Search Results

Search found 4084 results on 164 pages for 'r tree'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 8/164 | < Previous Page | 4 5 6 7 8 9 10 11 12 13 14 15  | Next Page >

  • Friday Fun: Christmas Tree Light Up

    - by Asian Angel
    Another week has thankfully passed by, so it is time to take a break and have some fun. This week’s game tests your ability to light up the whole Christmas tree…can you figure out the correct wiring configuration? Christmas Tree Light Up The object of the game is simple…light up all of the bulbs on the Christmas tree. While the game may look quick and easy at first you will need to do some thinking and experimenting to come up with the correct wiring configuration. The instructions are very simple…just click on any of the wiring sections or bulbs to rotate them. Keep in mind that you may have to click a few times to line the wiring sections or bulbs up as desired since the rotation is always clockwise. Note: You will need use all of the wiring sections available to completely light the tree up. Each time you will be presented with a different starting setup coming from your power source. Time to hook up the lights! Note: It is recommended that you disable the sound for the game since the “rotation” sounds can be slightly irritating. A nice start but there are still a lot of bulbs to light up. Getting closer… Almost there…only two more bulbs to light up. Success! Have fun playing! Play Christmas Tree Light Up Latest Features How-To Geek ETC The 50 Best Registry Hacks that Make Windows Better The How-To Geek Holiday Gift Guide (Geeky Stuff We Like) LCD? LED? Plasma? The How-To Geek Guide to HDTV Technology The How-To Geek Guide to Learning Photoshop, Part 8: Filters Improve Digital Photography by Calibrating Your Monitor Our Favorite Tech: What We’re Thankful For at How-To Geek Settle into Orbit with the Voyage Theme for Chrome and Iron Awesome Safari Compass Icons Set Escape from the Exploding Planet Wallpaper Move Your Tumblr Blog to WordPress Pytask is an Easy to Use To-Do List Manager for Your Ubuntu System Snowy Christmas House Personas Theme for Firefox

    Read the article

  • Display root node of Hierarchical Tree using ADF - EJB DC

    - by arul.wilson(at)oracle.com
    Displaying Employee (HR schema) records in Hierarchical Tree can be achieved in ADF-BC by creating custom VO and a Viewlink for displaying root node. This can be more easily done using  EJB-DC by just introducing a NamedQuery to get the root node.Here you go to get this scenario working.Create DB connection based on HR schema.Create Entity Bean from Employees Table.Add custom NamedQuery to Employees.java bean, this named query is responsible for fetching the root node (King in this example). @NamedQueries({  @NamedQuery(name = "Employees.findAll", query = "select o from Employees o"),  @NamedQuery(name = "Employees.findRootEmp", query = "select p from Employees p where p.employees is null")}) Create Stateless Session Bean and expose the Named Queries through the Session Facade.Create Datacontrol from SessionBean local interface.Create jspx page in ViewController project.Drop employeesFindRootEmp from Data Controls Palette as ADF Tree.Add employeesList as Tree level rule.Run page to see the hierarchical tree with root node as 'King'

    Read the article

  • Binary Search Tree Implementation

    - by Gabe
    I've searched the forum, and tried to implement the code in the threads I found. But I've been working on this real simple program since about 10am, and can't solve the seg. faults for the life of me. Any ideas on what I'm doing wrong would be greatly appreciated. BST.h (All the implementation problems should be in here.) #ifndef BST_H_ #define BST_H_ #include <stdexcept> #include <iostream> #include "btnode.h" using namespace std; /* A class to represent a templated binary search tree. */ template <typename T> class BST { private: //pointer to the root node in the tree BTNode<T>* root; public: //default constructor to make an empty tree BST(); /* You have to document these 4 functions */ void insert(T value); bool search(const T& value) const; bool search(BTNode<T>* node, const T& value) const; void printInOrder() const; void remove(const T& value); //function to print out a visual representation //of the tree (not just print the tree's values //on a single line) void print() const; private: //recursive helper function for "print()" void print(BTNode<T>* node,int depth) const; }; /* Default constructor to make an empty tree */ template <typename T> BST<T>::BST() { root = NULL; } template <typename T> void BST<T>::insert(T value) { BTNode<T>* newNode = new BTNode<T>(value); cout << newNode->data; if(root == NULL) { root = newNode; return; } BTNode<T>* current = new BTNode<T>(NULL); current = root; current->data = root->data; while(true) { if(current->left == NULL && current->right == NULL) break; if(current->right != NULL && current->left != NULL) { if(newNode->data > current->data) current = current->right; else if(newNode->data < current->data) current = current->left; } else if(current->right != NULL && current->left == NULL) { if(newNode->data < current->data) break; else if(newNode->data > current->data) current = current->right; } else if(current->right == NULL && current->left != NULL) { if(newNode->data > current->data) break; else if(newNode->data < current->data) current = current->left; } } if(current->data > newNode->data) current->left = newNode; else current->right = newNode; return; } //public helper function template <typename T> bool BST<T>::search(const T& value) const { return(search(root,value)); //start at the root } //recursive function template <typename T> bool BST<T>::search(BTNode<T>* node, const T& value) const { if(node == NULL || node->data == value) return(node != NULL); //found or couldn't find value else if(value < node->data) return search(node->left,value); //search left subtree else return search(node->right,value); //search right subtree } template <typename T> void BST<T>::printInOrder() const { //print out the value's in the tree in order // //You may need to use this function as a helper //and create a second recursive function //(see "print()" for an example) } template <typename T> void BST<T>::remove(const T& value) { if(root == NULL) { cout << "Tree is empty. No removal. "<<endl; return; } if(!search(value)) { cout << "Value is not in the tree. No removal." << endl; return; } BTNode<T>* current; BTNode<T>* parent; current = root; parent->left = NULL; parent->right = NULL; cout << root->left << "LEFT " << root->right << "RIGHT " << endl; cout << root->data << " ROOT" << endl; cout << current->data << "CURRENT BEFORE" << endl; while(current != NULL) { cout << "INTkhkjhbljkhblkjhlk " << endl; if(current->data == value) break; else if(value > current->data) { parent = current; current = current->right; } else { parent = current; current = current->left; } } cout << current->data << "CURRENT AFTER" << endl; // 3 cases : //We're looking at a leaf node if(current->left == NULL && current->right == NULL) // It's a leaf { if(parent->left == current) parent->left = NULL; else parent->right = NULL; delete current; cout << "The value " << value << " was removed." << endl; return; } // Node with single child if((current->left == NULL && current->right != NULL) || (current->left != NULL && current->right == NULL)) { if(current->left == NULL && current->right != NULL) { if(parent->left == current) { parent->left = current->right; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->right; cout << "The value " << value << " was removed." << endl; delete current; } } else // left child present, no right child { if(parent->left == current) { parent->left = current->left; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->left; cout << "The value " << value << " was removed." << endl; delete current; } } return; } //Node with 2 children - Replace node with smallest value in right subtree. if (current->left != NULL && current->right != NULL) { BTNode<T>* check; check = current->right; if((check->left == NULL) && (check->right == NULL)) { current = check; delete check; current->right = NULL; cout << "The value " << value << " was removed." << endl; } else // right child has children { //if the node's right child has a left child; Move all the way down left to locate smallest element if((current->right)->left != NULL) { BTNode<T>* leftCurrent; BTNode<T>* leftParent; leftParent = current->right; leftCurrent = (current->right)->left; while(leftCurrent->left != NULL) { leftParent = leftCurrent; leftCurrent = leftCurrent->left; } current->data = leftCurrent->data; delete leftCurrent; leftParent->left = NULL; cout << "The value " << value << " was removed." << endl; } else { BTNode<T>* temp; temp = current->right; current->data = temp->data; current->right = temp->right; delete temp; cout << "The value " << value << " was removed." << endl; } } return; } } /* Print out the values in the tree and their relationships visually. Sample output: 22 18 15 10 9 5 3 1 */ template <typename T> void BST<T>::print() const { print(root,0); } template <typename T> void BST<T>::print(BTNode<T>* node,int depth) const { if(node == NULL) { std::cout << std::endl; return; } print(node->right,depth+1); for(int i=0; i < depth; i++) { std::cout << "\t"; } std::cout << node->data << std::endl; print(node->left,depth+1); } #endif main.cpp #include "bst.h" #include <iostream> using namespace std; int main() { BST<int> tree; cout << endl << "LAB #13 - BINARY SEARCH TREE PROGRAM" << endl; cout << "----------------------------------------------------------" << endl; // Insert. cout << endl << "INSERT TESTS" << endl; // No duplicates allowed. tree.insert(0); tree.insert(5); tree.insert(15); tree.insert(25); tree.insert(20); // Search. cout << endl << "SEARCH TESTS" << endl; int x = 0; int y = 1; if(tree.search(x)) cout << "The value " << x << " is on the tree." << endl; else cout << "The value " << x << " is NOT on the tree." << endl; if(tree.search(y)) cout << "The value " << y << " is on the tree." << endl; else cout << "The value " << y << " is NOT on the tree." << endl; // Removal. cout << endl << "REMOVAL TESTS" << endl; tree.remove(0); tree.remove(1); tree.remove(20); // Print. cout << endl << "PRINTED DIAGRAM OF BINARY SEARCH TREE" << endl; cout << "----------------------------------------------------------" << endl; tree.print(); cout << endl << "Program terminated. Goodbye." << endl << endl; } BTNode.h #ifndef BTNODE_H_ #define BTNODE_H_ #include <iostream> /* A class to represent a node in a binary search tree. */ template <typename T> class BTNode { public: //constructor BTNode(T d); //the node's data value T data; //pointer to the node's left child BTNode<T>* left; //pointer to the node's right child BTNode<T>* right; }; /* Simple constructor. Sets the data value of the BTNode to "d" and defaults its left and right child pointers to NULL. */ template <typename T> BTNode<T>::BTNode(T d) : left(NULL), right(NULL) { data = d; } #endif Thanks.

    Read the article

  • Tortoise SVN tree conflict with myself

    - by Jesse Pepper
    Has anyone had the experience of moving a file in tortoise and committing successfully, only to later commit a different change and be told of a tree conflict where: the file in its original location has been deleted, but in tortoise is marked as missing the file in its new location is there, but marked as already added. (I use tortoise SVN, and we have client and server 1.60) Nobody else changed either the directory or the file (according to svn log). Why is this happening? Is there a way to avoid it happening? If it does happen, is there a more elegant way of fixing the problem than by deleting the whole folder and updating again?

    Read the article

  • B-Tree Revision

    - by stan
    Hi, If we are looking for line intersections (horizontal and vertical lines only) and we have n lines with half of them vertical and no intersections then Sorting the list of line end points on y value will take N log N using mergesort Each insert delete and search of our data structue (assuming its a b-tree) will be < log n so the total search time will be N log N What am i missing here, if the time to sort using mergesort takes a time of N log N and insert and delete takes a time of < log n are we dropping the constant factor to give an overal time of N log N. If not then how comes < log n goes missing in total ONotation run time? Thanks

    Read the article

  • (External) Java library for creating Tree structure ?

    - by suVasH.....
    I am planning to implement a tree structure where every node has two children and a parent along with various other node properties (and I'd want to do this in Java ) Now, the way to it probably is to create the node such that it links to other nodes ( linked list trick ), but I was wondering if there is any good external library to handle all this low level stuff. ( for eg. the ease of stl::vector vs array in C++ ). I've heard of JDots, but still since i haven't started (and haven't programmed a lot in Java), I'd rather hear out before I begin.

    Read the article

  • Speech.Recognition GrammarBuilder/Choices Tree Structure

    - by user2210179
    In playing around with C#'s Speech Recognition, I've stumbled across a road block in the creation of an effective GrammerBuilder with Choices (more specifically, Choices of Choices). IE considering the following logical commands. One solution would to "hard code" every combination of Speech lines and add them to a GrammarBuilder (ie "SET LEFT COLOR RED" and "SET RIGHT CLEAR", however, this would quickly max out the limit of 1024, especially when dealing with number combinations. Another solution would to Append all 'columns' as "Choices" (and filter out incorrect paths upon 'recognition', however this seems like it's processor heavy and unnecessary. The middle ground, seems like the best path - with Choices of Choices - like a tree structure on a GrammarBuilder - however I'm not sure how to proceed. Any suggestions?

    Read the article

  • Unix tree convert to recursive php array

    - by Fordnox
    I have a response from remote server like this: /home/computer/Downloads |-- /home/computer/Downloads/Apple | `-- /home/computer/Downloads/Apple/Pad |-- /home/computer/Downloads/Empty_Folder `-- /home/computer/Downloads/Subfolder |-- /home/computer/Downloads/Subfolder/Empty `-- /home/computer/Downloads/Subfolder/SubSubFolder `-- /home/computer/Downloads/Subfolder/SubSubFolder/Test this is the output for command computer@athome:$ tree -df --noreport -L 5 /home/computer/Downloads/ I would like to parse this string to recursive php array or object, something like this. I would show only part of result to get the idea. array( 'title' => '/home/computer/Downloads', 'children' => array( 0 => array( 'title' => '/home/computer/Downloads/Apple', 'children' => array( ... ) ) ); Response from server can change according to scanned directory. Can someone help me write this function. Please note that this is response from remote server and php functions can not scan any remote dir.

    Read the article

  • WPF component for 2D tree diagram

    - by pdm2011
    I'm looking for a well-documented, supported WPF component that provides an API for visualisation of 2D tree diagrams. Ideally something easy to use, customisable (i.e. supports various flavours of nodes and splines) and preferably with automated layout control. Tools that look good so far are GoXam (http://www.nwoods.com/components/silverlight-wpf/goxam-overview.htm) and yFiles WPF (http://www.yworks.com/en/products_yfileswpf_about.html). Just wondering if anyone has experience with either of these, or can recommend an alternative? Thanks!

    Read the article

  • Recursive Binary Search Tree Insert

    - by Nick Sinklier
    So this is my first java program, but I've done c++ for a few years. I wrote what I think should work, but in fact it does not. So I had a stipulation of having to write a method for this call: tree.insertNode(value); where value is an int. I wanted to write it recursively, for obvious reasons, so I had to do a work around: public void insertNode(int key) { Node temp = new Node(key); if(root == null) root = temp; else insertNode(temp); } public void insertNode(Node temp) { if(root == null) root = temp; else if(temp.getKey() <= root.getKey()) insertNode(root.getLeft()); else insertNode(root.getRight()); } Thanks for any advice.

    Read the article

  • A B+tree simple implementation in C

    - by initpy
    Hi guys, I'm working on a fun project where I need a simple key/value store that uses B+Trees. I studied them some years ago, and to be honest, I don't want to reinvent the wheel, so I'm looking for a simple implementation in C of b+tree that I can just include in my project. I know of sqlite's, dbm's and tokyocabinet's ones but they're a little too "complicated" for my needs. Is there any (even pedagogical) work on this you can refer me to? Do you have some code to share? Thanks a lot!

    Read the article

  • Flex: change item Style on certain Tree based ItemRenderers

    - by Markus
    Hi Everybody, I have a question concerning Tree items. I want to show where a drop action would be placed... The item will be placed in between two existing elements. So what I want to do is, to take the upper item and draw a line underneath it. But I struggling to address the itemRenderer... I have the index for the itemrenderer, but I dont get a instance of that object. Any help is appreciated! Markus

    Read the article

  • In a binary search Tree

    - by user1044800
    In a binary search tree that takes a simple object.....when creating the getter and setter methods for the left, right, and parent. do I a do a null pointer? as in this=this or do I create the object in each method? Code bellow... This is my code: public void setParent(Person parent) { parent = new Person( parent.getName(), parent.getWeight()); //or is the parent supposed to be a null pointer ???? This is the code it came from: public void setParent(Node parent) { this.parent = parent; } Their code takes a node from the node class...my set parent is taking a person object from my person class.....

    Read the article

  • Stuck on solving the Minimal Spanning Tree problem.

    - by kunjaan
    I have reduced my problem to finding the minimal spanning tree in the graph. But I want to have one more constraint which is that the total degree for each vertex shouldnt exceed a certain constant factor. How do I model my problem? Is MST the wrong path? Do you know any algorithms that will help me? One more problem: My graph has duplicate edge weights so is there a way to count the number of unique MSTs? Are there algorithms that do this? Thank You. Edit: By degree, I mean the total number of edges connecting the vertex. By duplicate edge weight I mean that two edges have the same weight.

    Read the article

  • movedown method not saving new position - cakephp tree

    - by Ryan
    Hi everyone, I am experiencing a problem that has popped up recently and is causing quite a bit of trouble for our system. The app we have relies on using the movedown method to organize content, but as of late it has stopped working and began to generate the following warning: Warning (2): array_values() [<a href='function.array-values'>function.array-values</a>]: The argument should be an array in [/usr/local/home/cake/cake_0_2_9/cake/libs/model/behaviors/tree.php, line 459] The line being referenced: list($node) = array_values($Model->find('first', array( 'conditions' => array($scope, $Model->escapeField() => $id), 'fields' => array($Model->primaryKey, $left, $right, $parent), 'recursive' => $recursive ))); The line calling the method: $this->movedown($id,abs((int)$position)); I have exhausted every idea I could come up with. Has anyone else crossed this issue before? Any help, or pointing in a direction would be much appreciated!

    Read the article

  • Java : Count even values in a Binary Search Tree recursively

    - by user307682
    Hi, I need to find out how many even values are contained in a binary tree. this is my code. private int countEven(BSTNode root){ if ((root == null)|| (root.value%2==1)) return 0; return 1+ countEven(root.left) + countEven(root.right); } this i just coded as i do not have a way to test this out. I'm not able to test it out at the moment but need an answer so badly. any help is deeply appreciated.

    Read the article

  • Generate syntax tree for simple math operations

    - by M28
    I am trying to generate a syntax tree, for a given string with simple math operators (+, -, *, /, and parenthesis). Given the string "1 + 2 * 3": It should return an array like this: ["+", [1, ["*", [2,3] ] ] ] I made a function to transform "1 + 2 * 3" in [1,"+",2,"*",3]. The problem is: I have no idea to give priority to certain operations. My code is: function isNumber(ch){ switch (ch) { case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9': case '.': return true; break; default: return false; break; } } function generateSyntaxTree(text){ if (typeof text != 'string') return []; var code = text.replace(new RegExp("[ \t\r\n\v\f]", "gm"), ""); var codeArray = []; var syntaxTree = []; // Put it in its on scope (function(){ var lastPos = 0; var wasNum = false; for (var i = 0; i < code.length; i++) { var cChar = code[i]; if (isNumber(cChar)) { if (!wasNum) { if (i != 0) { codeArray.push(code.slice(lastPos, i)); } lastPos = i; wasNum = true; } } else { if (wasNum) { var n = Number(code.slice(lastPos, i)); if (isNaN(n)) { throw new Error("Invalid Number"); return []; } else { codeArray.push(n); } wasNum = false; lastPos = i; } } } if (wasNum) { var n = Number(code.slice(lastPos, code.length)); if (isNaN(n)) { throw new Error("Invalid Number"); return []; } else { codeArray.push(n); } } })(); // At this moment, codeArray = [1,"+",2,"*",3] return syntaxTree; } alert('Returned: ' + generateSyntaxTree("1 + 2 * 3"));

    Read the article

  • How can a B-tree node be represented?

    - by chronodekar
    We're learning B-trees in class and have been asked to implement them in code. The teacher has left choice of programming language to us and I want to try and do it in C#. My problem is that the following structure is illegal in C#, unsafe struct BtreeNode { int key_num; // The number of keys in a node int[] key; // Array of keys bool leaf; // Is it a leaf node or not? BtreeNode*[] c; // Pointers to next nodes } Specifically, one is not allowed to create a pointer to point to the structure itself. Is there some work-around or alternate approach I could use? I'm fairly certain that there MUST be a way to do this within the managed code, but I can't figure it out. EDIT: Eric's answer pointed me in the right direction. Here's what I ended up using, class BtreeNode { public List<BtreeNode> children; // The child nodes public static int MinDeg; // The Minimum Degree of the tree public bool IsLeaf { get; set; } // Is the current node a leaf or not? public List<int> key; // The list of keys ... }

    Read the article

  • OutOfMemoryError creating a tree recursively?

    - by Alexander Khaos Greenstein
    root = new TreeNode(N); constructTree(N, root); private void constructTree(int N, TreeNode node) { if (N > 0) { node.setLeft(new TreeNode(N-1)); constructTree(N-1, node.getLeft()); } if (N > 1) { node.setMiddle(new TreeNode(N-2)); constructTree(N-2, node.getMiddle()); } if (N > 2) { node.setRight(new TreeNode(N-3)); constructTree(N-3, node.getRight()); } Assume N is the root number, and the three will create a left middle right node of N-1, N-2, N-3. EX: 5 / | \ 4 3 2 /|\ 3 2 1 etc. My GameNode class has the following variables: private int number; private GameNode left, middle, right; Whenever I construct a tree with an integer greater than 28, I get a OutOfMemoryError. Is my recursive method just incredibly inefficient or is this natural? Thanks!

    Read the article

  • How to reload Ext.tree.TreePanel on demand?

    - by Sergei Stolyarov
    I want to create Ext.tree.TreePanel component and periodically load content from the external URl. So I've written something like new Ext.tree.TreePanel({ root: { nodeType: 'async', text: 'asdasd', draggable: false, id: 'folders-tree-root' }, loader: new Ext.tree.TreeLoader() }); And now I want to reload this tree, so I write: tree.loader.dataUrl = 'folders-sample.json'; tree.root.reload(); And nothing happens. add: The only way I've found is set some invalid value to dataUrl param on TreeLoader creation: new Ext.tree.TreePanel({ root: { nodeType: 'async', text: 'asdasd', draggable: false, id: 'folders-tree-root' }, loader: new Ext.tree.TreeLoader(dataUrl: 'something') });

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 4 5 6 7 8 9 10 11 12 13 14 15  | Next Page >