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  • Should I convert overlong UTF-8 strings to their shortest normal form?

    - by Grant McLean
    I've just been reworking my Encoding::FixLatin Perl module to handle overlong UTF-8 byte sequences and convert them to the shortest normal form. My question is quite simply "is this a bad idea"? A number of sources (including this RFC) suggest that any over-long UTF-8 should be treated as an error and rejected. They caution against "naive implementations" and leave me with the impression that these things are inherently unsafe. Since the whole purpose of my module is to clean up messy data files with mixed encodings and convert them to nice clean utf8, this seems like just one more thing I can clean up so the application layer doesn't have to deal with it. My code does not concern itself with any semantic meaning the resulting characters might have, it simply converts them into a normalised form. Am I missing something. Is there a hidden danger I haven't considered?

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  • What is CSS? what should be the shortest and appropriate answer of this question? [closed]

    - by metal-gear-solid
    What is CSS? what should be the shortest and appropriate answer of this question? How simple we can explain , What is CSS ,just in words? mine answer is this Stands for "Cascading Style Sheet." Cascading style sheets are used to format the layout of Web pages. They can be used to define text styles, table sizes, and other aspects of Web pages that previously could only be defined in a page's HTML. Is it ok or it can be more perfect or shorten?

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  • Shortest distance between points on a toroidally wrapped (x- and y- wrapping) map?

    - by mstksg
    I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean rectangle, but when a point moves to the right boundary, it will appear at the left boundary (at the same y value), given by x_new = x_old % width Basically, points are plotted based on: (x_new, y_new) = ( x_old % width, y_old % height) Think Pac Man -- walking off one edge of the screen will make you appear on the opposite edge. What's the best way to calculate the shortest distance between two points? The typical implementation suggests a large distance for points on opposite corners of the map, when in reality, the real wrapped distance is very close. The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula. But that would involve many checks, calculations, operations -- some that I feel might be unnecessary. Is there a better way?

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  • Should I convert overly-long UTF-8 strings to their shortest normal form?

    - by Grant McLean
    I've just been reworking my Encoding::FixLatin Perl module to handle overly-long UTF-8 byte sequences and convert them to the shortest normal form. My question is quite simply "is this a bad idea"? A number of sources (including this RFC) suggest that any over-long UTF-8 should be treated as an error and rejected. They caution against "naive implementations" and leave me with the impression that these things are inherently unsafe. Since the whole purpose of my module is to clean up messy data files with mixed encodings and convert them to nice clean utf8, this seems like just one more thing I can clean up so the application layer doesn't have to deal with it. My code does not concern itself with any semantic meaning the resulting characters might have, it simply converts them into a normalised form. Am I missing something. Is there a hidden danger I haven't considered?

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  • GRAPH PROBLEM: find an algorithm to determine the shortest path from one point to another in a recta

    - by newba
    I'm getting such an headache trying to elaborate an appropriate algorithm to go from a START position to a EXIT position in a maze. For what is worth, the maze is rectangular, maxsize 500x500 and, in theory, is resolvable by DFS with some branch and bound techniques ... 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 Output: 5 1 4 2 Explanation: Our agent looses energy every time he gives a step and he can only move UP, DOWN, LEFT and RIGHT. Also, if the agent arrives with a remaining energy of zero or less, he dies, so we print something like "Impossible". So, in the input 10 is the initial agent's energy, 3 4 is the START position (i.e. column 3, line 4) and we have a maze 7x6. Think this as a kind of labyrinth, in which I want to find the exit that gives the agent a better remaining energy (shortest path). In case there are paths which lead to the same remaining energy, we choose the one which has the small number of steps, of course. I need to know if a DFS to a maze 500x500 in the worst case is feasible with these limitations and how to do it, storing the remaining energy in each step and the number of steps taken so far. The output means the agent arrived with remaining energy= 5 to the exit pos 1 4 in 2 steps. If we look carefully, in this maze it's also possible to exit at pos 3 1 (column 3, row 1) with the same energy but with 3 steps, so we choose the better one. With these in mind, can someone help me some code or pseudo-code? I have troubles working this around with a 2D array and how to store the remaining energy, the path (or number of steps taken)....

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  • What's the shortest way to post a cropped screenshot on the web?

    - by Borek
    If I want to send someone a piece of my screen this is what I currently do: PrtScr or Alt+PrtScr Open Paint.NET Paste the screen shot Make a selection Crop image to selection Save image to some location - and remember it! Go to some image hosting site (there are plenty of them in the days of Twitter) Click their "Browse" button Browse for the image if I happened to remember the location where I stored it :) Upload the image and obtain the link which I can share This is simply too many steps. I don't usually mind doing steps 1 to 5 but especially steps 6 and 9 are annoying. Jing is pretty close to what I'm looking for but I find their horrible URLs unbearable. If there was something with similar functionality but better or pluggable hosting, that would be great.

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  • How to recalculate all-pairs shorthest paths on-line if nodes are getting removed?

    - by Pavel Shved
    Latest news about underground bombing made me curious about the following problem. Assume we have a weighted undirected graph, nodes of which are sometimes removed. The problem is to re-calculate shortest paths between all pairs of nodes fast after such removals. With a simple modification of Floyd-Warshall algorithm we can calculate shortest paths between all pairs. These paths may be stored in a table, where shortest[i][j] contains the index of the next node on the shortest path between i and j (or NULL value if there's no path). The algorithm requires O(n³) time to build the table, and eacho query shortest(i,j) takes O(1). Unfortunately, we should re-run this algorithm after each removal. The other alternative is to run graph search on each query. This way each removal takes zero time to update an auxiliary structure (because there's none), but each query takes O(E) time. What algorithm can be used to "balance" the query and update time for all-pairs shortest-paths problem when nodes of the graph are being removed?

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  • A* navigational mesh path finding

    - by theguywholikeslinux
    So I've been making this top down 2D java game in this framework called Greenfoot [1] and I've been working on the AI for the guys you are gonna fight. I want them to be able to move around the world realistically so I soon realized, amongst a couple of other things, I would need some kind of pathfinding. I have made two A* prototypes. One is grid based and then I made one that works with waypoints so now I need to work out a way to get from a 2d "map" of the obstacles/buildings to a graph of nodes that I can make a path from. The actual pathfinding seems fine, just my open and closed lists could use a more efficient data structure, but I'll get to that if and when I need to. I intend to use a navigational mesh for all the reasons out lined in this post on ai-blog.net [2]. However, the problem I have faced is that what A* thinks is the shortest path from the polygon centres/edges is not necessarily the shortest path if you travel through any part of the node. To get a better idea you can see the question I asked on stackoverflow [3]. I got a good answer concerning a visibility graph. I have since purchased the book (Computational Geometry: Algorithms and Applications [4]) and read further into the topic, however I am still in favour of a navigational mesh (See "Managing Complexity" [5] from Amit’s Notes about Path-Finding [6]). (As a side note, maybe I could possibly use Theta* to convert multiple waypoints into one straight line if the first and last are not obscured. Or each time I move back check to the waypoint before last to see if I can go straight from that to this) So basically what I want is a navigational mesh where once I have put it through a funnel algorithm (e.g. this one from Digesting Duck [7]) I will get the true shortest path, rather than get one that is the shortest path following node to node only, but not the actual shortest given that you can go through some polygons and skip nodes/edges. Oh and I also want to know how you suggest storing the information concerning the polygons. For the waypoint prototype example I made I just had each node as an object and stored a list of all the other nodes you could travel to from that node, I'm guessing that won't work with polygons? and how to I tell if a polygon is open/traversable or if it is a solid object? How do I store which nodes make up the polygon? Finally, for the record: I do want to programme this by myself from scratch even though there are already other solutions available and I don't intend to be (re) using this code in anything other than this game so it does not matter that it will inevitably be poor quality. http://greenfoot.org http://www.ai-blog.net/archives/000152.html http://stackoverflow.com/q/7585515/ http://www.cs.uu.nl/geobook/ http://theory.stanford.edu/~amitp/GameProgramming/MapRepresentations.html http://theory.stanford.edu/~amitp/GameProgramming/ http://digestingduck.blogspot.com/2010/03/simple-stupid-funnel-algorithm.html

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  • Shortest way of attaching multiple DOM elements to one event handler using Jquery?

    - by Camsoft
    I need to attach a single event handler to multiple DOM elements using jQuery. I'm currently achieving this using jQuery's each() method. See below: $([elm1, elm2, elm3]).each(function() { this.click(eventHandler); }); I was wondering if there is a way to do this without using the each method and closure. I thought I would be able to do something like: $(elm1, elm2, elm3).click(eventHandler); This kind of statement works using Prototype.js but not in jQuery.The reason I ask the question is because after using both Prototype.js and jQuery I have found that jQuery requires simpler statements to achieve the same tasks in almost every area so I assume there must be a better way of doing this?

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  • Shortest command to calculate the sum of a column of output on Unix?

    - by Andrew
    I'm sure there is a quick and easy way to calculate the sum of a column of values on Unix systems (using something like awk or xargs perhaps), but writing a shell script to parse the rows line by line is the only thing that comes to mind at the moment. For example, what's the simplest way to modify the command below to compute and display the total for the SEGSZ column (70300)? ipcs -mb | head -6 IPC status from /dev/kmem as of Mon Nov 17 08:58:17 2008 T ID KEY MODE OWNER GROUP SEGSZ Shared Memory: m 0 0x411c322e --rw-rw-rw- root root 348 m 1 0x4e0c0002 --rw-rw-rw- root root 61760 m 2 0x412013f5 --rw-rw-rw- root root 8192

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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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  • Finding direction of travel in a world with wrapped edges

    - by crazy
    I need to find the shortest distance direction from one point in my 2D world to another point where the edges are wrapped (like asteroids etc). I know how to find the shortest distance but am struggling to find which direction it's in. The shortest distance is given by: int rows = MapY; int cols = MapX; int d1 = abs(S.Y - T.Y); int d2 = abs(S.X - T.X); int dr = min(d1, rows-d1); int dc = min(d2, cols-d2); double dist = sqrt((double)(dr*dr + dc*dc)); Example of the world : : T : :--------------:--------- : : : S : : : : : : T : : : :--------------: In the diagram the edges are shown with : and -. I've shown a wrapped repeat of the world at the top right too. I want to find the direction in degrees from S to T. So the shortest distance is to the top right repeat of T. but how do I calculate the direction in degreed from S to the repeated T in the top right? I know the positions of both S and T but I suppose I need to find the position of the repeated T however there more than 1. The worlds coordinates system starts at 0,0 at the top left and 0 degrees for the direction could start at West. It seems like this shouldn’t be too hard but I haven’t been able to work out a solution. I hope somone can help? Any websites would be appreciated.

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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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  • Inconsistent movement / line-of-sight around obstacles on a hexagonal grid

    - by Darq
    In a roguelike game I've been working on, one of my core design goals has been to allow the player to "Play the game, not the grid." In essence, I want the player's positioning to be tactical because of elements in the game world, not simply because some grid tiles are more advantageous than others, in relation to enemies. I am fine with world geometry not being realistic, but it needs to be consistent. In this process I have ran into most of the common problems (Square tiles? Diagonal movement, LOS, corner cases, etc.) and have moved to a hexagonal tile grid. For the most part this has been great, and I've not had too many inconsistencies. Recently however I have been stumped by the following: Points A and B are both distance 4 from the player (red lines). Line-of-sight to both are blocked by walls (black tiles). However, due to the hexagonal grid, A can be reached in 4 moves, whereas B requires 5 moves (blue lines). On a hex grid, "shortest path" seems divorced from "direct path", there may be multiple shortest paths to any point, but there is only one direct path (or two in some situations). This is fine, geometry need not be realistic. However this also seems inconsistent, similar obstacles are more effective in some positions than in others. A player running away from an enemy should be able to run in any direction, increasing the distance between the two actors. However when placing obstacles or traps between themselves and enemies, the player is best served by running in one of the six directions that don't have multiple shortest paths. Is there a way to rationalise this? Am I missing something that makes this behaviour consistent? Or is there a way to make this behaviour consistent? I am most certainly over-thinking this, but as it is one of my goals, I should do it due diligence.

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  • [python]: path between two nodes

    - by www.yegorov-p.ru
    I'm using networkx to work with graphs. I have pretty large graph (it's near 200 nodes in it) and I try to find all possible paths between two nodes. But, as I understand, networkx can find only shortest path. How can I get not just shortest path, but all possible paths?

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  • physics game programming box2d - orientating a turret-like object using torques

    - by egarcia
    This is a problem I hit when trying to implement a game using the LÖVE engine, which covers box2d with Lua scripting. The objective is simple: A turret-like object (seen from the top, on a 2D environment) needs to orientate itself so it points to a target. The turret is on the x,y coordinates, and the target is on tx, ty. We can consider that x,y are fixed, but tx, ty tend to vary from one instant to the other (i.e. they would be the mouse cursor). The turret has a rotor that can apply a rotational force (torque) on any given moment, clockwise or counter-clockwise. The magnitude of that force has an upper limit called maxTorque. The turret also has certain rotational inertia, which acts for angular movement the same way mass acts for linear movement. There's no friction of any kind, so the turret will keep spinning if it has an angular velocity. The turret has a small AI function that re-evaluates its orientation to verify that it points to the right direction, and activates the rotator. This happens every dt (~60 times per second). It looks like this right now: function Turret:update(dt) local x,y = self:getPositon() local tx,ty = self:getTarget() local maxTorque = self:getMaxTorque() -- max force of the turret rotor local inertia = self:getInertia() -- the rotational inertia local w = self:getAngularVelocity() -- current angular velocity of the turret local angle = self:getAngle() -- the angle the turret is facing currently -- the angle of the like that links the turret center with the target local targetAngle = math.atan2(oy-y,ox-x) local differenceAngle = _normalizeAngle(targetAngle - angle) if(differenceAngle <= math.pi) then -- counter-clockwise is the shortest path self:applyTorque(maxTorque) else -- clockwise is the shortest path self:applyTorque(-maxTorque) end end ... it fails. Let me explain with two illustrative situations: The turret "oscillates" around the targetAngle. If the target is "right behind the turret, just a little clock-wise", the turret will start applying clockwise torques, and keep applying them until the instant in which it surpasses the target angle. At that moment it will start applying torques on the opposite direction. But it will have gained a significant angular velocity, so it will keep going clockwise for some time... until the target will be "just behind, but a bit counter-clockwise". And it will start again. So the turret will oscillate or even go in round circles. I think that my turret should start applying torques in the "opposite direction of the shortest path" before it reaches the target angle (like a car braking before stopping). Intuitively, I think the turret should "start applying torques on the opposite direction of the shortest path when it is about half-way to the target objective". My intuition tells me that it has something to do with the angular velocity. And then there's the fact that the target is mobile - I don't know if I should take that into account somehow or just ignore it. How do I calculate when the turret must "start braking"?

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  • I am trying to build a list of limitations of all graph algorithms

    - by Jack
    Single Source shortest Path Dijkstra's - directed and undirected - works only for positive edge weights - cycles ?? Bellman Ford - directed - no cycles should exist All source shortest path Floyd Warshall - no info Minimum Spanning Tree ( no info about edge weights or nature of graph or cycles) Kruskal's Prim's - undirected Baruvka's

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  • Am I right about the differences between Floyd-Warshall, Dijkstra's and Bellman-Ford algorithms?

    - by Programming Noob
    I've been studying the three and I'm stating my inferences from them below. Could someone tell me if I have understood them accurately enough or not? Thank you. Dijkstra's algorithm is used only when you have a single source and you want to know the smallest path from one node to another, but fails in cases like this Floyd-Warshall's algorithm is used when any of all the nodes can be a source, so you want the shortest distance to reach any destination node from any source node. This only fails when there are negative cycles (this is the most important one. I mean, this is the one I'm least sure about:) 3.Bellman-Ford is used like Dijkstra's, when there is only one source. This can handle negative weights and its working is the same as Floyd-Warshall's except for one source, right? If you need to have a look, the corresponding algorithms are (courtesy Wikipedia): Bellman-Ford: procedure BellmanFord(list vertices, list edges, vertex source) // This implementation takes in a graph, represented as lists of vertices // and edges, and modifies the vertices so that their distance and // predecessor attributes store the shortest paths. // Step 1: initialize graph for each vertex v in vertices: if v is source then v.distance := 0 else v.distance := infinity v.predecessor := null // Step 2: relax edges repeatedly for i from 1 to size(vertices)-1: for each edge uv in edges: // uv is the edge from u to v u := uv.source v := uv.destination if u.distance + uv.weight < v.distance: v.distance := u.distance + uv.weight v.predecessor := u // Step 3: check for negative-weight cycles for each edge uv in edges: u := uv.source v := uv.destination if u.distance + uv.weight < v.distance: error "Graph contains a negative-weight cycle" Dijkstra: 1 function Dijkstra(Graph, source): 2 for each vertex v in Graph: // Initializations 3 dist[v] := infinity ; // Unknown distance function from 4 // source to v 5 previous[v] := undefined ; // Previous node in optimal path 6 // from source 7 8 dist[source] := 0 ; // Distance from source to source 9 Q := the set of all nodes in Graph ; // All nodes in the graph are 10 // unoptimized - thus are in Q 11 while Q is not empty: // The main loop 12 u := vertex in Q with smallest distance in dist[] ; // Start node in first case 13 if dist[u] = infinity: 14 break ; // all remaining vertices are 15 // inaccessible from source 16 17 remove u from Q ; 18 for each neighbor v of u: // where v has not yet been 19 removed from Q. 20 alt := dist[u] + dist_between(u, v) ; 21 if alt < dist[v]: // Relax (u,v,a) 22 dist[v] := alt ; 23 previous[v] := u ; 24 decrease-key v in Q; // Reorder v in the Queue 25 return dist; Floyd-Warshall: 1 /* Assume a function edgeCost(i,j) which returns the cost of the edge from i to j 2 (infinity if there is none). 3 Also assume that n is the number of vertices and edgeCost(i,i) = 0 4 */ 5 6 int path[][]; 7 /* A 2-dimensional matrix. At each step in the algorithm, path[i][j] is the shortest path 8 from i to j using intermediate vertices (1..k-1). Each path[i][j] is initialized to 9 edgeCost(i,j). 10 */ 11 12 procedure FloydWarshall () 13 for k := 1 to n 14 for i := 1 to n 15 for j := 1 to n 16 path[i][j] = min ( path[i][j], path[i][k]+path[k][j] );

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  • Reading graph inputs for a programming puzzle and then solving it

    - by Vrashabh
    I just took a programming competition question and I absolutely bombed it. I had trouble right at the beginning itself from reading the input set. The question was basically a variant of this puzzle http://codercharts.com/puzzle/evacuation-plan but also had an hour component in the first line(say 3 hours after start of evacuation). It reads like this This puzzle is a tribute to all the people who suffered from the earthquake in Japan. The goal of this puzzle is, given a network of road and locations, to determine the maximum number of people that can be evacuated. The people must be evacuated from evacuation points to rescue points. The list of road and the number of people they can carry per hour is provided. Input Specifications Your program must accept one and only one command line argument: the input file. The input file is formatted as follows: the first line contains 4 integers n r s t n is the number of locations (each location is given by a number from 0 to n-1) r is the number of roads s is the number of locations to be evacuated from (evacuation points) t is the number of locations where people must be evacuated to (rescue points) the second line contains s integers giving the locations of the evacuation points the third line contains t integers giving the locations of the rescue points the r following lines contain to the road definitions. Each road is defined by 3 integers l1 l2 width where l1 and l2 are the locations connected by the road (roads are one-way) and width is the number of people per hour that can fit on the road Now look at the sample input set 5 5 1 2 3 0 3 4 0 1 10 0 2 5 1 2 4 1 3 5 2 4 10 The 3 in the first line is the additional component and is defined as the number of hours since the resuce has started which is 3 in this case. Now my solution was to use Dijisktras algorithm to find the shortest path between each of the rescue and evac nodes. Now my problem started with how to read the input set. I read the first line in python and stored the values in variables. But then I did not know how to store the values of the distance between the nodes and what DS to use and how to input it to say a standard implementation of dijikstras algorithm. So my question is two fold 1.) How do I take the input of such problems? - I have faced this problem in quite a few competitions recently and I hope I can get a simple code snippet or an explanation in java or python to read the data input set in such a way that I can input it as a graph to graph algorithms like dijikstra and floyd/warshall. Also a solution to the above problem would also help. 2.) How to solve this puzzle? My algorithm was: Find shortest path between evac points (in the above example it is 14 from 0 to 3) Multiply it by number of hours to get maximal number of saves Also the answer given for the variant for the input set was 24 which I dont understand. Can someone explain that also. UPDATE: I get how the answer is 14 in the given problem link - it seems to be just the shortest path between node 0 and 3. But with the 3 hour component how is the answer 24 UPDATE I get how it is 24 - its a complete graph traversal at every hour and this is how I solve it Hour 1 Node 0 to Node 1 - 10 people Node 0 to Node 2- 5 people TotalRescueCount=0 Node 1=10 Node 2= 5 Hour 2 Node 1 to Node 3 = 5(Rescued) Node 2 to Node 4 = 5(Rescued) Node 0 to Node 1 = 10 Node 0 to Node 2 = 5 Node 1 to Node 2 = 4 TotalRescueCount = 10 Node 1 = 10 Node 2= 5+4 = 9 Hour 3 Node 1 to Node 3 = 5(Rescued) Node 2 to Node 4 = 5+4 = 9(Rescued) TotalRescueCount = 9+5+10 = 24 It hard enough for this case , for multiple evac and rescue points how in the world would I write a pgm for this ?

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