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• Get the Latest on MySQL Enterprise Edition

  - by monica.kumar

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text-indent:-.25in; font-family:Symbol;}ol {margin-bottom:0in;}ul {margin-bottom:0in;}--Oracle just announced MySQL 5.5 Enterprise Edition. MySQL Enterprise Edition is a comprehensive subscription that includes:- MySQL Database- MySQL Enterprise Backup- MySQL Enterprise Monitor- MySQL Workbench- Oracle Premier Support; 24x7, WorldwideNew in this release is the addition of MySQL Enterprise Backup and MySQL Workbench along with enhancements in MySQL Enterprise Monitor. Recent integration with MyOracle Support allows MySQL customers to access the same support infrastructure used for Oracle Database customers. Joint MySQL and Oracle customers can experience faster problem resolution by using a common technical support interface. Supporting multiple operating systems, including Linux and Windows, MySQL Enterprise Edition can enable customers to achieve up to 90 percent TCO savingsover Microsoft SQL Server. See what Booking.com is saying:“With more than 50 million unique monthly visitors, performance and uptime are our first priorities,” said Bert Lindner, Senior Systems Architect, Booking.com. “The MySQL Enterprise Monitor is an essential tool to monitor, tune and manage our many MySQL instances. It allows us to zoom in quickly on the right areas, so we can spend our time and resources where it matters.” Read the press release for detailson technology enhancements.

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• Ubuntu, the family album

  <b>Pourquoi pas:</b> "A few days before the release of the new Ubuntu, here's a guided tour through the Ubuntu family album with some annotations telling my story with the different versions."

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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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• Soluto’s New Quick Question Button Makes Family Tech Support Simple

  - by Jason Fitzpatrick

  Soluto, a computer and boot management tool, now features a Quick Question button that allows the people you help out to easily click a button and send you both a short message and a screenshot of the problem. Any time your friend or family member presses F8, Soluto will take a screenshot of the screen, the Task Manager history, and a note from the user highlighting what issue they’re experiencing, and then email it all to you. After reviewing the email you can easily login to Soluto to remotely manage your friend’s computer and help with the problem. For more information about Soluto you can check out our previous reviews of the service here and here, or just hit up the link below to read more and take Soluto for a test drive. Soluto is a free service (for the first 5 computers), Windows only. Introducing Quick Question [The Soluto Blog] Java is Insecure and Awful, It’s Time to Disable It, and Here’s How What Are the Windows A: and B: Drives Used For? HTG Explains: What is DNS?

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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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• Balanced Quad Tree

  - by Marcos Perez

  Hi, how can I build a balanced quad tree. Thanks, Marcos

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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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• java polymorphic binary search tree

  - by ashraf

  how can i implement a polymorphic binary search tree (that makes use of EmptyTree and NonEmptyTree) without using downcasting or class checking?

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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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• mx:Tree not dispatching "itemClick" event when click on icon

  - by Xshare

  I have a flex tree that worked perfectly fine when we set the defaultLeafIcon={null} and the folderClosedIcon and folderOpenIcon to {null}. We decided to put the icons back in and took out the nulls. Now they show up fine, but if you click on the icon instead of the label or the rest of the row, it seems to change the selected item, shows the highlight around the new item, but doesn't dispatch the ItemClick event. This makes it really hard to know that the tree's selected item has changed! The weird part is that once you have clicked on the icon once and it looked like the selectedItem changed (or at least it applied that style), if you click the same icon again, it will actually fire the itemClick event. if you click any other icon, it does the same thing again, switching the selectedItem and styling that row, but not firing the itemClick event. Any ideas? Thanks. (This is in flex 4 btw)

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• how to use a tree data structure in C#

  - by matti

  I found an implementation for a tree at this SO question. Unfortunately I don't know how to use it. Also I made a change to it since LinkedList does not have Add method: delegate void TreeVisitor<T>(T nodeData); class NTree<T> { T data; List<NTree<T>> children; public NTree(T data) { this.data = data; children = new List<NTree<T>>(); } public void AddChild(T data) { children.Add(new NTree<T>(data)); } public NTree<T> GetChild(int i) { return children[i]; } public void Traverse(NTree<T> node, TreeVisitor<T> visitor) { visitor(node.data); foreach (NTree<T> kid in node.children) Traverse(kid, visitor); } } I have class named tTable and I want to store it's children and their grandchildren (...) in this tree. My need is to find immediate children and not traverse entire tree. I also might need to find children with some criteria. Let's say tTable has only name and I want to find children with names matching some criteria. tTables constructor gives name a value according to int-value (somehow). How do I use Traverse (write delegate) if I have code like this; int i = 0; Dictionary<string, NTree<tTable>> tableTreeByRootTableName = new Dictionary<string, NTree<tTable>>(); tTable aTable = new tTable(i++); tableTreeByRootTableName[aTable.Name] = new NTree(aTable); tableTreeByRootTableName[aTable.Name].AddChild(new tTable(i++)); tableTreeByRootTableName[aTable.Name].AddChild(new tTable(i++)); tableTreeByRootTableName[aTable.Name].GetChild(1).AddChild(new tTable(i++));

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• flex: move item around in a tree control

  - by Markus

  Hi everybody, I have a tree control and I want to give the user the ability that he can move up and down the element he just selected with a up and a downbutton. The tree gets generated from XML. I managed to insert the selected item a second time at a other place, with the following code: var parentXML:XML = XML(containerTree.selectedItem).parent(); var upperItem:XML = topContainer.source[containerTree.selectedIndex-1]; parentXML.insertChildBefore(upperItem,XML(containerTree.selectedItem)); but then I have the item there twice in the List. How can I remove to reinsert it? Thanks for Hints! Markus

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• PHP SPL RecursiveDirectoryIterator RecursiveIteratorIterator retrieving the full tree

  - by kmunky

  how can i retrieve the full directory tree using SPL ?

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• How to use R-Tree for plotting large number of map markers on google maps

  - by Eeyore

  After searching SO and multiple articles I haven't found a solution to my problem. What I am trying to achieve is to load 20,000 markers on Google Maps. R-Tree seems like a good approach but it's only helpful when searching for points within the visible part of the map. When the map is zoomed out it will return all of the points and...crash the browser. There is also the problem with dragging the map and at the end of dragging re-running the query. I would like to know how I can use R-Tree and be able to achieve the all of the above.

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• Height of a binary tree

  - by Programmer

  Consider the following code: public int heightOfBinaryTree(Node node) { if (node == null) { return 0; } else { return 1 + Math.max(heightOfBinaryTree(node.left), heightOfBinaryTree(node.right)); } } I want to know the logical reasoning behind this code. How did people come up with it? Does some have an inductive proof? Moreover, I thought of just doing a BFS with the root of the binary tree as the argument to get the height of the binary tree. Is the previous approach better than mine?Why?

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• Maximum number of descendants in a B-tree

  - by Phenom

  If you know the order of a B-tree, how do you figure out the maximum number of descendants from a page?

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• Selecting a different parent in Family Safety filters for Windows 8?

  - by Zhaph - Ben Duguid

  I've got Family Safety up and running nicely on a Windows 8 Surface, with two parent accounts and two child accounts. When one of the child accounts reaches its time limit, the user can "Ask a parent for more time". However, the next dialog doesn't allow the parents to choose which parent account to use - it always comes up with my account, and my wife can't log in to authorise. How can I allow the her account to allow more time/authorise sites etc? She can use the same Live login on our Windows 7 computers to control these settings, and is listed as a Parent on the Family Safety website.

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• Getting a table's values into a tree

  - by Jason

  So, I have a table like such: id|root|kw1|kw2|kw3|kw4|kw5|name 1| A| B| C| D| E| F|fileA 2| A| B| | | | |fileB 3| B| C| D| E| | |fileC 4| A| B| | | | |fileD (several hundred rows...) And I need to get it into a tree like the following: *A *B -fileB -fileD *C *D *E *F -fileA *B *C *D *E -fileC I'm pretty sure the table is laid out poorly but it's what I have to live with. I've read a little about Adjacency List Model & Modified Preorder Tree Traversal but I don't think my data is laid out correctly. I think this requires a recursive function, but I'm not at all sure how to go about that. I'm open to any ideas of how to get this done even if it means extracting the data into a new table just to process this. Are there any good options available to me or any good ways to do this? (Examples are a bonus of course)

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• Is there any tool which can show the call tree for SQL stored procedures

  - by DBZ_A

  I have a huge SQL script which i need to analyse. It would be really helpful if i could find a tool which can generate a call tree; ie, to see which all procedures are called from a particular procedure. a perl based example is here, http://sqlblog.com/blogs/linchi_shea/archive/2009/10/23/find-the-complete-call-tree-for-a-stored-procedure.aspx but i need a tool to analyse the text file (.sql file), not the procedure stored in the database. due to some reasons i will not be able to create the whole set of procedures in the database and use the above mentioned tool. please respond if you have come across any ide/tool with this feature.

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• dojo dgrid tree, subrows in wrong position

  - by Ventura

  I have a dgrid, working with tree column plugin. Every time that the user click on the tree, I call the server, catch the subrows(json) and bind it. But when it happens, these subrows are show in wrong position, like the image bellow. The most strange is when I change the pagination, after go back to first page, the subrows stay on the correct place. (please, tell me if is possible to understand my english, then I can try to improve the text) My dgrid code: var CustomGrid = declare([OnDemandGrid, Keyboard, Selection, Pagination]); var grid = new CustomGrid({ columns: [ selector({label: "#", disabled: function(object){ return object.type == 'DOCx'; }}, "radio"), {label:'Id', field:'id', sortable: false}, tree({label: "Title", field:"title", sortable: true, indentWidth:20, allowDuplicates:true}), //{label:'Title', field:'title', sortable: false}, {label:'Count', field:'count', sortable: false} ], store: this.memoryStore, collapseOnRefresh:true, pagingLinks: false, pagingTextBox: true, firstLastArrows: true, pageSizeOptions: [10, 15, 25], selectionMode: "single", // for Selection; only select a single row at a time cellNavigation: false // for Keyboard; allow only row-level keyboard navigation }, "grid"); My memory store: loadMemoryStore: function(items){ this.memoryStore = Observable(new Memory({ data: items, getChildren: function(parent, options){ return this.query({parent: parent.id}, options); }, mayHaveChildren: function(parent){ return (parent.count != 0) && (parent.type != 'DOC'); } })); }, This moment I am binding the subrows: success: function(data){ for(var i=0; i<data.report.length; i++){ this.memoryStore.put({id:data.report[i].id, title:data.report[i].created, type:'DOC', parent:this.designId}); } }, I was thinking, maybe every moment that I bind the subrows, I could do like a refresh on the grid, maybe works. I think that the pagination does the same thing. Thanks. edit: I forgot the question. Well, How can I correct this bug? If The refresh in dgrid works. How can I do it? Other thing that I was thinking, maybe my getChildren is wrong, but I could not identify it. thanks again.

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