if } // end while if (found == true) break; } // end for return found; } } } Just wondering if I may be doing something wrong somewhere or something.

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