Search Results

Search found 16892 results on 676 pages for 'linear search'.

Page 47/676 | < Previous Page | 43 44 45 46 47 48 49 50 51 52 53 54  | Next Page >

  • Transform between two 3d cartesian coordinate systems

    - by Pris
    I'd like to know how to get the rotation matrix for the transformation from one cartesian coordinate system (X,Y,Z) to another one (X',Y',Z'). Both systems are defined with three orthogonal vectors as one would expect. No scaling or translation occurs. I'm using OpenSceneGraph and it offers a Matrix convenience class, if it makes finding the matrix easier: http://www.openscenegraph.org/documentation/OpenSceneGraphReferenceDocs/a00403.html.

    Read the article

  • How to implement a simple bullet trajectory

    - by AirieFenix
    I searched and searched and although it's a fair simple question, I don't find the proper answer but general ideas (which I already have). I have a top-down game and I want to implement a gun which shoots bullets that follow a simple path (no physics nor change of trajectory, just go from A to B thing). a: vector of the position of the gun/player. b: vector of the mouse position (cross-hair). w: the vector of the bullet's trajectory. So, w=b-a. And the position of the bullet = [x=x0+speed*time*normalized w.x , y=y0+speed*time * normalized w.y]. I have the constructor: public Shot(int shipX, int shipY, int mouseX, int mouseY) { //I get mouse with Gdx.input.getX()/getY() ... this.shotTime = TimeUtils.millis(); this.posX = shipX; this.posY = shipY; //I used aVector = aVector.nor() here before but for some reason didn't work float tmp = (float) (Math.pow(mouseX-shipX, 2) + Math.pow(mouseY-shipY, 2)); tmp = (float) Math.sqrt(Math.abs(tmp)); this.vecX = (mouseX-shipX)/tmp; this.vecY = (mouseY-shipY)/tmp; } And here I update the position and draw the shot: public void drawShot(SpriteBatch batch) { this.lifeTime = TimeUtils.millis() - this.shotTime; //position = positionBefore + v*t this.posX = this.posX + this.vecX*this.lifeTime*speed*Gdx.graphics.getDeltaTime(); this.posY = this.posY + this.vecY*this.lifeTime*speed*Gdx.graphics.getDeltaTime(); ... } Now, the behavior of the bullet seems very awkward, not going exactly where my mouse is (it's like the mouse is 30px off) and with a random speed. I know I probably need to open the old algebra book from college but I'd like somebody says if I'm in the right direction (or points me to it); if it's a calculation problem, a code problem or both. Also, is it possible that Gdx.input.getX() gives me non-precise position? Because when I draw the cross-hair it also draws off the cursor position. Sorry for the long post and sorry if it's a very basic question. Thanks!

    Read the article

  • Create Auto Customization Criteria OAF Search Page

    - by PRajkumar
    1. Create a New Workspace and Project Right click Workspaces and click create new OAworkspace and name it as PRajkumarCustSearch. Automatically a new OA Project will also be created. Name the project as CustSearchDemo and package as prajkumar.oracle.apps.fnd.custsearchdemo   2. Create a New Application Module (AM) Right Click on CustSearchDemo > New > ADF Business Components > Application Module Name -- CustSearchAM Package -- prajkumar.oracle.apps.fnd.custsearchdemo.server   3. Enable Passivation for the Root UI Application Module (AM) Right Click on CustSearchAM > Edit SearchAM > Custom Properties > Name – RETENTION_LEVEL Value – MANAGE_STATE Click add > Apply > OK   4. Create Test Table and insert data some data in it (For Testing Purpose)   CREATE TABLE xx_custsearch_demo (   -- ---------------------     -- Data Columns     -- ---------------------     column1                  VARCHAR2(100),     column2                  VARCHAR2(100),     column3                  VARCHAR2(100),     column4                  VARCHAR2(100),     -- ---------------------     -- Who Columns     -- ---------------------     last_update_date    DATE         NOT NULL,     last_updated_by     NUMBER   NOT NULL,     creation_date          DATE         NOT NULL,     created_by               NUMBER   NOT NULL,     last_update_login   NUMBER  );   INSERT INTO xx_custsearch_demo VALUES('v1','v2','v3','v4',SYSDATE,0,SYSDATE,0,0); INSERT INTO xx_custsearch_demo VALUES('v1','v3','v4','v5',SYSDATE,0,SYSDATE,0,0); INSERT INTO xx_custsearch_demo VALUES('v2','v3','v4','v5',SYSDATE,0,SYSDATE,0,0); INSERT INTO xx_custsearch_demo VALUES('v3','v4','v5','v6',SYSDATE,0,SYSDATE,0,0); Now we have 4 records in our custom table   5. Create a New Entity Object (EO) Right click on SearchDemo > New > ADF Business Components > Entity Object Name – CustSearchEO Package -- prajkumar.oracle.apps.fnd.custsearchdemo.schema.server Database Objects -- XX_CUSTSEARCH_DEMO   Note – By default ROWID will be the primary key if we will not make any column to be primary key   Check the Accessors, Create Method, Validation Method and Remove Method   6. Create a New View Object (VO) Right click on CustSearchDemo > New > ADF Business Components > View Object Name -- CustSearchVO Package -- prajkumar.oracle.apps.fnd.custsearchdemo.server   In Step2 in Entity Page select CustSearchEO and shuttle them to selected list   In Step3 in Attributes Window select columns Column1, Column2, Column3, Column4, and shuttle them to selected list   In Java page deselect Generate Java file for View Object Class: CustSearchVOImpl and Select Generate Java File for View Row Class: CustSearchVORowImpl   7. Add Your View Object to Root UI Application Module Select Right click on CustSearchAM > Application Modules > Data Model Select CustSearchVO and shuttle to Data Model list   8. Create a New Page Right click on CustSearchDemo > New > Web Tier > OA Components > Page Name -- CustSearchPG Package -- prajkumar.oracle.apps.fnd.custsearchdemo.webui   9. Select the CustSearchPG and go to the strcuture pane where a default region has been created   10. Select region1 and set the following properties: ID -- PageLayoutRN Region Style -- PageLayout AM Definition -- prajkumar.oracle.apps.fnd.custsearchdemo.server.CustSearchAM Window Title – AutoCustomize Search Page Window Title – AutoCustomization Search Page Auto Footer -- True   11. Add a Query Bean to Your Page Right click on PageLayoutRN > New > Region Select new region region1 and set following properties ID – QueryRN Region Style – query Construction Mode – autoCustomizationCriteria Include Simple Panel – False Include Views Panel – False Include Advanced Panel – False   12. Create a New Region of style table Right Click on QueryRN > New > Region Using Wizard Application Module – prajkumar.oracle.apps.fnd.custsearchdemo.server.CustSearchAM Available View Usages – CustSearchVO1   In Step2 in Region Properties set following properties Region ID – CustSearchTable Region Style – Table   In Step3 in View Attributes shuttle all the items (Column1, Column2, Column3, Column4) available in “Available View Attributes” to Selected View Attributes: In Step4 in Region Items page set style to “messageStyledText” for all items   13. Select CustSearchTable in Structure Panel and set property Width to 100%   14. Include Simple Search Panel Right Click on QueryRN > New > simpleSearchPanel Automatically region2 (header Region) and region1 (MessageComponentLayout Region) created Set Following Properties for region2 Id – SimpleSearchHeader Text -- Simple Search   15. Now right click on message Component Layout Region (SimpleSearchMappings) and create two message text input beans and set the below properties to each   Message TextInputBean1 Id – SearchColumn1 Search Allowed – True Data Type – VARCHAR2 Maximum Length – CSS Class – OraFieldText Prompt – Column1   Message TextInputBean2 Id – SearchColumn2 Search Allowed -- True Data Type – VARCHAR2 Maximum Length – 100 CSS Class – OraFieldText Prompt – Column2   16. Now Right Click on query Components and create simple Search Mappings. Then automatically SimpleSearchMappings and QueryCriteriaMap1 created   17.  Now select the QueryCriteriaMap1 and set the below properties Id – SearchColumn1Map Search Item – SearchColumn1 Result Item – Column1   18. Now again right click on simpleSearchMappings -> New -> queryCriteriaMap, and then set the below properties Id – SearchColumn2Map Search Item – SearchColumn2 Result Item – Column2   19. Congratulation you have successfully finished Auto Customization Search page. Run Your CustSearchPG page and Test Your Work            

    Read the article

  • 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.

    Read the article

  • Make an object slide around an obstacle

    - by Isaiah
    I have path areas set up in a game I'm making for canvas/html5 and have got it working to keep the player within these areas. I have a function isOut(boundary, x, y) that returns true if the point is outside the boundary. What I do is check only the new position x/y separately with the corresponding old position x/y. Then if each one is out I assign them the past value from the frame before. The old positions are kept in a variable from a closure I made. like this: opos = [x,y];//old position npos = [x,y];//new position if(isOut(bound, npos[0], opos[1])){ npos[0] = opos[0]; //assign it the old x position } if(isOut(bound, opos[0], npos[1])){ npos[1] = opos[1]; //assign it the old y position } It looks nice and works good at certain angles, but if your boundary has diagonal regions it results in jittery motion. What's happening is the y pos exits the area while x doesn't and continues pushing the player to the side, once it has moved the player to the side a bit the player can move forward and then the y exits again and the whole process repeats. Anyone know how I may be able to achieve a smoother slide? I have access to the player's velocity vector, the angle, and the speed(when used with the angle). I can move the play with either angle/speed or x/yvelocities as I've built in backups to translate one to the other if either have been altered manually.

    Read the article

  • How can I test if a point lies between two parallel lines?

    - by Harold
    In the game I'm designing there is a blast that shoots out from an origin point towards the direction of the mouse. The width of this blast is always going to be the same. Along the bottom of the screen (what's currently) squares move about which should be effected by the blast that the player controls. Currently I am trying to work out a way to discover if the corners of these squares are within the blast's two bounding lines. I thought the best way to do this would be to rotate the corners of the square around an origin point as if the blast were completely horizontal and see if the Y values of the corners were less than or equal to the width of the blast which would mean that they lie within the effected region, but I can't work out

    Read the article

  • Moving objects colliding when using unalligned collision avoidance (steering)

    - by James Bedford
    I'm having trouble with unaligned collision avoidance for what I think is a rare case. I have set two objects to move towards each other but with a slight offset, so one of the objects is moving slightly upwards, and one of the objects is moving slightly downwards. In my unaligned collision avoidance steering algorithm I'm finding the points on the object's forward line and the other object's forward line where these two lines are the closest. If these closest points are within a collision avoidance distance, and if the distance between them is smaller than the two radii of the two object's bounding spheres, then the objects should steer away in the appropriate direction. The problem is that for my case, the closest points on the lines are calculated to be really far away from the actual collision point. This is because the two forward lines for each object are moving away from each other as the objects pass. The problem is that because of this, no steering takes place, and the two objects partially collide. Does anyone have any suggestions as to how I can correctly calculate the point of collision? Perhaps by somehow taking into account the size of the two objects?

    Read the article

  • Best way to implement a simple bullet trajectory

    - by AirieFenix
    I searched and searched and although it's a fair simple question, I don't find the proper answer but general ideas (which I already have). I have a top-down game and I want to implement a gun which shoots bullets that follow a simple path (no physics nor change of trajectory, just go from A to B thing). a: vector of the position of the gun/player. b: vector of the mouse position (cross-hair). w: the vector of the bullet's trajectory. So, w=b-a. And the position of the bullet = [x=x0+speed*time*normalized w.x , y=y0+speed*time * normalized w.y]. I have the constructor: public Shot(int shipX, int shipY, int mouseX, int mouseY) { //I get mouse with Gdx.input.getX()/getY() ... this.shotTime = TimeUtils.millis(); this.posX = shipX; this.posY = shipY; //I used aVector = aVector.nor() here before but for some reason didn't work float tmp = (float) (Math.pow(mouseX-shipX, 2) + Math.pow(mouseY-shipY, 2)); tmp = (float) Math.sqrt(Math.abs(tmp)); this.vecX = (mouseX-shipX)/tmp; this.vecY = (mouseY-shipY)/tmp; } And here I update the position and draw the shot: public void drawShot(SpriteBatch batch) { this.lifeTime = TimeUtils.millis() - this.shotTime; //position = positionBefore + v*t this.posX = this.posX + this.vecX*this.lifeTime*speed*Gdx.graphics.getDeltaTime(); this.posY = this.posY + this.vecY*this.lifeTime*speed*Gdx.graphics.getDeltaTime(); ... } Now, the behavior of the bullet seems very awkward, not going exactly where my mouse is (it's like the mouse is 30px off) and with a random speed. I know I probably need to open the old algebra book from college but I'd like somebody says if I'm in the right direction (or points me to it); if it's a calculation problem, a code problem or both. Also, is it possible that Gdx.input.getX() gives me non-precise position? Because when I draw the cross-hair it also draws off the cursor position. Sorry for the long post and sorry if it's a very basic question. Thanks!

    Read the article

  • Understanding dot notation

    - by Starkers
    Here's my interpretation of dot notation: a = [2,6] b = [1,4] c = [0,8] a . b . c = (2*6)+(1*4)+(0*8) = 12 + 4 + 0 = 16 What is the significance of 16? Apparently it's a scalar. Am I right in thinking that a scalar is the number we times a unit vector by to get a vector that has a scaled up magnitude but the same direction as the unit vector? So again, what is the relevance of 16? When is it used? It's not the magnitude of all the vectors added up. The magnitude of all of them is calculated as follows: sqrt( ax * ax + ay * ay ) + sqrt( bx * bx + by * by ) + sqrt( cx * cx + cy * cy) sqrt( 2 * 2 + 6 * 6 ) + sqrt( 1 * 1 + 4 * 4 ) + sqrt( 0 * 0 + 8 * 8) sqrt( 4 + 36 ) + sqrt( 1 + 16 ) + sqrt( 0 + 64) sqrt( 40 ) + sqrt( 17 ) + sqrt( 64) 6.3 + 4.1 + 8 10.4 + 8 18.4 So I don't really get this diagram: Attempting with sensible numbers: a = [1,0] b = [4,3] a . b = (1*0) + (4*3) = 0 + 12 = 12 So what exactly is a . b describing here? The magnitude of that vector? Because that isn't right: the 'a.b' vector = [4,0] sqrt( x*x + y*y ) sqrt( 4*4 + 0*0 ) sqrt( 16 + 0 ) 4 So what is 12 describing?

    Read the article

  • Lerping to a center point while in motion

    - by Fibericon
    I have an enemy that initially flies in a circular motion, while facing away from the center point. This is how I achieve that: position.Y = (float)(Math.Cos(timeAlive * MathHelper.PiOver4) * radius + origin.Y); position.X = (float)(Math.Sin(timeAlive * MathHelper.PiOver4) * radius + origin.X); if (timeAlive < 5) { angle = (float)Math.Atan((0 - position.X) / (0 - position.Y)); if (0 < position.Y) RotationMatrix = Matrix.CreateRotationX(MathHelper.PiOver2) * Matrix.CreateRotationZ(-1 * angle); else RotationMatrix = Matrix.CreateRotationX(MathHelper.PiOver2) * Matrix.CreateRotationZ(MathHelper.Pi - angle); } That part works just fine. After five seconds of this, I want the enemy to turn inward, facing the center point. However, I've been trying to lerp to that point, since I don't want it to simply jump to the new rotation. Here's my code for trying to do that: else { float newAngle = -1 * (float)Math.Atan((0 - position.X) / (0 - position.Y)); angle = MathHelper.Lerp(angle, newAngle, (float)gameTime.ElapsedGameTime.Milliseconds / 1000); if (0 < position.Y) RotationMatrix = Matrix.CreateRotationX(MathHelper.PiOver2) * Matrix.CreateRotationZ(MathHelper.Pi - angle); else RotationMatrix = Matrix.CreateRotationX(MathHelper.PiOver2) * Matrix.CreateRotationZ(-1 * angle); } That doesn't work so fine. It seems like it's going to at first, but then it just sort of skips around. How can I achieve what I want here?

    Read the article

  • How to get xy coordinates along a given path

    - by netbrain
    Say i have two points (x,y), (0,0) and (10,10). Now i wan´t to get coordinates along the line by stepping through values of x and y. I thought i solved it with the following functions: fy = startY + (x - startX) * ((destY-startY)/(destX-startX)); fx = (y + startY) / ((destY-startY)/(destX-startX)) + startX; taken from http://en.wikipedia.org/wiki/Linear_interpolation However, it seems that im getting a problem when destX and startX is the same value, so you get division by zero. Is there a better way of getting coordinates along a line when knowing the start and endpoint of the line?

    Read the article

  • Trouble with Collada bones

    - by KyleT
    I have a Collada file with a rigged mesh. I've read the node tags in the library_visual_scenes tag and extracted the matrix for each node and stored everything in a hierarchical bone structure. My Matrix container is "row major", so I'd store the first float of a matrix tag in the 1st row, 1st column, the second in the 1st row, 2nd column, etc. From what I gather this is the Bind Pose Matrix. After that I went through the tag and extracted the float array in the source tag of the skin tag of the controller for the mesh. I stored each matrix from this float array in their corresponding Bone as the Inverse Bind Matrix. I also extracted the bind-shape-matrix and stored it. Now I'd like to draw the skeleton with OpenGL to see if everything is working correctly before I go about skinning. I iterate once over my bones and multiply a bone's Bind Pose Matrix by it's parents and store that. After that I iterate again over the bones and multiply the result of the previous matrix multiplication by the Inverse Bind Matrix and then by the Bind Shape Matrix. The results look something like this: [0.2, 9.2, 5.8, 1.2 ] [4.6, -3.3, -0.2, -0.1 ] [-1.8, 0.2, -4.2, -3.9 ] [0, 0, 0, 1 ] I've had to go to various sources to get the little understanding of Collada I have and books about 3d transform matricies can get pretty intense. I've hit a brick wall and if you could please read through this and see if there is something I'm doing wrong, and how I'd go about getting an X,Y,Z to draw a point for each of these joints once I've calculated the final transform, I'd really appreciate it.

    Read the article

  • Matrices: Arrays or separate member variables?

    - by bjz
    I'm teaching myself 3D maths and in the process building my own rudimentary engine (of sorts). I was wondering what would be the best way to structure my matrix class. There are a few options: Separate member variables: struct Mat4 { float m11, m12, m13, m14, m21, m22, m23, m24, m31, m32, m33, m34, m41, m42, m43, m44; // methods } A multi-dimensional array: struct Mat4 { float[4][4] m; // methods } An array of vectors struct Mat4 { Vec4[4] m; // methods } I'm guessing there would be positives and negatives to each. From 3D Math Primer for Graphics and Game Development, 2nd Edition p.155: Matrices use 1-based indices, so the first row and column are numbered 1. For example, a12 (read “a one two,” not “a twelve”) is the element in the first row, second column. Notice that this is different from programming languages such as C++ and Java, which use 0-based array indices. A matrix does not have a column 0 or row 0. This difference in indexing can cause some confusion if matrices are stored using an actual array data type. For this reason, it’s common for classes that store small, fixed size matrices of the type used for geometric purposes to give each element its own named member variable, such as float a11, instead of using the language’s native array support with something like float elem[3][3]. So that's one vote for method one. Is this really the accepted way to do things? It seems rather unwieldy if the only benefit would be sticking with the conventional math notation.

    Read the article

  • Portal View/Projection Matrix near plane

    - by melak47
    For RenderToTexture/Camera based portal rendering, the basics seems simple enough. However, with a free camera, most of the time it is going to be looking at such portals at an angle: Now a regular near clipping plane will not always work here, it will either intersect with the wall the portal is sitting on, or possibly with objects in front of the wall. The desired near clipping plane would be aligned like the portal, producing a view volume more like this: or this in 3D: So here is my question: How does one construct or "truncate" a view/projection matrix to achieve such an off-camera-normal (near) clipping plane?

    Read the article

  • Mouse pointer position to screen space

    - by Ylisar
    If I have a mouse pointer position in pixels of canvas, I can easily convert it to the -1..1 range for both X & Y by lerping by dividing with canvas dimensions. However, the problem is what I should put in Z & W if I want my screen space position to be on the near plane? The step afterwards would be for me to multiply by the inverse of view-projection to take me to world space, where I easily can construct a ray from the cameras world space position.

    Read the article

  • ASP.NET MVC search route

    - by sahina
    I setup a search route: routes.MapRoute( "Search", "Search/{q}", new { controller = "Search", action = "Index" } ); The search form has an input box and a button. I want the search with a GET as below. <% using(Html.BeginForm("Index", "Search", FormMethod.Get)) {%> <%:Html.TextBox("q")%> <span class="query-button"> <input type="submit" value="select" /></span> <% } %> </div> The action on the SearchController is: public ActionResult Index(string q) { // search logic here return View(new SearchResult(q)); } The URL becomes like this: http://localhost:19502/search?q=mvc+is+great But I want the search to be like: http://localhost:19502/search/mvc+is+great How do I setup the route or the Html.BeginForm

    Read the article

  • Duplicate GET request from multiple IPs - can anyone explain this?

    - by dwq
    We've seen a pattern in our webserver access logs which we're having problem explaining. A GET request appears in the access log which is a legitimate, but private, url as part of normal e-commerce website use (by private, we mean there is a unique key in a url form variable generated specifically for that customer session). Then a few seconds later we get hit with an identical request maybe 10-15 times within the space of a second. The duplicate requests are all from different IP addresses. The UserAgent for the duplicates are all the same (but different from the original request). The reverse DNS lookup on the IPs for all the duplicates requests resolve to the same large hosting company. Can anyone think of a scenario what would explain this? EDIT 1 Here's an example that's probably anonymised beyond being any actual use, but it might give an idea of the sort of pattern we're seeing (it's from a search query as they sometimes get duplicated too): xx.xx.xx.xx - - [21/Jun/2013:21:42:57 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "http://www.ourdomain.com/index.html" "Mozilla/5.0 (compatible; MSIE 10.0; Windows NT 6.1; WOW64; Trident/6.0)" xx.xx.xx.xx - - [21/Jun/2013:21:43:03 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:03 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" xx.xx.xx.xx - - [21/Jun/2013:21:43:04 +0100] "GET /search.html?search=widget&Submit=Search HTTP/1.0" 200 5475 "" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_6_7) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.91 Safari/534.30" UPDATE 2 Sometimes it is part of a checkout flow that's duplicated to I'd think twitter is unlikely.

    Read the article

  • Use a vector to index a matrix without linear index

    - by David_G
    G'day, I'm trying to find a way to use a vector of [x,y] points to index from a large matrix in MATLAB. Usually, I would convert the subscript points to the linear index of the matrix.(for eg. Use a vector as an index to a matrix in MATLab) However, the matrix is 4-dimensional, and I want to take all of the elements of the 3rd and 4th dimensions that have the same 1st and 2nd dimension. Let me hopefully demonstrate with an example: Matrix = nan(4,4,2,2); % where the dimensions are (x,y,depth,time) Matrix(1,2,:,:) = 999; % note that this value could change in depth (3rd dim) and time (4th time) Matrix(3,4,:,:) = 888; % note that this value could change in depth (3rd dim) and time (4th time) Matrix(4,4,:,:) = 124; Now, I want to be able to index with the subscripts (1,2) and (3,4), etc and return not only the 999 and 888 which exist in Matrix(:,:,1,1) but the contents which exist at Matrix(:,:,1,2),Matrix(:,:,2,1) and Matrix(:,:,2,2), and so on (IRL, the dimensions of Matrix might be more like size(Matrix) = (300 250 30 200) I don't want to use linear indices because I would like the results to be in a similar vector fashion. For example, I would like a result which is something like: ans(time=1) 999 888 124 999 888 124 ans(time=2) etc etc etc etc etc etc I'd also like to add that due to the size of the matrix I'm dealing with, speed is an issue here - thus why I'd like to use subscript indices to index to the data. I should also mention that (unlike this question: Accessing values using subscripts without using sub2ind) since I want all the information stored in the extra dimensions, 3 and 4, of the i and jth indices, I don't think that a slightly faster version of sub2ind still would not cut it..

    Read the article

  • How to partition bits in a bit array with less than linear time

    - by SiLent SoNG
    This is an interview question I faced recently. Given an array of 1 and 0, find a way to partition the bits in place so that 0's are grouped together, and 1's are grouped together. It does not matter whether 1's are ahead of 0's or 0's are ahead of 1's. An example input is 101010101, and output is either 111110000 or 000011111. Solve the problem in less than linear time. Make the problem simpler. The input is an integer array, with each element either 1 or 0. Output is the same integer array with integers partitioned well. To me, this is an easy question if it can be solved in O(N). My approach is to use two pointers, starting from both ends of the array. Increases and decreases each pointer; if it does not point to the correct integer, swap the two. int * start = array; int * end = array + length - 1; while (start < end) { // Assume 0 always at the end if (*end == 0) { --end; continue; } // Assume 1 always at the beginning if (*start == 1) { ++start; continue; } swap(*start, *end); } However, the interview insists there is a sub-linear solution. This makes me thinking hard but still not get an answer. Can anyone help on this interview question?

    Read the article

  • Linear Search with Jagged Array?

    - by Nerathas
    Hello, I have the following program that creates 100 random elements trough a array. Those 100 random value's are unique, and every value only gets displayed once. Although with the linear search it keeps looking up the entire array. How would i be able to get a Jagged Array into this, so it only "scans" the remaining places left? (assuming i keep the table at 100 max elements, so if one random value is generated the array holds 99 elements with linear search scans and on...) I assume i would have to implent the jagged array somewhere in the FoundLinearInArray? Hopefully this made any sence. Regards. private int ValidNumber(int[] T, int X, int Range) { Random RndInt = new Random(); do { X = RndInt.Next(1, Range + 1); } while (FoundLinearInArray(T, X)); return X; }/*ValidNumber*/ private bool FoundLinearInArray(int[] A, int X) { byte I = 0; while ((I < A.Length) && (A[I] != X)) { I++; } return (I < A.Length); }/*FoundInArray*/ public void FillArray(int[] T, int Range) { for (byte I = 0; I < T.Length; I++) { T[I] = ValidNumber(T, I, Range); } }/*FillArray*/

    Read the article

  • Admob in xml not showing in Linear

    - by NoobMe
    i am implementing admob on my app it appears when the parent is in relative layout but i must not use the alignparentbottom so i am changing it to linear but it doesnt show when i change it to linear.. any tips? help? thanks in advance here it is in xml: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="match_parent" android:layout_height="match_parent" android:orientation="vertical" > <RelativeLayout android:id="@+id/banner_holder" android:layout_width="match_parent" android:layout_height="wrap_content" > <ImageView android:id="@+id/offline_banner" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_centerInParent="true" android:background="@color/black" android:src="@drawable/offline_banner" /> <com.google.ads.AdView xmlns:ads="http://schemas.android.com/apk/lib/com.google.ads" android:id="@+id/adView" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_centerInParent="true" ads:adSize="SMART_BANNER" ads:adUnitId="@string/unit_id" ads:loadAdOnCreate="true" /> </RelativeLayout> <FrameLayout android:id="@+id/fragmentContainer" android:layout_width="match_parent" android:layout_height="wrap_content" /> </LinearLayout> i want the admob to be at the bottom part of the screen without using the alignparentbottom of relative layout thanks~

    Read the article

  • Can not search my company howto blog site anylonger in Sharepoint

    - by Worldunix
    I have a Howto company Blog site that i post to for my clients to access for help. For some reason it has stopped letting anyone search on it. I can search for Mysites or users. But when you drop down the tab to search: This Site: "blog site name" you get the following reply: No results matching your search were found. Check your spelling. Are the words in your query spelled correctly? Try using synonyms. Maybe what you're looking for uses slightly different words. Make your search more general. Try more general terms in place of specific ones. Try your search in a different scope. Different scopes can have different results. I have tried the following command: from the Index server net stop osearch net start osearch iisreset /noforce But still not able to search a local blog site I can only search for users and Sites. please help Don

    Read the article

  • Can not search my company howto blog site anylonger in Sharepoint

    - by Worldunix
    I have a Howto company Blog site that i post to for my clients to access for help. For some reason it has stopped letting anyone search on it. I can search for Mysites or users. But when you drop down the tab to search: This Site: "blog site name" you get the following reply: No results matching your search were found. Check your spelling. Are the words in your query spelled correctly? Try using synonyms. Maybe what you're looking for uses slightly different words. Make your search more general. Try more general terms in place of specific ones. Try your search in a different scope. Different scopes can have different results. I have tried the following command: from the Index server net stop osearch net start osearch iisreset /noforce But still not able to search a local blog site I can only search for users and Sites. please help Don

    Read the article

  • Removing malware of a particular kind

    - by Cyclone
    I need to remove some malware from my computer. It is a trojan, and very annoying. It blocks access to Google and search sites. The trojan, with its name spelled out on each line cause it seems to block sites when i reference it in a url, is a r t (some text to mess it up) e m (more text i s First off, what is it, what does it do? Second, why can't I access google or yahoo or any other search sites at all? Third, can it be removed via McAffee? It says it quarantined it when I scanned I found a suspicious process "c"s"r"s"s".exe and it will not let me terminate it, and this is what Mcaffee says it is. Why on earth isn't Mcaffee getting rid of it? I even blocked internet access for this program. Thanks so much, I get kinda freaked out with things like this... Here is my entire Hosts file: 127.0.0.1 go.mail.ru 127.0.0.1 nova.rambler.ru 127.0.0.1 google.ad 127.0.0.1 www.google.ad 127.0.0.1 google.ae 127.0.0.1 www.google.ae 127.0.0.1 google.am 127.0.0.1 www.google.am 127.0.0.1 google.com.ar 127.0.0.1 www.google.com.ar 127.0.0.1 google.as 127.0.0.1 www.google.as 127.0.0.1 google.at 127.0.0.1 www.google.at 127.0.0.1 google.com.au 127.0.0.1 www.google.com.au 127.0.0.1 google.az 127.0.0.1 www.google.az 127.0.0.1 google.ba 127.0.0.1 www.google.ba 127.0.0.1 google.be 127.0.0.1 www.google.be 127.0.0.1 google.bg 127.0.0.1 www.google.bg 127.0.0.1 google.bs 127.0.0.1 www.google.bs 127.0.0.1 google.com.by 127.0.0.1 www.google.com.by 127.0.0.1 google.ca 127.0.0.1 www.google.ca 127.0.0.1 google.ch 127.0.0.1 www.google.ch 127.0.0.1 google.cn 127.0.0.1 www.google.cn 127.0.0.1 google.cz 127.0.0.1 www.google.cz 127.0.0.1 google.de 127.0.0.1 www.google.de 127.0.0.1 google.dk 127.0.0.1 www.google.dk 127.0.0.1 google.ee 127.0.0.1 www.google.ee 127.0.0.1 google.es 127.0.0.1 www.google.es 127.0.0.1 google.fi 127.0.0.1 www.google.fi 127.0.0.1 google.fr 127.0.0.1 www.google.fr 127.0.0.1 google.gr 127.0.0.1 www.google.gr 127.0.0.1 google.com.hk 127.0.0.1 www.google.com.hk 127.0.0.1 google.hr 127.0.0.1 www.google.hr 127.0.0.1 google.hu 127.0.0.1 www.google.hu 127.0.0.1 google.ie 127.0.0.1 www.google.ie 127.0.0.1 google.co.il 127.0.0.1 www.google.co.il 127.0.0.1 google.co.in 127.0.0.1 www.google.co.in 127.0.0.1 google.is 127.0.0.1 www.google.is 127.0.0.1 google.it 127.0.0.1 www.google.it 127.0.0.1 google.co.jp 127.0.0.1 www.google.co.jp 127.0.0.1 google.kg 127.0.0.1 www.google.kg 127.0.0.1 google.co.kr 127.0.0.1 www.google.co.kr 127.0.0.1 google.li 127.0.0.1 www.google.li 127.0.0.1 google.lt 127.0.0.1 www.google.lt 127.0.0.1 google.lu 127.0.0.1 www.google.lu 127.0.0.1 google.lv 127.0.0.1 www.google.lv 127.0.0.1 google.md 127.0.0.1 www.google.md 127.0.0.1 google.com.mx 127.0.0.1 www.google.com.mx 127.0.0.1 google.nl 127.0.0.1 www.google.nl 127.0.0.1 google.no 127.0.0.1 www.google.no 127.0.0.1 google.co.nz 127.0.0.1 www.google.co.nz 127.0.0.1 google.com.pe 127.0.0.1 www.google.com.pe 127.0.0.1 google.com.ph 127.0.0.1 www.google.com.ph 127.0.0.1 google.pl 127.0.0.1 www.google.pl 127.0.0.1 google.pt 127.0.0.1 www.google.pt 127.0.0.1 google.ro 127.0.0.1 www.google.ro 127.0.0.1 google.ru 127.0.0.1 www.google.ru 127.0.0.1 google.com.ru 127.0.0.1 www.google.com.ru 127.0.0.1 google.com.sa 127.0.0.1 www.google.com.sa 127.0.0.1 google.se 127.0.0.1 www.google.se 127.0.0.1 google.com.sg 127.0.0.1 www.google.com.sg 127.0.0.1 google.si 127.0.0.1 www.google.si 127.0.0.1 google.sk 127.0.0.1 www.google.sk 127.0.0.1 google.co.th 127.0.0.1 www.google.co.th 127.0.0.1 google.com.tj 127.0.0.1 www.google.com.tj 127.0.0.1 google.tm 127.0.0.1 www.google.tm 127.0.0.1 google.com.tr 127.0.0.1 www.google.com.tr 127.0.0.1 google.com.tw 127.0.0.1 www.google.com.tw 127.0.0.1 google.com.ua 127.0.0.1 www.google.com.ua 127.0.0.1 google.co.uk 127.0.0.1 www.google.co.uk 127.0.0.1 google.co.vi 127.0.0.1 www.google.co.vi 127.0.0.1 google.com 127.0.0.1 www.google.com 127.0.0.1 google.us 127.0.0.1 www.google.us 127.0.0.1 google.com.pl 127.0.0.1 www.google.com.pl 127.0.0.1 google.co.hu 127.0.0.1 www.google.co.hu 127.0.0.1 google.ge 127.0.0.1 www.google.ge 127.0.0.1 google.kz 127.0.0.1 www.google.kz 127.0.0.1 google.co.uz 127.0.0.1 www.google.co.uz 127.0.0.1 bing.com 127.0.0.1 www.bing.com 127.0.0.1 search.yahoo.com 127.0.0.1 ca.search.yahoo.com 127.0.0.1 ar.search.yahoo.com 127.0.0.1 cl.search.yahoo.com 127.0.0.1 co.search.yahoo.com 127.0.0.1 mx.search.yahoo.com 127.0.0.1 espanol.search.yahoo.com 127.0.0.1 qc.search.yahoo.com 127.0.0.1 ve.search.yahoo.com 127.0.0.1 pe.search.yahoo.com 127.0.0.1 at.search.yahoo.com 127.0.0.1 ct.search.yahoo.com 127.0.0.1 dk.search.yahoo.com 127.0.0.1 fi.search.yahoo.com 127.0.0.1 fr.search.yahoo.com 127.0.0.1 de.search.yahoo.com 127.0.0.1 it.search.yahoo.com 127.0.0.1 nl.search.yahoo.com 127.0.0.1 no.search.yahoo.com 127.0.0.1 ru.search.yahoo.com 127.0.0.1 es.search.yahoo.com 127.0.0.1 se.search.yahoo.com 127.0.0.1 ch.search.yahoo.com 127.0.0.1 uk.search.yahoo.com 127.0.0.1 asia.search.yahoo.com 127.0.0.1 au.search.yahoo.com 127.0.0.1 one.cn.yahoo.com 127.0.0.1 hk.search.yahoo.com 127.0.0.1 in.search.yahoo.com 127.0.0.1 id.search.yahoo.com 127.0.0.1 search.yahoo.co.jp 127.0.0.1 kr.search.yahoo.com 127.0.0.1 malaysia.search.yahoo.com 127.0.0.1 nz.search.yahoo.com 127.0.0.1 ph.search.yahoo.com 127.0.0.1 sg.search.yahoo.com 127.0.0.1 tw.search.yahoo.com 127.0.0.1 th.search.yahoo.com 127.0.0.1 vn.search.yahoo.com 127.0.0.1 images.google.com 127.0.0.1 images.google.ca 127.0.0.1 images.google.co.uk 127.0.0.1 news.google.com 127.0.0.1 news.google.ca 127.0.0.1 news.google.co.uk 127.0.0.1 video.google.com 127.0.0.1 video.google.ca 127.0.0.1 video.google.co.uk 127.0.0.1 blogsearch.google.com 127.0.0.1 blogsearch.google.ca 127.0.0.1 blogsearch.google.co.uk 127.0.0.1 searchservice.myspace.com 127.0.0.1 ask.com 127.0.0.1 www.ask.com 127.0.0.1 search.aol.com 127.0.0.1 search.netscape.com 127.0.0.1 yandex.ru 127.0.0.1 www.yandex.ru 127.0.0.1 yandex.ua 127.0.0.1 www.yandex.ua 127.0.0.1 search.about.com 127.0.0.1 www.verizon.net 127.0.0.1 verizon.net

    Read the article

< Previous Page | 43 44 45 46 47 48 49 50 51 52 53 54  | Next Page >