Search Results

Search found 33788 results on 1352 pages for 'codeit right'.

Page 63/1352 | < Previous Page | 59 60 61 62 63 64 65 66 67 68 69 70  | Next Page >

  • I want to write a program to control the square moving by using WINAPI

    - by code_new
    This is the code as attempted so far: #include <windows.h> LRESULT CALLBACK WndProc (HWND, UINT, WPARAM, LPARAM) ; int WINAPI WinMain (HINSTANCE hInstance, HINSTANCE hPrevInstance, PSTR szCmdLine, int iCmdShow) { static TCHAR szAppname[] = TEXT ("win0") ; WNDCLASS wndclass ; MSG msg ; HWND hwnd ; wndclass.style = CS_HREDRAW | CS_VREDRAW ; wndclass.cbWndExtra = 0 ; wndclass.cbClsExtra = 0 ; wndclass.lpfnWndProc = WndProc ; wndclass.lpszClassName = szAppname ; wndclass.lpszMenuName = NULL ; wndclass.hbrBackground = (HBRUSH) GetStockObject (WHITE_BRUSH) ; wndclass.hCursor = LoadCursor (NULL, IDC_ARROW) ; wndclass.hIcon = LoadIcon (NULL, IDI_APPLICATION) ; wndclass.hInstance = hInstance ; if (!RegisterClass (&wndclass)) { MessageBox (NULL, TEXT ("Register fail"), szAppname, 0) ; return 0 ; } hwnd = CreateWindow ( szAppname, TEXT ("mywin"), WS_OVERLAPPEDWINDOW, CW_USEDEFAULT, CW_USEDEFAULT, CW_USEDEFAULT, CW_USEDEFAULT, NULL, NULL, hInstance, NULL) ; ShowWindow (hwnd, iCmdShow) ; UpdateWindow (hwnd) ; while (GetMessage (&msg, NULL, 0, 0)) { TranslateMessage (&msg) ; DispatchMessage (&msg) ; } return msg.wParam ; } LRESULT CALLBACK WndProc (HWND hwnd, UINT message, WPARAM wParam, LPARAM lParam) { static int cxClient, cyClient, Left, Top, Right, Down ; PAINTSTRUCT ps ; HDC hdc ; RECT rect ; Right = 20 ; Down = 20 ; switch (message) { case WM_SIZE : cxClient = LOWORD (lParam) ; cyClient = HIWORD (lParam) ; return 0 ; case WM_PAINT : hdc = BeginPaint (hwnd, &ps) ; SetRect (&rect, Left, Top, Right, Down) ; FillRect (hdc, &rect, CreateSolidBrush (RGB (100, 100, 100))) ; EndPaint (hwnd, &ps) ; return 0 ; case WM_KEYDOWN : InvalidateRect (hwnd, &rect, TRUE) ; switch (wParam) { case VK_UP : if (Top - 20 < 0) { Top = 0 ; Down = 20 ; } else { Top -= 20 ; Down -= 20 ; } SendMessage (hwnd, WM_PAINT, wParam, lParam) ; break ; case VK_DOWN : if (Down + 20 > cyClient) { Down = cyClient ; Top = Down - 20 ; } else { Down += 20 ; Top += 20 ; }SendMessage (hwnd, WM_PAINT, wParam, lParam) ; break ; case VK_LEFT : if (Left - 20 < 0) { Left = 0 ; Right = 20 ; } else { Left -= 20 ; Right -= 20 ; }SendMessage (hwnd, WM_PAINT, wParam, lParam) ; break ; case VK_RIGHT : if (Right + 20 > cxClient) { Right = cxClient ; Left = Right - 20 ; } else { Right += 20 ; Left += 20 ; }SendMessage (hwnd, WM_PAINT, wParam, lParam) ; break ; default : break ; } return 0 ; case WM_DESTROY : PostQuitMessage (0) ; return 0 ; } return DefWindowProc (hwnd, message, wParam, lParam); } I considered that I didn't deal the message well, so I can't control it.

    Read the article

  • Parallelism in .NET – Part 11, Divide and Conquer via Parallel.Invoke

    - by Reed
    Many algorithms are easily written to work via recursion.  For example, most data-oriented tasks where a tree of data must be processed are much more easily handled by starting at the root, and recursively “walking” the tree.  Some algorithms work this way on flat data structures, such as arrays, as well.  This is a form of divide and conquer: an algorithm design which is based around breaking up a set of work recursively, “dividing” the total work in each recursive step, and “conquering” the work when the remaining work is small enough to be solved easily. Recursive algorithms, especially ones based on a form of divide and conquer, are often a very good candidate for parallelization. This is apparent from a common sense standpoint.  Since we’re dividing up the total work in the algorithm, we have an obvious, built-in partitioning scheme.  Once partitioned, the data can be worked upon independently, so there is good, clean isolation of data. Implementing this type of algorithm is fairly simple.  The Parallel class in .NET 4 includes a method suited for this type of operation: Parallel.Invoke.  This method works by taking any number of delegates defined as an Action, and operating them all in parallel.  The method returns when every delegate has completed: Parallel.Invoke( () => { Console.WriteLine("Action 1 executing in thread {0}", Thread.CurrentThread.ManagedThreadId); }, () => { Console.WriteLine("Action 2 executing in thread {0}", Thread.CurrentThread.ManagedThreadId); }, () => { Console.WriteLine("Action 3 executing in thread {0}", Thread.CurrentThread.ManagedThreadId); } ); .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } Running this simple example demonstrates the ease of using this method.  For example, on my system, I get three separate thread IDs when running the above code.  By allowing any number of delegates to be executed directly, concurrently, the Parallel.Invoke method provides us an easy way to parallelize any algorithm based on divide and conquer.  We can divide our work in each step, and execute each task in parallel, recursively. For example, suppose we wanted to implement our own quicksort routine.  The quicksort algorithm can be designed based on divide and conquer.  In each iteration, we pick a pivot point, and use that to partition the total array.  We swap the elements around the pivot, then recursively sort the lists on each side of the pivot.  For example, let’s look at this simple, sequential implementation of quicksort: public static void QuickSort<T>(T[] array) where T : IComparable<T> { QuickSortInternal(array, 0, array.Length - 1); } private static void QuickSortInternal<T>(T[] array, int left, int right) where T : IComparable<T> { if (left >= right) { return; } SwapElements(array, left, (left + right) / 2); int last = left; for (int current = left + 1; current <= right; ++current) { if (array[current].CompareTo(array[left]) < 0) { ++last; SwapElements(array, last, current); } } SwapElements(array, left, last); QuickSortInternal(array, left, last - 1); QuickSortInternal(array, last + 1, right); } static void SwapElements<T>(T[] array, int i, int j) { T temp = array[i]; array[i] = array[j]; array[j] = temp; } Here, we implement the quicksort algorithm in a very common, divide and conquer approach.  Running this against the built-in Array.Sort routine shows that we get the exact same answers (although the framework’s sort routine is slightly faster).  On my system, for example, I can use framework’s sort to sort ten million random doubles in about 7.3s, and this implementation takes about 9.3s on average. Looking at this routine, though, there is a clear opportunity to parallelize.  At the end of QuickSortInternal, we recursively call into QuickSortInternal with each partition of the array after the pivot is chosen.  This can be rewritten to use Parallel.Invoke by simply changing it to: // Code above is unchanged... SwapElements(array, left, last); Parallel.Invoke( () => QuickSortInternal(array, left, last - 1), () => QuickSortInternal(array, last + 1, right) ); } This routine will now run in parallel.  When executing, we now see the CPU usage across all cores spike while it executes.  However, there is a significant problem here – by parallelizing this routine, we took it from an execution time of 9.3s to an execution time of approximately 14 seconds!  We’re using more resources as seen in the CPU usage, but the overall result is a dramatic slowdown in overall processing time. This occurs because parallelization adds overhead.  Each time we split this array, we spawn two new tasks to parallelize this algorithm!  This is far, far too many tasks for our cores to operate upon at a single time.  In effect, we’re “over-parallelizing” this routine.  This is a common problem when working with divide and conquer algorithms, and leads to an important observation: When parallelizing a recursive routine, take special care not to add more tasks than necessary to fully utilize your system. This can be done with a few different approaches, in this case.  Typically, the way to handle this is to stop parallelizing the routine at a certain point, and revert back to the serial approach.  Since the first few recursions will all still be parallelized, our “deeper” recursive tasks will be running in parallel, and can take full advantage of the machine.  This also dramatically reduces the overhead added by parallelizing, since we’re only adding overhead for the first few recursive calls.  There are two basic approaches we can take here.  The first approach would be to look at the total work size, and if it’s smaller than a specific threshold, revert to our serial implementation.  In this case, we could just check right-left, and if it’s under a threshold, call the methods directly instead of using Parallel.Invoke. The second approach is to track how “deep” in the “tree” we are currently at, and if we are below some number of levels, stop parallelizing.  This approach is a more general-purpose approach, since it works on routines which parse trees as well as routines working off of a single array, but may not work as well if a poor partitioning strategy is chosen or the tree is not balanced evenly. This can be written very easily.  If we pass a maxDepth parameter into our internal routine, we can restrict the amount of times we parallelize by changing the recursive call to: // Code above is unchanged... SwapElements(array, left, last); if (maxDepth < 1) { QuickSortInternal(array, left, last - 1, maxDepth); QuickSortInternal(array, last + 1, right, maxDepth); } else { --maxDepth; Parallel.Invoke( () => QuickSortInternal(array, left, last - 1, maxDepth), () => QuickSortInternal(array, last + 1, right, maxDepth)); } We no longer allow this to parallelize indefinitely – only to a specific depth, at which time we revert to a serial implementation.  By starting the routine with a maxDepth equal to Environment.ProcessorCount, we can restrict the total amount of parallel operations significantly, but still provide adequate work for each processing core. With this final change, my timings are much better.  On average, I get the following timings: Framework via Array.Sort: 7.3 seconds Serial Quicksort Implementation: 9.3 seconds Naive Parallel Implementation: 14 seconds Parallel Implementation Restricting Depth: 4.7 seconds Finally, we are now faster than the framework’s Array.Sort implementation.

    Read the article

  • Create and Consume WCF service using Visual Studio 2010

    - by sreejukg
    In this article I am going to demonstrate how to create a WCF service, that can be hosted inside IIS and a windows application that consume the WCF service. To support service oriented architecture, Microsoft developed the programming model named Windows Communication Foundation (WCF). ASMX was the prior version from Microsoft, was completely based on XML and .Net framework continues to support ASMX web services in future versions also. While ASMX web services was the first step towards the service oriented architecture, Microsoft has made a big step forward by introducing WCF. An overview of planning for WCF can be found from this link http://msdn.microsoft.com/en-us/library/ff649584.aspx . The following are the important differences between WCF and ASMX from an asp.net developer point of view. 1. ASMX web services are easy to write, configure and consume 2. ASMX web services are only hosted in IIS 3. ASMX web services can only use http 4. WCF, can be hosted inside IIS, windows service, console application, WAS(Windows Process Activation Service) etc 5. WCF can be used with HTTP, TCP/IP, MSMQ and other protocols. The detailed difference between ASMX web service and WCF can be found here. http://msdn.microsoft.com/en-us/library/cc304771.aspx Though WCF is a bigger step for future, Visual Studio makes it simpler to create, publish and consume the WCF service. In this demonstration, I am going to create a service named SayHello that accepts 2 parameters such as name and language code. The service will return a hello to user name that corresponds to the language. So the proposed service usage is as follows. Caller: SayHello(“Sreeju”, “en”) -> return value -> Hello Sreeju Caller: SayHello(“???”, “ar”) -> return value -> ????? ??? Caller: SayHello(“Sreeju”, “es”) - > return value -> Hola Sreeju Note: calling an automated translation service is not the intention of this article. If you are interested, you can find bing translator API and can use in your application. http://www.microsofttranslator.com/dev/ So Let us start First I am going to create a Service Application that offer the SayHello Service. Open Visual Studio 2010, Go to File -> New Project, from your preferred language from the templates section select WCF, select WCF service application as the project type, give the project a name(I named it as HelloService), click ok so that visual studio will create the project for you. In this demonstration, I have used C# as the programming language. Visual studio will create the necessary files for you to start with. By default it will create a service with name Service1.svc and there will be an interface named IService.cs. The screenshot for the project in solution explorer is as follows Since I want to demonstrate how to create new service, I deleted Service1.Svc and IService1.cs files from the project by right click the file and select delete. Now in the project there is no service available, I am going to create one. From the solution explorer, right click the project, select Add -> New Item Add new item dialog will appear to you. Select WCF service from the list, give the name as HelloService.svc, and click on the Add button. Now Visual studio will create 2 files with name IHelloService.cs and HelloService.svc. These files are basically the service definition (IHelloService.cs) and the service implementation (HelloService.svc). Let us examine the IHelloService interface. The code state that IHelloService is the service definition and it provides an operation/method (similar to web method in ASMX web services) named DoWork(). Any WCF service will have a definition file as an Interface that defines the service. Let us see what is inside HelloService.svc The code illustrated is implementing the interface IHelloService. The code is self-explanatory; the HelloService class needs to implement all the methods defined in the Service Definition. Let me do the service as I require. Open IHelloService.cs in visual studio, and delete the DoWork() method and add a definition for SayHello(), do not forget to add OperationContract attribute to the method. The modified IHelloService.cs will look as follows Now implement the SayHello method in the HelloService.svc.cs file. Here I wrote the code for SayHello method as follows. I am done with the service. Now you can build and run the service by clicking f5 (or selecting start debugging from the debug menu). Visual studio will host the service in give you a client to test it. The screenshot is as follows. In the left pane, it shows the services available in the server and in right side you can invoke the service. To test the service sayHello, double click on it from the above window. It will ask you to enter the parameters and click on the invoke button. See a sample output below. Now I have done with the service. The next step is to write a service client. Creating a consumer application involves 2 steps. One generating the class and configuration file corresponds to the service. Create a project that utilizes the generated class and configuration file. First I am going to generate the class and configuration file. There is a great tool available with Visual Studio named svcutil.exe, this tool will create the necessary class and configuration files for you. Read the documentation for the svcutil.exe here http://msdn.microsoft.com/en-us/library/aa347733.aspx . Open Visual studio command prompt, you can find it under Start Menu -> All Programs -> Visual Studio 2010 -> Visual Studio Tools -> Visual Studio command prompt Make sure the service is in running state in visual studio. Note the url for the service(from the running window, you can right click and choose copy address). Now from the command prompt, enter the svcutil.exe command as follows. I have mentioned the url and the /d switch – for the directory to store the output files(In this case d:\temp). If you are using windows drive(in my case it is c: ) , make sure you open the command prompt with run as administrator option, otherwise you will get permission error(Only in windows 7 or windows vista). The tool has created 2 files, HelloService.cs and output.config. Now the next step is to create a new project and use the created files and consume the service. Let us do that now. I am going to add a console application to the current solution. Right click solution name in the solution explorer, right click, Add-> New Project Under Visual C#, select console application, give the project a name, I named it TestService Now navigate to d:\temp where I generated the files with the svcutil.exe. Rename output.config to app.config. Next step is to add both files (d:\temp\helloservice.cs and app.config) to the files. In the solution explorer, right click the project, Add -> Add existing item, browse to the d:\temp folder, select the 2 files as mentioned before, click on the add button. Now you need to add a reference to the System.ServiceModel to the project. From solution explorer, right click the references under testservice project, select Add reference. In the Add reference dialog, select the .Net tab, select System.ServiceModel, and click ok Now open program.cs by double clicking on it and add the code to consume the web service to the main method. The modified file looks as follows Right click the testservice project and set as startup project. Click f5 to run the project. See the sample output as follows Publishing WCF service under IIS is similar to publishing ASP.Net application. Publish the application to a folder using Visual studio publishing feature, create a virtual directory and create it as an application. Don’t forget to set the application pool to use ASP.Net version 4. One last thing you need to check is the app.config file you have added to the solution. See the element client under ServiceModel element. There is an endpoint element with address attribute that points to the published service URL. If you permanently host the service under IIS, you can simply change the address parameter to the corresponding one and your application will consume the service. You have seen how easily you can build/consume WCF service. If you need the solution in zipped format, please post your email below.

    Read the article

  • Point inside Oriented Bounding Box?

    - by Milo
    I have an OBB2D class based on SAT. This is my point in OBB method: public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } Here is the rest of the class; the parts that pertain: public class OBB2D { private Vector2D projVec = new Vector2D(); private static Vector2D projAVec = new Vector2D(); private static Vector2D projBVec = new Vector2D(); private static Vector2D tempNormal = new Vector2D(); private Vector2D deltaVec = new Vector2D(); private ArrayList<Vector2D> collisionPoints = new ArrayList<Vector2D>(); // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(float centerx, float centery, float w, float h, float angle) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(centerx,centery,w,h,angle); } public OBB2D(float left, float top, float width, float height) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(left + (width / 2), top + (height / 2),width,height,0.0f); } public void set(float centerx,float centery,float w, float h,float angle) { float vxx = (float)Math.cos(angle); float vxy = (float)Math.sin(angle); float vyx = (float)-Math.sin(angle); float vyy = (float)Math.cos(angle); vxx *= w / 2; vxy *= (w / 2); vyx *= (h / 2); vyy *= (h / 2); corner[0].x = centerx - vxx - vyx; corner[0].y = centery - vxy - vyy; corner[1].x = centerx + vxx - vyx; corner[1].y = centery + vxy - vyy; corner[2].x = centerx + vxx + vyx; corner[2].y = centery + vxy + vyy; corner[3].x = centerx - vxx + vyx; corner[3].y = centery - vxy + vyy; this.center.x = centerx; this.center.y = centery; this.angle = angle; computeAxes(); extents.x = w / 2; extents.y = h / 2; computeBoundingRect(); } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0].x = corner[1].x - corner[0].x; axis[0].y = corner[1].y - corner[0].y; axis[1].x = corner[3].x - corner[0].x; axis[1].y = corner[3].y - corner[0].y; // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { float l = axis[a].length(); float ll = l * l; axis[a].x = axis[a].x / ll; axis[a].y = axis[a].y / ll; origin[a] = corner[0].dot(axis[a]); } } public void computeBoundingRect() { boundingRect.left = JMath.min(JMath.min(corner[0].x, corner[3].x), JMath.min(corner[1].x, corner[2].x)); boundingRect.top = JMath.min(JMath.min(corner[0].y, corner[1].y),JMath.min(corner[2].y, corner[3].y)); boundingRect.right = JMath.max(JMath.max(corner[1].x, corner[2].x), JMath.max(corner[0].x, corner[3].x)); boundingRect.bottom = JMath.max(JMath.max(corner[2].y, corner[3].y),JMath.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(rect.centerX(),rect.centerY(),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } public void moveTo(float centerx, float centery) { float cx,cy; cx = center.x; cy = center.y; deltaVec.x = centerx - cx; deltaVec.y = centery - cy; for (int c = 0; c < 4; ++c) { corner[c].x += deltaVec.x; corner[c].y += deltaVec.y; } boundingRect.left += deltaVec.x; boundingRect.top += deltaVec.y; boundingRect.right += deltaVec.x; boundingRect.bottom += deltaVec.y; this.center.x = centerx; this.center.y = centery; computeAxes(); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center.x,center.y,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center.x,center.y,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } public static float distance(float ax, float ay,float bx, float by) { if (ax < bx) return bx - ay; else return ax - by; } public Vector2D project(float ax, float ay) { projVec.x = Float.MAX_VALUE; projVec.y = Float.MIN_VALUE; for (int i = 0; i < corner.length; ++i) { float dot = Vector2D.dot(corner[i].x,corner[i].y,ax,ay); projVec.x = JMath.min(dot, projVec.x); projVec.y = JMath.max(dot, projVec.y); } return projVec; } public Vector2D getCorner(int c) { return corner[c]; } public int getNumCorners() { return corner.length; } public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } public ArrayList<Vector2D> getCollsionPoints(OBB2D b) { collisionPoints.clear(); for(int i = 0; i < corner.length; ++i) { if(b.pointInside(corner[i])) { collisionPoints.add(corner[i]); } } for(int i = 0; i < b.corner.length; ++i) { if(pointInside(b.corner[i])) { collisionPoints.add(b.corner[i]); } } return collisionPoints; } }; What could be wrong? When I getCollisionPoints for 2 OBBs I know are penetrating, it returns no points. Thanks

    Read the article

  • Point of contact of 2 OBBs?

    - by Milo
    I'm working on the physics for my GTA2-like game so I can learn more about game physics. The collision detection and resolution are working great. I'm now just unsure how to compute the point of contact when I hit a wall. Here is my OBB class: public class OBB2D { private Vector2D projVec = new Vector2D(); private static Vector2D projAVec = new Vector2D(); private static Vector2D projBVec = new Vector2D(); private static Vector2D tempNormal = new Vector2D(); private Vector2D deltaVec = new Vector2D(); // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(float centerx, float centery, float w, float h, float angle) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(centerx,centery,w,h,angle); } public OBB2D(float left, float top, float width, float height) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(left + (width / 2), top + (height / 2),width,height,0.0f); } public void set(float centerx,float centery,float w, float h,float angle) { float vxx = (float)Math.cos(angle); float vxy = (float)Math.sin(angle); float vyx = (float)-Math.sin(angle); float vyy = (float)Math.cos(angle); vxx *= w / 2; vxy *= (w / 2); vyx *= (h / 2); vyy *= (h / 2); corner[0].x = centerx - vxx - vyx; corner[0].y = centery - vxy - vyy; corner[1].x = centerx + vxx - vyx; corner[1].y = centery + vxy - vyy; corner[2].x = centerx + vxx + vyx; corner[2].y = centery + vxy + vyy; corner[3].x = centerx - vxx + vyx; corner[3].y = centery - vxy + vyy; this.center.x = centerx; this.center.y = centery; this.angle = angle; computeAxes(); extents.x = w / 2; extents.y = h / 2; computeBoundingRect(); } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0].x = corner[1].x - corner[0].x; axis[0].y = corner[1].y - corner[0].y; axis[1].x = corner[3].x - corner[0].x; axis[1].y = corner[3].y - corner[0].y; // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { float l = axis[a].length(); float ll = l * l; axis[a].x = axis[a].x / ll; axis[a].y = axis[a].y / ll; origin[a] = corner[0].dot(axis[a]); } } public void computeBoundingRect() { boundingRect.left = JMath.min(JMath.min(corner[0].x, corner[3].x), JMath.min(corner[1].x, corner[2].x)); boundingRect.top = JMath.min(JMath.min(corner[0].y, corner[1].y),JMath.min(corner[2].y, corner[3].y)); boundingRect.right = JMath.max(JMath.max(corner[1].x, corner[2].x), JMath.max(corner[0].x, corner[3].x)); boundingRect.bottom = JMath.max(JMath.max(corner[2].y, corner[3].y),JMath.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(rect.centerX(),rect.centerY(),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } public void moveTo(float centerx, float centery) { float cx,cy; cx = center.x; cy = center.y; deltaVec.x = centerx - cx; deltaVec.y = centery - cy; for (int c = 0; c < 4; ++c) { corner[c].x += deltaVec.x; corner[c].y += deltaVec.y; } boundingRect.left += deltaVec.x; boundingRect.top += deltaVec.y; boundingRect.right += deltaVec.x; boundingRect.bottom += deltaVec.y; this.center.x = centerx; this.center.y = centery; computeAxes(); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center.x,center.y,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center.x,center.y,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } public static float distance(float ax, float ay,float bx, float by) { if (ax < bx) return bx - ay; else return ax - by; } public Vector2D project(float ax, float ay) { projVec.x = Float.MAX_VALUE; projVec.y = Float.MIN_VALUE; for (int i = 0; i < corner.length; ++i) { float dot = Vector2D.dot(corner[i].x,corner[i].y,ax,ay); projVec.x = JMath.min(dot, projVec.x); projVec.y = JMath.max(dot, projVec.y); } return projVec; } public Vector2D getCorner(int c) { return corner[c]; } public int getNumCorners() { return corner.length; } public static float collisionResponse(OBB2D a, OBB2D b, Vector2D outNormal) { float depth = Float.MAX_VALUE; for (int i = 0; i < a.getNumCorners() + b.getNumCorners(); ++i) { Vector2D edgeA; Vector2D edgeB; if(i >= a.getNumCorners()) { edgeA = b.getCorner((i + b.getNumCorners() - 1) % b.getNumCorners()); edgeB = b.getCorner(i % b.getNumCorners()); } else { edgeA = a.getCorner((i + a.getNumCorners() - 1) % a.getNumCorners()); edgeB = a.getCorner(i % a.getNumCorners()); } tempNormal.x = edgeB.x -edgeA.x; tempNormal.y = edgeB.y - edgeA.y; tempNormal.normalize(); projAVec.equals(a.project(tempNormal.x,tempNormal.y)); projBVec.equals(b.project(tempNormal.x,tempNormal.y)); float distance = OBB2D.distance(projAVec.x, projAVec.y,projBVec.x,projBVec.y); if (distance > 0.0f) { return 0.0f; } else { float d = Math.abs(distance); if (d < depth) { depth = d; outNormal.equals(tempNormal); } } } float dx,dy; dx = b.getCenter().x - a.getCenter().x; dy = b.getCenter().y - a.getCenter().y; float dot = Vector2D.dot(dx,dy,outNormal.x,outNormal.y); if(dot > 0) { outNormal.x = -outNormal.x; outNormal.y = -outNormal.y; } return depth; } public Vector2D getMoveDeltaVec() { return deltaVec; } }; Thanks!

    Read the article

  • Snake Game Help

    - by MuhammadA
    I am making a snake game and learning XNA at the same time. I have 3 classes : Game.cs, Snake.cs and Apple.cs My problem is more of a conceptual problem, I want to know which class is really responsible for ... detecting collision of snake head on apple/itself/wall? which class should increase the snakes speed, size? It seems to me that however much I try and put the snake stuff into snake.cs that game.cs has to know a lot about the snake, like : -- I want to increase the score depending on size of snake, the score variable is inside game.cs, which means now I have to ask the snake its size on every hit of the apple... seems a bit unclean all this highly coupled code. or -- I DO NOT want to place the apple under the snake... now the apple suddenly has to know about all the snake parts, my head hurts when I think of that. Maybe there should be some sort of AppleLayer.cs class that should know about the snake... Whats the best approach in such a simple scenario? Any tips welcome. Game.cs : using System; using System.Collections.Generic; using System.Linq; using Microsoft.Xna.Framework; using Microsoft.Xna.Framework.Audio; using Microsoft.Xna.Framework.Content; using Microsoft.Xna.Framework.GamerServices; using Microsoft.Xna.Framework.Graphics; using Microsoft.Xna.Framework.Input; using Microsoft.Xna.Framework.Media; using Microsoft.Xna.Framework.Design; namespace Snakez { public enum CurrentGameState { Playing, Paused, NotPlaying } public class Game1 : Microsoft.Xna.Framework.Game { private GraphicsDeviceManager _graphics; private SpriteBatch _spriteBatch; private readonly Color _niceGreenColour = new Color(167, 255, 124); private KeyboardState _oldKeyboardState; private SpriteFont _scoreFont; private SoundEffect _biteSound, _crashSound; private Vector2 _scoreLocation = new Vector2(10, 10); private Apple _apple; private Snake _snake; private int _score = 0; private int _speed = 1; public Game1() { _graphics = new GraphicsDeviceManager(this); Content.RootDirectory = "Content"; } /// <summary> /// Allows the game to perform any initialization it needs to before starting to run. /// This is where it can query for any required services and load any non-graphic /// related content. Calling base.Initialize will enumerate through any components /// and initialize them as well. /// </summary> protected override void Initialize() { base.Initialize(); } /// <summary> /// LoadContent will be called once per game and is the place to load /// all of your content. /// </summary> protected override void LoadContent() { _spriteBatch = new SpriteBatch(GraphicsDevice); _scoreFont = Content.Load<SpriteFont>("Score"); _apple = new Apple(800, 480, Content.Load<Texture2D>("Apple")); _snake = new Snake(Content.Load<Texture2D>("BodyBlock")); _biteSound = Content.Load<SoundEffect>("Bite"); _crashSound = Content.Load<SoundEffect>("Crash"); } /// <summary> /// UnloadContent will be called once per game and is the place to unload /// all content. /// </summary> protected override void UnloadContent() { Content.Unload(); } /// <summary> /// Allows the game to run logic such as updating the world, /// checking for collisions, gathering input, and playing audio. /// </summary> /// <param name="gameTime">Provides a snapshot of timing values.</param> protected override void Update(GameTime gameTime) { KeyboardState newKeyboardState = Keyboard.GetState(); if (newKeyboardState.IsKeyDown(Keys.Escape)) { this.Exit(); // Allows the game to exit } else if (newKeyboardState.IsKeyDown(Keys.Up) && !_oldKeyboardState.IsKeyDown(Keys.Up)) { _snake.SetDirection(Direction.Up); } else if (newKeyboardState.IsKeyDown(Keys.Down) && !_oldKeyboardState.IsKeyDown(Keys.Down)) { _snake.SetDirection(Direction.Down); } else if (newKeyboardState.IsKeyDown(Keys.Left) && !_oldKeyboardState.IsKeyDown(Keys.Left)) { _snake.SetDirection(Direction.Left); } else if (newKeyboardState.IsKeyDown(Keys.Right) && !_oldKeyboardState.IsKeyDown(Keys.Right)) { _snake.SetDirection(Direction.Right); } _oldKeyboardState = newKeyboardState; _snake.Update(); if (_snake.IsEating(_apple)) { _biteSound.Play(); _score += 10; _apple.Place(); } base.Update(gameTime); } /// <summary> /// This is called when the game should draw itself. /// </summary> /// <param name="gameTime">Provides a snapshot of timing values.</param> protected override void Draw(GameTime gameTime) { GraphicsDevice.Clear(_niceGreenColour); float frameRate = 1 / (float)gameTime.ElapsedGameTime.TotalSeconds; _spriteBatch.Begin(); _spriteBatch.DrawString(_scoreFont, "Score : " + _score, _scoreLocation, Color.Red); _apple.Draw(_spriteBatch); _snake.Draw(_spriteBatch); _spriteBatch.End(); base.Draw(gameTime); } } } Snake.cs : using System; using System.Collections.Generic; using System.Linq; using System.Text; using Microsoft.Xna.Framework.Graphics; using Microsoft.Xna.Framework; namespace Snakez { public enum Direction { Up, Down, Left, Right } public class Snake { private List<Rectangle> _parts; private readonly Texture2D _bodyBlock; private readonly int _startX = 160; private readonly int _startY = 120; private int _moveDelay = 100; private DateTime _lastUpdatedAt; private Direction _direction; private Rectangle _lastTail; public Snake(Texture2D bodyBlock) { _bodyBlock = bodyBlock; _parts = new List<Rectangle>(); _parts.Add(new Rectangle(_startX, _startY, _bodyBlock.Width, _bodyBlock.Height)); _parts.Add(new Rectangle(_startX + bodyBlock.Width, _startY, _bodyBlock.Width, _bodyBlock.Height)); _parts.Add(new Rectangle(_startX + (bodyBlock.Width) * 2, _startY, _bodyBlock.Width, _bodyBlock.Height)); _parts.Add(new Rectangle(_startX + (bodyBlock.Width) * 3, _startY, _bodyBlock.Width, _bodyBlock.Height)); _direction = Direction.Right; _lastUpdatedAt = DateTime.Now; } public void Draw(SpriteBatch spriteBatch) { foreach (var p in _parts) { spriteBatch.Draw(_bodyBlock, new Vector2(p.X, p.Y), Color.White); } } public void Update() { if (DateTime.Now.Subtract(_lastUpdatedAt).TotalMilliseconds > _moveDelay) { //DateTime.Now.Ticks _lastTail = _parts.First(); _parts.Remove(_lastTail); /* add new head in right direction */ var lastHead = _parts.Last(); var newHead = new Rectangle(0, 0, _bodyBlock.Width, _bodyBlock.Height); switch (_direction) { case Direction.Up: newHead.X = lastHead.X; newHead.Y = lastHead.Y - _bodyBlock.Width; break; case Direction.Down: newHead.X = lastHead.X; newHead.Y = lastHead.Y + _bodyBlock.Width; break; case Direction.Left: newHead.X = lastHead.X - _bodyBlock.Width; newHead.Y = lastHead.Y; break; case Direction.Right: newHead.X = lastHead.X + _bodyBlock.Width; newHead.Y = lastHead.Y; break; } _parts.Add(newHead); _lastUpdatedAt = DateTime.Now; } } public void SetDirection(Direction newDirection) { if (_direction == Direction.Up && newDirection == Direction.Down) { return; } else if (_direction == Direction.Down && newDirection == Direction.Up) { return; } else if (_direction == Direction.Left && newDirection == Direction.Right) { return; } else if (_direction == Direction.Right && newDirection == Direction.Left) { return; } _direction = newDirection; } public bool IsEating(Apple apple) { if (_parts.Last().Intersects(apple.Location)) { GrowBiggerAndFaster(); return true; } return false; } private void GrowBiggerAndFaster() { _parts.Insert(0, _lastTail); _moveDelay -= (_moveDelay / 100)*2; } } } Apple.cs : using System; using System.Collections.Generic; using System.Linq; using System.Text; using Microsoft.Xna.Framework.Graphics; using Microsoft.Xna.Framework; namespace Snakez { public class Apple { private readonly int _maxWidth, _maxHeight; private readonly Texture2D _texture; private readonly Random random = new Random(); public Rectangle Location { get; private set; } public Apple(int screenWidth, int screenHeight, Texture2D texture) { _maxWidth = (screenWidth + 1) - texture.Width; _maxHeight = (screenHeight + 1) - texture.Height; _texture = texture; Place(); } public void Place() { Location = GetRandomLocation(_maxWidth, _maxHeight); } private Rectangle GetRandomLocation(int maxWidth, int maxHeight) { // x and y -- multiple of 20 int x = random.Next(1, maxWidth); var leftOver = x % 20; x = x - leftOver; int y = random.Next(1, maxHeight); leftOver = y % 20; y = y - leftOver; return new Rectangle(x, y, _texture.Width, _texture.Height); } public void Draw(SpriteBatch spriteBatch) { spriteBatch.Draw(_texture, Location, Color.White); } } }

    Read the article

  • Creating an ASP.NET report using Visual Studio 2010 - Part 2

    - by rajbk
    We continue building our report in this three part series. Creating an ASP.NET report using Visual Studio 2010 - Part 1 Creating an ASP.NET report using Visual Studio 2010 - Part 3 Creating the Client Report Definition file (RDLC) Add a folder called “RDLC”. This will hold our RDLC report.   Right click on the RDLC folder, select “Add new item..” and add an “RDLC” name of “Products”. We will use the “Report Wizard” to walk us through the steps of creating the RDLC.   In the next dialog, give the dataset a name called “ProductDataSet”. Change the data source to “NorthwindReports.DAL” and select “ProductRepository(GetProductsProjected)”. The fields that are returned from the method are shown on the right. Click next.   Drag and drop the ProductName, CategoryName, UnitPrice and Discontinued into the Values container. Note that you can create much more complex grouping using this UI. Click Next.   Most of the selections on this screen are grayed out because we did not choose a grouping in the previous screen. Click next. Choose a style for your report. Click next. The report graphic design surface is now visible. Right click on the report and add a page header and page footer. With the report design surface active, drag and drop a TextBox from the tool box to the page header. Drag one more textbox to the page header. We will use the text boxes to add some header text as shown in the next figure. You can change the font size and other properties of the textboxes using the formatting tool bar (marked in red). You can also resize the columns by moving your cursor in between columns and dragging. Adding Expressions Add two more text boxes to the page footer. We will use these to add the time the report was generated and page numbers. Right click on the first textbox in the page footer and select “Expression”. Add the following expression for the print date (note the = sign at the left of the expression in the dialog below) "© Northwind Traders " & Format(Now(),"MM/dd/yyyy hh:mm tt") Right click on the second text box and add the following for the page count.   Globals.PageNumber & " of " & Globals.TotalPages Formatting the page footer is complete.   We are now going to format the “Unit Price” column so it displays the number in currency format.  Right click on the [UnitPrice] column (not header) and select “Text Box Properties..” Under “Number”, select “Currency”. Hit OK. Adding a chart With the design surface active, go to the toolbox and drag and drop a chart control. You will need to move the product list table down first to make space for the chart contorl. The document can also be resized by dragging on the corner or at the page header/footer separator. In the next dialog, pick the first chart type. This can be changed later if needed. Click OK. The chart gets added to the design surface.   Click on the blue bars in the chart (not legend). This will bring up drop locations for dropping the fields. Drag and drop the UnitPrice and CategoryName into the top (y axis) and bottom (x axis) as shown below. This will give us the total unit prices for a given category. That is the best I could come up with as far as what report to render, sorry :-) Delete the legend area to get more screen estate. Resize the chart to your liking. Change the header, x axis and y axis text by double clicking on those areas. We made it this far. Let’s impress the client by adding a gradient to the bar graph :-) Right click on the blue bar and select “Series properties”. Under “Fill”, add a color and secondary color and select the Gradient style. We are done designing our report. In the next section you will see how to add the report to the report viewer control, bind to the data and make it refresh when the filter criteria are changed.   Creating an ASP.NET report using Visual Studio 2010 - Part 3

    Read the article

  • Move penetrating OBB out of another OBB to resolve collision

    - by Milo
    I'm working on collision resolution for my game. I just need a good way to get an object out of another object if it gets stuck. In this case a car. Here is a typical scenario. The red car is in the green object. How do I correctly get it out so the car can slide along the edge of the object as it should. I tried: if(buildings.size() > 0) { Entity e = buildings.get(0); Vector2D vel = new Vector2D(); vel.x = vehicle.getVelocity().x; vel.y = vehicle.getVelocity().y; vel.normalize(); while(vehicle.getRect().overlaps(e.getRect())) { vehicle.setCenter(vehicle.getCenterX() - vel.x * 0.1f, vehicle.getCenterY() - vel.y * 0.1f); } colided = true; } But that does not work too well. Is there some sort of vector I could calculate to use as the vector to move the car away from the object? Thanks Here is my OBB2D class: public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

    Read the article

  • Finding the normal of OBB face with an OBB penetrating

    - by Milo
    Below is an illustration: I have an OBB in an OBB (see below for OBB2D code if needed). What I need to determine is, what face it is in, and what direction do I point the normal? The goal is to get the OBB out of the OBB so the normal needs to face outward of the OBB. How could I go about: Finding what face the line is penetrating given the 4 corners of the OBB and the class below: if we define dx=x2-x1 and dy=y2-y1, then the normals are (-dy, dx) and (dy, -dx). Which normal points outward of the OBB? Thanks public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

    Read the article

  • Java: initialization problem with private-final-int-value and empty constructor

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; InitInt(){} public static void main(String[] args) { // I don't want to assign any value. // just initialize it, how? InitInt test = new InitInt(); System.out.println(test.getRight()); // later assiging a value } public int getRight(){return right;} } Initialization problem with Constructor InitInt{ // Still the error, "may not be initialized" // How to initialise it? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } }

    Read the article

  • jQuery to get innerHTML not working on a HTMLFontElement object...

    - by Matt
    I have jQuery to select all font elements that are children of the element with id="right" within the html stored in the var html... when I do an alert to see how many elements it gets: alert($("#right > font", html).length); it gives me an alert of: 5 but when I try any of the following, I don't get any alerts... alert($("#right > font", html)[0].html()); alert($("#right > font", html)[0].text()); alert($("#right > font", html)[0].val()); Any Ideas? Thanks, Matt

    Read the article

  • Off center projection

    - by N0xus
    I'm trying to implement the code that was freely given by a very kind developer at the following link: http://forum.unity3d.com/threads/142383-Code-sample-Off-Center-Projection-Code-for-VR-CAVE-or-just-for-fun Right now, all I'm trying to do is bring it in on one camera, but I have a few issues. My class, looks as follows: using UnityEngine; using System.Collections; public class PerspectiveOffCenter : MonoBehaviour { // Use this for initialization void Start () { } // Update is called once per frame void Update () { } public static Matrix4x4 GeneralizedPerspectiveProjection(Vector3 pa, Vector3 pb, Vector3 pc, Vector3 pe, float near, float far) { Vector3 va, vb, vc; Vector3 vr, vu, vn; float left, right, bottom, top, eyedistance; Matrix4x4 transformMatrix; Matrix4x4 projectionM; Matrix4x4 eyeTranslateM; Matrix4x4 finalProjection; ///Calculate the orthonormal for the screen (the screen coordinate system vr = pb - pa; vr.Normalize(); vu = pc - pa; vu.Normalize(); vn = Vector3.Cross(vr, vu); vn.Normalize(); //Calculate the vector from eye (pe) to screen corners (pa, pb, pc) va = pa-pe; vb = pb-pe; vc = pc-pe; //Get the distance;; from the eye to the screen plane eyedistance = -(Vector3.Dot(va, vn)); //Get the varaibles for the off center projection left = (Vector3.Dot(vr, va)*near)/eyedistance; right = (Vector3.Dot(vr, vb)*near)/eyedistance; bottom = (Vector3.Dot(vu, va)*near)/eyedistance; top = (Vector3.Dot(vu, vc)*near)/eyedistance; //Get this projection projectionM = PerspectiveOffCenter(left, right, bottom, top, near, far); //Fill in the transform matrix transformMatrix = new Matrix4x4(); transformMatrix[0, 0] = vr.x; transformMatrix[0, 1] = vr.y; transformMatrix[0, 2] = vr.z; transformMatrix[0, 3] = 0; transformMatrix[1, 0] = vu.x; transformMatrix[1, 1] = vu.y; transformMatrix[1, 2] = vu.z; transformMatrix[1, 3] = 0; transformMatrix[2, 0] = vn.x; transformMatrix[2, 1] = vn.y; transformMatrix[2, 2] = vn.z; transformMatrix[2, 3] = 0; transformMatrix[3, 0] = 0; transformMatrix[3, 1] = 0; transformMatrix[3, 2] = 0; transformMatrix[3, 3] = 1; //Now for the eye transform eyeTranslateM = new Matrix4x4(); eyeTranslateM[0, 0] = 1; eyeTranslateM[0, 1] = 0; eyeTranslateM[0, 2] = 0; eyeTranslateM[0, 3] = -pe.x; eyeTranslateM[1, 0] = 0; eyeTranslateM[1, 1] = 1; eyeTranslateM[1, 2] = 0; eyeTranslateM[1, 3] = -pe.y; eyeTranslateM[2, 0] = 0; eyeTranslateM[2, 1] = 0; eyeTranslateM[2, 2] = 1; eyeTranslateM[2, 3] = -pe.z; eyeTranslateM[3, 0] = 0; eyeTranslateM[3, 1] = 0; eyeTranslateM[3, 2] = 0; eyeTranslateM[3, 3] = 1f; //Multiply all together finalProjection = new Matrix4x4(); finalProjection = Matrix4x4.identity * projectionM*transformMatrix*eyeTranslateM; //finally return return finalProjection; } // Update is called once per frame public void FixedUpdate () { Camera cam = camera; //calculate projection Matrix4x4 genProjection = GeneralizedPerspectiveProjection( new Vector3(0,1,0), new Vector3(1,1,0), new Vector3(0,0,0), new Vector3(0,0,0), cam.nearClipPlane, cam.farClipPlane); //(BottomLeftCorner, BottomRightCorner, TopLeftCorner, trackerPosition, cam.nearClipPlane, cam.farClipPlane); cam.projectionMatrix = genProjection; } } My error lies in projectionM = PerspectiveOffCenter(left, right, bottom, top, near, far); The debugger states: Expression denotes a `type', where a 'variable', 'value' or 'method group' was expected. Thus, I changed the line to read: projectionM = new PerspectiveOffCenter(left, right, bottom, top, near, far); But then the error is changed to: The type 'PerspectiveOffCenter' does not contain a constructor that takes '6' arguments. For reasons that are obvious. So, finally, I changed the line to read: projectionM = new GeneralizedPerspectiveProjection(left, right, bottom, top, near, far); And the error I get is: is a 'method' but a 'type' was expected. With this last error, I'm not sure what it is I should do / missing. Can anyone see what it is that I'm missing to fix this error?

    Read the article

  • Java: design problem with final-value and empty constructor

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; // Design Problem? // I feel the initialization problem is just due to bad style. InitInt(){} InitInt{ // Still the error, "may not be initialized" // How to initialise it? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } } public static void main(String[] args) { InitInt test = new InitInt(); System.out.println(test.getRight()); } public int getRight(){return right;} } Partial Solutions and Suggestions use "this" to access methods in the class, instead of creating empty constructor change final to non-final with final field value: initialize all final values in every constructor remove the empty constructor, keep your code simple and clean

    Read the article

  • How to occupy all the space in a div when working with min-height header / footer

    - by javacoder
    I believe this is a beginner's CSS question. I am utilizing the method described in http://www.xs4all.nl/~peterned/examples/csslayout1.html to fix a header to the top and a footer to the bottom. What I'd like to achieve now is two columns inside the content div. A left one of 200px and a right one that takes up the rest of the width. Unfortunately, I can't get the left and right divs to display correctly: they just don't grow vertically, and if I make the right div "width: 100%" it positions itself underneath the left one. What is the trick to make the left and right div take up all the space within the content div? The layout1.css is the original one. I just added two entries: #left and #right layout1.css: /** * 100% height layout with header and footer * ---------------------------------------------- * Feel free to copy/use/change/improve */ html,body { margin: 0; padding: 0; height: 100%; /* needed for container min-height */ background: gray; font-family: arial, sans-serif; font-size: small; color: #666; } h1 { font: 1.5em georgia, serif; margin: 0.5em 0; } h2 { font: 1.25em georgia, serif; margin: 0 0 0.5em; } h1,h2,a { color: orange; } p { line-height: 1.5; margin: 0 0 1em; } div#container { position: relative; /* needed for footer positioning*/ margin: 0 auto; /* center, not in IE5 */ width: 750px; background: #f0f0f0; height: auto !important; /* real browsers */ height: 100%; /* IE6: treaded as min-height*/ min-height: 100%; /* real browsers */ } div#header { padding: 1em; background: #ddd url("../csslayout.gif") 98% 10px no-repeat; border-bottom: 6px double gray; } div#header p { font-style: italic; font-size: 1.1em; margin: 0; } div#content { padding: 1em 1em 5em; /* bottom padding for footer */ } div#content p { text-align: justify; padding: 0 1em; } div#footer { position: absolute; width: 100%; bottom: 0; /* stick to bottom */ background: #ddd; border-top: 6px double gray; } div#footer p { padding: 1em; margin: 0; } // added the following: div#left { border: 1px solid red; width: 200px; float: left; min-height: 100%; height: 100%; padding-left: 10px; margin-right: 10px; } div#right { border: 1px solid blue; float: left; min-height: 100%; height: 100%; padding-left: 10px; margin-right: 10px; } layout.html: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>CSS Layout - 100% height</title> <link rel="stylesheet" type="text/css" href="layout1.css" /> </head> <body> <div id="container"> <div id="header"> <h1>header</h1> </div> <div id="content"> <div id="left"> left column </div> <div id="right"> right column </div> </div> <div id="footer"> <p> footer </p> </div> </div> </body>

    Read the article

  • Java: how to initialize private final int value with if-else in constructor?

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; InitInt(){} public static void main(String[] args) { // I don't want to assign any value. // just initialize it, how? InitInt test = new InitInt(); System.out.println(test.getRight()); // later assiging a value } public int getRight(){return right;} } Initialization problem with Constructor, due to if-else -loop InitInt{ // Still the error, "may not be initialized" // How to initialise it, without removing if-else? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } }

    Read the article

  • Top 10 Browser Productivity Tips

    - by Renso
    Originally posted on: http://geekswithblogs.net/renso/archive/2013/10/14/top-10-browser-productivity-tips.aspxYou don’t have to be a geek to be a productive browser user. The tips below have been selected by actions users take most of the time to navigate a web-site but use long-standing keyboard or mouse actions to get them done, when there are keyboard short-cuts you can use instead. Since you hands are already on the keyboard it is almost always faster to sue a keyboard shortcut to get something done that you usually used the mouse for. For example right-clicking on something to copy it and then doing to same for pasting something is very time consuming, keyboard shortcuts have been created that simplify the task. All it takes are a few memory brain cells to remember them. Here are the tips, in no particular order:   Tip 1 Hold down the spacebar on your keyboard to page to the end of your web page rather than using your mouse. This is really a slow way of doing it. If you want to page one page at a time, hit the spacebar once, and again to page again. But if you want to page all the way to the end of the web page simply hit Ctrl+End (that is hold down the Ctrl key and hit the End button on your keyboard). To get to the top of your web page, simply hit Ctrl + Home to go all the way to the top of your web page. Tip 2 Where are my downloads? Some folks run downloads again-and-again because they do not know where the last one went and they do not see the popup, or browser note on their web page in the footer, etc. Simply hit Ctrl+J. Works in most browsers. Tip 3 Selecting a US state from a drop down box. Don’t use the mouse, takes just way too long to scroll. When you tab to the drop down box or click on it with your mouse, simply hit the first character of the state and it will be selected. For Texas for example hit the letter “T” twice on your keyboard to get to it. The same concept can be applied to any drop down box that is alphabetical or numerically sorted. Tip 4 Fixing spelling errors. All modern-day browsers support this now. You see the red wavy lines underscoring a word, yes it is a spelling error. How do you fix it? Don’t overtype it or try and fix it manually, fist right-click on it and a list of suggestions comes up. If it does not show up, like my name “Renso” and you know how to spell your name as in this example, look further down the list of options (the little window popup that appears when you right click) and you should see an option to “Add to Dictionary”. Be warned, when you add it, it only adds it to the browser you’re using’s dictionary. If you use Google Chrome, Firefox and IE, each one will have their own list. Tip 5 So you have trouble seeing the text on the screen. Or you are looking at a photo, for example in Facebook. You want to zoom in to read better or zoom into a photo a bit more. Hit Ctrl++ (hold down Ctrl key and hit the plus key – actually it’s the equal key but it is easier to remember that it is plus for bigger). Hit the minus to zoom out. Now you can’t remember what the original size was since you were so excited to hit it 20 times, or was that 21… Simply hit Ctrl+0 (that is zero) and it will reset it to the default. Tip 6 So you closed a couple of tabs in your browser. Suddenly you remember something you wanted to double-check something on one of the tabs, you cannot remember the URL ad the tab is gone forever, or is it? Simply hit Ctrl+Shift+t and it will bring back your tabs one by one each time you click the T. This has also been a great way for me to quickly close some tabs because I don’t want my boss to see I’m shopping and then hitting Ctrl+Shift+t to quickly get it back and complete my check-put and purchase. Or, for parents, when you walk into your daughter’s room and you see she quickly clicks and closes a window/tab in here browser. Not to worry my little darling, daddy will Ctrl+Shift+t and see what boys on Facebook you were talking too… Tip 7 The web browser is frozen on your PC/Laptop/Whatever, in this example it may be your Internet Explorer browser. I don’t mention Firefox or Chrome here because it probably never happens in their world. You cannot close it, it won’t respond to anything you have done s far except for the next step you are about to take, which is throw your two-day old coffee on your keyboard. This happens especially on sites that want to force you to complete a purchase order. Hit Ctrl+Alt+Del on your keyboard on any version of windows, select TASK MANAGER. In the  First Tab, which is the Process Tab, look for the item in question. In this example you should see Internet Explorer. Right-click it and select “End Task”. It will force the thread out of memory and terminate that process. You can of course do this with any program running under your account. Tip 8 This is a personal favorite of mine. To select words in the paragraph without using the mouse. You don’t want to select one character at a time like when you use the Ctrl+arrows as it can be very slow if you want to select a lot of text. You also want to select whole words. Simply use the Ctrl+Shift_arrow (right or left depending which direction you want to go. Tip 9 I was a bit reluctant to add this one, but being in the professional services industry still come across many-a-folk that simply can’t copy-and-paste them-all text or images that reside on them screens, y’all. Ctrl+c to copy and Ctrl+v to paste it. Works a lot faster than using the mouse. You may be asking: “Well why in the devil did they not use Ctrl+p for paste…. because that is for printing. This is of course not limited to the browser world, it applies to almost any piece of software running on PC or Mac. Go try it on an image on your browser, right-click it and select copy. Open a word document and Ctrl+v to paste the image in there. Please consider copyright laws. Tip 10 Getting rid of annoying ads. Now this only works when you load a web page, meaning when you get back to the same page later you will have to do this again and you will need to learn a tool to do it, WELL WORTH IT. For example, I use GrooveShark to listen to music but I don’t like the ads they show. Install a tool like Firebug for Firefox or use the Ctrl+Shift+I on Chrome to bring up the developer toolbar. Shows at the bottom of the page. With Firefox, once you have installed Firebug as an add-on, a yellow bug should appear on the top right-hand-side of your browser, click on it to display the developer toolbar. You will need to learn how to use it, but once you know how to select an item/section on the window (usually just right-click the add you don’t want to see and select “Inspect Element”, the developer toolbar will appear (if not already there)) and then simply hit delete and it will remove the add from the screen. If you don’t know HTML you may need to play with it a bit, but once you understand how it works can open up a whole new world for you on how web pages actually work. If you can think of any others that have saved you a ton of time please let me know so I can add them to a top 99 list.

    Read the article

  • liquid CSS issues - rtl, floating and scrollers

    - by Rani
    hi I want to build a site that will have these restrictions: RTL direction vertical scroll on the right side whole page is floating to the right page has 2 columns the right (main) column has min width the right (main) column has table inside it that can expend in its data and get wider making all other data in the column expend to the same width as well the sidebar should be on the left side but still floating to the right of the main div it should fit low resolution so the page will be able to add horizontal scroll if needed should work in all major browsers don't use table for constructing the page Can someone help or direct me? Thanks Rani

    Read the article

  • 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.

    Read the article

  • Displaying UserControls based on the type a TreeView selection is bound to

    - by Ray Wenderlich
    I am making an app in WPF in a style similar to Windows Explorer, with a TreeView on the left and a pane on the right. I want the contents of the right pane to change depending on the type of the selected element in the TreeView. For example, say the top level in the Tree View contains objects of class "A", and if you expand the "A" object you'll see a list of "B" objects as children of the "A" object. If the "A" object is selected, I want the right pane to show a user control for "A", and if "B" is selected I want the right pane to show a user control for "B". I've currently got this working by: setting up the TreeView with one HierarchialDataTemplate per type adding all the UserControls to the right pane, but collapsed implementing SelectedItemChanged on the TreeView, and setting the appropriate usercontrol to visible and the others to collapsed. However, I'm sure there's a better/more elegant way to switch out the views based on the type the selection is bound to, perhaps by making more use of data binding... any ideas?

    Read the article

  • Placing an background image with padding in h2 tag

    - by Cedar Jensen
    I want to create a headline (h2) with an image at the right-most area of the bounding box. I have the layout almost right except I can't push the image a little bit to the right of the element's bounding box -- how would I tweak my css so it is displayed correctly? I'm trying to do something like this: [{someHeadLineText}{dynamic space }{image}{5px space}] where the [] indicate the total available width of my content. Html: <div class="primaryHeader"> <h2>News</h2> </div> Css: .primaryHeader h2 { background-color: green; /* the header looks like a box */ color: black; background: transparent url(../images/edit.png) no-repeat right center; border: 1px solid red; } I am placing the image to the right of my h2 element and centered vertically -- but how do I adjust the placement of the background image?

    Read the article

  • How do implement a breadth first traversal?

    - by not looking for answer
    //This is what I have. I thought pre-order was the same and mixed it up with depth first! import java.util.LinkedList; import java.util.Queue; public class Exercise25_1 { public static void main(String[] args) { BinaryTree tree = new BinaryTree(new Integer[] {10, 5, 15, 12, 4, 8 }); System.out.print("\nInorder: "); tree.inorder(); System.out.print("\nPreorder: "); tree.preorder(); System.out.print("\nPostorder: "); tree.postorder(); //call the breadth method to test it System.out.print("\nBreadthFirst:"); tree.breadth(); } } class BinaryTree { private TreeNode root; /** Create a default binary tree */ public BinaryTree() { } /** Create a binary tree from an array of objects */ public BinaryTree(Object[] objects) { for (int i = 0; i < objects.length; i++) { insert(objects[i]); } } /** Search element o in this binary tree */ public boolean search(Object o) { return search(o, root); } public boolean search(Object o, TreeNode root) { if (root == null) { return false; } if (root.element.equals(o)) { return true; } else { return search(o, root.left) || search(o, root.right); } } /** Return the number of nodes in this binary tree */ public int size() { return size(root); } public int size(TreeNode root) { if (root == null) { return 0; } else { return 1 + size(root.left) + size(root.right); } } /** Return the depth of this binary tree. Depth is the * number of the nodes in the longest path of the tree */ public int depth() { return depth(root); } public int depth(TreeNode root) { if (root == null) { return 0; } else { return 1 + Math.max(depth(root.left), depth(root.right)); } } /** Insert element o into the binary tree * Return true if the element is inserted successfully */ public boolean insert(Object o) { if (root == null) { root = new TreeNode(o); // Create a new root } else { // Locate the parent node TreeNode parent = null; TreeNode current = root; while (current != null) { if (((Comparable)o).compareTo(current.element) < 0) { parent = current; current = current.left; } else if (((Comparable)o).compareTo(current.element) > 0) { parent = current; current = current.right; } else { return false; // Duplicate node not inserted } } // Create the new node and attach it to the parent node if (((Comparable)o).compareTo(parent.element) < 0) { parent.left = new TreeNode(o); } else { parent.right = new TreeNode(o); } } return true; // Element inserted } public void breadth() { breadth(root); } // Implement this method to produce a breadth first // search traversal public void breadth(TreeNode root){ if (root == null) return; System.out.print(root.element + " "); breadth(root.left); breadth(root.right); } /** Inorder traversal */ public void inorder() { inorder(root); } /** Inorder traversal from a subtree */ private void inorder(TreeNode root) { if (root == null) { return; } inorder(root.left); System.out.print(root.element + " "); inorder(root.right); } /** Postorder traversal */ public void postorder() { postorder(root); } /** Postorder traversal from a subtree */ private void postorder(TreeNode root) { if (root == null) { return; } postorder(root.left); postorder(root.right); System.out.print(root.element + " "); } /** Preorder traversal */ public void preorder() { preorder(root); } /** Preorder traversal from a subtree */ private void preorder(TreeNode root) { if (root == null) { return; } System.out.print(root.element + " "); preorder(root.left); preorder(root.right); } /** Inner class tree node */ private class TreeNode { Object element; TreeNode left; TreeNode right; public TreeNode(Object o) { element = o; } } }

    Read the article

  • Java: design problem with private-final-int-value and empty constructor

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; //DUE to new Klowledge: Design Problem //I think having an empty constructor like this // is an design problem, shall I remove it? What do you think? // When to use an empty constructor? InitInt(){} public static void main(String[] args) { InitInt test = new InitInt(); System.out.println(test.getRight()); } public int getRight(){return right;} } Initialization problem with Constructor InitInt{ // Still the error, "may not be initialized" // How to initialise it? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } }

    Read the article

  • make tree in scheme

    - by ???
    (define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (define (mktree order items_list) (cond ((= (length items_list) 1) (make-tree (car items_list) '() '())) (else (insert2 order (car items_list) (mktree order (cdr items_list)))))) (define (insert2 order x t) (cond ((null? t) (make-tree x '() '())) ((order x (entry t)) (make-tree (entry t) (insert2 order x (left-branch t)) (right-branch t))) ((order (entry t) x ) (make-tree (entry t) (left-branch t) (insert2 order x (right-branch t)))) (else t))) The result is: (mktree (lambda (x y) (< x y)) (list 7 3 5 1 9 11)) (11 (9 (1 () (5 (3 () ()) (7 () ()))) ()) ()) But I'm trying to get: (7 (3 (1 () ()) (5 () ())) (9 () (11 () ()))) Where is the problem?

    Read the article

  • How do I make the footer stretch vertically downward align to footer.

    - by text
    I am new to web design using tableless and I'm having problem positioning some elements on my page.. Here's the sample html: http://christianruado.comuf.com/sample.html As you can see from the screen shots I want my right div to be vertically stretched down to the same level of my footer and position my bottom element to the lowest part of the right container. #right { float:right; width: 19%; background:#FF3300; margin-left:2px; height: 100%; } #right .bottom { bottom:0; display:block; background-color:#FFCCFF; height:30px; }

    Read the article

  • In 3D camera math, calculate what Z depth is pixel unity for a given FOV

    - by badweasel
    I am working in iOS and OpenGL ES 2.0. Through trial and error I've figured out a frustum to where at a specific z depth pixels drawn are 1 to 1 with my source textures. So 1 pixel in my texture is 1 pixel on the screen. For 2d games this is good. Of course it means that I also factor in things like the size of the quad and the size of the texture. For example if my sprite is a quad 32x32 pixels. The quad size is 3.2 units wide and tall. And the texcoords are 32 / the size of the texture wide and tall. Then the frustum is: matrixFrustum(-(float)backingWidth/frustumScale,(float)backingWidth/frustumScale, -(float)backingHeight/frustumScale, (float)backingHeight/frustumScale, 40, 1000, mProjection); Where frustumScale is 800 for a retina screen. Then at a distance of 800 from camera the sprite is pixel for pixel the same as photoshop. For 3d games sometimes I still want to be able to do this. But depending on the scene I sometimes need the FOV to be different things. I'm looking for a way to figure out what Z depth will achieve this same pixel unity for a given FOV. For this my mProjection is set using: matrixPerspective(cameraFOV, near, far, (float)backingWidth / (float)backingHeight, mProjection); With testing I found that at an FOV of 45.0 a Z of 38.5 is very close to pixel unity. And at an FOV of 30.0 a Z of 59.5 is about right. But how can I calculate a value that is spot on? Here's my matrixPerspecitve code: void matrixPerspective(float angle, float near, float far, float aspect, mat4 m) { //float size = near * tanf(angle / 360.0 * M_PI); float size = near * tanf(degreesToRadians(angle) / 2.0); float left = -size, right = size, bottom = -size / aspect, top = size / aspect; // Unused values in perspective formula. m[1] = m[2] = m[3] = m[4] = 0; m[6] = m[7] = m[12] = m[13] = m[15] = 0; // Perspective formula. m[0] = 2 * near / (right - left); m[5] = 2 * near / (top - bottom); m[8] = (right + left) / (right - left); m[9] = (top + bottom) / (top - bottom); m[10] = -(far + near) / (far - near); m[11] = -1; m[14] = -(2 * far * near) / (far - near); } And my mView is set using: lookAtMatrix(cameraPos, camLookAt, camUpVector, mView); * UPDATE * I'm going to leave this here in case anyone has a different solution, can explain how they do it, or why this works. This is what I figured out. In my system I use a 10th scale unit to pixels on non-retina displays and a 20th scale on retina displays. The iPhone is 640 pixels wide on retina and 320 pixels wide on non-retina (obsolete). So if I want something to be the full screen width I divide by 20 to get the OpenGL unit width. Then divide that by 2 to get the left and right unit position. Something 32 units wide centered on the screen goes from -16 to +16. Believe it or not I have an excel spreadsheet do all this math for me and output all the vertex data for my sprite sheet. It's an arbitrary thing I made up to do .1 units = 1 non-retina pixel or 2 retina pixels. I could have made it .01 units = 2 pixels and someday I might switch to that. But for now it's the other. So the width of the screen in units is 32.0, and that means the left most pixel is at -16.0 and the right most is at 16.0. After messing a bit I figured out that if I take the [0] value of an identity modelViewProjection matrix and multiply it by 16 I get the depth required to get 1:1 pixels. I don't know why. I don't know if the 16 is related to the screen size or just a lucky guess. But I did a test where I placed a sprite at that calculated depth and varied the FOV through all the valid values and the object stays steady on screen with 1:1 pixels. So now I'm just calculating the unityDepth that way. If someone gives me a better answer I'll checkmark it.

    Read the article

< Previous Page | 59 60 61 62 63 64 65 66 67 68 69 70  | Next Page >