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  • "Distance" between colours in PHP

    - by Phil
    I'm looking for a function that can accurately represent the distance between two colours as a number or something. For example I am looking to have an array of HEX values or RGB arrays and I want to find the most similar colour in the array for a given colour eg. I pass a function a RGB value and the 'closest' colour in the array is returned

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  • Efficiently generate numpy array from list comprehension output?

    - by shootingstars
    Is there a more efficient way than using numpy.asarray() to generate an array from output in the form of a list? This appears to be copying everything in memory, which doesn't seem like it would be that efficient with very large arrays. (Updated) Example: import numpy as np a1 = np.array([1,2,3,4,5,6,7,8,9,10]) # pretend this has thousands of elements a2 = np.array([3,7,8]) results = np.asarray([np.amax(np.where(a1 > element)) for element in a2])

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  • fill array with binary numbers

    - by davit-datuashvili
    hi, first of all this is not homework!! my question is from book: Algorithms in C++ third edition by robert sedgewick question is: there is given array of size n by 2^n(two dimensional) and we should fill it by binary numbers with bits size exactly n or for example n=5 so result will be 00001 00010 00011 00100 00101 00110 00111 and so on we should put this sequence of bits into arrays please help me

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  • Variable Length Array

    - by corinaf
    Hello, I would like to know how a variable length array is managed (what extra variables or data structures are kept on the stack in order to have variable length arrays). Thanks a lot.

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  • C/C++ variable length automatic array performance

    - by aaa
    hello. Is there significant cpu/memory overhead associated with using automatic arrays with g++/Intel on 64-bit x86 linux platform? int function(int N) { double array[N]; overhead compared to allocating array before hand (assuming function is called multiple times) overhead compared to using new overhead compared to using malloc range of N maybe from 1kb to 16kb roughly, stack overrun is not a problem Thank you

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  • Is there a static code analyzer [like Lint] for PHP files?

    - by eswald
    Is there a static code analyzer for PHP files? The binary itself can check for syntax errors, but I'm looking for something that does more, like unused variable assignments, arrays that are assigned into without being initialized first, and possibly code style warnings. Open-source programs would be preferred, but we might convince the company to pay for something if it's highly recommended.

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  • Can't re-mount existing RAID10 on Ubuntu

    - by Zoran
    I saw similar questions, but didn't find what solution to my problem. After power-cut, one of RAID10 (4 disks were) appears to be malfunctioning. I make tha array active one, but can not mount it. Always the same error: mount: you must specify the filesystem type So, here is what I have when type mdadm --detail /dev/md0 /dev/md0: Version : 00.90.03 Creation Time : Tue Sep 1 11:00:40 2009 Raid Level : raid10 Array Size : 1465148928 (1397.27 GiB 1500.31 GB) Used Dev Size : 732574464 (698.64 GiB 750.16 GB) Raid Devices : 4 Total Devices : 3 Preferred Minor : 0 Persistence : Superblock is persistent Update Time : Mon Jun 11 09:54:27 2012 State : clean, degraded Active Devices : 3 Working Devices : 3 Failed Devices : 0 Spare Devices : 0 Layout : near=2, far=1 Chunk Size : 64K UUID : 1a02e789:c34377a1:2e29483d:f114274d Events : 0.166 Number Major Minor RaidDevice State 0 8 16 0 active sync /dev/sdb 1 0 0 1 removed 2 8 48 2 active sync /dev/sdd 3 8 64 3 active sync /dev/sde At the /etc/mdadm/mdadm.conf I have by default, scan all partitions (/proc/partitions) for MD superblocks. alternatively, specify devices to scan, using wildcards if desired. DEVICE partitions auto-create devices with Debian standard permissions CREATE owner=root group=disk mode=0660 auto=yes automatically tag new arrays as belonging to the local system HOMEHOST <system> instruct the monitoring daemon where to send mail alerts MAILADDR root definitions of existing MD arrays ARRAY /dev/md0 level=raid10 num-devices=4 UUID=1a02e789:c34377a1:2e29483d:f114274d ARRAY /dev/md1 level=raid1 num-devices=2 UUID=9b592be7:c6a2052f:2e29483d:f114274d This file was auto-generated... So, my question is, how can I mount md0 array (md1 has been mounted without problem) in order to preserve existing data? One more thing, fdisk -l command gives the following result: Disk /dev/sdb: 750.1 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0x660a6799 Device Boot Start End Blocks Id System /dev/sdb1 * 1 88217 708603021 83 Linux /dev/sdb2 88218 91201 23968980 5 Extended /dev/sdb5 88218 91201 23968948+ 82 Linux swap / Solaris Disk /dev/sdc: 750.1 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0x0008f8ae Device Boot Start End Blocks Id System /dev/sdc1 1 88217 708603021 83 Linux /dev/sdc2 88218 91201 23968980 5 Extended /dev/sdc5 88218 91201 23968948+ 82 Linux swap / Solaris Disk /dev/sdd: 750.1 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0x4be1abdb Device Boot Start End Blocks Id System Disk /dev/sde: 750.1 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0xa4d5632e Device Boot Start End Blocks Id System Disk /dev/sdf: 750.1 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0xdacb141c Device Boot Start End Blocks Id System Disk /dev/sdg: 750.1 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0xdacb141c Device Boot Start End Blocks Id System Disk /dev/md1: 750.1 GB, 750156251136 bytes 2 heads, 4 sectors/track, 183143616 cylinders Units = cylinders of 8 * 512 = 4096 bytes Disk identifier: 0xdacb141c Device Boot Start End Blocks Id System Warning: ignoring extra data in partition table 5 Warning: ignoring extra data in partition table 5 Warning: ignoring extra data in partition table 5 Warning: invalid flag 0x7b6e of partition table 5 will be corrected by w(rite) Disk /dev/md0: 1500.3 GB, 1500312502272 bytes 255 heads, 63 sectors/track, 182402 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Disk identifier: 0x660a6799 Device Boot Start End Blocks Id System /dev/md0p1 * 1 88217 708603021 83 Linux /dev/md0p2 88218 91201 23968980 5 Extended /dev/md0p5 ? 121767 155317 269488144 20 Unknown And one more thing. When using mdadm --examine command, here ise result: mdadm -v --examine --scan /dev/sdb /dev/sdc /dev/sdd /dev/sde /dev/sdf /dev/sd ARRAY /dev/md1 level=raid1 num-devices=2 UUID=9b592be7:c6a2052f:2e29483d:f114274d devices=/dev/sdf ARRAY /dev/md0 level=raid10 num-devices=4 UUID=1a02e789:c34377a1:2e29483d:f114274d devices=/dev/sdb,/dev/sdc,/dev/sdd,/dev/sde md0 has 3 devices which are active. Can someone instruct me how to solve this issue? If it is possible, I would like not to removing faulty HDD. Please advise

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  • RAID controller dropping the wrong drive

    - by bramp
    I've been having an issue with 3ware 9500S-8 RAID 10, and I have contracted their tech support, but I wanted to hear the serverfault community's recommendations. Firstly, all my data is backuped and secure, so I don't mind blowing my RAID away if I have to. But let me describe the problem I've been seeing. A month ago, disk 6 dropped out of the RAID. It is mirrored with disk 7, so I wasn't that bothered. I went to the data centre and replaced it. When I got back to the office, I noticed that disk 6 will still not in the RAID, and in fact the controller was show the name of the old drive still. A week later I went back and replace the drive again, thinking I might have swapped in a bad drive. Still the same problem. I decided to reboot the machine, to see if that would "force" the controller into seeing the new drive. It did, and a rebuild started to happen (from disk 7). Eventually both drives were showing as good. A week later, the MySQL database has flagged the database is corrupt, and is unable to repair it. I don't know what has gone wrong, but I suspected this 6-7 pair. At this point I noticed that the RAID had constantly been verifying itself, over and over. Regardless of this I began to rebuild the database, which took about 19 hours. It's a big database. Near the end of the repair, the RAID controller told me it had dropped disk 7, and that some data was most likely corrupted. I contacted LSI tech support, and they very promptly started to help me. I mentioned that drive 7 had been dropped. They suspect that drive 7 was always at fault, and drive 6 had always been good. I want to know how often a RAID controller would drop the wrong drive (in this case dropping drive 6 a month ago, instead of 7). I foolishly didn't run smartctl on the drives before I started swapping them out. I just assumed the RAID controller knew what it was talking about. I think my plan of action is to replace drive 7, rebuild the array from scratch, double check smartctl on ALL the disks, and then start restoring my data again. I would appreciate anyone's input on what the correct procedure for swapping drives is, and how often failures like this happen. If anyone would like more information then I'd be happy to provide it. thanks in advance. Oh some more information. I'm running CentOS 5.3, with two RAID arrays, a simple RAID 1 for the OS, and RAID 10 for the database. Both arrays are on different controllers. The RAID 10 is made of 10 identical ST3640323AS drives, until I swapped in a SAMSUNG HD103SJ last month.

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  • [OpenGL ES - Android] Better way to generate tiles

    - by Inoe
    Hi ! I'll start by saying that i'm REALLY new to OpenGL ES (I started yesterday =), but I do have some Java and other languages experience. I've looked a lot of tutorials, of course Nehe's ones and my work is mainly based on that. As a test, I started creating a "tile generator" in order to create a small Zelda-like game (just moving a dude in a textured square would be awsome :p). So far, I have achieved a working tile generator, I define a char map[][] array to store wich tile is on : private char[][] map = { {0, 0, 20, 11, 11, 11, 11, 4, 0, 0}, {0, 20, 16, 12, 12, 12, 12, 7, 4, 0}, {20, 16, 17, 13, 13, 13, 13, 9, 7, 4}, {21, 24, 18, 14, 14, 14, 14, 8, 5, 1}, {21, 22, 25, 15, 15, 15, 15, 6, 2, 1}, {21, 22, 23, 0, 0, 0, 0, 3, 2, 1}, {21, 22, 23, 0, 0, 0, 0, 3, 2, 1}, {26, 0, 0, 0, 0, 0, 0, 3, 2, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1} }; It's working but I'm no happy with it, I'm sure there is a beter way to do those things : 1) Loading Textures : I create an ugly looking array containing the tiles I want to use on that map : private int[] textures = { R.drawable.herbe, //0 R.drawable.murdroite_haut, //1 R.drawable.murdroite_milieu, //2 R.drawable.murdroite_bas, //3 R.drawable.angledroitehaut_haut, //4 R.drawable.angledroitehaut_milieu, //5 }; (I cutted this on purpose, I currently load 27 tiles) All of theses are stored in the drawable folder, each one is a 16*16 tile. I then use this array to generate the textures and store them in a HashMap for a later use : int[] tmp_tex = new int[textures.length]; gl.glGenTextures(textures.length, tmp_tex, 0); texturesgen = tmp_tex; //Store the generated names in texturesgen for(int i=0; i < textures.length; i++) { //Bitmap bmp = BitmapFactory.decodeResource(context.getResources(), textures[i]); InputStream is = context.getResources().openRawResource(textures[i]); Bitmap bitmap = null; try { //BitmapFactory is an Android graphics utility for images bitmap = BitmapFactory.decodeStream(is); } finally { //Always clear and close try { is.close(); is = null; } catch (IOException e) { } } // Get a new texture name // Load it up this.textureMap.put(new Integer(textures[i]),new Integer(i)); int tex = tmp_tex[i]; gl.glBindTexture(GL10.GL_TEXTURE_2D, tex); //Create Nearest Filtered Texture gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_MIN_FILTER, GL10.GL_NEAREST); gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_MAG_FILTER, GL10.GL_LINEAR); //Different possible texture parameters, e.g. GL10.GL_CLAMP_TO_EDGE gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_WRAP_S, GL10.GL_REPEAT); gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_WRAP_T, GL10.GL_REPEAT); //Use the Android GLUtils to specify a two-dimensional texture image from our bitmap GLUtils.texImage2D(GL10.GL_TEXTURE_2D, 0, bitmap, 0); bitmap.recycle(); } I'm quite sure there is a better way to handle that... I just was unable to figure it. If someone has an idea, i'm all ears. 2) Drawing the tiles What I did was create a single square and a single texture map : /** The initial vertex definition */ private float vertices[] = { -1.0f, -1.0f, 0.0f, //Bottom Left 1.0f, -1.0f, 0.0f, //Bottom Right -1.0f, 1.0f, 0.0f, //Top Left 1.0f, 1.0f, 0.0f //Top Right }; private float texture[] = { //Mapping coordinates for the vertices 0.0f, 1.0f, 1.0f, 1.0f, 0.0f, 0.0f, 1.0f, 0.0f }; Then, in my draw function, I loop through the map to define the texture to use (after pointing to and enabling the buffers) : for(int y = 0; y < Y; y++){ for(int x = 0; x < X; x++){ tile = map[y][x]; try { //Get the texture from the HashMap int textureid = ((Integer) this.textureMap.get(new Integer(textures[tile]))).intValue(); gl.glBindTexture(GL10.GL_TEXTURE_2D, this.texturesgen[textureid]); } catch(Exception e) { return; } //Draw the vertices as triangle strip gl.glDrawArrays(GL10.GL_TRIANGLE_STRIP, 0, vertices.length / 3); gl.glTranslatef(2.0f, 0.0f, 0.0f); //A square takes 2x so I move +2x before drawing the next tile } gl.glTranslatef(-(float)(2*X), -2.0f, 0.0f); //Go back to the begining of the map X-wise and move 2y down before drawing the next line } This works great by I really think that on a 1000*1000 or more map, it will be lagging as hell (as a reminder, this is a typical Zelda world map : http://vgmaps.com/Atlas/SuperNES/LegendOfZelda-ALinkToThePast-LightWorld.png ). I've read things about Vertex Buffer Object and DisplayList but I couldn't find a good tutorial and nodoby seems to be OK on wich one is the best / has the better support (T1 and Nexus One are ages away). I think that's it, I've putted a lot of code but I think it helps. Thanks in advance !

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  • Translation of clustering problem to graph theory language

    - by honk
    I have a rectangular planar grid, with each cell assigned some integer weight. I am looking for an algorithm to identify clusters of 3 to 6 adjacent cells with higher-than-average weight. These blobs should have approximately circular shape. For my case the average weight of the cells not containing a cluster is around 6, and that for cells containing a cluster is around 6+4, i.e. there is a "background weight" somewhere around 6. The weights fluctuate with a Poisson statistic. For small background greedy or seeded algorithms perform pretty well, but this breaks down if my cluster cells have weights close to fluctuations in the background. Also, I cannot do a brute-force search looping through all possible setups because my grid is large (something like 1000x1000). I have the impression there might exist ways to tackle this in graph theory. I heard of vertex-covers and cliques, but am not sure how to best translate my problem into their language.

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  • OpenGL ES Simple Undo Last Drawing

    - by Erika
    Hi Everyone, I am trying to figure out how to implement a simple "undo" of last drawing action on the iPhone screen. I draw by first preparing the frame buffer: [EAGLContext setCurrentContext:context]; glBindFramebufferOES(GL_FRAMEBUFFER_OES, viewFramebuffer); I then prepare the vertex array and draw this way: glVertexPointer(2, GL_FLOAT, 0, vertexBuffer); glDrawArrays(GL_POINTS, 0, vertexCount); glBindRenderbufferOES(GL_RENDERBUFFER_OES, viewRenderbuffer); [context presentRenderbuffer:GL_RENDERBUFFER_OES]; How do I simple undo this last action? There has to be a way to save previous state or an built-in OpenGL ES function, I would think. Thanks

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  • problem with my texture coordinates on a square.

    - by Evan Kimia
    Im very new to OpenGL ES, and have been doing a tutorial to build a square. The square is made, and now im trying to map a 256 by 256 image onto it. The problem is, im only seeing a very zoomed in portion of this bitmap; Im fairly certain my texture coords are whats wrong here. Thanks! package se.jayway.opengl.tutorial; import java.io.IOException; import java.io.InputStream; import java.nio.ByteBuffer; import java.nio.ByteOrder; import java.nio.FloatBuffer; import java.nio.ShortBuffer; import javax.microedition.khronos.opengles.GL10; import android.content.Context; import android.graphics.Bitmap; import android.graphics.BitmapFactory; import android.opengl.GLUtils; public class Square { // Our vertices. private float vertices[] = { -1.0f, 1.0f, 0.0f, // 0, Top Left -1.0f, -1.0f, 0.0f, // 1, Bottom Left 1.0f, -1.0f, 0.0f, // 2, Bottom Right 1.0f, 1.0f, 0.0f, // 3, Top Right }; //Our texture. private float texture[] = { 0.0f, 0.0f, 0.0f, 0.0f, 1.0f, 0.0f, 1.0f, 1.0f, 0.0f, 1.0f, 0.0f, 0.0f, }; // The order we like to connect them. private short[] indices = { 0, 1, 2, 0, 2, 3 }; // Our vertex buffer. private FloatBuffer vertexBuffer; // Our index buffer. private ShortBuffer indexBuffer; //texture buffer. private FloatBuffer textureBuffer; //Our texture pointer. private int[] textures = new int[1]; public Square() { // a float is 4 bytes, therefore we multiply the number if // vertices with 4. ByteBuffer vbb = ByteBuffer.allocateDirect(vertices.length * 4); vbb.order(ByteOrder.nativeOrder()); vertexBuffer = vbb.asFloatBuffer(); vertexBuffer.put(vertices); vertexBuffer.position(0); // a float is 4 bytes, therefore we multiply the number of // vertices with 4. ByteBuffer tbb = ByteBuffer.allocateDirect(texture.length * 4); vbb.order(ByteOrder.nativeOrder()); textureBuffer = tbb.asFloatBuffer(); textureBuffer.put(texture); textureBuffer.position(0); // short is 2 bytes, therefore we multiply the number if // vertices with 2. ByteBuffer ibb = ByteBuffer.allocateDirect(indices.length * 2); ibb.order(ByteOrder.nativeOrder()); indexBuffer = ibb.asShortBuffer(); indexBuffer.put(indices); indexBuffer.position(0); } /** * This function draws our square on screen. * @param gl */ public void draw(GL10 gl) { // Counter-clockwise winding. gl.glFrontFace(GL10.GL_CCW); // Enable face culling. gl.glEnable(GL10.GL_CULL_FACE); // What faces to remove with the face culling. gl.glCullFace(GL10.GL_BACK); //Bind our only previously generated texture in this case gl.glBindTexture(GL10.GL_TEXTURE_2D, textures[0]); // Enabled the vertices buffer for writing and to be used during // rendering. gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); //Enable texture buffer array gl.glEnableClientState(GL10.GL_TEXTURE_COORD_ARRAY); // Specifies the location and data format of an array of vertex // coordinates to use when rendering. gl.glVertexPointer(3, GL10.GL_FLOAT, 0, vertexBuffer); gl.glTexCoordPointer(2, GL10.GL_FLOAT, 0, textureBuffer); gl.glDrawElements(GL10.GL_TRIANGLES, indices.length, GL10.GL_UNSIGNED_SHORT, indexBuffer); // Disable the vertices buffer. gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); //Disable the texture buffer. gl.glDisableClientState(GL10.GL_TEXTURE_COORD_ARRAY); // Disable face culling. gl.glDisable(GL10.GL_CULL_FACE); } /** * Load the textures * * @param gl - The GL Context * @param context - The Activity context */ public void loadGLTexture(GL10 gl, Context context) { //Get the texture from the Android resource directory InputStream is = context.getResources().openRawResource(R.drawable.test); Bitmap bitmap = null; try { //BitmapFactory is an Android graphics utility for images bitmap = BitmapFactory.decodeStream(is); } finally { //Always clear and close try { is.close(); is = null; } catch (IOException e) { } } //Generate one texture pointer... gl.glGenTextures(1, textures, 0); //...and bind it to our array gl.glBindTexture(GL10.GL_TEXTURE_2D, textures[0]); //Create Nearest Filtered Texture gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_MIN_FILTER, GL10.GL_NEAREST); gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_MAG_FILTER, GL10.GL_LINEAR); //Different possible texture parameters, e.g. GL10.GL_CLAMP_TO_EDGE gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_WRAP_S, GL10.GL_REPEAT); gl.glTexParameterf(GL10.GL_TEXTURE_2D, GL10.GL_TEXTURE_WRAP_T, GL10.GL_REPEAT); //Use the Android GLUtils to specify a two-dimensional texture image from our bitmap GLUtils.texImage2D(GL10.GL_TEXTURE_2D, 0, bitmap, 0); //Clean up bitmap.recycle(); } }

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  • Format of compiled directx9 shader files?

    - by JB
    Is the format of compiled pixel and vertex shader object files as produced by fxc.exe documented anywhere either officially or unofficially? I'd like to be able to read the constant name to register assignments from the shader files. I know that the effects framework in D3DX can do this, but I need to avoid using D3DX as it may not be installed on user's machines and I don't need it for anything else so I want to avoid them having to run the directx update. If the effects framework can do it, then so can I if I can find out the file format but I can' seem to find it documented anywhere.

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  • Finding the centeroid of a polygon?

    - by user146780
    I have tried: for each vertex, add to total, divide by number of verities to get center. I'v also tried: Find the topmost, bottommost - get midpoint... find leftmost, rightmost, find midpoint. Both of these did not return the perfect center because I'm relying on the center to scale a polygon. I want to scale my polygons so I may put a border around them. What is the best way to find the centroid of a polygon given that the polygon may be concave, convex and have many many sides of various lengths. Thanks

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  • Interpolating 2d data that is piecewise constant on faces

    - by celil
    I have an irregular mesh which is described by two variables - a faces array that stores the indices of the vertices that constitute each face, and a verts array that stores the coordinates of each vertex. I also have a function that is assumed to be piecewise constant over each face, and it is stored in the form of an array of values per face. I am looking for a way to construct a function f from this data. Something along the following lines: faces = [[0,1,2], [1,2,3], [2,3,4] ...] verts = [[0,0], [0,1], [1,0], [1,1],....] vals = [0.0, 1.0, 0.5, 3.0,....] f = interpolate(faces, verts, vals) f(0.2, 0.2) = 0.0 # point inside face [0,1,2] f(0.6, 0.6) = 1.0 # point inside face [1,2,3] The manual way of evaluating f(x,y) would be to find the corresponding face that the point x,y lies in, and return the value that is stored in that face. Is there a function that already implements this in scipy (or in matlab)?

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  • Dislpay List and transformation

    - by Gary
    Greetings! I have this question. Whenever, I enter a transformation (gltranslate, glrotate, glscale) within a display list, the transformation remains as a command within the display list. Everytime the display list is rendered, it will calculate all and over again. Is there a way, I can make an opengl transformation and the transformed vertex coordinates be stored permanently in a display list instead of transformation & intial coordinates? Hope my question makes sense. Thank you in advance. Gary

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  • PyOpenGL: glVertexPointer() offset problem

    - by SurvivalMachine
    My vertices are interleaved in a numpy array (dtype = float32) like this: ... tu, tv, nx, ny, nz, vx, vy, vz, ... When rendering, I'm calling gl*Pointer() like this (I have enabled the arrays before): stride = (2 + 3 + 3) * 4 glTexCoordPointer( 2, GL_FLOAT, stride, self.vertArray ) glNormalPointer( GL_FLOAT, stride, self.vertArray + 2 ) glVertexPointer( 3, GL_FLOAT, stride, self.vertArray + 5 ) glDrawElements( GL_TRIANGLES, len( self.indices ), GL_UNSIGNED_SHORT, self.indices ) The result is that nothing renders. However, if I organize my array so that the vertex position is the first element ( ... vx, vy, vz, tu, tv, nx, ny, nz, ... ) I get correct positions for vertices while rendering but texture coords and normals aren't rendered correctly. This leads me to believe that I'm not setting the pointer offset right. How should I set it? I'm using almost the exact same code in my other app in C++ and it works.

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  • Graph coloring Algorithm

    - by Amitd
    From wiki In its simplest form, it is a way of coloring the vertices of a graph such that no two adjacent vertices share the same color; this is called a vertex coloring. Similarly, an edge coloring assigns a color to each edge so that no two adjacent edges share the same color, and a face coloring of a planar graph assigns a color to each face or region so that no two faces that share a boundary have the same color. Given 'n' colors and m vertices, how easily can a graph coloring algorithm be implemented? Lan

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  • Simple reduction (NP completeness)

    - by Allen
    hey guys I'm looking for a means to prove that the bicriteria shortest path problem is np complete. That is, given a graph with lengths and weights, I need to know if a there exists a path in the graph from s to t with total length <= L and weight <= W. I know that i must take an NP complete problem and reduce it to this one. We have at our disposal the following problems to choose from: 3-SAT, independent set, vertex cover, hamiltonian cycle, and 3-dimensional matching. Any ideas on which may be viable? thanks

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  • Finding the heaviest length-constrained path in a weighted Binary Tree

    - by Hristo
    UPDATE I worked out an algorithm that I think runs in O(n*k) running time. Below is the pseudo-code: routine heaviestKPath( T, k ) // create 2D matrix with n rows and k columns with each element = -8 // we make it size k+1 because the 0th column must be all 0s for a later // function to work properly and simplicity in our algorithm matrix = new array[ T.getVertexCount() ][ k + 1 ] (-8); // set all elements in the first column of this matrix = 0 matrix[ n ][ 0 ] = 0; // fill our matrix by traversing the tree traverseToFillMatrix( T.root, k ); // consider a path that would arc over a node globalMaxWeight = -8; findArcs( T.root, k ); return globalMaxWeight end routine // node = the current node; k = the path length; node.lc = node’s left child; // node.rc = node’s right child; node.idx = node’s index (row) in the matrix; // node.lc.wt/node.rc.wt = weight of the edge to left/right child; routine traverseToFillMatrix( node, k ) if (node == null) return; traverseToFillMatrix(node.lc, k ); // recurse left traverseToFillMatrix(node.rc, k ); // recurse right // in the case that a left/right child doesn’t exist, or both, // let’s assume the code is smart enough to handle these cases matrix[ node.idx ][ 1 ] = max( node.lc.wt, node.rc.wt ); for i = 2 to k { // max returns the heavier of the 2 paths matrix[node.idx][i] = max( matrix[node.lc.idx][i-1] + node.lc.wt, matrix[node.rc.idx][i-1] + node.rc.wt); } end routine // node = the current node, k = the path length routine findArcs( node, k ) if (node == null) return; nodeMax = matrix[node.idx][k]; longPath = path[node.idx][k]; i = 1; j = k-1; while ( i+j == k AND i < k ) { left = node.lc.wt + matrix[node.lc.idx][i-1]; right = node.rc.wt + matrix[node.rc.idx][j-1]; if ( left + right > nodeMax ) { nodeMax = left + right; } i++; j--; } // if this node’s max weight is larger than the global max weight, update if ( globalMaxWeight < nodeMax ) { globalMaxWeight = nodeMax; } findArcs( node.lc, k ); // recurse left findArcs( node.rc, k ); // recurse right end routine Let me know what you think. Feedback is welcome. I think have come up with two naive algorithms that find the heaviest length-constrained path in a weighted Binary Tree. Firstly, the description of the algorithm is as follows: given an n-vertex Binary Tree with weighted edges and some value k, find the heaviest path of length k. For both algorithms, I'll need a reference to all vertices so I'll just do a simple traversal of the Tree to have a reference to all vertices, with each vertex having a reference to its left, right, and parent nodes in the tree. Algorithm 1 For this algorithm, I'm basically planning on running DFS from each node in the Tree, with consideration to the fixed path length. In addition, since the path I'm looking for has the potential of going from left subtree to root to right subtree, I will have to consider 3 choices at each node. But this will result in a O(n*3^k) algorithm and I don't like that. Algorithm 2 I'm essentially thinking about using a modified version of Dijkstra's Algorithm in order to consider a fixed path length. Since I'm looking for heaviest and Dijkstra's Algorithm finds the lightest, I'm planning on negating all edge weights before starting the traversal. Actually... this doesn't make sense since I'd have to run Dijkstra's on each node and that doesn't seem very efficient much better than the above algorithm. So I guess my main questions are several. Firstly, do the algorithms I've described above solve the problem at hand? I'm not totally certain the Dijkstra's version will work as Dijkstra's is meant for positive edge values. Now, I am sure there exist more clever/efficient algorithms for this... what is a better algorithm? I've read about "Using spine decompositions to efficiently solve the length-constrained heaviest path problem for trees" but that is really complicated and I don't understand it at all. Are there other algorithms that tackle this problem, maybe not as efficiently as spine decomposition but easier to understand? Thanks.

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  • Drawing Directed Acyclic Graphs: Minimizing edge crossing?

    - by Robert Fraser
    Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple without graph drawing algorithms such as Efficient Sugimiya. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of: EDIT: As the picture suggests, a vertex's inputs are always on top and outputs are always below, which is another barrier to just pasting in an existing layout algorithm.

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