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  • Eigenvector computation using OpenCV

    - by Andriyev
    Hi I have this matrix A, representing similarities of pixel intensities of an image. For example: Consider a 10 x 10 image. Matrix A in this case would be of dimension 100 x 100, and element A(i,j) would have a value in the range 0 to 1, representing the similarity of pixel i to j in terms of intensity. I am using OpenCV for image processing and the development environment is C on Linux. Objective is to compute the Eigenvectors of matrix A and I have used the following approach: static CvMat mat, *eigenVec, *eigenVal; static double A[100][100]={}, Ain1D[10000]={}; int cnt=0; //Converting matrix A into a one dimensional array //Reason: That is how cvMat requires it for(i = 0;i < affnDim;i++){ for(j = 0;j < affnDim;j++){ Ain1D[cnt++] = A[i][j]; } } mat = cvMat(100, 100, CV_32FC1, Ain1D); cvEigenVV(&mat, eigenVec, eigenVal, 1e-300); for(i=0;i < 100;i++){ val1 = cvmGet(eigenVal,i,0); //Fetching Eigen Value for(j=0;j < 100;j++){ matX[i][j] = cvmGet(eigenVec,i,j); //Fetching each component of Eigenvector i } } Problem: After execution I get nearly all components of all the Eigenvectors to be zero. I tried different images and also tried populating A with random values between 0 and 1, but the same result. Few of the top eigenvalues returned look like the following: 9805401476911479666115491135488.000000 -9805401476911479666115491135488.000000 -89222871725331592641813413888.000000 89222862280598626902522986496.000000 5255391142666987110400.000000 I am now thinking on the lines of using cvSVD() which performs singular value decomposition of real floating-point matrix and might yield me the eigenvectors. But before that I thought of asking it here. Is there anything absurd in my current approach? Am I using the right API i.e. cvEigenVV() for the right input matrix (my matrix A is a floating point matrix)? cheers

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  • problems with scrolling a java TextArea

    - by Jonathan
    All, I am running into an issue using JTextArea and JScrollPane. For some reason the scroll pane appears to not recognize the last line in the document, and will only scroll down to the line before it. The scroll bar does not even change to a state where I can slide it until the lines in the document are two greater than the number of lines the textArea shows (it should happen as soon as it is one greater). Has anyone run into this before? What would be a good solution (I want to avoid having to add an extra 'blank' line to the end of the document, which I would have to remove every time I add a new line)? Here is how I instantiate the TextArea and ScrollPane: JFrame frame = new JFrame("Java Chat Program"); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); Container pane = frame.getContentPane(); if (!(pane.getLayout() instanceof BorderLayout)) { System.err.println("Error: UI Container does not implement BorderLayout."); System.exit(-1); } textArea = new JTextArea(); textArea.setPreferredSize(new Dimension(500, 100)); textArea.setEditable(false); textArea.setLineWrap(true); textArea.setWrapStyleWord(true); JScrollPane scroller = new JScrollPane(textArea); scroller.setVerticalScrollBarPolicy(ScrollPaneConstants.VERTICAL_SCROLLBAR_ALWAYS); pane.add(scroller, BorderLayout.CENTER); Here is the method I use to add a new line to textArea: public void println(String a) { textArea.append(" "+a+"\n"); textArea.setCaretPosition(textArea.getDocument().getLength()); } Thanks for your help, Jonathan EDIT: Also, as a side note, with the current code I have to manually scroll down. I assumed that setCaretPosition(doc.getLength()) in the println(line) method would automatically set the page to the bottom after a line is entered... Should that be the case, or do I need to do something differently?

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  • How to avoid reallocation using the STL (C++)

    - by Tue Christensen
    This question is derived of the topic: http://stackoverflow.com/questions/2280655/vector-reserve-c I am using a datastructur of the type vector<vector<vector<double> > >. It is not possible to know the size of each of these vector (except the outer one) before items (doubles) are added. I can get an approximate size (upper bound) on the number of items in each "dimension". A solution with the shared pointers might be the way to go, but I would like to try a solution where the vector<vector<vector<double> > > simply has .reserve()'ed enough space (or in some other way has allocated enough memory). Will A.reserve(500) (assumming 500 is the size or, alternatively an upper bound on the size) be enough to hold "2D" vectors of large size, say [1000][10000]? The reason for my question is mainly because I cannot see any way of reasonably estimating the size of the interior of A at the time of .reserve(500). An example of my question: vector A; A.reserve(500+1); vector temp2; vector temp1 (666,666); for(int i=0;i<500;i++) { A.push_back(temp2); for(int j=0; j< 10000;j++) { A.back().push_back(temp1); } } Will this ensure that no reallocation is done for A? If temp2.reserve(100000) and temp1.reserve(1000) where added at creation will this ensure no reallocation at all will occur at all? In the above please disregard the fact that memory could be wasted due to conservative .reserve() calls. Thank you all in advance!

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  • Problem with using APACHE-POI to convert PPT to Image

    - by SpawnCxy
    Hi all, I got a problem when I try to use Apache POI project to convert my PPT to Images.My code as follows: FileInputStream is = new FileInputStream("test.ppt"); SlideShow ppt = new SlideShow(is); is.close(); Dimension pgsize = ppt.getPageSize(); Slide[] slide = ppt.getSlides(); for (int i = 0; i < slide.length; i++) { BufferedImage img = new BufferedImage(pgsize.width, pgsize.height, BufferedImage.TYPE_INT_RGB); Graphics2D graphics = img.createGraphics(); //clear the drawing area graphics.setPaint(Color.white); graphics.fill(new Rectangle2D.Float(0, 0, pgsize.width, pgsize.height)); //render slide[i].draw(graphics); //save the output FileOutputStream out = new FileOutputStream("slide-" + (i+1) + ".png"); javax.imageio.ImageIO.write(img, "png", out); out.close(); It works fine except that all Chinese words are converted to some squares.The png image I got is like following image: Then how can I fix this?Thanks in advance!

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  • C# program that generates html pages - limit image dimensions on page

    - by Professor Mustard
    I have a C# program that generates a large number of html pages, based on various bits of data and images that I have stored on the file system. The html itself works just fine, but the images can vary greatly in their dimensions. I need a way to ensure that a given image won't exceed a certain size on the page. The simplest way to accomplish this would be through the html itself... if there was some kind of "maxwidth" or "maxheight" property I could set in the html, or maybe a way to force the image to fit inside a table cell (if I used something like this, I'd have to be sure that the non-offending dimension would automatically scale with the one that's being reduced). The problem is, I don't know much about this "fine tuning" kind of stuff in html, and it seems to be a tough thing to Google around for (most of the solutions involve some sort of html specialization; I'm just using plain html). Alternatively, I could determine the width and height of each image at runtime by examining the image in C#, and then setting width/height values in the html if the image's dimensions exceed a certain value. The problem here is that is seems incredibly inefficient to load an entire image into memory, just to get its dimensions. I would need a way to "peek" at an image and just get its size (it could be bmp, jpg, gif or png). Any recommendations for either approach would be greatly appreciated.

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  • Attempting to find a formula for tessellating rectangles onto a board, where middle square can't be

    - by timemirror
    I'm working on a spatial stacking problem... at the moment I'm trying to solve in 2D but will eventually have to make this work in 3D. I divide up space into n x n squares around a central block, therefore n is always odd... and I'm trying to find the number of locations that a rectangle of any dimension less than n x n (eg 1x1, 1x2, 2x2 etc) can be placed, where the middle square is not available. So far I've got this.. total number of rectangles = ((n^2 + n)^2 ) / 4 ..also the total number of squares = (n (n+1) (2n+1)) / 6 However I'm stuck in working out a formula to find how many of those locations are impossible as the middle square would be occupied. So for example: [] [] [] [] [x] [] [] [] [] 3 x 3 board... with 8 possible locations for storing stuff as mid square is in use. I can use 1x1 shapes, 1x2 shapes, 2x1, 3x1, etc... Formula gives me the number of rectangles as: (9+3)^2 / 4 = 144/4 = 36 stacking locations However as the middle square is unoccupiable these can not all be realized. By hand I can see that these are impossible options: 1x1 shapes = 1 impossible (mid square) 2x1 shapes = 4 impossible (anything which uses mid square) 3x1 = 2 impossible 2x2 = 4 impossible etc Total impossible combinations = 16 Therefore the solution I'm after is 36-16 = 20 possible rectangular stacking locations on a 3x3 board. I've coded this in C# to solve it through trial and error, but I'm really after a formula as I want to solve for massive values of n, and also to eventually make this 3D. Can anyone point me to any formulas for these kind of spatial / tessellation problem? Also any idea on how to take the total rectangle formula into 3D very welcome! Thanks!

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  • How to create offline OLAP cube in C#?

    - by jimmyjoe
    I have a problem with creating an offline OLAP cube from C# using following code: using (var connection = new OleDbConnection()) { connection.ConnectionString = "Provider=MSOLAP; Initial Catalog=[OCWCube]; Data Source=C:\\temp\\test.cub; CreateCube=CREATE CUBE [OCWCube] ( DIMENSION [NAME], LEVEL [Wszystkie] TYPE ALL, LEVEL [NAME], MEASURE [Liczba DESCRIPTIO] FUNCTION COUNT ); InsertInto=INSERT INTO OCWCube([Liczba DESCRIPTIO], [NAME].[NAME]) OPTIONS ATTEMPT_ANALYSIS SELECT Planners.DESCRIPTIO, Planners.NAME FROM Planners Planners; Source_DSN=\"CollatingSequence=ASCII;DefaultDir=c:\\temp;Deleted=1;Driver={Microsoft dBase Driver (*.dbf)};DriverId=277;FIL=dBase IV;MaxBufferSize=2048;MaxScanRows=8;PageTimeout=600;SafeTransactions=0;Statistics=0;Threads=3;UserCommitSync=Yes;\";Mode=Write;UseExistingFile=True"; try { connection.Open(); } catch (OleDbException e) { Console.WriteLine(e); } } I keep on getting the following exception: "Multiple-step operation generated errors. Check each OLE database status value. No action was taken." I took the connection string literally from OQY file generated by Excel. I had to add "Mode=Write" section, otherwise I was getting another exception ("file may be in use"). What is wrong with the connection string? How to diagnose the error? Somebody please guide me...

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  • Background image in a JFrame.

    - by thepandaatemyface
    Hi, This question has been asked a lot but everywhere the answers fall short. I can get a JFrame to display a background image just fine by extending JPanel and overriding paintComponent, like so: class BackgroundPanel extends JPanel { private ImageIcon imageIcon; public BackgroundPanel() { this.imageIcon = Icons.getIcon("foo"); } @Override protected void paintComponent(Graphics g) { super.paintComponent(g); g.drawImage(imageIcon.getImage(), 0,0,imageIcon.getIconWidth(),imageIcon.getIconHeight(),this); } } But now, how do you add a component on top of that background? When I go JFrame w = new JFrame() ; Container cp = w.getContentPane(); cp.setLayout(null); BackgroundPanel bg = new BackgroundPanel(); cp.add(bg); JPanel b = new JPanel(); b.setSize(new Dimension(30, 40)); b.setBackground(Color.red); cp.add(b); w.pack() w.setVisible(true) It shows the little red square (or any other component) and not the background, but when I remove cp.setLayout(null);, the background shows up but not my other component. I'm guessing this has something to do with the paintComponent not being called by the null LayoutManager, but I'm not at all familiar with how LayoutManagers work (this is a project for college and the assignment specifically says not to use a LayoutManager) When i make the image the background has to display null (and so, transparant (??)) the red square shows up so it might be that the background is actually above my other components) Does anyone anyone have any ideas? Thanks

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  • Suggest a good method with least lookup time complexity

    - by Amrish
    I have a structure which has 3 identifier fields and one value field. I have a list of these objects. To give an analogy, the identifier fields are like the primary keys to the object. These 3 fields uniquely identify an object. Class { int a1; int a2; int a3; int value; }; I would be having a list of say 1000 object of this datatype. I need to check for specific values of these identity key values by passing values of a1, a2 and a3 to a lookup function which would check if any object with those specific values of a1, a2 and a3 is present and returns that value. What is the most effective way to implement this to achieve a best lookup time? One solution I could think of is to have a 3 dimensional matrix of length say 1000 and populate the value in it. This has a lookup time of O(1). But the disadvantages are. 1. I need to know the length of array. 2. For higher identity fields (say 20), then I will need a 20 dimension matrix which would be an overkill on the memory. For my actual implementation, I have 23 identity fields. Can you suggest a good way to store this data which would give me the best look up time?

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  • MDX equivalent to SQL subqueries with aggregation

    - by James Lampe
    I'm new to MDX and trying to solve the following problem. Investigated calculated members, subselects, scope statements, etc but can't quite get it to do what I want. Let's say I'm trying to come up with the MDX equivalent to the following SQL query: SELECT SUM(netMarketValue) net, SUM(CASE WHEN netMarketValue > 0 THEN netMarketValue ELSE 0 END) assets, SUM(CASE WHEN netMarketValue < 0 THEN netMarketValue ELSE 0 END) liabilities, SUM(ABS(netMarketValue)) gross someEntity1 FROM ( SELECT SUM(marketValue) netMarketValue, someEntity1, someEntity2 FROM <some set of tables> GROUP BY someEntity1, someEntity2) t GROUP BY someEntity1 In other words, I have an account ledger where I hide internal offsetting transactions (within someEntity2), then calculate assets & liabilities after aggregating them by someEntity2. Then I want to see the grand total of those assets & liabilities aggregated by the bigger entity, someEntity1. In my MDX schema I'd presumably have a cube with dimensions for someEntity1 & someEntity2, and marketValue would be my fact table/measure. I suppose i could create another DSV that did what my subquery does (calculating net), and simply create a cube with that as my measure dimension, but I wonder if there is a better way. I'd rather not have 2 cubes (one for these net calculations and another to go to a lower level of granularity for other use cases), since it will be a lot of duplicate info in my database. These will be very large cubes.

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  • This is more a matlab/math brain teaser than a question

    - by gd047
    Here is the setup. No assumptions for the values I am using. n=2; % dimension of vectors x and (square) matrix P r=2; % number of x vectors and P matrices x1 = [3;5] x2 = [9;6] x = cat(2,x1,x2) P1 = [6,11;15,-1] P2 = [2,21;-2,3] P(:,1)=P1(:) P(:,2)=P2(:) modePr = [-.4;16] TransPr=[5.9,0.1;20.2,-4.8] pred_modePr = TransPr'*modePr MixPr = TransPr.*(modePr*(pred_modePr.^(-1))') x0 = x*MixPr Then it was time to apply the following formula to get myP , where µij is MixPr. I used this code to get it: myP=zeros(n*n,r); Ptables(:,:,1)=P1; Ptables(:,:,2)=P2; for j=1:r for i = 1:r; temp = MixPr(i,j)*(Ptables(:,:,i) + ... (x(:,i)-x0(:,j))*(x(:,i)-x0(:,j))'); myP(:,j)= myP(:,j) + temp(:); end end Some brilliant guy proposed this formula as another way to produce myP for j=1:r xk1=x(:,j); PP=xk1*xk1'; PP0(:,j)=PP(:); xk1=x0(:,j); PP=xk1*xk1'; PP1(:,j)=PP(:); end myP = (P+PP0)*MixPr-PP1 I tried to formulate the equality between the two methods and seems to be this one. To make things easier, I ignored from both methods the summation of matrix P. where the first part denotes the formula that I used, while the second comes from his code snippet. Do you think this is an obvious equality? If yes, ignore all the above and just try to explain why. I could only start from the LHS, and after some algebra I think I proved it equals to the RHS. However I can't see how did he (or she) think of it in the first place.

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  • How to get image's coordinate on JPanel

    - by Jessy
    This question is related to my previous question http://stackoverflow.com/questions/2376027/how-to-generate-cartesian-coordinate-x-y-from-gridbaglayout I have successfully get the coordinate of each pictures, however when I checked the coordinate through (System.out.println) and the placement of the images on the screen, it seems to be wrong. e.g. if on the screen it was obvious that the x point of the first picture is on cell 2 which is on coordinate of 20, but the program shows x=1. Here is part of the code: public Grid (){ setPreferredSize(new Dimension(600,600)); .... setLayout(new GridBagLayout()); GridBagConstraints gc = new GridBagConstraints(); gc.weightx = 1d; gc.weighty = 1d; gc.insets = new Insets(0, 0, 0, 0);//top, left, bottom, and right gc.fill = GridBagConstraints.BOTH; JLabel[][] label = new JLabel[ROWS][COLS]; Random rand = new Random(); // fill the panel with labels for (int i=0;i<IMAGES;i++){ ImageIcon icon = createImageIcon("myPics.jpg"); int r, c; do{ //pick random cell which is empty r = (int)Math.floor(Math.random() * ROWS); c = (int)Math.floor(Math.random() * COLS); } while (label[r][c]!=null); //randomly scale the images int x = rand.nextInt(50)+30; int y = rand.nextInt(50)+30; Image image = icon.getImage().getScaledInstance(x,y, Image.SCALE_SMOOTH); icon.setImage(image); JLabel lbl = new JLabel(icon); // Instantiate GUI components gc.gridx = r; gc.gridy = c; add(lbl, gc); //add(component, constraintObj); label[r][c] = lbl; } I checked the coordinate through this code: Component[] components = getComponents(); for (Component component : components) { System.out.println(component.getBounds()); }

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  • Getting problem in collision detection in Java Game

    - by chetans
    Hi I am developing Spaceship Game in which i am getting problem in collision detection of moving images Game has a spaceship and number of asteroids(obstacles) i want to detect the collision between them How can i do this?`package Game; import java.applet.Applet; import java.awt.Color; import java.awt.Dimension; import java.awt.Graphics; import java.awt.Image; import java.awt.MediaTracker; import java.awt.event.KeyEvent; import java.awt.event.KeyListener; import java.net.MalformedURLException; import java.net.URL; public class ThreadInApplet extends Applet implements KeyListener { private static final long serialVersionUID = 1L; Image[] asteroidImage; Image spaceshipImage; int[] XPosObst,YPosObst; int numberOfObstacles=0,XPosOfSpaceship,YPosOfSpaceship; int spaceButtnCntr=0,noOfObstaclesLevel=20; boolean gameStart=false,collideUp=false,collideDown=false,collideLeft=false,collideRight=false; private Image offScreenImage; private Dimension offScreenSize,d; private Graphics offScreenGraphics; int speedObstacles=1; String spaceshipImagePath="images/spaceship.png",obstacleImagepath="images/asteroid.png"; String buttonToStart="Press Space to start"; public void init() { try { asteroidImage=new Image[noOfObstaclesLevel]; XPosObst=new int[noOfObstaclesLevel]; YPosObst=new int[noOfObstaclesLevel]; XPosOfSpaceship=getWidth()/2-35; YPosOfSpaceship=getHeight()-100; spaceshipImage=getImage(new URL(getCodeBase(),spaceshipImagePath)); for(int i=0;i<noOfObstaclesLevel;i++) { asteroidImage[i]=getImage(new URL(getCodeBase(),obstacleImagepath)); XPosObst[i]=(int) (Math.random()*700); YPosObst[i]=0; } MediaTracker tracker = new MediaTracker (this); for(int i=0;i<noOfObstaclesLevel;i++) { tracker.addImage (asteroidImage[i], 0); } } catch (MalformedURLException e) { e.printStackTrace(); } setBackground(Color.black); addKeyListener(this); } public void paint(Graphics g) { g.setColor(Color.white); if(gameStart==false) { g.drawString(buttonToStart, (getWidth()/2)-60, getHeight()/2); } g.drawString("HEADfitted Solutions Pvt.Ltd.", (getWidth()/2)-80, getHeight()-20); for(int n=0;n<numberOfObstacles;n++) { if(n>0) g.drawImage(asteroidImage[n],XPosObst[n],YPosObst[n],this); } g.drawImage(spaceshipImage,XPosOfSpaceship,YPosOfSpaceship,this); } @SuppressWarnings("deprecation") public void update(Graphics g) { d = size(); if((offScreenImage == null) || (d.width != offScreenSize.width) || (d.height != offScreenSize.height)) { offScreenImage = createImage(d.width, d.height); offScreenSize = d; offScreenGraphics = offScreenImage.getGraphics(); } offScreenGraphics.clearRect(0, 0, d.width, d.height); paint(offScreenGraphics); g.drawImage(offScreenImage, 0, 0, null); } public void keyReleased(KeyEvent arg0){} public void keyTyped(KeyEvent arg0) {} Thread mainThread=new Thread() { synchronized public void run () { try { //System.out.println("in main thread"); if (gameStart==true) { moveObstacles.start(); if(collide()==false) { createObsThread.start(); } } } catch (Exception e) { e.printStackTrace(); } } }; Thread createObsThread=new Thread() { synchronized public void run () { if (spaceButtnCntr==1) { if (collide()==false) { for(int g=0;g<noOfObstaclesLevel;g++) { try { sleep(1000); } catch (InterruptedException e) { e.printStackTrace(); } numberOfObstacles++; } } } } }; Thread moveObstacles=new Thread() // Moving Obstacle images downwards after every 10 ms { synchronized public void run () { while(YPosObst[19]!=600) { if (collide()==false) { //createObsThread.start(); for(int l=0;l } repaint(); try { sleep(10); } catch (InterruptedException e) { e.printStackTrace(); } } } } }; public void keyPressed(KeyEvent e) { if(e.getKeyCode()==32) { gameStart=true; spaceButtnCntr++; if (spaceButtnCntr==1) { mainThread.start(); } } if(gameStart==true) { if(e.getKeyCode()==37 && collideLeft==false)//Spaceship movement left { new Thread () { synchronized public void run () { XPosOfSpaceship-=10; repaint(); } }.start(); } if(e.getKeyCode()==38 && collideUp==false)//Spaceship movement up { new Thread () { synchronized public void run () { YPosOfSpaceship-=10; repaint(); } }.start(); } if(e.getKeyCode()==39 && collideRight==false)//Spaceship movement right { new Thread () { synchronized public void run () { XPosOfSpaceship+=10; repaint(); } }.start(); } if(e.getKeyCode()==40 && collideDown==false)//Spaceship movement down { new Thread () { synchronized public void run () { YPosOfSpaceship+=10; repaint(); } }.start(); } } } /*public boolean collide() { int x0, y0, w0, h0, x2, y2, w2, h2; x0=XPosOfSpaceship; y0=YPosOfSpaceship; h0=spaceshipImage.getHeight(null); w0=spaceshipImage.getWidth(null); for(int i=0;i<20;i++) { x2=XPosObst[i]; y2=YPosObst[i]; h2=asteroidImage[i].getHeight(null); w2=asteroidImage[i].getWidth(null); if ((x0 > (x2 + w2)) || ((x0 + w0) < x2)) return false; System.out.println(x2+" "+y2+" "+h2+" "+w2); if ((y0 > (y2 + h2)) || ((y0 + h0) < y2)) return false; } return true; }*/ public boolean collide() { int x1,y1,x2,y2,x3,y3,x4,y4; //coordinates of obstacles int a1,b1,a2,b2,a3,b3,a4,b4; //coordinates of spaceship a1 =XPosOfSpaceship; b1=YPosOfSpaceship; a2=a1+spaceshipImage.getWidth(this); b2=b1; a3=a1; b3=b1+spaceshipImage.getHeight(this); a4=a2; b4=b3; for(int a=0;a if(x1>=a1 && x1<=a2 && x1<=b3 && x1>=b1) return (true); if(x2>=a1 && x2<=a2 && x2<=b3 && x2>=b1) return(true); //********checking asteroid touch spaceship from up direction******** if(y3==b1 && x4>=a1 && x4<=a2) { collideUp = true; return(true); } if(y3==b1 && x3>=a1 && x3<=a2) { collideUp = true; return(true); } //********checking asteroid touch spaceship from left direction****** if(x2==a1 && y4>=b1 && y4<=b3) { collideLeft=true; return(true); } if(x2==a1 && y2>=b1 && y2<=b3) { collideLeft=true; return(true); } //********checking asteroid touch spaceship from right direction***** if(x1==a2 && y3>=b2 && y3<=b4) { collideRight=true; return(true); } if(x1==a2 && y1>=b2 && y1<=b4) { collideRight=true; return(true); } //********checking asteroid touch spaceship from down direction***** if(y1==b3 && x2>=a3 && x2<=a4) { collideDown=true; return(true); } if(y1==b3 && x1>=a3 && x1<=a4) { collideDown=true; return(true); } else { collideUp=false; collideDown=false; collideLeft=false; collideRight=false; } } return(false); } } `

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  • Code Golf: Easter Spiral

    - by friol
    What's more appropriate than a Spiral for Easter Code Golf sessions? Well, I guess almost anything. The Challenge The shortest code by character count to display a nice ASCII Spiral made of asterisks ('*'). Input is a single number, R, that will be the x-size of the Spiral. The other dimension (y) is always R-2. The program can assume R to be always odd and = 5. Some examples: Input 7 Output ******* * * * *** * * * * ***** * Input 9 Output ********* * * * ***** * * * * * * *** * * * * * ******* * Input 11 Output *********** * * * ******* * * * * * * * *** * * * * * * * * ***** * * * * * ********* * Code count includes input/output (i.e., full program). Any language is permitted. My easily beatable 303 chars long Python example: import sys; d=int(sys.argv[1]); a=[d*[' '] for i in range(d-2)]; r=[0,-1,0,1]; x=d-1;y=x-2;z=0;pz=d-2;v=2; while d>2: while v>0: while pz>0: a[y][x]='*'; pz-=1; if pz>0: x+=r[z]; y+=r[(z+1)%4]; z=(z+1)%4; pz=d; v-=1; v=2;d-=2;pz=d; for w in a: print ''.join(w); Now, enter the Spiral...

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  • Haskell Linear Algebra Matrix Library for Arbitrary Element Types

    - by Johannes Weiß
    I'm looking for a Haskell linear algebra library that has the following features: Matrix multiplication Matrix addition Matrix transposition Rank calculation Matrix inversion is a plus and has the following properties: arbitrary element (scalar) types (in particular element types that are not Storable instances). My elements are an instance of Num, additionally the multiplicative inverse can be calculated. The elements mathematically form a finite field (??2256). That should be enough to implement the features mentioned above. arbitrary matrix sizes (I'll probably need something like 100x100, but the matrix sizes will depend on the user's input so it should not be limited by anything else but the memory or the computational power available) as fast as possible, but I'm aware that a library for arbitrary elements will probably not perform like a C/Fortran library that does the work (interfaced via FFI) because of the indirection of arbitrary (non Int, Double or similar) types. At least one pointer gets dereferenced when an element is touched (written in Haskell, this is not a real requirement for me, but since my elements are no Storable instances the library has to be written in Haskell) I already tried very hard and evaluated everything that looked promising (most of the libraries on Hackage directly state that they wont work for me). In particular I wrote test code using: hmatrix, assumes Storable elements Vec, but the documentation states: Low Dimension : Although the dimensionality is limited only by what GHC will handle, the library is meant for 2,3 and 4 dimensions. For general linear algebra, check out the excellent hmatrix library and blas bindings I looked into the code and the documentation of many more libraries but nothing seems to suit my needs :-(. Update Since there seems to be nothing, I started a project on GitHub which aims to develop such a library. The current state is very minimalistic, not optimized for speed at all and only the most basic functions have tests and therefore should work. But should you be interested in using or helping out developing it: Contact me (you'll find my mail address on my web site) or send pull requests.

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  • Opengl-es draw an .obj file, but how?

    - by lacas
    I d like to parse an .obj file. My parser is working good, but my displaying is not good. Obj file is here my code is: public ObjModelParser parse() { long startTime = System.currentTimeMillis(); InputStream fileIn = resources.openRawResource(resourceID); BufferedReader buffer = new BufferedReader(new InputStreamReader(fileIn)); String line=""; Log.e("model loader", "Start parsing object " + resourceID); try { while ((line = buffer.readLine()) != null) { StringTokenizer parts = new StringTokenizer(line, " "); int numTokens = parts.countTokens(); if (numTokens == 0) continue; String part = parts.nextToken(); if (part.equals(VERTEX)) { Log.e("v ", line); vertices.add(Float.parseFloat(parts.nextToken())); vertices.add(Float.parseFloat(parts.nextToken())); vertices.add(Float.parseFloat(parts.nextToken())); .... and my displaying code is: draw that model with TRIANGLE_STRIP and gl.glDrawArrays(rendermode, 0, coords.length/dimension); What is the mistake here? edited: file here to show what is my good coords from my program for a cube, and what is from .obj file, that never show Thanks, Leslie

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  • C# Taking a element off each time (stack)

    - by Sef
    Greetings, Have a question considering a program that stimulates a stack.(not using any build in stack features or any such) stack2= 1 2 3 4 5 //single dimension array of 5 elements By calling up a method "pop" the stack should look like the following: Basically taking a element off each time the stack is being "called" up again. stack2= 1 2 3 4 0 stack2= 1 2 3 0 0 stack2= 1 2 0 0 0 stack2= 1 0 0 0 0 stack2= 0 0 0 0 0 - for (int i = 1; i <= 6; i++) { number= TryPop(s2); //use number ShowStack(s2, "s2"); } Basically I already have code that fills my array with values (trough a push method). The pop method should basically take the last value and place it on 0. Then calls up the next stack and place the following on 0 (like shown above in stack2). The current pop method that keeps track of the top index (0 elements = 0 top, 1 element = 1 top etc..). Already includes a underflow warning if this goes on 0 or below (which is correct). public int Pop() { if(top <= 0) { throw new Exception("Stack underflow..."); } else { for (int j = tabel.Length - 1; j >= 0; j--) { //...Really not sure what to do here. } } return number; }/*Pop*/ Since in the other class I already have a loop (for loop shown above) that simulates 6 times the s2 stack. (first stack: 1 2 3 4 0, second stack 1 2 3 0 0 and so on.) How exactly do I take a element off each time? Either I have the entire display on 0 or the 0 in the wrong places / out of index errors. Thanks in advance!

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  • How to display image within cell boundary

    - by Jessy
    How can I place an image within the cell boundary? I mean without taking the space of other cell? In the code below, random cells were selected to display images. One image in one cell. The problem is that, the image seems to take other cells as well. ... setPreferredSize(new Dimension(600,600)); final int ROWS = 6; final int COLS = 6; final int IMAGES = 10; setLayout(new GridBagLayout()); GridBagConstraints gc = new GridBagConstraints(); gc.weightx = 1d; gc.weighty = 1d; gc.insets = new Insets(0, 0, 0, 0);//top, left, bottom, and right gc.fill = GridBagConstraints.NONE; JLabel[][] label = new JLabel[ROWS][COLS]; Random rand = new Random(); // fill the panel with labels for (int i=0;i<IMAGES;i++){ ImageIcon icon = createImageIcon("myImage.jpg"); int r, c; do{ //pick random cell which is empty to avoid overlap image in the same cell r = (int)Math.floor(Math.random() * ROWS); c = (int)Math.floor(Math.random() * COLS); } while (label[r][c]!=null); //scale the image int x = rand.nextInt(20)+30; int y = rand.nextInt(20)+30; Image image = icon.getImage().getScaledInstance(x,y, Image.SCALE_SMOOTH); icon.setImage(image); JLabel lbl = new JLabel(icon); gc.gridx = r; gc.gridy = c; add(lbl, gc); //add image to the cell label[r][c] = lbl; }

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  • How to display image within cell boundary

    - by Jessy
    How can I place an image within the cell boundary? I mean without taking the space of other cell? In the code below, random cells were selected to display images. One image in one cell. The problem is that, the image seems to take other cells as well. ... setPreferredSize(new Dimension(600,600)); final int ROWS = 6; final int COLS = 6; final int IMAGES = 10; setLayout(new GridBagLayout()); GridBagConstraints gc = new GridBagConstraints(); gc.weightx = 1d; gc.weighty = 1d; gc.insets = new Insets(0, 0, 0, 0);//top, left, bottom, and right gc.fill = GridBagConstraints.NONE; JLabel[][] label = new JLabel[ROWS][COLS]; Random rand = new Random(); // fill the panel with labels for (int i=0;i<IMAGES;i++){ ImageIcon icon = createImageIcon("myImage.jpg"); int r, c; do{ //pick random cell which is empty to avoid overlap image in the same cell r = (int)Math.floor(Math.random() * ROWS); c = (int)Math.floor(Math.random() * COLS); } while (label[r][c]!=null); //scale the image int x = rand.nextInt(20)+30; int y = rand.nextInt(20)+30; Image image = icon.getImage().getScaledInstance(x,y, Image.SCALE_SMOOTH); icon.setImage(image); JLabel lbl = new JLabel(icon); gc.gridx = r; gc.gridy = c; add(lbl, gc); //add image to the cell label[r][c] = lbl; }

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  • JEditorPanes, Preferred Size, and printing HTML

    - by Ryan Elkins
    I'm trying to print some HTML directly, without displaying anything to the user. It works currently (somewhat) using a custom JEditorPane that implements Printable. The problem I'm having is that it always wants to use a preferred size of 582px x 560px. If I manually change the size using something like setSize(x,y) it will change the size of the pane, put the content renders at preferred size, not actual size (so it's still 582x560). I can scale it up to fit the page, but it's basically just an enlarged version where the images are all pixelated and the layout is wrong (based on the smaller window size). Inside the print method of my Printable JEditorPane I used this to try and get the size: javax.swing.JWindow wnd = new javax.swing.JWindow(); wnd.setContentPane(this); wnd.setSize(1024,1584); wnd.pack(); Dimension d = wnd.getPreferredSize(); With or without that setSize and/or pack methods on the JWindow the preferred size always comes back as 582x560. I do have control over the html that I'm trying to print but I'd rather not have to rewrite all of that to scale it down so it will print correctly at full size (when scaled up).

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  • BorderLayout problem with JSplitPane after adding JToolbar (Java)

    - by Alex Cheng
    Hello all. Problem: My program layout is fine, as below before I add JToolbar to BorderLayout.PAGE_START Here's a screenshot before JToolbar is added: Here's how it looked like after adding JToolbar: May I know what did I do wrong? Here's the code I used: //Create the text pane and configure it. textPane = new JTextPane(); -snipped code- JScrollPane scrollPane = new JScrollPane(textPane); scrollPane.setPreferredSize(new Dimension(300, 300)); //Create the text area for the status log and configure it. changeLog = new JTextArea(5, 30); changeLog.setEditable(false); JScrollPane scrollPaneForLog = new JScrollPane(changeLog); //Create a split pane for the change log and the text area. JSplitPane splitPane = new JSplitPane( JSplitPane.VERTICAL_SPLIT, scrollPane, scrollPaneForLog); splitPane.setOneTouchExpandable(true); //Create the status area. JPanel statusPane = new JPanel(new GridLayout(1, 1)); CaretListenerLabel caretListenerLabel = new CaretListenerLabel("Caret Status"); statusPane.add(caretListenerLabel); //Create the toolbar JToolBar toolBar = new JToolBar(); -snipped code- //Add the components. getContentPane().add(toolBar, BorderLayout.PAGE_START); getContentPane().add(splitPane, BorderLayout.CENTER); getContentPane().add(statusPane, BorderLayout.PAGE_END); //Set up the menu bar. actions = createActionTable(textPane); JMenu editMenu = createEditMenu(); JMenu styleMenu = createStyleMenu(); JMenuBar mb = new JMenuBar(); mb.add(editMenu); mb.add(styleMenu); setJMenuBar(mb); Please help, I'm new to GUI Building, and I don't feel like using Netbeans to drag and drop the UI for me... Thank you in advance.

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  • LWUIT HtmlComponent

    - by Pavan Kumar Ragi
    I've an Html String: String html="<p><img border=\"1\" align=\"left\" width=\"150\" vspace=\"2\" hspace=\"2\" height=\"159\" src=/"tmdbuserfiles/ntr-vv-vinayak-pics.jpg\" alt=\"Prithvi II, ballistic missile, DRDO, armed forces,Chandipur, Balasore district, Odisha State\" />The Strategic Forces Command of the armed forces successfully flight-tested the surface-to-surface Prithvi II missile from Chandipur in Balasore </P>"; I want to display the text as well as Image on my LWUIT Form Screen,For my Requirement I've used the below code: public class LwuitMidlet extends MIDlet { public void startApp() { Display.init(this); Form f = new Form("Hello, LWUIT!"); String html="<p><img border=\"1\" align=\"left\" width=\"150\" vspace=\"2\" hspace=\"2\" height=\"159\" src=www.teluguone.com/tmdbuserfiles/ntr-vv-vinayak-pics.jpg\" alt=\"Prithvi II, ballistic missile, DRDO, armed forces,Chandipur, Balasore district, Odisha State\" />The Strategic Forces Command of the armed forces successfully flight-tested the surface-to-surface Prithvi II missile from Chandipur in Balasore </P>"; HTMLComponent com=new HTMLComponent(); com.setPreferredSize(new Dimension(300,300)); com.setHTML(html, null, null, false); com.setShowImages(true); //com.setHTML(image, null, null, false); f.addComponent(com); f.show(); } public void pauseApp() { } public void destroyApp(boolean unconditional) { } } If i use the above code,I'm able to display only the text,but i'm not able to display the image,I've tested my app on Nokia SDK 2.O and SDK 3.0.5 can any one help me?

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  • how to reuse the area in a container when any component is removed from it?

    - by vybhav
    hi! i hav a panel and i m trying to remove labels from it which were added to it during run-time. but when labels are removed succesfully i m not able to use the space,left by that label,again to add any label to it. thanks in anticipation of the solution. here is the relevant code snippet: to add label to the panel: JLabel jl = new JLabel(); jl.setOpaque(true); jl.setIcon(new ImageIcon("D:/Project/router2.jpg")); jl.setBounds(x, y,jl.getPreferredSize().width,jl.getPreferredSize().height); for(Component c :lcomponent) { flag=true; Rectangle r4=c.getBounds(); int x1=(int) r4.getX(); int y1=(int) r4.getY(); Rectangle r5 = new Rectangle(new Point(x1-60, y1-60),new Dimension(170,170)); if(r5.contains(p)){ //To ensure that two labels do not overlap or are too close flag = false; // to each other break; }} if(flag) { p2.add(jl); //p2 is a panel Component c2 = p2.getComponentAt(x,y); p2.repaint(); lcomponent.add(c2); //lcomponent is an ArrayList<Component> to store all the labels added to the panel } 2.to remove the label: p2.remove(); p2.repaint();

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  • Object oriented design of game in Java: How to handle a party of NPCs?

    - by Arvanem
    Hi folks, I'm making a very simple 2D RPG in Java. My goal is to do this in as simple code as possible. Stripped down to basics, my class structure at the moment is like this: Physical objects have an x and y dimension. Roaming objects are physical objects that can move(). Humanoid objects are roaming objects that have inventories of GameItems. The Player is a singleton humanoid object that can hire up to 4 NPC Humanoids to join his or her party, and do other actions, such as fight non-humanoid objects. NPC Humanoids can be hired by the Player object to join his or her party, and once hired can fight for the Player. So far I have given the Player class a "party" ArrayList of NPC Humanoids, and the NPC Humanoids class a "hired" Boolean. However, my fight method is clunky, using an if to check the party size before implementing combat, e.g. public class Player extends Humanoids { private ArrayList<Humanoids> party; // GETTERS AND SETTERS for party here //... public void fightEnemy(Enemy eneObj) { if (this.getParty().size() == 0) // Do combat without party issues else if (this.getParty().size() == 1) // Do combat with party of 1 else if (this.getParty().size() == 2) // Do combat with party of 2 // etc. My question is, thinking in object oriented design, am I on the right track to do this in as simple code as possible? Is there a better way?

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  • TCL How to read , extract and count the occurent in .txt file (Current Directory)

    - by Passion
    Hi Folks, I am beginner to scripting and vigorously learning TCL for the development of embedded system. I have to Search for the files with only .txt format in the current directory, count the number of cases of each different "Interface # nnnn" string in .txt file, where nnnn is a four or 32 digits max hexadecimal no and o/p of a table of Interface number against occurrence. I am facing implementation issues while writing a script i.e, Unable to implement the data structure like Linked List, Two Dimensional array. I am rewriting a script using multi dimension array (Pass values into the arrays in and out of procedure) in TCL to scan through the every .txt file and search for the the string/regular expression ‘Interface # ’ to count and display the number of occurrences. If someone could help me to complete this part will be much appreciated. Search for only .txt extension files and obtain the size of the file Here is my piece of code for searching a .txt file in present directory set files [glob *.txt] if { [llength $files] > 0 } { puts "Files:" foreach f [lsort $files] { puts " [file size $f] - $f" } } else { puts "(no files)" } I reckon these are all the possible logical steps behind to complete it i) Once searched and find the .txt file then open all .txt files in read only mode ii) Create a array or list using the procedure (proc) Interface number to NULL and Interface count to zero 0 iii) Scan thro the .txt file and search for the string or regular expression "interface # iv) When a match found in .txt file, check the Interface Number and increment the count for the corresponding entry. Else add new element to the Interface Number list v) If there are no files return to the first directory My o/p is like follows Interface Frequency 123f 3 1232 4

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