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  • Graph Isomorphism > What kind of Graph is this?

    - by oodavid
    Essentially, this is a variation of Comparing Two Tree Structures, however I do not have "trees", but rather another type of graph. I need to know what kind of Graph I have in order to figure out if there's a Graph Isomorphism Special Case... As you can see, they are: Not Directed Not A Tree Cyclic Max 4 connections But I still don't know the correct terminology, nor the which Isomorphism algorithm to pursue, guidance appreciated.

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  • Finding most Important Node(s) in a Directed Graph

    - by Srikar Appal
    I have a large (˜ 20 million nodes) directed Graph with in-edges & out-edges. I want to figure out which parts of of the graph deserve the most attention. Often most of the graph is boring, or at least it is already well understood. The way I am defining "attention" is by the concept of "connectedness" i.e. How can i find the most connected node(s) in the graph? In what follows, One can assume that nodes by themselves have no score, the edges have no weight & they are either connected or not. This website suggest some pretty complicated procedures like n-dimensional space, Eigen Vectors, graph centrality concepts, pageRank etc. Is this problem that complex? Can I not do a simple Breadth-First Traversal of the entire graph where at each node I figure out a way to find the number of in-edges. The node with most in-edges is the most important node in the graph. Am I missing something here?

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  • How to use Boost 1.41.0 graph layout algorithmes

    - by daniil-k
    Hi I have problem using boost graph layout algorithmes. boost verision 1_41_0 mingw g++ 4.4.0. So there are issues I have encountered Can you suggest me with them? The function fruchterman_reingold_force_directed_layout isn't compiled. The kamada_kawai_spring_layout compiled but program crashed. Boost documentation to layout algorithms is wrong, sample to fruchterman_reingold_force_directed_layout isn't compiled. This is my example. To use function just uncomment one. String 60, 61, 63. #include <boost/config.hpp> #include <boost/graph/adjacency_list.hpp> #include <boost/graph/graph_utility.hpp> #include <boost/graph/simple_point.hpp> #include <boost/property_map/property_map.hpp> #include <boost/graph/circle_layout.hpp> #include <boost/graph/fruchterman_reingold.hpp> #include <boost/graph/kamada_kawai_spring_layout.hpp> #include <iostream> //typedef boost::square_topology<>::point_difference_type Point; typedef boost::square_topology<>::point_type Point; struct VertexProperties { std::size_t index; Point point; }; struct EdgeProperty { EdgeProperty(const std::size_t &w):weight(w) {} double weight; }; typedef boost::adjacency_list<boost::listS, boost::listS, boost::undirectedS, VertexProperties, EdgeProperty > Graph; typedef boost::property_map<Graph, std::size_t VertexProperties::*>::type VertexIndexPropertyMap; typedef boost::property_map<Graph, Point VertexProperties::*>::type PositionMap; typedef boost::property_map<Graph, double EdgeProperty::*>::type WeightPropertyMap; typedef boost::graph_traits<Graph>::vertex_descriptor VirtexDescriptor; int main() { Graph graph; VertexIndexPropertyMap vertexIdPropertyMap = boost::get(&VertexProperties::index, graph); for (int i = 0; i < 3; ++i) { VirtexDescriptor vd = boost::add_vertex(graph); vertexIdPropertyMap[vd] = i + 2; } boost::add_edge(boost::vertex(1, graph), boost::vertex(0, graph), EdgeProperty(5), graph); boost::add_edge(boost::vertex(2, graph), boost::vertex(0, graph), EdgeProperty(5), graph); std::cout << "Vertices\n"; boost::print_vertices(graph, vertexIdPropertyMap); std::cout << "Edges\n"; boost::print_edges(graph, vertexIdPropertyMap); PositionMap positionMap = boost::get(&VertexProperties::point, graph); WeightPropertyMap weightPropertyMap = boost::get(&EdgeProperty::weight, graph); boost::circle_graph_layout(graph, positionMap, 100); // boost::fruchterman_reingold_force_directed_layout(graph, positionMap, boost::square_topology<>()); boost::kamada_kawai_spring_layout(graph, positionMap, weightPropertyMap, boost::square_topology<>(), boost::side_length<double>(10), boost::layout_tolerance<>(), 1, vertexIdPropertyMap); std::cout << "Coordinates\n"; boost::graph_traits<Graph>::vertex_iterator i, end; for (boost::tie(i, end) = boost::vertices(graph); i != end; ++i) { std::cout << "ID: (" << vertexIdPropertyMap[*i] << ") x: " << positionMap[*i][0] << " y: " << positionMap[*i][1] << "\n"; } return 0; }

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  • Class instance clustering in object reference graph for multi-entries serialization

    - by Juh_
    My question is on the best way to cluster a graph of class instances (i.e. objects, the graph nodes) linked by object references (the -directed- edges of the graph) around specifically marked objects. To explain better my question, let me explain my motivation: I currently use a moderately complex system to serialize the data used in my projects: "marked" objects have a specific attributes which stores a "saving entry": the path to an associated file on disc (but it could be done for any storage type providing the suitable interface) Those object can then be serialized automatically (eg: obj.save()) The serialization of a marked object 'a' contains implicitly all objects 'b' for which 'a' has a reference to, directly s.t: a.b = b, or indirectly s.t.: a.c.b = b for some object 'c' This is very simple and basically define specific storage entries to specific objects. I have then "container" type objects that: can be serialized similarly (in fact their are or can-be "marked") they don't serialize in their storage entries the "marked" objects (with direct reference): if a and a.b are both marked, a.save() calls b.save() and stores a.b = storage_entry(b) So, if I serialize 'a', it will serialize automatically all objects that can be reached from 'a' through the object reference graph, possibly in multiples entries. That is what I want, and is usually provides the functionalities I need. However, it is very ad-hoc and there are some structural limitations to this approach: the multi-entry saving can only works through direct connections in "container" objects, and there are situations with undefined behavior such as if two "marked" objects 'a'and 'b' both have a reference to an unmarked object 'c'. In this case my system will stores 'c' in both 'a' and 'b' making an implicit copy which not only double the storage size, but also change the object reference graph after re-loading. I am thinking of generalizing the process. Apart for the practical questions on implementation (I am coding in python, and use Pickle to serialize my objects), there is a general question on the way to attach (cluster) unmarked objects to marked ones. So, my questions are: What are the important issues that should be considered? Basically why not just use any graph parsing algorithm with the "attach to last marked node" behavior. Is there any work done on this problem, practical or theoretical, that I should be aware of? Note: I added the tag graph-database because I think the answer might come from that fields, even if the question is not.

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  • Is finding graph minors without single node pinch points possible?

    - by Alturis
    Is it possible to robustly find all the graph minors within an arbitrary node graph where the pinch points are generally not single nodes? I have read some other posts on here about how to break up your graph into a Hamiltonian cycle and then from that find the graph minors but it seems to be such an algorithm would require that each "room" had "doorways" consisting of single nodes. To explain a bit more a visual aid is necessary. Lets say the nodes below are an example of the typical node graph. What I am looking for is a way to automatically find the different colored regions of the graph (or graph minors)

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  • Best Planar graph program

    - by brian
    In graph theory, a planar graph is a graph that can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their endpoints. What is the best open source program for drawing the planar graph with support of input nodes size and fixed drawing boundary region

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  • Understanding and Implementing a Force based graph layout algorithm

    - by zcourts
    I'm trying to implement a force base graph layout algorithm, based on http://en.wikipedia.org/wiki/Force-based_algorithms_(graph_drawing) My first attempt didn't work so I looked at http://blog.ivank.net/force-based-graph-drawing-in-javascript.html and https://github.com/dhotson/springy I changed my implementation based on what I thought I understood from those two but I haven't managed to get it right and I'm hoping someone can help? JavaScript isn't my strong point so be gentle... If you're wondering why write my own. In reality I have no real reason to write my own I'm just trying to understand how the algorithm is implemented. Especially in my first link, that demo is brilliant. This is what I've come up with //support function.bind - https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Function/bind#Compatibility if (!Function.prototype.bind) { Function.prototype.bind = function (oThis) { if (typeof this !== "function") { // closest thing possible to the ECMAScript 5 internal IsCallable function throw new TypeError("Function.prototype.bind - what is trying to be bound is not callable"); } var aArgs = Array.prototype.slice.call(arguments, 1), fToBind = this, fNOP = function () {}, fBound = function () { return fToBind.apply(this instanceof fNOP ? this : oThis || window, aArgs.concat(Array.prototype.slice.call(arguments))); }; fNOP.prototype = this.prototype; fBound.prototype = new fNOP(); return fBound; }; } (function() { var lastTime = 0; var vendors = ['ms', 'moz', 'webkit', 'o']; for(var x = 0; x < vendors.length && !window.requestAnimationFrame; ++x) { window.requestAnimationFrame = window[vendors[x]+'RequestAnimationFrame']; window.cancelAnimationFrame = window[vendors[x]+'CancelAnimationFrame'] || window[vendors[x]+'CancelRequestAnimationFrame']; } if (!window.requestAnimationFrame) window.requestAnimationFrame = function(callback, element) { var currTime = new Date().getTime(); var timeToCall = Math.max(0, 16 - (currTime - lastTime)); var id = window.setTimeout(function() { callback(currTime + timeToCall); }, timeToCall); lastTime = currTime + timeToCall; return id; }; if (!window.cancelAnimationFrame) window.cancelAnimationFrame = function(id) { clearTimeout(id); }; }()); function Graph(o){ this.options=o; this.vertices={}; this.edges={};//form {vertexID:{edgeID:edge}} } /** *Adds an edge to the graph. If the verticies in this edge are not already in the *graph then they are added */ Graph.prototype.addEdge=function(e){ //if vertex1 and vertex2 doesn't exist in this.vertices add them if(typeof(this.vertices[e.vertex1])==='undefined') this.vertices[e.vertex1]=new Vertex(e.vertex1); if(typeof(this.vertices[e.vertex2])==='undefined') this.vertices[e.vertex2]=new Vertex(e.vertex2); //add the edge if(typeof(this.edges[e.vertex1])==='undefined') this.edges[e.vertex1]={}; this.edges[e.vertex1][e.id]=e; } /** * Add a vertex to the graph. If a vertex with the same ID already exists then * the existing vertex's .data property is replaced with the @param v.data */ Graph.prototype.addVertex=function(v){ if(typeof(this.vertices[v.id])==='undefined') this.vertices[v.id]=v; else this.vertices[v.id].data=v.data; } function Vertex(id,data){ this.id=id; this.data=data?data:{}; //initialize to data.[x|y|z] or generate random number for each this.x = this.data.x?this.data.x:-100 + Math.random()*200; this.y = this.data.y?this.data.y:-100 + Math.random()*200; this.z = this.data.y?this.data.y:-100 + Math.random()*200; //set initial velocity to 0 this.velocity = new Point(0, 0, 0); this.mass=this.data.mass?this.data.mass:Math.random(); this.force=new Point(0,0,0); } function Edge(vertex1ID,vertex2ID){ vertex1ID=vertex1ID?vertex1ID:Math.random() vertex2ID=vertex2ID?vertex2ID:Math.random() this.id=vertex1ID+"->"+vertex2ID; this.vertex1=vertex1ID; this.vertex2=vertex2ID; } function Point(x, y, z) { this.x = x; this.y = y; this.z = z; } Point.prototype.plus=function(p){ this.x +=p.x this.y +=p.y this.z +=p.z } function ForceLayout(o){ this.repulsion = o.repulsion?o.repulsion:200; this.attraction = o.attraction?o.attraction:0.06; this.damping = o.damping?o.damping:0.9; this.graph = o.graph?o.graph:new Graph(); this.total_kinetic_energy =0; this.animationID=-1; } ForceLayout.prototype.draw=function(){ //vertex velocities initialized to (0,0,0) when a vertex is created //vertex positions initialized to random position when created cc=0; do{ this.total_kinetic_energy =0; //for each vertex for(var i in this.graph.vertices){ var thisNode=this.graph.vertices[i]; // running sum of total force on this particular node var netForce=new Point(0,0,0) //for each other node for(var j in this.graph.vertices){ if(thisNode!=this.graph.vertices[j]){ //net-force := net-force + Coulomb_repulsion( this_node, other_node ) netForce.plus(this.CoulombRepulsion( thisNode,this.graph.vertices[j])) } } //for each spring connected to this node for(var k in this.graph.edges[thisNode.id]){ //(this node, node its connected to) //pass id of this node and the node its connected to so hookesattraction //can update the force on both vertices and return that force to be //added to the net force this.HookesAttraction(thisNode.id, this.graph.edges[thisNode.id][k].vertex2 ) } // without damping, it moves forever // this_node.velocity := (this_node.velocity + timestep * net-force) * damping thisNode.velocity.x=(thisNode.velocity.x+thisNode.force.x)*this.damping; thisNode.velocity.y=(thisNode.velocity.y+thisNode.force.y)*this.damping; thisNode.velocity.z=(thisNode.velocity.z+thisNode.force.z)*this.damping; //this_node.position := this_node.position + timestep * this_node.velocity thisNode.x=thisNode.velocity.x; thisNode.y=thisNode.velocity.y; thisNode.z=thisNode.velocity.z; //normalize x,y,z??? //total_kinetic_energy := total_kinetic_energy + this_node.mass * (this_node.velocity)^2 this.total_kinetic_energy +=thisNode.mass*((thisNode.velocity.x+thisNode.velocity.y+thisNode.velocity.z)* (thisNode.velocity.x+thisNode.velocity.y+thisNode.velocity.z)) } cc+=1; }while(this.total_kinetic_energy >0.5) console.log(cc,this.total_kinetic_energy,this.graph) this.cancelAnimation(); } ForceLayout.prototype.HookesAttraction=function(v1ID,v2ID){ var a=this.graph.vertices[v1ID] var b=this.graph.vertices[v2ID] var force=new Point(this.attraction*(b.x - a.x),this.attraction*(b.y - a.y),this.attraction*(b.z - a.z)) // hook's attraction a.force.x += force.x; a.force.y += force.y; a.force.z += force.z; b.force.x += this.attraction*(a.x - b.x); b.force.y += this.attraction*(a.y - b.y); b.force.z += this.attraction*(a.z - b.z); return force; } ForceLayout.prototype.CoulombRepulsion=function(vertex1,vertex2){ //http://en.wikipedia.org/wiki/Coulomb's_law // distance squared = ((x1-x2)*(x1-x2)) + ((y1-y2)*(y1-y2)) + ((z1-z2)*(z1-z2)) var distanceSquared = ( (vertex1.x-vertex2.x)*(vertex1.x-vertex2.x)+ (vertex1.y-vertex2.y)*(vertex1.y-vertex2.y)+ (vertex1.z-vertex2.z)*(vertex1.z-vertex2.z) ); if(distanceSquared==0) distanceSquared = 0.001; var coul = this.repulsion / distanceSquared; return new Point(coul * (vertex1.x-vertex2.x),coul * (vertex1.y-vertex2.y), coul * (vertex1.z-vertex2.z)); } ForceLayout.prototype.animate=function(){ if(this.animating) this.animationID=requestAnimationFrame(this.animate.bind(this)); this.draw(); } ForceLayout.prototype.cancelAnimation=function(){ cancelAnimationFrame(this.animationID); this.animating=false; } ForceLayout.prototype.redraw=function(){ this.animating=true; this.animate(); } $(document).ready(function(){ var g= new Graph(); for(var i=0;i<=100;i++){ var v1=new Vertex(Math.random(), {}) var v2=new Vertex(Math.random(), {}) var e1= new Edge(v1.id,v2.id); g.addEdge(e1); } console.log(g); var l=new ForceLayout({ graph:g }); l.redraw(); });

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  • Need some help understanding this problem about maximizing graph connectivity

    - by Legend
    I was wondering if someone could help me understand this problem. I prepared a small diagram because it is much easier to explain it visually. Problem I am trying to solve: 1. Constructing the dependency graph Given the connectivity of the graph and a metric that determines how well a node depends on the other, order the dependencies. For instance, I could put in a few rules saying that node 3 depends on node 4 node 2 depends on node 3 node 3 depends on node 5 But because the final rule is not "valuable" (again based on the same metric), I will not add the rule to my system. 2. Execute the request order Once I built a dependency graph, execute the list in an order that maximizes the final connectivity. I am not sure if this is a really a problem but I somehow have a feeling that there might exist more than one order in which case, it is required to choose the best order. First and foremost, I am wondering if I constructed the problem correctly and if I should be aware of any corner cases. Secondly, is there a closely related algorithm that I can look at? Currently, I am thinking of something like Feedback Arc Set or the Secretary Problem but I am a little confused at the moment. Any suggestions? PS: I am a little confused about the problem myself so please don't flame on me for that. If any clarifications are needed, I will try to update the question.

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  • How to spread changes in oriented graph?

    - by joseph
    Hello. I have oriented graph. Graph can be strongly connected. Every vertix can have a set of anything, for example letters. The set is user editable. Every vertix makes intersection of sets in previous vertices (only one step back). But now, there is problem: When I update set of one vertex, the change should expand to all vertices and uptate their intersection of sets of previous vertices. How to do every vertex have correct intersection after update of any vertex? Restriction: algorithm must avoid to stick in infinity. Any idea how to solve it?

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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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  • box stacking in graph theory

    - by mozhdeh
    Please help me find a good solution for this problem. We have n boxes with 3 dimensions. We can orient them and we want to put them on top of another to have a maximun height. We can put a box on top of an other box, if 2 dimensions (width and lenght) are lower than the dimensions of the box below. For exapmle we have 3 dimensions w*D*h, we can show it in to (h*d,d*h,w*d,d*W,h*w,w*h) please help me to solve it in graph theory.

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  • Graph colouring algorithm: typical scheduling problem

    - by newba
    Hi, I'm training code problems like UvA and I have this one in which I have to, given a set of n exams and k students enrolled in the exams, find whether it is possible to schedule all exams in two time slots. Input Several test cases. Each one starts with a line containing 1 < n < 200 of different examinations to be scheduled. The 2nd line has the number of cases k in which there exist at least 1 student enrolled in 2 examinations. Then, k lines will follow, each containing 2 numbers that specify the pair of examinations for each case above. (An input with n = 0 will means end of the input and is not to be processed). Output: You have to decide whether the examination plan is possible or not for 2 time slots. Example: Input: 3 3 0 1 1 2 2 0 9 8 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 Ouput: NOT POSSIBLE. POSSIBLE. I think the general approach is graph colouring, but I'm really a newb and I may confess that I had some trouble understanding the problem. Anyway, I'm trying to do it and then submit it. Could someone please help me doing some code for this problem? I will have to handle and understand this algo now in order to use it later, over and over. I prefer C or C++, but if you want, Java is fine to me ;) Thanks in advance

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  • evaluating a code of a graph [migrated]

    - by mazen.r.f
    This is relatively a long code,if you have the tolerance and the will to find out how to make this code work then take a look please, i will appreciate your feed back. i have spent two days trying to come up with a code to represent a graph , then calculate the shortest path using dijkastra algorithm , but i am not able to get the right result , even the code runs without errors , but the result is not correct , always i am getting 0. briefly,i have three classes , Vertex, Edge, Graph , the Vertex class represents the nodes in the graph and it has id and carried ( which carry the weight of the links connected to it while using dijkastra algorithm ) and a vector of the ids belong to other nodes the path will go through before arriving to the node itself , this vector is named previous_nodes. the Edge class represents the edges in the graph it has two vertices ( one in each side ) and a wight ( the distance between the two vertices ). the Graph class represents the graph , it has two vectors one is the vertices included in this graph , and the other is the edges included in the graph. inside the class Graph there is a method its name shortest takes the sources node id and the destination and calculates the shortest path using dijkastra algorithm, and i think that it is the most important part of the code. my theory about the code is that i will create two vectors one for the vertices in the graph i will name it vertices and another vector its name is ver_out it will include the vertices out of calculation in the graph, also i will have two vectors of type Edge , one its name edges for all the edges in the graph and the other its name is track to contain temporarily the edges linked to the temporarily source node in every round , after the calculation of every round the vector track will be cleared. in main() i created five vertices and 10 edges to simulate a graph , the result of the shortest path supposedly to be 4 , but i am always getting 0 , that means i am having something wrong in my code , so if you are interesting in helping me find my mistake and how to make the code work , please take a look. the way shortest work is as follow at the beginning all the edges will be included in the vector edges , we select the edges related to the source and put them in the vector track , then we iterate through track and add the wight of every edge to the vertex (node ) related to it ( not the source vertex ) , then after we clear track and remove the source vertex from the vector vertices and select a new source , and start over again select the edges related to the new source , put them in track , iterate over edges in tack , adding the weights to the corresponding vertices then remove this vertex from the vector vertices, and clear track , and select a new source , and so on . here is the code. #include<iostream> #include<vector> #include <stdlib.h> // for rand() using namespace std; class Vertex { private: unsigned int id; // the name of the vertex unsigned int carried; // the weight a vertex may carry when calculating shortest path vector<unsigned int> previous_nodes; public: unsigned int get_id(){return id;}; unsigned int get_carried(){return carried;}; void set_id(unsigned int value) {id = value;}; void set_carried(unsigned int value) {carried = value;}; void previous_nodes_update(unsigned int val){previous_nodes.push_back(val);}; void previous_nodes_erase(unsigned int val){previous_nodes.erase(previous_nodes.begin() + val);}; Vertex(unsigned int init_val = 0, unsigned int init_carried = 0) :id (init_val), carried(init_carried) // constructor { } ~Vertex() {}; // destructor }; class Edge { private: Vertex first_vertex; // a vertex on one side of the edge Vertex second_vertex; // a vertex on the other side of the edge unsigned int weight; // the value of the edge ( or its weight ) public: unsigned int get_weight() {return weight;}; void set_weight(unsigned int value) {weight = value;}; Vertex get_ver_1(){return first_vertex;}; Vertex get_ver_2(){return second_vertex;}; void set_first_vertex(Vertex v1) {first_vertex = v1;}; void set_second_vertex(Vertex v2) {second_vertex = v2;}; Edge(const Vertex& vertex_1 = 0, const Vertex& vertex_2 = 0, unsigned int init_weight = 0) : first_vertex(vertex_1), second_vertex(vertex_2), weight(init_weight) { } ~Edge() {} ; // destructor }; class Graph { private: std::vector<Vertex> vertices; std::vector<Edge> edges; public: Graph(vector<Vertex> ver_vector, vector<Edge> edg_vector) : vertices(ver_vector), edges(edg_vector) { } ~Graph() {}; vector<Vertex> get_vertices(){return vertices;}; vector<Edge> get_edges(){return edges;}; void set_vertices(vector<Vertex> vector_value) {vertices = vector_value;}; void set_edges(vector<Edge> vector_ed_value) {edges = vector_ed_value;}; unsigned int shortest(unsigned int src, unsigned int dis) { vector<Vertex> ver_out; vector<Edge> track; for(unsigned int i = 0; i < edges.size(); ++i) { if((edges[i].get_ver_1().get_id() == vertices[src].get_id()) || (edges[i].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[i]); edges.erase(edges.begin()+i); } }; for(unsigned int i = 0; i < track.size(); ++i) { if(track[i].get_ver_1().get_id() != vertices[src].get_id()) { track[i].get_ver_1().set_carried((track[i].get_weight()) + track[i].get_ver_2().get_carried()); track[i].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else { track[i].get_ver_2().set_carried((track[i].get_weight()) + track[i].get_ver_1().get_carried()); track[i].get_ver_2().previous_nodes_update(vertices[src].get_id()); } } for(unsigned int i = 0; i < vertices.size(); ++i) if(vertices[i].get_id() == src) vertices.erase(vertices.begin() + i); // removing the sources vertex from the vertices vector ver_out.push_back (vertices[src]); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int i = 0; i < vertices.size(); ++i) if((vertices[i].get_carried() < vertices[src].get_carried()) && (vertices[i].get_id() != dis)) src = vertices[i].get_id(); //while(!edges.empty()) for(unsigned int round = 0; round < vertices.size(); ++round) { for(unsigned int k = 0; k < edges.size(); ++k) { if((edges[k].get_ver_1().get_id() == vertices[src].get_id()) || (edges[k].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[k]); edges.erase(edges.begin()+k); } }; for(unsigned int n = 0; n < track.size(); ++n) if((track[n].get_ver_1().get_id() != vertices[src].get_id()) && (track[n].get_ver_1().get_carried() > (track[n].get_ver_2().get_carried() + track[n].get_weight()))) { track[n].get_ver_1().set_carried((track[n].get_weight()) + track[n].get_ver_2().get_carried()); track[n].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else if(track[n].get_ver_2().get_carried() > (track[n].get_ver_1().get_carried() + track[n].get_weight())) { track[n].get_ver_2().set_carried((track[n].get_weight()) + track[n].get_ver_1().get_carried()); track[n].get_ver_2().previous_nodes_update(vertices[src].get_id()); } for(unsigned int t = 0; t < vertices.size(); ++t) if(vertices[t].get_id() == src) vertices.erase(vertices.begin() + t); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int tt = 0; tt < edges.size(); ++tt) { if(vertices[tt].get_carried() < vertices[src].get_carried()) { src = vertices[tt].get_id(); } } } return vertices[dis].get_carried(); } }; int main() { cout<< "Hello, This is a graph"<< endl; vector<Vertex> vers(5); vers[0].set_id(0); vers[1].set_id(1); vers[2].set_id(2); vers[3].set_id(3); vers[4].set_id(4); vector<Edge> eds(10); eds[0].set_first_vertex(vers[0]); eds[0].set_second_vertex(vers[1]); eds[0].set_weight(5); eds[1].set_first_vertex(vers[0]); eds[1].set_second_vertex(vers[2]); eds[1].set_weight(9); eds[2].set_first_vertex(vers[0]); eds[2].set_second_vertex(vers[3]); eds[2].set_weight(4); eds[3].set_first_vertex(vers[0]); eds[3].set_second_vertex(vers[4]); eds[3].set_weight(6); eds[4].set_first_vertex(vers[1]); eds[4].set_second_vertex(vers[2]); eds[4].set_weight(2); eds[5].set_first_vertex(vers[1]); eds[5].set_second_vertex(vers[3]); eds[5].set_weight(5); eds[6].set_first_vertex(vers[1]); eds[6].set_second_vertex(vers[4]); eds[6].set_weight(7); eds[7].set_first_vertex(vers[2]); eds[7].set_second_vertex(vers[3]); eds[7].set_weight(1); eds[8].set_first_vertex(vers[2]); eds[8].set_second_vertex(vers[4]); eds[8].set_weight(8); eds[9].set_first_vertex(vers[3]); eds[9].set_second_vertex(vers[4]); eds[9].set_weight(3); unsigned int path; Graph graf(vers, eds); path = graf.shortest(2, 4); cout<< path << endl; return 0; }

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  • Evaluating code for a graph [migrated]

    - by mazen.r.f
    This is relatively long code. Please take a look at this code if you are still willing to do so. I will appreciate your feedback. I have spent two days trying to come up with code to represent a graph, calculating the shortest path using Dijkstra's algorithm. But I am not able to get the right result, even though the code runs without errors. The result is not correct and I am always getting 0. I have three classes: Vertex, Edge, and Graph. The Vertex class represents the nodes in the graph and it has id and carried (which carry the weight of the links connected to it while using Dijkstra's algorithm) and a vector of the ids belong to other nodes the path will go through before arriving to the node itself. This vector is named previous_nodes. The Edge class represents the edges in the graph and has two vertices (one in each side) and a width (the distance between the two vertices). The Graph class represents the graph. It has two vectors, where one is the vertices included in this graph, and the other is the edges included in the graph. Inside the class Graph, there is a method named shortest() that takes the sources node id and the destination and calculates the shortest path using Dijkstra's algorithm. I think that it is the most important part of the code. My theory about the code is that I will create two vectors, one for the vertices in the graph named vertices, and another vector named ver_out (it will include the vertices out of calculation in the graph). I will also have two vectors of type Edge, where one is named edges (for all the edges in the graph), and the other is named track (to temporarily contain the edges linked to the temporary source node in every round). After the calculation of every round, the vector track will be cleared. In main(), I've created five vertices and 10 edges to simulate a graph. The result of the shortest path supposedly is 4, but I am always getting 0. That means I have something wrong in my code. If you are interesting in helping me find my mistake and making the code work, please take a look. The way shortest work is as follow: at the beginning, all the edges will be included in the vector edges. We select the edges related to the source and put them in the vector track, then we iterate through track and add the width of every edge to the vertex (node) related to it (not the source vertex). After that, we clear track and remove the source vertex from the vector vertices and select a new source. Then we start over again and select the edges related to the new source, put them in track, iterate over edges in track, adding the weights to the corresponding vertices, then remove this vertex from the vector vertices. Then clear track, and select a new source, and so on. #include<iostream> #include<vector> #include <stdlib.h> // for rand() using namespace std; class Vertex { private: unsigned int id; // the name of the vertex unsigned int carried; // the weight a vertex may carry when calculating shortest path vector<unsigned int> previous_nodes; public: unsigned int get_id(){return id;}; unsigned int get_carried(){return carried;}; void set_id(unsigned int value) {id = value;}; void set_carried(unsigned int value) {carried = value;}; void previous_nodes_update(unsigned int val){previous_nodes.push_back(val);}; void previous_nodes_erase(unsigned int val){previous_nodes.erase(previous_nodes.begin() + val);}; Vertex(unsigned int init_val = 0, unsigned int init_carried = 0) :id (init_val), carried(init_carried) // constructor { } ~Vertex() {}; // destructor }; class Edge { private: Vertex first_vertex; // a vertex on one side of the edge Vertex second_vertex; // a vertex on the other side of the edge unsigned int weight; // the value of the edge ( or its weight ) public: unsigned int get_weight() {return weight;}; void set_weight(unsigned int value) {weight = value;}; Vertex get_ver_1(){return first_vertex;}; Vertex get_ver_2(){return second_vertex;}; void set_first_vertex(Vertex v1) {first_vertex = v1;}; void set_second_vertex(Vertex v2) {second_vertex = v2;}; Edge(const Vertex& vertex_1 = 0, const Vertex& vertex_2 = 0, unsigned int init_weight = 0) : first_vertex(vertex_1), second_vertex(vertex_2), weight(init_weight) { } ~Edge() {} ; // destructor }; class Graph { private: std::vector<Vertex> vertices; std::vector<Edge> edges; public: Graph(vector<Vertex> ver_vector, vector<Edge> edg_vector) : vertices(ver_vector), edges(edg_vector) { } ~Graph() {}; vector<Vertex> get_vertices(){return vertices;}; vector<Edge> get_edges(){return edges;}; void set_vertices(vector<Vertex> vector_value) {vertices = vector_value;}; void set_edges(vector<Edge> vector_ed_value) {edges = vector_ed_value;}; unsigned int shortest(unsigned int src, unsigned int dis) { vector<Vertex> ver_out; vector<Edge> track; for(unsigned int i = 0; i < edges.size(); ++i) { if((edges[i].get_ver_1().get_id() == vertices[src].get_id()) || (edges[i].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[i]); edges.erase(edges.begin()+i); } }; for(unsigned int i = 0; i < track.size(); ++i) { if(track[i].get_ver_1().get_id() != vertices[src].get_id()) { track[i].get_ver_1().set_carried((track[i].get_weight()) + track[i].get_ver_2().get_carried()); track[i].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else { track[i].get_ver_2().set_carried((track[i].get_weight()) + track[i].get_ver_1().get_carried()); track[i].get_ver_2().previous_nodes_update(vertices[src].get_id()); } } for(unsigned int i = 0; i < vertices.size(); ++i) if(vertices[i].get_id() == src) vertices.erase(vertices.begin() + i); // removing the sources vertex from the vertices vector ver_out.push_back (vertices[src]); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int i = 0; i < vertices.size(); ++i) if((vertices[i].get_carried() < vertices[src].get_carried()) && (vertices[i].get_id() != dis)) src = vertices[i].get_id(); //while(!edges.empty()) for(unsigned int round = 0; round < vertices.size(); ++round) { for(unsigned int k = 0; k < edges.size(); ++k) { if((edges[k].get_ver_1().get_id() == vertices[src].get_id()) || (edges[k].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[k]); edges.erase(edges.begin()+k); } }; for(unsigned int n = 0; n < track.size(); ++n) if((track[n].get_ver_1().get_id() != vertices[src].get_id()) && (track[n].get_ver_1().get_carried() > (track[n].get_ver_2().get_carried() + track[n].get_weight()))) { track[n].get_ver_1().set_carried((track[n].get_weight()) + track[n].get_ver_2().get_carried()); track[n].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else if(track[n].get_ver_2().get_carried() > (track[n].get_ver_1().get_carried() + track[n].get_weight())) { track[n].get_ver_2().set_carried((track[n].get_weight()) + track[n].get_ver_1().get_carried()); track[n].get_ver_2().previous_nodes_update(vertices[src].get_id()); } for(unsigned int t = 0; t < vertices.size(); ++t) if(vertices[t].get_id() == src) vertices.erase(vertices.begin() + t); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int tt = 0; tt < edges.size(); ++tt) { if(vertices[tt].get_carried() < vertices[src].get_carried()) { src = vertices[tt].get_id(); } } } return vertices[dis].get_carried(); } }; int main() { cout<< "Hello, This is a graph"<< endl; vector<Vertex> vers(5); vers[0].set_id(0); vers[1].set_id(1); vers[2].set_id(2); vers[3].set_id(3); vers[4].set_id(4); vector<Edge> eds(10); eds[0].set_first_vertex(vers[0]); eds[0].set_second_vertex(vers[1]); eds[0].set_weight(5); eds[1].set_first_vertex(vers[0]); eds[1].set_second_vertex(vers[2]); eds[1].set_weight(9); eds[2].set_first_vertex(vers[0]); eds[2].set_second_vertex(vers[3]); eds[2].set_weight(4); eds[3].set_first_vertex(vers[0]); eds[3].set_second_vertex(vers[4]); eds[3].set_weight(6); eds[4].set_first_vertex(vers[1]); eds[4].set_second_vertex(vers[2]); eds[4].set_weight(2); eds[5].set_first_vertex(vers[1]); eds[5].set_second_vertex(vers[3]); eds[5].set_weight(5); eds[6].set_first_vertex(vers[1]); eds[6].set_second_vertex(vers[4]); eds[6].set_weight(7); eds[7].set_first_vertex(vers[2]); eds[7].set_second_vertex(vers[3]); eds[7].set_weight(1); eds[8].set_first_vertex(vers[2]); eds[8].set_second_vertex(vers[4]); eds[8].set_weight(8); eds[9].set_first_vertex(vers[3]); eds[9].set_second_vertex(vers[4]); eds[9].set_weight(3); unsigned int path; Graph graf(vers, eds); path = graf.shortest(2, 4); cout<< path << endl; return 0; }

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  • Help:Graph contest problem: maybe a modified Dijkstra or another alternative algorithm

    - by newba
    Hi you all, I'm trying to do this contest exercise about graphs: XPTO is an intrepid adventurer (a little too temerarious for his own good) who boasts about exploring every corner of the universe, no matter how inhospitable. In fact, he doesn't visit the planets where people can easily live in, he prefers those where only a madman would go with a very good reason (several millions of credits for instance). His latest exploit is trying to survive in Proxima III. The problem is that Proxima III suffers from storms of highly corrosive acids that destroy everything, including spacesuits that were especially designed to withstand corrosion. Our intrepid explorer was caught in a rectangular area in the middle of one of these storms. Fortunately, he has an instrument that is capable of measuring the exact concentration of acid on each sector and how much damage it does to his spacesuit. Now, he only needs to find out if he can escape the storm. Problem The problem consists of finding an escape route that will allow XPTOto escape the noxious storm. You are given the initial energy of the spacesuit, the size of the rectangular area and the damage that the spacesuit will suffer while standing in each sector. Your task is to find the exit sector, the number of steps necessary to reach it and the amount of energy his suit will have when he leaves the rectangular area. The escape route chosen should be the safest one (i.e., the one where his spacesuit will be the least damaged). Notice that Rodericus will perish if the energy of his suit reaches zero. In case there are more than one possible solutions, choose the one that uses the least number of steps. If there are at least two sectors with the same number of steps (X1, Y1) and (X2, Y2) then choose the first if X1 < X2 or if X1 = X2 and Y1 < Y2. Constraints 0 < E = 30000 the suit's starting energy 0 = W = 500 the rectangle's width 0 = H = 500 rectangle's height 0 < X < W the starting X position 0 < Y < H the starting Y position 0 = D = 10000 the damage sustained in each sector Input The first number given is the number of test cases. Each case will consist of a line with the integers E, X and Y. The following line will have the integers W and H. The following lines will hold the matrix containing the damage D the spacesuit will suffer whilst in the corresponding sector. Notice that, as is often the case for computer geeks, (1,1) corresponds to the upper left corner. Output If there is a solution, the output will be the remaining energy, the exit sector's X and Y coordinates and the number of steps of the route that will lead Rodericus to safety. In case there is no solution, the phrase Goodbye cruel world! will be written. Sample Input 3 40 3 3 7 8 12 11 12 11 3 12 12 12 11 11 12 2 1 13 11 11 12 2 13 2 14 10 11 13 3 2 1 12 10 11 13 13 11 12 13 12 12 11 13 11 13 12 13 12 12 11 11 11 11 13 13 10 10 13 11 12 8 3 4 7 6 4 3 3 2 2 3 2 2 5 2 2 2 3 3 2 1 2 2 3 2 2 4 3 3 2 2 4 1 3 1 4 3 2 3 1 2 2 3 3 0 3 4 10 3 4 7 6 3 3 1 2 2 1 0 2 2 2 4 2 2 5 2 2 1 3 0 2 2 2 2 1 3 3 4 2 3 4 4 3 1 1 3 1 2 2 4 2 2 1 Sample Output 12 5 1 8 Goodbye cruel world! 5 1 4 2 Basically, I think we have to do a modified Dijkstra, in which the distance between nodes is the suit's energy (and we have to subtract it instead of suming up like is normal with distances) and the steps are the ....steps made along the path. The pos with the bester binomial (Energy,num_Steps) is our "way out". Important : XPTO obviously can't move in diagonals, so we have to cut out this cases. I have many ideas, but I have such a problem implementing them... Could someone please help me thinking about this with some code or, at least, ideas? Am I totally wrong?

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  • C++: Error in Xcode; "Graph::Coordinate::Coordinate()", referenced from: ...

    - by Alexandstein
    In a program I am writing, I wrote for two classes (Coordinate, and Graph), with one of them taking the other as constructor arguments. When I try to compile it I get the following error for Graph.cpp: Undefined symbols: "Graph::Coordinate::Coordinate(double)", referenced from: Graph::Graph() in Graph.o Graph::Graph() in Graph.o "Graph::Coordinate::Coordinate()", referenced from: Graph::Graph(Graph::Coordinate, Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate, Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate)in Graph.o Graph::Graph(Graph::Coordinate)in Graph.o Graph::Graph() in Graph.o Graph::Graph() in Graph.o Graph::Graph() in Graph.o Graph::Graph() in Graph.o Graph::Graph() in Graph.o Graph::Graph() in Graph.o ld: symbol(s) not found collect2: ld returned 1 exit status I checked the code and couldn't find anything out of the ordinary. Here are the four class files: (Sorry if it's a lot of code to sift through.) Coordinate.h class Graph{ #include "Coordinate.h" public: Graph(); Graph(Coordinate); Graph(Coordinate, Coordinate); Graph(Coordinate, Coordinate, Coordinate); void setXSize(int); void setYSize(int); void setX(int); //int corresponds to coordinates 1, 2, or 3 void setY(int); void setZ(int); int getXSize(); int getYSize(); double getX(int); //int corresponds to coordinates 1, 2, or 3 double getY(int); double getZ(int); void outputGraph(); void animateGraph(); private: int xSize; int ySize; Coordinate coord1; Coordinate coord2; Coordinate coord3; }; Coordinate.cpp #include <iostream> #include "Coordinate.h" Coordinate::Coordinate() { xCoord = 1; yCoord = 1; zCoord = 1; xVel = 1; yVel = 1; zVel = 1; } Coordinate::Coordinate(double xCoo) { xCoord = xCoo; yCoord = 1; zCoord = 1; xVel = 1; yVel = 1; zVel = 1; } Coordinate::Coordinate(double xCoo,double yCoo) { xCoord = xCoo; yCoord = yCoo; zCoord = 1; xVel = 1; yVel = 1; zVel = 1; } Coordinate::Coordinate(double xCoo,double yCoo,double zCoo) { xCoord = xCoo; yCoord = yCoo; zCoord = zCoo; xVel = 1; yVel = 1; zVel = 1; } void Coordinate::setXCoord(double xCoo) { xCoord = xCoo; } void Coordinate::setYCoord(double yCoo) { yCoord = yCoo; } void Coordinate::setZCoord(double zCoo) { zCoord = zCoo; } void Coordinate::setXVel(double xVelo) { xVel = xVelo; } void Coordinate::setYVel(double yVelo) { yVel = yVelo; } void Coordinate::setZVel(double zVelo) { zVel = zVelo; } double Coordinate::getXCoord() { return xCoord; } double Coordinate::getYCoord() { return yCoord; } double Coordinate::getZCoord() { return zCoord; } double Coordinate::getXVel() { return xVel; } double Coordinate::GetYVel() { return yVel; } double Coordinate::GetZVel() { return zVel; } Graph.h class Graph{ #include "Coordinate.h" public: Graph(); Graph(Coordinate); Graph(Coordinate, Coordinate); Graph(Coordinate, Coordinate, Coordinate); void setXSize(int); void setYSize(int); void setX(int); //int corresponds to coordinates 1, 2, or 3 void setY(int); void setZ(int); int getXSize(); int getYSize(); double getX(int); //int corresponds to coordinates 1, 2, or 3 double getY(int); double getZ(int); void outputGraph(); void animateGraph(); private: int xSize; int ySize; Coordinate coord1; Coordinate coord2; Coordinate coord3; }; Graph.cpp #include "Graph.h" #include "Coordinate.h" #include <iostream> #include <ctime> using namespace std; Graph::Graph() { Coordinate coord1(0); } Graph::Graph(Coordinate cOne) { coord1 = cOne; xSize = 20; ySize = 20; } Graph::Graph(Coordinate cOne, Coordinate cTwo) { coord1 = cOne; coord2 = cTwo; xSize = 20; ySize = 20; } Graph::Graph(Coordinate cOne, Coordinate cTwo, Coordinate cThree) { coord1 = cOne; coord2 = cTwo; coord3 = cThree; xSize = 20; ySize = 20; } void Graph::setXSize(int size) { xSize = size; } void Graph::setYSize(int size) { ySize = size; } int Graph::getXSize() { return xSize; } int Graph::getYSize() { return ySize; } void Graph::outputGraph() { } void Graph::animateGraph() { } Thanks very much for any help!

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  • Given an XML which contains a representation of a graph, how to apply it DFS algorithm? [on hold]

    - by winston smith
    Given the followin XML which is a directed graph: <?xml version="1.0" encoding="iso-8859-1" ?> <!DOCTYPE graph PUBLIC "-//FC//DTD red//EN" "../dtd/graph.dtd"> <graph direct="1"> <vertex label="V0"/> <vertex label="V1"/> <vertex label="V2"/> <vertex label="V3"/> <vertex label="V4"/> <vertex label="V5"/> <edge source="V0" target="V1" weight="1"/> <edge source="V0" target="V4" weight="1"/> <edge source="V5" target="V2" weight="1"/> <edge source="V5" target="V4" weight="1"/> <edge source="V1" target="V2" weight="1"/> <edge source="V1" target="V3" weight="1"/> <edge source="V1" target="V4" weight="1"/> <edge source="V2" target="V3" weight="1"/> </graph> With this classes i parsed the graph and give it an adjacency list representation: import java.io.IOException; import java.util.HashSet; import java.util.LinkedList; import java.util.Collection; import java.util.Iterator; import java.util.logging.Level; import java.util.logging.Logger; import practica3.util.Disc; public class ParsingXML { public static void main(String[] args) { try { // TODO code application logic here Collection<Vertex> sources = new HashSet<Vertex>(); LinkedList<String> lines = Disc.readFile("xml/directed.xml"); for (String lin : lines) { int i = Disc.find(lin, "source=\""); String data = ""; if (i > 0 && i < lin.length()) { while (lin.charAt(i + 1) != '"') { data += lin.charAt(i + 1); i++; } Vertex v = new Vertex(); v.setName(data); v.setAdy(new HashSet<Vertex>()); sources.add(v); } } Iterator it = sources.iterator(); while (it.hasNext()) { Vertex ver = (Vertex) it.next(); Collection<Vertex> adyacencias = ver.getAdy(); LinkedList<String> ls = Disc.readFile("xml/graphs.xml"); for (String lin : ls) { int i = Disc.find(lin, "target=\""); String data = ""; if (lin.contains("source=\""+ver.getName())) { Vertex v = new Vertex(); if (i > 0 && i < lin.length()) { while (lin.charAt(i + 1) != '"') { data += lin.charAt(i + 1); i++; } v.setName(data); } i = Disc.find(lin, "weight=\""); data = ""; if (i > 0 && i < lin.length()) { while (lin.charAt(i + 1) != '"') { data += lin.charAt(i + 1); i++; } v.setWeight(Integer.parseInt(data)); } if (v.getName() != null) { adyacencias.add(v); } } } } for (Vertex vert : sources) { System.out.println(vert); System.out.println("adyacencias: " + vert.getAdy()); } } catch (IOException ex) { Logger.getLogger(ParsingXML.class.getName()).log(Level.SEVERE, null, ex); } } } This is another class: import java.util.Collection; import java.util.Objects; public class Vertex { private String name; private int weight; private Collection ady; public Collection getAdy() { return ady; } public void setAdy(Collection adyacencias) { this.ady = adyacencias; } public String getName() { return name; } public void setName(String nombre) { this.name = nombre; } public int getWeight() { return weight; } public void setWeight(int weight) { this.weight = weight; } @Override public int hashCode() { int hash = 7; hash = 43 * hash + Objects.hashCode(this.name); hash = 43 * hash + this.weight; return hash; } @Override public boolean equals(Object obj) { if (obj == null) { return false; } if (getClass() != obj.getClass()) { return false; } final Vertex other = (Vertex) obj; if (!Objects.equals(this.name, other.name)) { return false; } if (this.weight != other.weight) { return false; } return true; } @Override public String toString() { return "Vertice{" + "name=" + name + ", weight=" + weight + '}'; } } And finally: /** * * @author user */ /* -*-jde-*- */ /* <Disc.java> Contains the main argument*/ import java.io.*; import java.util.LinkedList; /** * Lectura y escritura de archivos en listas de cadenas * Ideal para el uso de las clases para gráficas. * * @author Peralta Santa Anna Victor Miguel * @since Julio 2011 */ public class Disc { /** * Metodo para lectura de un archivo * * @param fileName archivo que se va a leer * @return El archivo en representacion de lista de cadenas */ public static LinkedList<String> readFile(String fileName) throws IOException { BufferedReader file = new BufferedReader(new FileReader(fileName)); LinkedList<String> textlist = new LinkedList<String>(); while (file.ready()) { textlist.add(file.readLine().trim()); } file.close(); /* for(String linea:textlist){ if(linea.contains("source")){ //String generado = linea.replaceAll("<\\w+\\s+\"", ""); //System.out.println(generado); } }*/ return textlist; }//readFile public static int find(String linea,String palabra){ int i,j; boolean found = false; for(i=0,j=0;i<linea.length();i++){ if(linea.charAt(i)==palabra.charAt(j)){ j++; if(j==palabra.length()){ found = true; return i; } }else{ continue; } } if(!found){ i= -1; } return i; } /** * Metodo para la escritura de un archivo * * @param fileName archivo que se va a escribir * @param tofile la lista de cadenas que quedaran en el archivo * @param append el bit que dira si se anexa el contenido o se empieza de cero */ public static void writeFile(String fileName, LinkedList<String> tofile, boolean append) throws IOException { FileWriter file = new FileWriter(fileName, append); for (int i = 0; i < tofile.size(); i++) { file.write(tofile.get(i) + "\n"); } file.close(); }//writeFile /** * Metodo para escritura de un archivo * @param msg archivo que se va a escribir * @param tofile la cadena que quedaran en el archivo * @param append el bit que dira si se anexa el contenido o se empieza de cero */ public static void writeFile(String msg, String tofile, boolean append) throws IOException { FileWriter file = new FileWriter(msg, append); file.write(tofile); file.close(); }//writeFile }// I'm stuck on what can be the best way to given an adjacency list representation of the graph how to apply it Depth-first search algorithm. Any idea of how to aproach to complete the task?

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  • How to choose an integer linear programming solver ?

    - by Cassie
    Hi all, I am newbie for integer linear programming. I plan to use a integer linear programming solver to solve my combinatorial optimization problem. I am more familiar with C++/object oriented programming on an IDE. Now I am using NetBeans with Cygwin to write my applications most of time. May I ask if there is an easy use ILP solver for me? Or it depends on the problem I want to solve ? I am trying to do some resources mapping optimization. Please let me know if any further information is required. Thank you very much, Cassie.

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  • help in the Donalds B. Johnson's algorithm, i cannot understand the pseudo code (PART II)

    - by Pitelk
    Hi all , i cannot understand a certain part of the paper published by Donald Johnson about finding cycles (Circuits) in a graph. More specific i cannot understand what is the matrix Ak which is mentioned in the following line of the pseudo code : Ak:=adjacency structure of strong component K with least vertex in subgraph of G induced by {s,s+1,....n}; to make things worse some lines after is mentins " for i in Vk do " without declaring what the Vk is... As far i have understand we have the following: 1) in general, a strong component is a sub-graph of a graph, in which for every node of this sub-graph there is a path to any node of the sub-graph (in other words you can access any node of the sub-graph from any other node of the sub-graph) 2) a sub-graph induced by a list of nodes is a graph containing all these nodes plus all the edges connecting these nodes. in paper the mathematical definition is " F is a subgraph of G induced by W if W is subset of V and F = (W,{u,y)|u,y in W and (u,y) in E)}) where u,y are edges , E is the set of all the edges in the graph, W is a set of nodes. 3)in the code implementation the nodes are named by integer numbers 1 ... n. 4) I suspect that the Vk is the set of nodes of the strong component K. now to the question. Lets say we have a graph G= (V,E) with V = {1,2,3,4,5,6,7,8,9} which it can be divided into 3 strong components the SC1 = {1,4,7,8} SC2= {2,3,9} SC3 = {5,6} (and their edges) Can anybody give me an example for s =1, s= 2, s= 5 what if going to be the Vk and Ak according to the code? The pseudo code is in my previous question in http://stackoverflow.com/questions/2908575/help-in-the-donalds-b-johnsons-algorithm-i-cannot-understand-the-pseudo-code and the paper can be found at http://stackoverflow.com/questions/2908575/help-in-the-donalds-b-johnsons-algorithm-i-cannot-understand-the-pseudo-code thank you in advance

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  • Suggestions of the easiest algorithms for some Graph operations

    - by Nazgulled
    Hi, The deadline for this project is closing in very quickly and I don't have much time to deal with what it's left. So, instead of looking for the best (and probably more complicated/time consuming) algorithms, I'm looking for the easiest algorithms to implement a few operations on a Graph structure. The operations I'll need to do is as follows: List all users in the graph network given a distance X List all users in the graph network given a distance X and the type of relation Calculate the shortest path between 2 users on the graph network given a type of relation Calculate the maximum distance between 2 users on the graph network Calculate the most distant connected users on the graph network A few notes about my Graph implementation: The edge node has 2 properties, one is of type char and another int. They represent the type of relation and weight, respectively. The Graph is implemented with linked lists, for both the vertices and edges. I mean, each vertex points to the next one and each vertex also points to the head of a different linked list, the edges for that specific vertex. What I know about what I need to do: I don't know if this is the easiest as I said above, but for the shortest path between 2 users, I believe the Dijkstra algorithm is what people seem to recommend pretty often so I think I'm going with that. I've been searching and searching and I'm finding it hard to implement this algorithm, does anyone know of any tutorial or something easy to understand so I can implement this algorithm myself? If possible, with C source code examples, it would help a lot. I see many examples with math notations but that just confuses me even more. Do you think it would help if I "converted" the graph to an adjacency matrix to represent the links weight and relation type? Would it be easier to perform the algorithm on that instead of the linked lists? I could easily implement a function to do that conversion when needed. I'm saying this because I got the feeling it would be easier after reading a couple of pages about the subject, but I could be wrong. I don't have any ideas about the other 4 operations, suggestions?

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  • Graph navigation problem

    - by affan
    I have graph of components and relation between them. User open graph and want to navigate through the graph base on his choice. He start with root node and click expand button which reveal new component that is related to current component. The problem is with when use decide to collapse a node. I have to choose a sub-tree to hide and at same time leave graph in consistent state so that there is no expanded node with missing relation to another node in graph. Now in case of cyclic/loop between component i have difficult of choosing sub-tree. For simplicity i choose the order in which they were expanded. So if a A expand into B and C collapse A will hide the nodes and edge that it has created. Now consider flowing scenario. [-] mean expanded state and [+] mean not yet expanded. A is expanded to reveal B and C. And then B is expanded to reveal D. C is expanded which create a link between C and exiting node D and also create node E. Now user decide to collapse B. Since by order of expansion D is child of B it will collapse and hide D. This leave graph in inconsistent state as C is expanded with edge to D but D is not anymore there if i remove CD edge it will still be inconsistent. If i collapse C. And E is again a cyclic link e.g to B will produce the same problem. /-----B[-]-----\ A[-] D[+] \-----C[-]-----/ \ E[+] So guys any idea how can i solve this problem. User need to navigate through graph and should be able to collapse but i am stuck with problem of cyclic nodes in which case any of node in loop if collapse will leave graph in inconsistent state.

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  • algorithm to use to return a specific range of nodes in a directed graph

    - by GatesReign
    I have a class Graph with two lists types namely nodes and edges I have a function List<int> GetNodesInRange(Graph graph, int Range) when I get these parameters I need an algorithm that will go through the graph and return the list of nodes only as deep (the level) as the range. The algorithm should be able to accommodate large number of nodes and large ranges. Atop this, should I use a similar function List<int> GetNodesInRange(Graph graph, int Range, int selected) I want to be able to search outwards from it, to the number of nodes outwards (range) specified. So in the first function, I expect it to return the nodes placed in the blue box. The other function, if I pass the nodes as in the graph with a range of 1 and it starts at node 5, I want it to return the list of nodes that satisfy this criteria (placed in the orange box)

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  • Randomly generate directed graph on a grid

    - by Talon876
    I am trying to randomly generate a directed graph for the purpose of making a puzzle game similar to the ice sliding puzzles from Pokemon. This is essentially what I want to be able to randomly generate: http://bulbanews.bulbagarden.net/wiki/Crunching_the_numbers:_Graph_theory. I need to be able to limit the size of the graph in an x and y dimension. In the example given in the link, it would be restricted to an 8x4 grid. The problem I am running into is not randomly generating the graph, but randomly generating a graph, which I can properly map out in a 2d space, since I need something (like a rock) on the opposite side of a node, to make it visually make sense when you stop sliding. The problem with this is that sometimes the rock ends up in the path between two other nodes or possibly on another node itself, which causes the entire graph to become broken. After discussing the problem with a few people I know, we came to a couple of conclusions that may lead to a solution. Including the obstacles in the grid as part of the graph when constructing it. Start out with a fully filled grid and just draw a random path and delete out blocks that will make that path work. The problem then becomes figuring out which ones to delete to avoid introducing an additional, shorter path. We were also thinking a dynamic programming algorithm may be beneficial, though none of us are too skilled with creating dynamic programming algorithms from nothing. Any ideas or references about what this problem is officially called (if it's an official graph problem) would be most helpful. Here are some examples of what I have accomplished so far by just randomly placing blocks and generating the navigation graph from the chosen start/finish. The idea (as described in the previous link) is you start at the green S and want to get to the green F. You do this by moving up/down/left/right and you continue moving in the direction chosen until you hit a wall. In these pictures, grey is a wall, white is the floor, and the purple line is the minimum length from start to finish, and the black lines and grey dots represented possible paths. Here are some bad examples of randomly generated graphs: http://i.stack.imgur.com/9uaM6.png Here are some good examples of randomly generated (or hand tweaked) graphs: i.stack.imgur.com/uUGeL.png (can't post another link, sorry) I've also seemed to notice the more challenging ones when actually playing this as a puzzle are ones which have lots of high degree nodes along the minimum path.

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