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  • red-black tree - construction

    - by Chaitanya
    Recently, I have been going through search trees and I encountered red-black trees, the point confusing me is, In r-b tree, the root node should be black thats fine, now how will I decide whether the incoming node assumes red or black color. I have gone through the wiki article but have not found a solution for this. I might be wrong, but I would be happy if someone can guide me through the exact material. Thank you.

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  • Effects of changing a node in a binary tree

    - by eSKay
    Suppose I want to change the orange node in the following tree. So, the only other change I'll need to make is in the left pointer of the green node. The blue node will remain the same. Am I wrong somewhere? Because according to this article (that explains zippers), even the blue node needs to be changed. Similarly, in this picture (recolored) from the same article, why do we change the orange nodes at all (when we change x)?

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  • trying to build a web ui tree from scratch

    - by Susan
    I have seen yahoo ui using div for every node of the tree. Is this a good thing to do. Seems like every node including leaf node is heavily nested. I saw a div, table, tr, td being used for creating a node. Is this necessary. Is there a better way to do the same. I tried to us tr for every node. Are there any issues with this approach Thanks

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  • Get HTML tree with Css

    - by 2x2p1p
    Hi guys :| Css can apply styles in elements through one ID, class or pseudo selector, but i would like to get the HTML tree, something like javascript: http://codeviewer.org/view/code:e86 Im tired of assign IDs to millions and millions of elements, can somebody help me ? Thanks

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  • How may I create an horizontal view from a tree table

    - by Giuseppe Alfieri
    I want to create a view for each tree for each customer of this 3 table structured in this way: Fields: db_attrib.name, db_attrib.name, db_attrib.name until the last db_tree.id_child = 0 Values: db_attrib_values.value where db_attrib.id = db_attrib_value.id_attrib and so on for each child And so on for each db_tree.id This is the link to sql fiddle with the prepared structure: http://sqlfiddle.com/#!9/21516

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  • Populate Tree using data from ArrayCollection

    - by jtorrance
    Let's say I had an ArrayCollection like this: public var ac:ArrayCollection= new ArrayCollection([ {item:"dog", group:"Animal"}, {item:"orange", group:"Fruits"}, {item:"cat", group:"Animal"}, {item:"apple", group:"Fruits"} ]); How would I create a Tree component in Flex 3 that uses the groups as nodes, with the appropriate items listed under each node?

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  • Tree View in c# with recursive

    - by user188886
    hi there i have table in DB with 3 column = ParentId , Id , Text i wanna make Tree in C# by the Records that we have saved in this table i dont know how should i write my recursive method in c# please help me thanks

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  • Real world examples of tree structures

    - by Arec Barrwin
    I'm looking for some examples of tree structures that are used in commercial/free software projects, modern or old. I can see examples on wikipedia, but I am looking for more concrete examples and how they're used. For example primary keys in databases are (from what I've read) stored in BST structure or a variation of the BST (feel free to correct me on this) My question isn't limited Binary Search Trees (BSTs), it can include any variation such as red-black, AVL and so on.

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  • Weakly connected tree

    - by wow_22
    hello I have an algorithmic problem using a weakly connected tree T where w(T)=sum(w(e)) for each edge e,by w i declare weight and i have to prove that we can use prim and Kruskal algorithm while w(T)=max{w(e)} maximum between any edge e belongs at T (I proved that) but i have also to prove the same for w(T)=?(w(e)) while ? states product of all edges belongs at T i tried a lot to prove it but i did not came up with a result that proving or disapproving the use of prim ,kruskal any help will be more than appreciated thanks

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  • ANTLR expressions rewrite intermediate tree

    - by user313856
    For expressions like 3+4 I would like to use the value 7 in an intermediate representation tree. I cannot work out how to get the returns value into a rewrite rule. expression returns [int v]: etc. How do I get expression.v into WR? At the moment I get (+ 3 4), I want (7) |^( WRITE c=expression) - ^(WRINT ^(INTC ^($c)) the next step is to emit 7 in an assembler.

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  • Tree structured resource Authorization

    - by user323883
    I have portfolio table with portoflio_id and parent_portfolio_id and I have user table now some users may have access to all portfolios, or selective portfolios or depending on group, everything under a portfolio tree. can someone suggest a good schema or any existing framework

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  • What will this B-tree look like?

    - by Phenom
    The B-tree is of order 4, meaning that it can hold 4 pointers, and 3 keys. The following is inserted: A G I Y Since they can't all fit in one node, I know that the node will split. So I know there's going to be a root node with 2 child nodes after these things are inserted, but I don't know exactly what they'll look like.

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  • Binary Search Tree Contains Function

    - by Suede
    I am trying to write a "contains" function for a binary search tree. I receive the following error at compile "Unhandled exception at 0x77291CB3 (ntdll.dll) in BST.exe: 0xC00000FD: Stack overflow (parameters: 0x00000001, 0x001E2FFC)." The following is my code. struct Node { int data; Node* leftChild; Node* rightChild; Node() : leftChild(NULL), rightChild(NULL) {} }; struct BST { Node* root; BST() : root(NULL) {} void insert(int value); bool contains(int value); }; void BST::insert(int value) { Node* temp = new Node(); temp->data = value; if(root == NULL) { root = temp; return; } Node* current; current = root; Node* parent; parent = root; current = (temp->data < current->data ? (current->leftChild) : (current->rightChild) while(current != NULL) { parent = current; current = (temp->data < current->data) ? (current->leftChild) : (current->rightChild) } if(temp->data < parent->data) { parent->leftChild = temp; } if(temp->data > parent->data) { parent->rightChild = temp; } } bool BST::contains(int value) { Node* temp = new Node(); temp->data = value; Node* current; current = root; if(temp->data == current->data) { // base case for when node with value is found std::cout << "true" << std::endl; return true; } if(current == NULL) { // base case if BST is empty or if a leaf is reached before value is found std::cout << "false" << std::endl; return false; } else { // recursive step current = (temp->data < current->data) ? (current->leftChild) : (current->rightChild); return contains(temp->data); } } int main() { BST bst; bst.insert(5); bst.contains(4); system("pause"); } As it stands, I would insert a single node with value '5' and I would search the binary search tree for a node with value '4' - thus, I would expect the result to be false.

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  • Problem with dojo tree

    - by Ewout
    Hello, I'm trying to get the dojo tree widget working. It works with a small json object, but when i try it with a large json object it goes wrong. There is no error, just the root node. Is this a normal behavior? Is there a maximum of objects you can load? My json object contains around 800 entries. Thanks, Ewout

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  • What is this Hash-like/Tree-like Construct Called?

    - by viatropos
    I want to create a "Config" class that acts somewhere between a hash and a tree. It's just for storing global values, which can have a context. Here's how I use it: Config.get("root.parent.child_b") #=> "value" Here's what the class might look like: class Construct def get(path) # split path by "." # search tree for nodes end def set(key, value) # split path by "." # create tree node if necessary # set tree value end def tree { :root => { :parent => { :child_a => "value", :child_b => "another value" }, :another_parent => { :something => { :nesting => "goes on and on" } } } } end end Is there a name for this kind of thing, somewhere between Hash and Tree (not a Computer Science major)? Basically a hash-like interface to a tree.

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  • Trace Your Browser’s Roots on the Browser Family Tree [Infographic]

    - by ETC
    The world of browsers is far more diverse than a glance at the big four browsers might lead you to believe. Check out the roots of your browser in the Browser Family Tree. You’re likely aware of mainstream browsers like Internet Explorer, Firefox, Chrome, and Opera, but do you know where they came from? That many of them share a common forefather? Not only that but what about lesser known browsers like Tamaya and OmniWeb? The browser family tree is a diverse thing. Hit up the link below to check out the full Browser Family Tree. Browser Family Tree [Wikipedia via Hotlinks] Latest Features How-To Geek ETC Macs Don’t Make You Creative! So Why Do Artists Really Love Apple? MacX DVD Ripper Pro is Free for How-To Geek Readers (Time Limited!) HTG Explains: What’s a Solid State Drive and What Do I Need to Know? How to Get Amazing Color from Photos in Photoshop, GIMP, and Paint.NET Learn To Adjust Contrast Like a Pro in Photoshop, GIMP, and Paint.NET Have You Ever Wondered How Your Operating System Got Its Name? Sync Blocker Stops iTunes from Automatically Syncing The Journey to the Mystical Forest [Wallpaper] Trace Your Browser’s Roots on the Browser Family Tree [Infographic] Save Files Directly from Your Browser to the Cloud in Chrome and Iron The Steve Jobs Chronicles – Charlie and the Apple Factory [Video] Google Chrome Updates; Faster, Cleaner Menus, Encrypted Password Syncing, and More

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  • 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

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  • 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'

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  • Preffered lambda syntax?

    - by Roger Alsing
    I'm playing around a bit with my own C like DSL grammar and would like some oppinions. I've reserved the use of "(...)" for invocations. eg: foo(1,2); My grammar supports "trailing closures" , pretty much like Ruby's blocks that can be passed as the last argument of an invocation. Currently my grammar support trailing closures like this: foo(1,2) { //parameterless closure passed as the last argument to foo } or foo(1,2) [x] { //closure with one argument (x) passed as the last argument to foo print (x); } The reason why I use [args] instead of (args) is that (args) is ambigious: foo(1,2) (x) { } There is no way in this case to tell if foo expects 3 arguments (int,int,closure(x)) or if foo expects 2 arguments and returns a closure with one argument(int,int) - closure(x) So thats pretty much the reason why I use [] as for now. I could change this to something like: foo(1,2) : (x) { } or foo(1,2) (x) -> { } So the actual question is, what do you think looks best? [...] is somewhat wrist unfriendly. let x = [a,b] { } Ideas?

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  • Preferred lambda syntax?

    - by Roger Alsing
    I'm playing around a bit with my own C like DSL grammar and would like some oppinions. I've reserved the use of "(...)" for invocations. eg: foo(1,2); My grammar supports "trailing closures" , pretty much like Ruby's blocks that can be passed as the last argument of an invocation. Currently my grammar support trailing closures like this: foo(1,2) { //parameterless closure passed as the last argument to foo } or foo(1,2) [x] { //closure with one argument (x) passed as the last argument to foo print (x); } The reason why I use [args] instead of (args) is that (args) is ambigious: foo(1,2) (x) { } There is no way in this case to tell if foo expects 3 arguments (int,int,closure(x)) or if foo expects 2 arguments and returns a closure with one argument(int,int) - closure(x) So thats pretty much the reason why I use [] as for now. I could change this to something like: foo(1,2) : (x) { } or foo(1,2) (x) -> { } So the actual question is, what do you think looks best? [...] is somewhat wrist unfriendly. let x = [a,b] { } Ideas?

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