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  • Return parent of node in Binary Tree

    - by user188995
    I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs. //Assume that nodes are represented by numbers from 1...n where 1=root and even //nos.=left child and odd nos=right child. public int parent(Node node){ if (node % 2 == 0){ if (root.left==node) return root; else return parent(root.left); } //same case for right } But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right. If it's the child or if the node that was queried else, recurses it with the child.

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  • .NET binary serialization conditionally without ISerializable

    - by SillyWhy
    I got 2 classes, for example: public class A { private B b; ... } public class B { ... } I need to serialize an object A using BinaryFormatter. When remoting it shall include the field b, but not when serialize to file. Here is what I added: [Serializable] public class A : MarshalByRefObject { private B b; [OnSerializing] private void OnSerializing(StreamingContext context) { if (context.State == StreamingContextStates.File) { this.b = null; } } ... } [Serializable] public class B : MarshalByRefObject { ... } I think this is a bad design because if another class C also contains B, in class C we must add the duplicate OnSerializing() logic as in A. Class B should decide what to do, not class A or C. I don't want to use ISerializable interface because there are too many variables in class B have to be added to SerializationInfo. I can create a SerializationSurrogate for class B, which perform nothing in GetObjectData() & SetObjectData(), then use it when serializing to file. However the same maintenance issue because whoever modify class B can't notice what going to happen during serialization & the existence of SerializationSurrogate. Is there a better alternative?

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  • findNode in binary search tree

    - by Weadadada Awda
    Does this look right? I mean I am trying to implement the delete function. Node* BST::findNode(int tofind) { Node* node = new Node; node = root; while (node != NULL) { if (node->val == tofind) { return node; } else if (tofind < node->val) { node = node->left; } else { node = node->right; } } } Here is the delete, it's not even close to done but, void BST::Delete(int todelete) { // bool found = false; Node* toDelete = new Node(); toDelete=findNode(todelete); if(toDelete->val!=NULL) { cout << toDelete->val << endl; } } This causes a segmentation fault just running that, any ideas?

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  • Write to a binary file?

    - by rick irby
    Here is data structure w/ variables: struct Part_record { char id_no[3]; int qoh; string desc; double price: }; --- (Using "cin" to input data) --- Part_record null_part = {" ", 0," ",0.0}; --- --- file.seekg( -(long)sizeof(Part_record), ios::cur); file.write( ( char *)&part, sizeof(Part_record) ); The three variables, qoh, Id_no & price, write out correctly, but the "desc" variable is not right. Do I need to initialize Part_record some other way? It should be 20 characters in length. If you have enough info here, pls share your advice,thanks.

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  • Error inserting data in binary tree

    - by chepe263
    I copied this code (in spanish) http://www.elrincondelc.com/nuevorincon/index.php?pag=codigos&id=4 and wrote a new one. This is my code: #include <cstdlib> #include <conio.h> #include <iostream> using namespace std; struct nodoarbol { int dato; struct nodoarbol *izq; struct nodoarbol *der; }; typedef nodoarbol Nodo; typedef Nodo *Arbol; void insertar(Arbol *, int); void inorden(Arbol); void postorden(Arbol); void preorden(Arbol); void insertar(Arbol *raiz, int nuevo){ if (*raiz==NULL){ *raiz = (Nodo *)malloc(sizeof(Nodo)); if (*raiz != NULL){ (*raiz)->dato=nuevo; (*raiz)->der=NULL; (*raiz)->izq=NULL; } else{ cout<<"No hay memoria suficiente u ocurrio un error"; } } else{ if (nuevo < (*raiz)->dato) insertar( &((*raiz)->izq), nuevo ); else if (nuevo > (*raiz)->dato) insertar(&((*raiz)->der), nuevo); } }//inseertar void inorden(Arbol raiz){ if (raiz != NULL){ inorden(raiz->izq); cout << raiz->dato << " "; inorden(raiz->der); } } void preorden(Arbol raiz){ if (raiz != NULL){ cout<< raiz->dato << " "; preorden(raiz->izq); preorden(raiz->der); } } void postorden(Arbol raiz){ if (raiz!=NULL){ postorden(raiz->izq); postorden(raiz->der); cout<<raiz->dato<<" "; } } int main() { int i; i=0; int val; Arbol raiz = NULL; for (i=0; i<10; i++){ cout<<"Inserte un numero"; cin>>val; insertar( (raiz), val); } cout<<"\nPreorden\n"; preorden(raiz); cout<<"\nIneorden\n"; inorden(raiz); cout<<"\nPostorden\n"; postorden(raiz); return 0; } I'm using netbeans 7.1.1, mingw32 compiler This is the output: make[2]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' make[1]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' main.cpp: In function 'int main()': main.cpp:110:30: error: cannot convert 'Arbol {aka nodoarbol*}' to 'Nodo** {aka nodoarbol**}' for argument '1' to 'void insertar(Nodo**, int)' make[2]: *** [build/Release/MinGW-Windows/main.o] Error 1 make[1]: *** [.build-conf] Error 2 make: *** [.build-impl] Error 2 BUILD FAILED (exit value 2, total time: 11s) I don't understand what's wrong since i just copied the code (and rewrite it to my own code). I'm really good in php, asp.net (vb) and other languages but c is a headche for me. I've been struggling with this problem for about an hour. Could somebody tell me what could it be?

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  • efficient video format/codec for sparse & binary blob tracking

    - by user391339
    I am working on a blob tracking project and have many high-definition videos that I would like to reduce in size for storage and downstream tracking/shape-analysis. I want to use a lossless method that takes advantage of the black and white nature of the video as well as the fact that not much is moving between individual frames. The videos are quite sparse, with 5 to 10 b&w blobs per frame occupying <30% of the space in total, with each blob moving <5-10% of the field of view between frames and not changing shape too much between 2-3 frames. I will work in Python, Matlab, or LabView for this project, and could use a batch utility if available. It may be worthwhile to export the files as compressed image stacks if a proper video format can't be found. What are the pros and cons of this? A video codec uses correlations between neighboring frames, so it should be more efficient, but not if the wrong one is chosen or if it is improperly configured.

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  • Binary Search Tree can't delete the root

    - by Ali Zahr
    Everything is working fine in this function, but the problem is that I can't delete the root, I couldn't figure out what's the bug here.I've traced the "else part" it works fine until the return, it returns the old value I don't know why. Plz Help! node *removeNode(node *Root, int key) { node *tmp = new node; if(key > Root->value) Root->right = removeNode(Root->right,key); else if(key < Root->value) Root->left = removeNode(Root->left, key); else if(Root->left != NULL && Root->right != NULL) { node *minNode = findNode(Root->right); Root->value = minNode->value; Root->right = removeNode(Root->right,Root->value); } else { tmp = Root; if(Root->left == NULL) Root = Root->right; else if(Root->right == NULL) Root = Root->left; delete tmp; } return Root; }

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  • Better way to download a binary file?

    - by geoff
    I have a site where a user can download a file. Some files are extremely large (the largest being 323 MB). When I test it to try and download this file I get an out of memory exception. The only way I know to download the file is below. The reason I'm using the code below is because the URL is encoded and I can't let the user link directly to the file. Is there another way to download this file without having to read the whole thing into a byte array? FileStream fs = new FileStream(context.Server.MapPath(url), FileMode.Open, FileAccess.Read); BinaryReader br = new BinaryReader(fs); long numBytes = new FileInfo(context.Server.MapPath(url)).Length; byte[] bytes = br.ReadBytes((int) numBytes); string filename = Path.GetFileName(url); context.Response.Buffer = true; context.Response.Charset = ""; context.Response.Cache.SetCacheability(HttpCacheability.NoCache); context.Response.ContentType = "application/x-rar-compressed"; context.Response.AddHeader("content-disposition", "attachment;filename=" + filename); context.Response.BinaryWrite(bytes); context.Response.Flush(); context.Response.End();

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  • How many posibilities on a binary ?

    - by Val
    in hexadecimal "10 10 10 10" system you have 0-255 posibilities right? in total 256 different posibilities as there are 8 1s and 0s. how many different posibilities would i get? if i had 10 digits. instead of 8? or how would i calculate that in php ?

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  • Problems in Binary Search Tree

    - by user2782324
    This is my first ever trial at implementing the BST, and I am unable to get it done. Please help The problem is that When I delete the node if the node is in the right subtree from the root or if its a right child in the left subtree, then it works fine. But if the node is in the left subtree from root and its any left child, then it does not get deleted. Can someone show me what mistake am I doing?? the markedNode here gets allocated to the parent node of the node to be deleted. the minValueNode here gets allocated to a node whose left value child is the smallest value and it will be used to replace the value to be deleted. package DataStructures; class Node { int value; Node rightNode; Node leftNode; } class BST { Node rootOfTree = null; public void insertintoBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { Node newNode = new Node(); newNode.value = value; rootOfTree = newNode; newNode.rightNode = null; newNode.leftNode = null; } else { while (true) { if (value >= markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.rightNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.leftNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } } } } public void searchBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { System.out.println("Element Not Found"); } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { System.out.println("Element Not Found"); break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { System.out.println("Element Not Found"); break; } } if (value == markedNode.value) { System.out.println("Element Found"); break; } } } } public void deleteFromBST(int value) { Node markedNode = rootOfTree; Node minValueNode = null; if (rootOfTree == null) { System.out.println("Element Not Found"); return; } if (rootOfTree.value == value) { if (rootOfTree.leftNode == null && rootOfTree.rightNode == null) { rootOfTree = null; return; } else if (rootOfTree.leftNode == null ^ rootOfTree.rightNode == null) { if (rootOfTree.rightNode != null) { rootOfTree = rootOfTree.rightNode; return; } else { rootOfTree = rootOfTree.leftNode; return; } } else { minValueNode = rootOfTree.rightNode; if (minValueNode.leftNode == null) { rootOfTree.rightNode.leftNode = rootOfTree.leftNode; rootOfTree = rootOfTree.rightNode; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node rootOfTree.value = minValueNode.leftNode.value; // The value has been swapped if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { break; } else { markedNode = markedNode.rightNode; } } else { System.out.println("Element Not Found"); return; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { break; } else { markedNode = markedNode.leftNode; } } else { System.out.println("Element Not Found"); return; } } } // Parent of the required element found // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { if (markedNode.rightNode.rightNode == null && markedNode.rightNode.leftNode == null) { markedNode.rightNode = null; return; } else if (markedNode.rightNode.rightNode == null ^ markedNode.rightNode.leftNode == null) { if (markedNode.rightNode.rightNode != null) { markedNode.rightNode = markedNode.rightNode.rightNode; return; } else { markedNode.rightNode = markedNode.rightNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.rightNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { if (markedNode.leftNode.rightNode == null && markedNode.leftNode.leftNode == null) { markedNode.leftNode = null; return; } else if (markedNode.leftNode.rightNode == null ^ markedNode.leftNode.leftNode == null) { if (markedNode.leftNode.rightNode != null) { markedNode.leftNode = markedNode.leftNode.rightNode; return; } else { markedNode.leftNode = markedNode.leftNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.leftNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// } } } } } } public class BSTImplementation { public static void main(String[] args) { BST newBst = new BST(); newBst.insertintoBST(19); newBst.insertintoBST(13); newBst.insertintoBST(10); newBst.insertintoBST(20); newBst.insertintoBST(5); newBst.insertintoBST(23); newBst.insertintoBST(28); newBst.insertintoBST(16); newBst.insertintoBST(27); newBst.insertintoBST(9); newBst.insertintoBST(4); newBst.insertintoBST(22); newBst.insertintoBST(17); newBst.insertintoBST(30); newBst.insertintoBST(40); newBst.deleteFromBST(5); newBst.deleteFromBST(4); newBst.deleteFromBST(9); newBst.deleteFromBST(10); newBst.deleteFromBST(13); newBst.deleteFromBST(16); newBst.deleteFromBST(17); newBst.searchBST(5); newBst.searchBST(4); newBst.searchBST(9); newBst.searchBST(10); newBst.searchBST(13); newBst.searchBST(16); newBst.searchBST(17); System.out.println(); newBst.deleteFromBST(20); newBst.deleteFromBST(23); newBst.deleteFromBST(27); newBst.deleteFromBST(28); newBst.deleteFromBST(30); newBst.deleteFromBST(40); newBst.searchBST(20); newBst.searchBST(23); newBst.searchBST(27); newBst.searchBST(28); newBst.searchBST(30); newBst.searchBST(40); } }

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  • Command to find the source package of a binary?

    - by Delan Azabani
    I know there's a which command, that echoes the full name of a binary (e.g. which sh). However, I'm fairly sure there's a command that echoes the package that provides a particular binary. Is there such a command? If so, what is it? I'd like to be able to run this: commandName ls and get coreutils for example.

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  • Ensure Future Browser Compatibility - SharePoint Branding

    - by KunaalKapoor
    HTML and Future Internet Explorer Compatibility with SharePointAs new versions of Internet Explorer are released, the way HTML is rendered by the browser could change over time. To address the possibility of changes, Microsoft uses the X-UA-Compatible META tag that targets HTML markup to a specific version of Internet Explorer. The default SharePoint 2010 master pages are set to force current and future versions of Internet Explorer to render HTML in Internet Explorer 8 mode like the following markup:<meta http-equiv="X-UA-Compatibile" content="IE=IE8" />The Adventure Works Travel HTML includes the META tag to help ensure future Internet Explorer versions will display the SharePoint HTML properly.For more information about the Internet Explorer Standards Mode, see META Tags and Locking in Future Compatibility.

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  • --log-slave-updates is OFF but updates received from master are still logged to slave binary log?

    - by quanta
    MySQL version 5.5.14 According to the document, by the default, slave does not log to its binary log any updates that are received from a master server. Here are my config. on the slave: # egrep 'bin|slave' /etc/my.cnf relay-log=mysqld-relay-bin log-bin = /var/log/mysql/mysql-bin binlog-format=MIXED sync_binlog = 1 log-bin-trust-function-creators = 1 mysql> show global variables like 'log_slave%'; +-------------------+-------+ | Variable_name | Value | +-------------------+-------+ | log_slave_updates | OFF | +-------------------+-------+ 1 row in set (0.01 sec) mysql> select @@log_slave_updates; +---------------------+ | @@log_slave_updates | +---------------------+ | 0 | +---------------------+ 1 row in set (0.00 sec) but slave still logs the updates that are received from a master to its binary logs, let's see the file size: -rw-rw---- 1 mysql mysql 37M Apr 1 01:00 /var/log/mysql/mysql-bin.001256 -rw-rw---- 1 mysql mysql 25M Apr 2 01:00 /var/log/mysql/mysql-bin.001257 -rw-rw---- 1 mysql mysql 46M Apr 3 01:00 /var/log/mysql/mysql-bin.001258 -rw-rw---- 1 mysql mysql 115M Apr 4 01:00 /var/log/mysql/mysql-bin.001259 -rw-rw---- 1 mysql mysql 105M Apr 4 18:54 /var/log/mysql/mysql-bin.001260 and the sample query when reading these binary files with mysqlbinlog utility: #120404 19:08:57 server id 3 end_log_pos 110324763 Query thread_id=382435 exec_time=0 error_code=0 SET TIMESTAMP=1333541337/*!*/; INSERT INTO norep_SplitValues VALUES ( NAME_CONST('cur_string',_utf8'118212' COLLATE 'utf8_general_ci')) /*!*/; # at 110324763 Did I miss something?

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  • Understanding Application binary interface (ABI)

    - by Tim
    I am trying to understand the concept of Application binary interface (ABI). From The Linux Kernel Primer: An ABI is a set of conventions that allows a linker to combine separately compiled modules into one unit without recompilation, such as calling conventions, machine interface, and operating-system interface. Among other things, an ABI defines the binary interface between these units. ... The benefits of conforming to an ABI are that it allows linking object files compiled by different compilers. From Wikipedia: an application binary interface (ABI) describes the low-level interface between an application (or any type of) program and the operating system or another application. ABIs cover details such as data type, size, and alignment; the calling convention, which controls how functions' arguments are passed and return values retrieved; the system call numbers and how an application should make system calls to the operating system; and in the case of a complete operating system ABI, the binary format of object files, program libraries and so on. I was wondering whether ABI depends on both the instruction set and the OS. Are the two all that ABI depends on? What kinds of role does ABI play in different stages of compilation: preprocessing, conversion of code from C to Assembly, conversion of code from Assembly to Machine code, and linking? From the first quote above, it seems to me that ABI is needed for only linking stage, not the other stages. Is it correct? When is ABI needed to be considered? Is ABI needed to be considered during programming in C, Assembly or other languages? If yes, how are ABI and API different? Or is it only for linker or compiler? Is ABI specified for/in machine code, Assembly language, and/or of C?

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  • Collaborative Whiteboard using WebSocket in GlassFish 4 - Text/JSON and Binary/ArrayBuffer Data Transfer (TOTD #189)

    - by arungupta
    This blog has published a few blogs on using JSR 356 Reference Implementation (Tyrus) as its integrated in GlassFish 4 promoted builds. TOTD #183: Getting Started with WebSocket in GlassFish TOTD #184: Logging WebSocket Frames using Chrome Developer Tools, Net-internals and Wireshark TOTD #185: Processing Text and Binary (Blob, ArrayBuffer, ArrayBufferView) Payload in WebSocket TOTD #186: Custom Text and Binary Payloads using WebSocket One of the typical usecase for WebSocket is online collaborative games. This Tip Of The Day (TOTD) explains a sample that can be used to build such games easily. The application is a collaborative whiteboard where different shapes can be drawn in multiple colors. The shapes drawn on one browser are automatically drawn on all other peer browsers that are connected to the same endpoint. The shape, color, and coordinates of the image are transfered using a JSON structure. A browser may opt-out of sharing the figures. Alternatively any browser can send a snapshot of their existing whiteboard to all other browsers. Take a look at this video to understand how the application work and the underlying code. The complete sample code can be downloaded here. The code behind the application is also explained below. The web page (index.jsp) has a HTML5 Canvas as shown: <canvas id="myCanvas" width="150" height="150" style="border:1px solid #000000;"></canvas> And some radio buttons to choose the color and shape. By default, the shape, color, and coordinates of any figure drawn on the canvas are put in a JSON structure and sent as a message to the WebSocket endpoint. The JSON structure looks like: { "shape": "square", "color": "#FF0000", "coords": { "x": 31.59999942779541, "y": 49.91999053955078 }} The endpoint definition looks like: @WebSocketEndpoint(value = "websocket",encoders = {FigureDecoderEncoder.class},decoders = {FigureDecoderEncoder.class})public class Whiteboard { As you can see, the endpoint has decoder and encoder registered that decodes JSON to a Figure (a POJO class) and vice versa respectively. The decode method looks like: public Figure decode(String string) throws DecodeException { try { JSONObject jsonObject = new JSONObject(string); return new Figure(jsonObject); } catch (JSONException ex) { throw new DecodeException("Error parsing JSON", ex.getMessage(), ex.fillInStackTrace()); }} And the encode method looks like: public String encode(Figure figure) throws EncodeException { return figure.getJson().toString();} FigureDecoderEncoder implements both decoder and encoder functionality but thats purely for convenience. But the recommended design pattern is to keep them in separate classes. In certain cases, you may even need only one of them. On the client-side, the Canvas is initialized as: var canvas = document.getElementById("myCanvas");var context = canvas.getContext("2d");canvas.addEventListener("click", defineImage, false); The defineImage method constructs the JSON structure as shown above and sends it to the endpoint using websocket.send(). An instant snapshot of the canvas is sent using binary transfer with WebSocket. The WebSocket is initialized as: var wsUri = "ws://localhost:8080/whiteboard/websocket";var websocket = new WebSocket(wsUri);websocket.binaryType = "arraybuffer"; The important part is to set the binaryType property of WebSocket to arraybuffer. This ensures that any binary transfers using WebSocket are done using ArrayBuffer as the default type seem to be blob. The actual binary data transfer is done using the following: var image = context.getImageData(0, 0, canvas.width, canvas.height);var buffer = new ArrayBuffer(image.data.length);var bytes = new Uint8Array(buffer);for (var i=0; i<bytes.length; i++) { bytes[i] = image.data[i];}websocket.send(bytes); This comprehensive sample shows the following features of JSR 356 API: Annotation-driven endpoints Send/receive text and binary payload in WebSocket Encoders/decoders for custom text payload In addition, it also shows how images can be captured and drawn using HTML5 Canvas in a JSP. How could this be turned in to an online game ? Imagine drawing a Tic-tac-toe board on the canvas with two players playing and others watching. Then you can build access rights and controls within the application itself. Instead of sending a snapshot of the canvas on demand, a new peer joining the game could be automatically transferred the current state as well. Do you want to build this game ? I built a similar game a few years ago. Do somebody want to rewrite the game using WebSocket APIs ? :-) Many thanks to Jitu and Akshay for helping through the WebSocket internals! Here are some references for you: JSR 356: Java API for WebSocket - Specification (Early Draft) and Implementation (already integrated in GlassFish 4 promoted builds) Subsequent blogs will discuss the following topics (not necessary in that order) ... Error handling Interface-driven WebSocket endpoint Java client API Client and Server configuration Security Subprotocols Extensions Other topics from the API

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  • [BST] Deletion procedure

    - by Metz
    Hi all. Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. Thank you. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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  • Browser Compatablitiy, Support, Modern Browsers and older Versions.

    - by smoop
    This is a question to all web-developers working in the industry more than a few years. Today the server guy at my work told me that his browser didnt render a google font I used on a site properly, so I checked his browser - it was firefox 3.5. (google font Raleway).. This made me think.. I know for some of our sites its a requirement to support IE6 (larger corperate sites) but for the rest (personal sites) should I be looking at developing for all incrments of firefox, chrome, IE and safari???? Does your company still support IE6? Does your company support previous versions of modern browsers or just the latest ones? and finally, has anyone else had this problem with FF 3.5 and Google Hosted font Raleway..(the font displays extra extra thing so the font is almost unreadable) If anyones interested I found a working solution to my font problem here: http://www.jshsolutions.net/google-webfonts-cross-browser-fix-howto/ Sam

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  • Launch file:// from Firefox/Chrome

    - by JuniorFlip
    I am looking to be able to launch a network file on our local intranet using FF or Chrome Currently the link work good in IE <a href="\\Start\Of\My\Network\file.xlsx">View Report</a>&nbsp; but in FF it show <a href="http://mydomain.com/\\Start\Of\My\Network\file.xlsx">View Report</a>&nbsp; is there a way to get the link to render properly...Just a simple click from a href tag.

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  • Why is C++ backward compatibility important / necessary?

    - by Giorgio
    As far as understand it is a well-established opinion within the C++ community that C is an obsolete language that was useful 20 years ago but cannot support many modern good programming practices, or even encourages bad practices; certain features that were typical of C++ (C with classes) during the nineties are also obsolete and considered bad practice in modern C++ (e.g., new and delete should be replaced by smart pointer primitives). In view of this, I often wonder why backward compatibility with C and obsolete C++ features is still considered important: to my knowledge there is no 100% compatibility, but most of C and C++ are contained in C++11 as a subset. Of course, there is a lot of legacy code and libraries (possibly containing templates) that are written using a previous standard of the language and which still need to be maintained or used in connection with new code. Nevertheless, maybe it would still be possible to drop obsolete C and C++ features (e.g. the mentioned new / delete) from a future C++ standard so that it is impossible to use them in new code. In this way, old and dangerous programming practices would be quickly banned from new code, and modern, better programming practices would be enforced by the compiler. Legacy code could still be maintained using separate compilation (having C alongside C++ source files is already a common practice). Developers would have to choose between one compiler supporting the old-style C++ that was common during the nineties and a compiler supporting the modern C++? style (the question mark indicates a future, hypothetical revision). Only mixing the two styles would be forbidden. Would this be a viable strategy for encouraging the adoption of modern C++ practices? Are there conceptual reasons or technical problems (e.g. compiling existing templates) that make such a change undesirable or even impossible? Has such a development been proposed in the C++ community. If there has been some extended discussion on the topic, is there any material on-line?

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  • Browser Compatibility

    There are five major browsers available today across the many operating system platforms, with over twenty minor browsers tagging along. With such diversity, the biggest problems a web programmer faces is browser compatibility.

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