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  • How to add a new node to a dijit.Tree

    - by Larry Bergman
    I want to add a new node to a dijit.ree as a sibling of the currently selected node. I've found sample code (I'm new to dojo) that adds a new item to the tree using the newItem method of ItemFileWriteStore, but the new item always appears at the bottom of the tree. How would I add to the store at a specified position, in particular the position corresponding to the current selection? Pointers to sample code would be welcome :) Thanks, Larry

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  • Behavior Tree Implementations

    - by Hamza Yerlikaya
    I am looking for behavior tree implementations in any language, I would like to learn more about how they are implemented and used so can roll my own but I could only find one Owyl, unfortunately, it does not contain examples of how it is used. Any one know any other open source ones that I can browse through the code see some examples of how they are used etc? EDIT: Behavior tree is the name of the data structure.

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  • drawing hierarchical tree with orthogonal lines

    - by user267530
    Hi I need to work on drawing a hierarchical tree structure with orthogonal lines(straight rectangular connecting lines) between root and children ( like the following: http://lab.kapit.fr/display/visualizationlayouts/Hierarchical+Tree+layout ). I want to know if there are any open source examples of the algorithm of drawing trees like that so that I can implement the same algorithm in actionscript. Thanks Palash

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  • Setting dijit.Tree cookie for all pages

    - by peirix
    I'm using the same dijit.Tree view over several pages in our application, and I'd like to have the cookie saved for the server name, instead of the folder name. Right now I've got 3 pages and 3 cookies, which each hold their own information on the state of the Tree, which is kinda annoying. Any ways to accomplish this? The only thing I've found on cookies in the API, is that I can set the cookieName and turn cookies on/off.

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  • Flex Tree leaf-element highlighting while drag&drop

    - by sani4xl
    Hi, i have TileList from which i'm dragging some stuff(image) to Tree (something like dragging sounds into playlist in iTunes), but when i can drop this stuff, i see only underline, this mean i can drop it only under or above some leaf-element in that Tree. How can i force it to hide this black underline and highlight leaf-element to which i wanna drop my stuff. Thanks

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  • C++ R - tree implementation wanted

    - by Kotti
    Hi, Does anyone know good and simple to use in production code R-tree (actually, any implementations - R*, R+ or PR-tree would be great)? It doesn't matter if it is a template or library implementation, but some implementations that google found look very disappointing... Thanks in advance.

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  • Tree data structure gems compared?

    - by huug
    I want to you use a tree structure for my navigation. I was thinking about using Ancestry, but then I found this article about 7 plugins for providing a tree structure to your models. What are the pros/cons for each plugin/gem and above all: which one do you recommend? Tnx!

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  • Estimating the size of a tree

    - by Full Decent
    I'd like to estimate the number of leaves in a large tree structure for which I can't visit every node exhaustively. Is this algorithm appropriate? Does it have a name? Also, please pedant if I am using any terms improperly. sum_trials = 0 num_trials = 0 WHILE time_is_not_up bits = 0 ptr = tree.root WHILE count(ptr.children) > 0 bits += log2(count(ptr.children)) ptr = ptr.children[rand()%count(ptr.children)] sum_trials += bits num_trials++ estimated_tree_size = 2^(sum_trials/num_trials)

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  • POV Christmas Tree Is a Holiday-Themed DIY Electronics Project

    - by Jason Fitzpatrick
    If you’re looking for an electronics project with a bit of holiday cheer, this clever POV Christmas tree combines LEDs, motors, and a simple vision hack to create a glowing Christmas tree. POV (or Persistence Of Vision) hacks rely on your visual circuit’s lag time. By taking advantage of that lag POV displays can create the illusion of shapes and words where there are none. In the case of this Christmas tree hack a spinning set of LED lights creates the illusion of a Christmas tree when, in reality, there is just a few LEDs suspended in space by wire. It’s not a beginner level project by any means but it is a great way to practice surface mounting electronics and polish up your PCB making skills. Hit up the link below for the full tutorial. POV Christmas Tree [Instructables] HTG Explains: Do You Really Need to Defrag Your PC? Use Amazon’s Barcode Scanner to Easily Buy Anything from Your Phone How To Migrate Windows 7 to a Solid State Drive

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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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  • dijit tree and focus node

    - by user220836
    Hello, I cannot get focusNode() or expandNode() get working. I also tried switching back to dojo 1.32 and even 1.3, no difference to 1.4. And I debugged with firebug, the node is a valid tree node and no errors occur but the node wont get focused. Help is VERY appreciated! <head> <script type="text/javascript"> dojo.declare("itcTree",[dijit.Tree], { focusNodeX : function(/* string */ id) { var node=this._itemNodesMap[id]; this.focusNode(node); } }); </script> </head> <body class="tundra"> <div dojoType="dojo.data.ItemFileReadStore" jsId="continentStore" url="countries.json"> </div> <div dojoType="dijit.tree.ForestStoreModel" jsId="continentModel" store="continentStore" query="{type:'continent'}" rootId="continentRoot" rootLabel="Continents" childrenAttrs="children"> </div> <div dojoType="itcTree" id="mytree" model="continentModel" openOnClick="true"> <script type="dojo/method" event="onClick" args="item"> dijit.byId('mytree').focusNodeX('AF'); </script> </div> <p> <button onclick="dijit.byId('mytree').focusNode('DE');">klick</button> </p> </body>

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  • Parsing an arithmetic expression and building a tree from it in Java

    - by ChocolateBear
    Hi, I needed some help with creating custom trees given an arithmetic expression. Say, for example, you input this arithmetic expression: (5+2)*7 The result tree should look like: * / \ + 7 / \ 5 2 I have some custom classes to represent the different types of nodes, i.e. PlusOp, LeafInt, etc. I don't need to evaluate the expression, just create the tree, so I can perform other functions on it later. Additionally, the negative operator '-' can only have one child, and to represent '5-2', you must input it as 5 + (-2). Some validation on the expression would be required to ensure each type of operator has the correct the no. of arguments/children, each opening bracket is accompanied by a closing bracket. Also, I should probably mention my friend has already written code which converts the input string into a stack of tokens, if that's going to be helpful for this. I'd appreciate any help at all. Thanks :) (I read that you can write a grammar and use antlr/JavaCC, etc. to create the parse tree, but I'm not familiar with these tools or with writing grammars, so if that's your solution, I'd be grateful if you could provide some helpful tutorials/links for them.)

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  • How to create a Binary Tree from a General Tree?

    - by mno4k
    I have to solve the following constructor for a BinaryTree class in java: BinaryTree(GeneralTree<T> aTree) This method should create a BinaryTree (bt) from a General Tree (gt) as follows: Every Vertex from gt will be represented as a leaf in bt. If gt is a leaf, then bt will be a leaf with the same value as gt If gt is not a leaf, then bt will be constructed as an empty root, a left subTree (lt) and a right subTree (lr). Lt is a stric binary tree created from the oldest subtree of gt (the left-most subtree) and lr is a stric binary tree created from gt without its left-most subtree. The frist part is trivial enough, but the second one is giving me some trouble. I've gotten this far: public BinaryTree(GeneralTree<T> aTree){ if (aTree.isLeaf()){ root= new BinaryNode<T>(aTree.getRootData()); }else{ root= new BinaryNode<T>(null); // empty root LinkedList<GeneralTree<T>> childs = aTree.getChilds(); // Childs of the GT are implemented as a LinkedList of SubTrees child.begin(); //start iteration trough list BinaryTree<T> lt = new BinaryTree<T>(childs.element(0)); // first element = left-most child this.addLeftChild(lt); aTree.DeleteChild(hijos.elemento(0)); BinaryTree<T> lr = new BinaryTree<T>(aTree); this.addRightChild(lr); } } Is this the right way? If not, can you think of a better way to solve this? Thank you!

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  • Saving tree-structures in Databases

    - by Nina Null
    Hello everyone. I use Hibernate/Spring and a MySQL Database for my data management. Currently I display a tree-structure in a JTable. A tree can have several branches, in turn a branch can have several branches (up to nine levels) again, or having leaves. Lately I have performanceproblemes, as soon as I want to create new branches on deeper levels. At this time a branch has a foreign key to its parent. The domainobject has access to its parent by calling getParent(), which returns the parent-branch. The deeper the level, the longer it takes to create a new branch. Microbenchmark results for creating a new branch are like: Level 1: 32 ms. Level 3: 80 ms. Level 9: 232 ms. Obviously the level (which means the number of parents) is responsible for this. So I wanted to ask, if there are any appendages to work around this kind of problem. I don’t understand why Hibernate needs to know about the whole object tree (all parents until the root) while creating a new branch. But as far as I know this can be the only reason for the delay while creating a new branch, because a branch doesn’t have any other relations to any other objects. I would be very thankful for any workarounds or suggestions. greets, jambusa

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  • Permuting a binary tree without the use of lists

    - by Banang
    I need to find an algorithm for generating every possible permutation of a binary tree, and need to do so without using lists (this is because the tree itself carries semantics and restraints that cannot be translated into lists). I've found an algorithm that works for trees with the height of three or less, but whenever I get to greater hights, I loose one set of possible permutations per height added. Each node carries information about its original state, so that one node can determine if all possible permutations have been tried for that node. Also, the node carries information on weather or not it's been 'swapped', i.e. if it has seen all possible permutations of it's subtree. The tree is left-centered, meaning that the right node should always (except in some cases that I don't need to cover for this algorithm) be a leaf node, while the left node is always either a leaf or a branch. The algorithm I'm using at the moment can be described sort of like this: if the left child node has been swapped swap my right node with the left child nodes right node set the left child node as 'unswapped' if the current node is back to its original state swap my right node with the lowest left nodes' right node swap the lowest left nodes two childnodes set my left node as 'unswapped' set my left chilnode to use this as it's original state set this node as swapped return null return this; else if the left child has not been swapped if the result of trying to permute left child is null return the permutation of this node else return the permutation of the left child node if this node has a left node and a right node that are both leaves swap them set this node to be 'swapped' The desired behaviour of the algoritm would be something like this: branch / | branch 3 / | branch 2 / | 0 1 branch / | branch 3 / | branch 2 / | 1 0 <-- first swap branch / | branch 3 / | branch 1 <-- second swap / | 2 0 branch / | branch 3 / | branch 1 / | 0 2 <-- third swap branch / | branch 3 / | branch 0 <-- fourth swap / | 1 2 and so on... Sorry for the ridiculisly long and waddly explanation, would really, really apreciate any sort of help you guys could offer me. Thanks a bunch!

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  • Delete on a very deep tree

    - by Kathoz
    I am building a suffix trie (unfortunately, no time to properly implement a suffix tree) for a 10 character set. The strings I wish to parse are going to be rather long (up to 1M characters). The tree is constructed without any problems, however, I run into some when I try to free the memory after being done with it. In particularly, if I set up my constructor and destructor to be as such (where CNode.child is a pointer to an array of 10 pointers to other CNodes, and count is a simple unsigned int): CNode::CNode(){ count = 0; child = new CNode* [10]; memset(child, 0, sizeof(CNode*) * 10); } CNode::~CNode(){ for (int i=0; i<10; i++) delete child[i]; } I get a stack overflow when trying to delete the root node. I might be wrong, but I am fairly certain that this is due to too many destructor calls (each destructor calls up to 10 other destructors). I know this is suboptimal both space, and time-wise, however, this is supposed to be a quick-and-dirty solution to a the repeated substring problem. tl;dr: how would one go about freeing the memory occupied by a very deep tree? Thank you for your time.

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  • EBS 12.0 Minimum Requirements for Extended Support Finalized

    - by Steven Chan
    Oracle E-Business Suite Release 12.0 will transition from Premier Support to Extended Support on February 1, 2012.  New EBS 12.0 patches will be created and tested during Extended Support against the minimum patching baseline documented in our E-Business Suite Error Correction Support Policy (Note 1195034.1).These new technical requirements have now been finalized.  To be eligible for Extended Support, all EBS 12.0 customers must apply the EBS 12.0.6 Release Update Pack, technology stack infrastructure updates, and updates for EBS products if they're shared or fully-installed.  The complete set of minimum EBS 12.0 baseline requirements are listed here:E-Business Suite Error Correction Support Policy (Note 1195034.1)

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  • How to determine if binary tree is balanced?

    - by user69514
    It's been a while from those school years. Got a job as IT specialist at a hospital. Trying to move to do some actual programming now. I'm working on binary trees now, and I was wondering what would be the best way to determine if the tree is height-balanced. I was thinking of something along this: public boolean isBalanced(Node root){ if(root==null){ return true; //tree is empty } else{ int lh = root.left.height(); int rh = root.right.height(); if(lh - rh > 1 || rh - lh > 1){ return false; } } return true; } Is this a good implementation? or am I missing something?

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  • Tree data structure in iphone with coredata

    - by ebabchick
    Hi, I am looking to store a tree structure in CoreData for iphone. Can someone give me some tips on the best way to go about this? Basically I want to have a bunch of folders that people can dive into in a table view, and I want to have leaves of the tree be photos. The thing is, I want to be able to let the user edit these folders on-the-fly and add folders and content (photos) at their discretion. I'm relatively new to CoreData. Any help would be appreciated. Thanks.

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  • Searching over a templated tree

    - by floatingfrisbee
    So I have 2 interfaces: A node that can have children public interface INode { IEnumeration<INode> Children { get; } void AddChild(INode node); } And a derived "Data Node" that can have data associated with it public interface IDataNode<DataType> : INode { DataType Data; IDataNode<DataType> FindNode(DataType dt); } Keep in mind that each node in the tree could have a different data type associated with it as its Data (because the INode.AddChild function just takes the base INode) Here is the implementation of the IDataNode interface: internal class DataNode<DataType> : IDataNode<DataType> { List<INode> m_Children; DataNode(DataType dt) { Data = dt; } public IEnumerable<INode> Children { get { return m_Children; } } public void AddChild(INode node) { if (null == m_Children) m_Children = new List<INode>(); m_Children.Add(node); } public DataType Data { get; private set; } Question is how do I implement the FindNode function without knowing what kinds of DataType I will encounter in the tree? public IDataNode<DataType> FindNode(DataType dt) { throw new NotImplementedException(); } } As you can imagine something like this will not work out public IDataNode<DataType> FindNode(DataType dt) { IDataNode<DataType> result = null; foreach (var child in Children) { if (child is IDataNode<DataType>) { var datachild = child as IDataNode<DataType>; if (datachild.Data.Equals(dt)) { result = child as IDataNode<DataType>; break; } } else { // What?? } } return result; } Is my only option to do this when I know what kinds of DataType a particular tree I use will have? Maybe I am going about this in the wrong way, so any tips are appreciated. Thanks!

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