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  • What is stable as Ubuntu in kernel space, but also cutting-edge as Fedora in user land?

    - by HRJ
    I am a long time Fedora user (since Fedora 6). Previously I have used Gentoo (for 2 years) and Slackware (for 5 years). The thing I liked about Fedora is frequency of package updates + great community. But lately I have noticed that Fedora is becoming too cutting-edge, nay, bleeding-edge. They changed the DNS client to be strict, without any warning, which broke some of their own packages for two Fedora releases. More critically, their LVM modules are not compatible across Fedora 12, 13, 14 (sometimes). Ubuntu is nicely polished but seems too stable for my liking. Some of the user-space applications are two major version numbers behind (even in testing or unstable or whatever they call it). Is there any Linux distro that has the stability of Ubuntu in kernel space and the bleeding edge in applications (especially harmless applications like, say, Stellarium)?

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  • Returning a CSS element

    - by TMP
    Is there a way to return a CSS element? I was using Adobe Edge and adding some of my own code in their code tab, but in order to create boundaries I would need to keep track of margin-top or margin-left. The following code works to move the element "woo" but I'm not sure how to call the elements to add something like "|| sym.$("woo").css({"margin-left"0px"}) to the move left code. //Move RIGHT if (e.which == 39) { sym.$("woo").css({"margin-left":"+=10px"}); } //Move UP else if (e.which == 38) { sym.$("woo").css({"margin-top":"-=10px"}); } //Move Left else if (e.which == 37) { sym.$("woo").css({"margin-left":"-=10px"}); } //Move DOWN else if (e.which == 40) { sym.$("woo").css({"margin-top":"+=10px"}); } EDIT: I changed the Left if statement to the following: else if (e.which == 37 && ($("woo").css("margin-left")>0)) { It seems to be working to some extent except for now it won't move left at all! I tried doing <0 too in case I was screwing up a sign but it won't let me move the element left either.

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  • Use Case diagrams as a requirements gathering tool for new functionality - particularly in systems t

    - by drelihan
    Hi Folks I'm interested in persuing the idea of using Use Case Diagrams as a tool for collecting user requirements. However, it will be for new features as opposed to developing a system from scratch. Also, the system only has a small level of user interaction - most of the actors will be external systems. I want to know what people's experiances have been with using this method of gathering requirements. How did your customers respond to the change and was it positive? Did it just not work for anybody? Thanks,

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  • How to detect if breaking an edge will make a graph disjoint?

    - by the_graph_guy
    I have a graph that starts off with a single, root node. Nodes are added one by one to the graph. At node creation time, they have to be linked either to the root node, or to another node, by a single edge. Edges can also be created and deleted (one by one, between any two nodes). Nodes can be deleted one at a time. Node and edge creation, deletion operations can happen in any arbitrary order. OK, so here's my question: When an edge is deleted, is it possible do determine, in constant time (i.e. with an O(1) algorithm), if doing this will divide the graph into two disjoint subgraphs? If it will, then which side of the edge will the root node belong? I'm willing to maintain, within reasonable limits, any additional data structure that can facilitate the derivation of this information. Maybe it is not possible to do it in O(1), if so any pointers to literature will be appreciated.

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  • output image is not displayed

    - by gerry chocolatos
    so, im doing an image detection which i have to process on each red, green, n blue element to get the edge map and combine them become one to show the output. but it doesnt show my output image would anyone pls be kind enough to help me? here is my code so far. //get the red element process_red = new int[width * height]; counter = 0; for(int i = 0; i < 256; i++) { for(int j = 0; j < 256; j++) { int clr = buff_red.getRGB(j, i); int red = (clr & 0x00ff0000) >> 16; red = (0xFF<<24)|(red<<16)|(red<<8)|red; process_red[counter] = red; counter++; } } //set threshold value for red element int threshold = 100; for (int x = 0; x < width; x++) { for (int y = 0; y < height; y++) { int bin = (buff_red.getRGB(x, y) & 0x000000ff); if (bin < threshold) bin = 0; else bin = 255; buff_red.setRGB(x,y, 0xff000000 | bin << 16 | bin << 8 | bin); } } and i do the same way for my green n blue elements. and then i wanted to get to combination of the three by doing it this way: //combine the three elements process_combine = new int[width * height]; counter = 0; for(int i = 0; i < 256; i++) { for(int j = 0; j < 256; j++) { int clr_a = buff_red.getRGB(j, i); int ar = clr_a & 0x000000ff; int clr_b = buff_green.getRGB(j, i); int bg = clr_b & 0x000000ff; int clr_c = buff_blue.getRGB(j, i); int cb = clr_b & 0x000000ff; int alpha = 0xff000000; int combine = alpha|(ar<<16)|(bg<<8)|cb; process_combine[counter] = combine; counter++; } } buff_rgb = new BufferedImage(width,height, BufferedImage.TYPE_INT_ARGB); Graphics rgb; rgb = buff_rgb.getGraphics(); rgb.drawImage(output_rgb, 0, 0, null); rgb.dispose(); repaint(); and to show the output whic is from the combining process, i use a draw method: g.drawImage(buff_rgb,800,100,this); but still it doesnt show the image. can anyone pls help me? ur help is really appreciated. thanks.

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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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  • Showing Egde Shaped Event Duration in StreamInsight using Debugger

    Whilst writing some courseware I wanted to be able to see the start and end times of Edge shaped events from within the debugger.  A quick recap on Edge events At the start of the event you do not know the end time and most probably cannot work it out or you should be using one of the other shapes. You enqueue an event (Start Edge) with the start time and payload of the event.  The end time of the event is set to infinity When you see the end edge come through, you enqueue another event (End Edge) with the previous start time and payload and restate the event’s end time.  This is the Retract Event All seems simple enough.  The problem is the debugger is a little shy about showing you what you need but you can get it to show you everything by also reading this article Here’s what I mean. Here is what the Event Debugger looks like by default when viewing 2 complete edge events.  Notice how all the end times are set to infinity   The above does not tell you for how long an event was valid.  I then add the “NewEndTime” column to the debugger output and there I can now see the duration of events.  You will see the Retract events (End Edge) have the same start time and payload as their respective start events (Start Edge)   You can follow the exact same logic when looking at Interval shape events.  They look a little different on the output adapter but using this article you can easily see what is happening.

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  • How to determine edges in an images optimally?

    - by SorinA.
    I recently was put in front of the problem of cropping and resizing images. I needed to crop the 'main content' of an image for example if i had an image similar to this: the result should be an image with the msn content without the white margins(left& right). I search on the X axis for the first and last color change and on the Y axis the same thing. The problem is that traversing the image line by line takes a while..for an image that is 2000x1600px it takes up to 2 seconds to return the CropRect = x1,y1,x2,y2 data. I tried to make for each coordinate a traversal and stop on the first value found but it didn't work in all test cases..sometimes the returned data wasn't the expected one and the duration of the operations was similar.. Any idea how to cut down the traversal time and discovery of the rectangle round the 'main content'? public static CropRect EdgeDetection(Bitmap Image, float Threshold) { CropRect cropRectangle = new CropRect(); int lowestX = 0; int lowestY = 0; int largestX = 0; int largestY = 0; lowestX = Image.Width; lowestY = Image.Height; //find the lowest X bound; for (int y = 0; y < Image.Height - 1; ++y) { for (int x = 0; x < Image.Width - 1; ++x) { Color currentColor = Image.GetPixel(x, y); Color tempXcolor = Image.GetPixel(x + 1, y); Color tempYColor = Image.GetPixel(x, y + 1); if ((Math.Sqrt(((currentColor.R - tempXcolor.R) * (currentColor.R - tempXcolor.R)) + ((currentColor.G - tempXcolor.G) * (currentColor.G - tempXcolor.G)) + ((currentColor.B - tempXcolor.B) * (currentColor.B - tempXcolor.B))) > Threshold)) { if (lowestX > x) lowestX = x; if (largestX < x) largestX = x; } if ((Math.Sqrt(((currentColor.R - tempYColor.R) * (currentColor.R - tempYColor.R)) + ((currentColor.G - tempYColor.G) * (currentColor.G - tempYColor.G)) + ((currentColor.B - tempYColor.B) * (currentColor.B - tempYColor.B))) > Threshold)) { if (lowestY > y) lowestY = y; if (largestY < y) largestY = y; } } } if (lowestX < Image.Width / 4) cropRectangle.X = lowestX - 3 > 0 ? lowestX - 3 : 0; else cropRectangle.X = 0; if (lowestY < Image.Height / 4) cropRectangle.Y = lowestY - 3 > 0 ? lowestY - 3 : 0; else cropRectangle.Y = 0; cropRectangle.Width = largestX - lowestX + 8 > Image.Width ? Image.Width : largestX - lowestX + 8; cropRectangle.Height = largestY + 8 > Image.Height ? Image.Height - lowestY : largestY - lowestY + 8; return cropRectangle; } }

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  • How to determine edges in an image optimally?

    - by SorinA.
    I recently was put in front of the problem of cropping and resizing images. I needed to crop the 'main content' of an image for example if i had an image similar to this: the result should be an image with the msn content without the white margins(left& right). I search on the X axis for the first and last color change and on the Y axis the same thing. The problem is that traversing the image line by line takes a while..for an image that is 2000x1600px it takes up to 2 seconds to return the CropRect = x1,y1,x2,y2 data. I tried to make for each coordinate a traversal and stop on the first value found but it didn't work in all test cases..sometimes the returned data wasn't the expected one and the duration of the operations was similar.. Any idea how to cut down the traversal time and discovery of the rectangle round the 'main content'? public static CropRect EdgeDetection(Bitmap Image, float Threshold) { CropRect cropRectangle = new CropRect(); int lowestX = 0; int lowestY = 0; int largestX = 0; int largestY = 0; lowestX = Image.Width; lowestY = Image.Height; //find the lowest X bound; for (int y = 0; y < Image.Height - 1; ++y) { for (int x = 0; x < Image.Width - 1; ++x) { Color currentColor = Image.GetPixel(x, y); Color tempXcolor = Image.GetPixel(x + 1, y); Color tempYColor = Image.GetPixel(x, y + 1); if ((Math.Sqrt(((currentColor.R - tempXcolor.R) * (currentColor.R - tempXcolor.R)) + ((currentColor.G - tempXcolor.G) * (currentColor.G - tempXcolor.G)) + ((currentColor.B - tempXcolor.B) * (currentColor.B - tempXcolor.B))) > Threshold)) { if (lowestX > x) lowestX = x; if (largestX < x) largestX = x; } if ((Math.Sqrt(((currentColor.R - tempYColor.R) * (currentColor.R - tempYColor.R)) + ((currentColor.G - tempYColor.G) * (currentColor.G - tempYColor.G)) + ((currentColor.B - tempYColor.B) * (currentColor.B - tempYColor.B))) > Threshold)) { if (lowestY > y) lowestY = y; if (largestY < y) largestY = y; } } } if (lowestX < Image.Width / 4) cropRectangle.X = lowestX - 3 > 0 ? lowestX - 3 : 0; else cropRectangle.X = 0; if (lowestY < Image.Height / 4) cropRectangle.Y = lowestY - 3 > 0 ? lowestY - 3 : 0; else cropRectangle.Y = 0; cropRectangle.Width = largestX - lowestX + 8 > Image.Width ? Image.Width : largestX - lowestX + 8; cropRectangle.Height = largestY + 8 > Image.Height ? Image.Height - lowestY : largestY - lowestY + 8; return cropRectangle; } }

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  • Should back end processes be included in use cases in requirements document?

    - by bizso09
    We're writing a requirements document for our client and need to include the use cases of the system. We're following this template: ID Description Actors Precondition Basic Steps Alternate Steps Exceptions Business validations/Rules Postconditions In the Basic Steps section, should we include steps that the system performs in the back end or should we only include steps that the user directly interacts with? Example: Basic Steps for Search 1: User goes to search page User enters term User presses search System matches search term with database entries System displays results vs Basic Steps for Search 2: User goes to search page User enters term User presses search System displays results

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  • XNA Diffuse Shader Issue. Edge lighting problem. Image Attached

    - by adtither
    As you can see in this image the diffuse shading is working correctly in some places but in other places such as the the bottom of the sphere you can see the squares/triangles of the mesh. Any idea what would be causing this? Let me know if you need anymore information related to code. I can upload my normals calculations and shader effect if required. EDIT: Here's a link to the shader I'm using http://pastebin.com/gymVc7CP Link to normals calculations: http://pastebin.com/KnMGdzHP Seems to be an issue with edge lighting. Can't seem to see where I'm going wrong with the normals calculations though.

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  • How to adjust TouchPad edge scroll area in Ubuntu?

    - by MikeVB
    I have an Acer Aspire that I have dual booting XP and Ubuntu. On the Windows side, the driver allows you to set how close your finger has to be to the edge of the TouchPad before it goes into scroll mode. In Ubuntu I don't have the option (at least in the GUI) to change the scroll area on the pad. Is there a conf file or other way to change this? I'm constantly getting into the scroll area during normal usage. I would like to leave it on without losing so much pad area to the scroll feature.

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  • Is there anyone with an Thinkpad Edge E320 with working Fn-Keys?

    - by Wauzl
    I recently bought a new Lenovo ThinkPad Edge E320. I'm having trouble with my Fn-Keys. The ones for Sound, Brightness and MUltimedia work but those that toggle Wifi/Bluetooth, Webcam and Microphone on and off won't. Fortunately the Wifi, Webcam and Mic are enabled by default. My problem is that I can't disable them (which would be nice to know that this works) and that I can't enable Bluetooth. (I checked that my machine has bluetooth support) Can't say if it would work in Windows as I purchased a machine without Windows. Here is my dmesg: http://pastebin.com/gAgYeDG7 Thanks for reading :)

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  • How to find optimal path visit every node with parallel workers complicated by dynamic edge costs?

    - by Aaron Anodide
    Say you have an acyclic directed graph with weighted edges and create N workers. My goal is to calculate the optimal way those workers can traverse the entire graph in parralel. However, edge costs may change along the way. Example: A -1-> B A -2-> C B -3-> C (if A has already been visited) B -5-> C (if A has not already been visited) Does what I describe lend itself to a standard algorithmic approach, or alternately can someone suggest if I'm looking at this in an inherently flawed way (i have an intuition I might be)?

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  • Creating an adjacency List for DFS

    - by user200081
    I'm having trouble creating a Depth First Search for my program. So far I have a class of edges and a class of regions. I want to store all the connected edges inside one node of my region. I can tell if something is connected by the getKey() function I have already implemented. If two edges have the same key, then they are connected. For the next region, I want to store another set of connected edges inside that region, etc etc. However, I am not fully understanding DFS and I'm having some trouble implementing it. I'm not sure when/where to call DFS again. Any help would be appreciated! class edge { private: int source, destination, length; int key; edge *next; public: getKey(){ return key; } } class region { edge *data; edge *next; region() { data = new edge(); next = NULL; } }; void runDFS(int i, edge **edge, int a) { region *head = new region(); aa[i]->visited == true;//mark the first vertex as true for(int v = 0; v < a; v++) { if(tem->edge[i].getKey() == tem->edge[v].getKey()) //if the edges of the vertex have the same root { if(head->data == NULL) { head->data = aa[i]; head->data->next == NULL; } //create an edge if(head->data) { head->data->next = aa[i]; head->data->next->next == NULL; }//if there is already a node connected to ti } if(aa[v]->visited == false) runDFS(v, edge, a); //call the DFS again } //for loop }

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  • How can I calculate a vertex normal for a hard edge?

    - by K.G.
    Here is a picture of a lovely polygon: Circled is a vertex, and numbered are its adjacent faces. I have calculated the normals of those faces as such (not yet normalized, 0-indexed): Vertex 1 normal 0: 0.000000 0.000000 -0.250000 Vertex 1 normal 1: 0.000000 0.000000 -0.250000 Vertex 1 normal 2: -0.250000 0.000000 0.000000 Vertex 1 normal 3: -0.250000 0.000000 0.000000 Vertex 1 normal 4: 0.250000 0.000000 0.000000 What I'm wondering is, how can I determine, taken as given that I want this vertex to represent a hard edge, whether its normal should be the normal of 1/2 or 3/4? My plan after I glanced at the sketch I used to put this together was "Ha! I'll just use whichever two faces have the same normal!" and now I see that there are two sets of two faces for which this is true. Is there a rule I can apply based on the face winding, angle of the adjacent edges, moon phase, coin flip, to consistently choose a normal direction for this box? For the record, all of the other polygons I plan to use will have their normals dictated in Maya, but after encountering this problem, it made me really curious.

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  • Is loose coupling w/o use cases an anti-pattern?

    - by dsimcha
    Loose coupling is, to some developers, the holy grail of well-engineered software. It's certainly a good thing when it makes code more flexible in the face of changes that are likely to occur in the foreseeable future, or avoids code duplication. On the other hand, efforts to loosely couple components increase the amount of indirection in a program, thus increasing its complexity, often making it more difficult to understand and often making it less efficient. Do you consider a focus on loose coupling without any use cases for the loose coupling (such as avoiding code duplication or planning for changes that are likely to occur in the foreseeable future) to be an anti-pattern? Can loose coupling fall under the umbrella of YAGNI?

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  • Which will be faster? Switching shaders or ignore that some cases don't need full code?

    - by PolGraphic
    I have two types of 2d objects: In first case (for about 70% of objects), I need that code in the shader: float2 texCoord = input.TexCoord + textureCoord.xy But in the second case I have to use: float2 texCoord = fmod(input.TexCoord, texCoordM.xy - textureCoord.xy) + textureCoord.xy I can use second code also for first case, but it will be a little slower (fmod is useless here, input.TexCoord will be always lower than textureCoord.xy - textureCoord.xy for sure). My question is, which way will be faster: Making two independent shaders for both types of rectangles, group rectangles by types and switch shaders during rendering. Make one shader and use some if statement. Make one shader and ignore that sometimes (70% of cases) I don't need to use fmod.

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  • Black screen with thinkpad edge e525, AMD radeon HD before and after installation, plymouth?

    - by carolien
    I have a new thinkpad edge e525 (which should work) and wanted to try ubuntu (first linux experience). With the ubuntu desktop-cd I had a blackscreen before installation, I could fix that with nomodeset. After the installation, when rebooting I got the black screen again. First I see some ubuntu colour with a green busted stripe. I have the same problem with the live boot, also tried mint live, the stripe gets pink. I installed xubuntu 64bit from an alternate cd and I didnt have problems until de reboot after the installation. Same problem, just a black screen with the green stripe than blackscreen. I tried several things: adding nomodeset to the bootmenu, but than I get: No connection to plymouth and it is stuck at checking the battery status. I tried to replace quiet splash with text. Didnt work either. actually one time I saw a blue booting image before that. I managed to get a root command though via the recovery boot. I didnt dare to just remove the plymouth package. Can I just delete plymouth or do I have to deinstall it as described in several ways. (Problem is, right now the notebook doesnt have a internet connection. And I dont know how to manage it with a usb stick?) Or do I have to install the AMD catalyst driver manually? (Again, how can I do that with an usb stick) I also tried this: Ubuntu hits a black screen after boot. Is there any solution without internet? And if I need internet which is the right one? Please explain step by step what I have to write and so on! Thank you! also tried: set gfxpayload=text and: just remove splash and adding vesafb.nonsense=1 to the grub editor, because I couldnt find it anywhere

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  • In what cases should I install (and configure) Postfix as a desktop user?

    - by Gonzalo
    Possible cases: 1) I plan to do Debian packaging (this case is the motivation since postfix gets installed as a dependency of some development packages, so it means that in such a case might be necessary). 2) I plan to use Evolution and a Internet provider mail account. 3) I plan to use gmail. Surely if I read Postfix documentation I may find the answer, but its huge and couldn't find it. In any case how (or where) should I find the answer to a question like that by myself? (I really tried)

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  • Anybody makes stylish PC cases that look like the Cobalt Cube?

    - by OverTheRainbow
    Hello I'm looking for computer cases that don't look like regular PC's in case customers would like to keep it on their desktop so it has to look stylish. I liked the Cobalt Cube: http://en.wikipedia.org/wiki/Cobalt_Qube Does someone know if some company makes cases that look the same (I know about Shuttle : they're too pricey), and possibly one that has a really quiet power supply? Thank you.

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  • Cutting Edge versus Just Average? Your SOA, Got BPM? by Mala Ramakrishnan

    - by JuergenKress
    Service Oriented Architecture (SOA) has completely transformed IT from the time it was introduced well over a decade ago. Organizations have been re-plumbing their infrastructure for reusability, efficiency and gain and succeeding with it. Best practices have emerged and people and technology have matured. We have got better at delivering on a stable platform on mission critical applications and services. Yet, there is this one secret that sets some SOA customers apart from the others. These companies grow and revolutionize their business and not just transform their IT infrastructure. The differences seem subtle for an untrained eye examining these organizations externally. And from within the company, it’s a bit like an ant sitting on an elephant, hard to differentiate between the IT trunk and business tail. What is it that some organizations do differently that makes them succeed beyond SOA? These organizations pull in business people more and more to weigh into their IT decisions. They wrench understanding process over services. They don’t settle easily when bridging business metrics and IT performance. They anguish over business requirements not translating seamlessly and quickly into IT. IT is not just an enabler but a pillar that revolutionizes their business. Okay, I’ll give it to you. These organizations layer Business Process Management (BPM) on top of their SOA. Think about lifeblood business processes in your own organizations. If you are Fedex, this would be shipping and handling. If you are Stanford Hospital, this would be patient case-management: from on-boarding through discharge and follow-up care. If you are Wells Fargo, this would be loan origination. Now think about how your SOA ties into your business process. Can you decouple your business processes from your SOA so that the two can transform and change independent of each other? Can you forecast success metrics for your business process, make the changes across the board and then look back over different periods of time to see if you are on track? Are your critical business processes entrenched in the minds of few experts in your organization or does everyone from the receptionist to your enterprise architect to your CEO understand what they can do to revolutionize it? Business Process Management is a superset of SOA. It is the process of getting your business to articulate business value and metrics and have it implemented in IT without any loss in translation. It is the act of extracting the business process from the minds of experts and IT applications in your organization and valuing them as assets for performance and gain. BPM is stepping outside your SOA and moving your organization to the next level of innovation. Oracle is accelerating BPM across industries with the latest launch. Join us to understand how BPM can give your organization a cutting edge over your SOA. SOA & BPM Partner Community For regular information on Oracle SOA Suite become a member in the SOA & BPM Partner Community for registration please visit www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Facebook Wiki Mix Forum Technorati Tags: SOA,BPM,SOA Community,Oracle SOA,Oracle BPM,Community,OPN,Jürgen Kress

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  • How can I draw an arrow at the edge of the screen pointing to an object that is off screen?

    - by Adam Henderson
    I am wishing to do what is described in this topic: http://www.allegro.cc/forums/print-thread/283220 I have attempted a variety of the methods mentioned here. First I tried to use the method described by Carrus85: Just take the ratio of the two triangle hypontenuses (doesn't matter which triagle you use for the other, I suggest point 1 and point 2 as the distance you calculate). This will give you the aspect ratio percentage of the triangle in the corner from the larger triangle. Then you simply multiply deltax by that value to get the x-coordinate offset, and deltay by that value to get the y-coordinate offset. But I could not find a way to calculate how far the object is away from the edge of the screen. I then tried using ray casting (which I have never done before) suggested by 23yrold3yrold: Fire a ray from the center of the screen to the offscreen object. Calculate where on the rectangle the ray intersects. There's your coordinates. I first calculated the hypotenuse of the triangle formed by the difference in x and y positions of the two points. I used this to create a unit vector along that line. I looped through that vector until either the x coordinate or the y coordinate was off the screen. The two current x and y values then form the x and y of the arrow. Here is the code for my ray casting method (written in C++ and Allegro 5) void renderArrows(Object* i) { float x1 = i->getX() + (i->getWidth() / 2); float y1 = i->getY() + (i->getHeight() / 2); float x2 = screenCentreX; float y2 = ScreenCentreY; float dx = x2 - x1; float dy = y2 - y1; float hypotSquared = (dx * dx) + (dy * dy); float hypot = sqrt(hypotSquared); float unitX = dx / hypot; float unitY = dy / hypot; float rayX = x2 - view->getViewportX(); float rayY = y2 - view->getViewportY(); float arrowX = 0; float arrowY = 0; bool posFound = false; while(posFound == false) { rayX += unitX; rayY += unitY; if(rayX <= 0 || rayX >= screenWidth || rayY <= 0 || rayY >= screenHeight) { arrowX = rayX; arrowY = rayY; posFound = true; } } al_draw_bitmap(sprite, arrowX - spriteWidth, arrowY - spriteHeight, 0); } This was relatively successful. Arrows are displayed in the bottom right section of the screen when objects are located above and left of the screen as if the locations of the where the arrows are drawn have been rotated 180 degrees around the center of the screen. I assumed this was due to the fact that when I was calculating the hypotenuse of the triangle, it would always be positive regardless of whether or not the difference in x or difference in y is negative. Thinking about it, ray casting does not seem like a good way of solving the problem (due to the fact that it involves using sqrt() and a large for loop). Any help finding a suitable solution would be greatly appreciated, Thanks Adam

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