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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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  • Hacker Disables More Than 100 Cars Remotely

    <b>Wired:</b> "More than 100 drivers in Austin, Texas found their cars disabled or the horns honking out of control, after an intruder ran amok in a web-based vehicle-immobilization system normally used to get the attention of consumers delinquent in their auto payments."

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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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  • 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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  • Deleting a node in a family tree

    - by user559142
    Hi, I'm trying to calclulate the best way to delete a node in a family tree. First, a little description of how the app works. My app makes the following assumption: Any node can only have one partner. That means that any child a single node has, it will also be the partner nodes child too. Therefore, step relations, divorces etc aren't compensated for. A node always has two parents - A mother and father cannot be added seperately. If the user doesn't know the details - the nodes attributes are set to a default value. Also any node can add parents, siblings, children to itself. Therefore in law relationships can be added. I have the following classes: FamilyMember String fName; String lName; String dob; String gender; FamilyMember mother, father, partner; ArrayListchildren; int index; int generation; void linkParents(); void linkPartner(); void addChild(); //gets & sets for fields Family ArrayListfamily; void addMember(); void removeMember(); FamilyMember getFamilyMember(index); ArrayListgetFamilyMembers(); FamilyTree Family family; void removeMember(); //need help void displayFamilyMembers(); void addFamilyMember(); void enterDetails(); void displayAncestors(); void displayDescendants(); void printDescendants(); FamilyMember findRootNode(); void sortGenerations(); void getRootGeneration(); I am having trouble with identifying the logic for removing a member. All other functions work fine. Has anyone developed a family tree app before who knows how to deal with removing various different nodes in the family "tree"? e.g. removing a leaf removing a leaf with partner (what if partner has parents etc) removing a parent It seems to be another recursive property but my head is swelling from over thought.

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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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  • Implementing an iterator over binary tree using C++ 11

    - by user1459339
    I would like to create an iterator over the binary tree so as to be able to use range-based for loop. I understand I ought to implement the begin() and end() function first. Begin should probably point to the root. According to the specification, however, the end() functions returns "the element following the last valid element". Which element (node) is that? Would it not be illegal to point to some "invalid" place? The other thing is the operator++. What is the best way to return "next" element in tree? I just need some advice to begin with this programming.

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  • Hacker Croll ??

    - by Haruto Kitano, CISSP-ISSJP
    ??????????????????????????????????? ???4?16??HP???HP Security DAY 2010????????????????ID????????????????????????2????????????????????????????????????????PCIDSS?????12??????????????????????????????????????????????????????????????????? ???????···

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  • Conditional Drag and Drop Operations in Flex/AS3 Tree

    - by user163757
    Good day everyone. I am currently working with a hierarchical tree structure in AS3/Flex, and want to enable drag and drop capabilities under certain conditions: Only parent/top level nodes can be moved Parent/top level nodes must remain at this level; they can not be moved to child nodes of other parent nodes Using the dragEnter event of the tree, I am able to handle condition 1 easily. private function onDragEnter(event:DragEvent):void { // only parent nodes (map layers) are moveable event.preventDefault(); if(toc.selectedItem.hasOwnProperty("layer")) DragManager.acceptDragDrop(event.target as UIComponent); else DragManager.showFeedback(DragManager.NONE); } Handling the second condition is proving to be a bit more difficult. I am pretty sure the dragOver event is the place for logic. I have been experimenting with calculateDropIndex, but that always gives me the index of the parent node, which doesn't help check if the potential drop location is acceptable or not. Below is some pseudo code of what I am looking to accomplish. private function onDragOver(e:DragEvent):void { // if potential drop location has parents // dont allow drop // else // allow drop } Can anyone provide advice how to implement this?

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  • Java Binary Tree. Priting InOrder traversal

    - by user69514
    I am having some problems printing an inOrder traversal of my binary tree. Even after inserting many items into the tree it's only printing 3 items. public class BinaryTree { private TreeNode root; private int size; public BinaryTree(){ this.size = 0; } public boolean insert(TreeNode node){ if( root == null) root = node; else{ TreeNode parent = null; TreeNode current = root; while( current != null){ if( node.getData().getValue().compareTo(current.getData().getValue()) <0){ parent = current; current = current.getLeft(); } else if( node.getData().getValue().compareTo(current.getData().getValue()) >0){ parent = current; current = current.getRight(); } else return false; if(node.getData().getValue().compareTo(parent.getData().getValue()) < 0) parent.setLeft(node); else parent.setRight(node); } } size++; return true; } /** * */ public void inOrder(){ inOrder(root); } private void inOrder(TreeNode root){ if( root.getLeft() !=null) this.inOrder(root.getLeft()); System.out.println(root.getData().getValue()); if( root.getRight() != null) this.inOrder(root.getRight()); } }

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  • Logic for family tree program

    - by david robers
    Hi All, I am creating a family tree program in Java, or at least trying to. I have developed several classes: Person - getters and setter for name gender age etc FamilyMember - extends Person getters and setters for setting arents and children Family - which consists of multiple family members and methods for adding removing members FamilyTree which is the main class for setting relationships. I have two main problems: 1) I need to set the relationships between people. Currently I am doing: FamilyMember A, FamilyMember B B.setMother(A); A.setChild(B); The example above is for setting a mother child relationship. This seems very clunky. Its getting very long winded to implement all relationships. Any ideas on how to implement multiple relationships in a less prodcedural way? 2) I have to be able to display the family tree. How can I do this? Are there any custom classes out there to make life easier? Thanks for your time...

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  • Creating subtree from tree which is represented in xml - python

    - by Jay
    Hi I have an XML (in the form of tree), I require to create sub-tree out of it. For ex: <a> <b> <c>Hello</c> <d> <e>Hi</e> </a> Subtree would be <root> <a> <b> <c>Hello</c> </b> </a> <a> <d> <e>Hi</e> </d> </a> </root> What is the best XML library in python to do it? Any algorithm that already does this would also be helpful. Note: the XML doc won't be that big, it will easily fit in memory.

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  • How to exclude rows where matching join is in an SQL tree

    - by Greg K
    Sorry for the poor title, I couldn't think how to concisely describe this problem. I have a set of items that should have a 1-to-1 relationship with an attribute. I have a query to return those rows where the data is wrong and this relationship has been broken (1-to-many). I'm gathering these rows to fix them and restore this 1-to-1 relationship. This is a theoretical simplification of my actual problem but I'll post example table schema here as it was requested. item table: +------------+------------+-----------+ | item_id | name | attr_id | +------------+------------+-----------+ | 1 | BMW 320d | 20 | | 1 | BMW 320d | 21 | | 2 | BMW 335i | 23 | | 2 | BMW 335i | 34 | +------------+------------+-----------+ attribute table: +---------+-----------------+------------+ | attr_id | value | parent_id | +---------+-----------------+------------+ | 20 | SE | 21 | | 21 | M Sport | 0 | | 23 | AC | 24 | | 24 | Climate control | 0 | .... | 34 | Leather seats | 0 | +---------+-----------------+------------+ A simple query to return items with more than one attribute. SELECT item_id, COUNT(DISTINCT(attr_id)) AS attributes FROM item GROUP BY item_id HAVING attributes > 1 This gets me a result set like so: +-----------+------------+ | item_id | attributes | +-----------+------------+ | 1 | 2 | | 2 | 2 | | 3 | 2 | -- etc. -- However, there's an exception. The attribute table can hold a tree structure, via parent links in the table. For certain rows, parent_id can hold the ID of another attribute. There's only one level to this tree. Example: +---------+-----------------+------------+ | attr_id | value | parent_id | +---------+-----------------+------------+ | 20 | SE | 21 | | 21 | M Sport | 0 | .... I do not want to retrieve items in my original query where, for a pair of associated attributes, they related like attributes 20 & 21. I do want to retrieve items where: the attributes have no parent for two or more attributes they are not related (e.g. attributes 23 & 34) Example result desired, just the item ID: +------------+ | item_id | +------------+ | 2 | +------------+ How can I join against attributes from items and exclude these rows? Do I use a temporary table or can I achieve this from a single query? Thanks.

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  • Getting tree construction with ANTLR

    - by prosseek
    As asked and answered in http://stackoverflow.com/questions/2999755/removing-left-recursion-in-antlr , I could remove the left recursion E - E + T|T T - T * F|F F - INT | ( E ) After left recursion removal, I get the following one E - TE' E' - null | + TE' T - FT' T' - null | * FT' Then, how to make the tree construction with the modified grammar? With the input 1+2, I want to have a tree ^('+' ^(INT 1) ^(INT 2)). Or similar. grammar T; options { output=AST; language=Python; ASTLabelType=CommonTree; } start : e - e ; e : t ep - ??? ; ep : | '+' t ep - ??? ; t : f tp - ??? ; tp : | '*' f tp - ??? ; f : INT | '(' e ')' - e ; INT : '0'..'9'+ ; WS: (' '|'\n'|'\r')+ {$channel=HIDDEN;} ;

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  • SVN X remains in tree-conflict

    - by Paul Knopf
    I am using VisualSVN (which uses Tortoise). I accidentally move a folder to a different location. When tries to move it back, SVN pukes with this error. It happened once before and I managed to do some random updates/commits, not knowing what I was doing and it was "fixed". I cannot pull the same magic again, so I need to know how to get my files and directory and of tree-conflict. Thanks!

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