Search Results

Search found 7638 results on 306 pages for 'binary 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 19/306 | < Previous Page | 15 16 17 18 19 20 21 22 23 24 25 26  | Next Page >

  • Express XPath as an expression tree

    - by 47d_
    If I have an XPath query like NodeA/NodeB[@WIDTH and not(@WIDTH="20")] | NodeC[@WIDTH and not(@WIDTH="20")]/NodeD Is there any API available to visualize this XPath query as a stack of atomic expressions, something like (following is generic) Get results of NodeA, call it "first set" Get results of NodeB from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for B" Get results of NodeC from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for C" Combine "node d for B" and "node d for C" I am trying to see if we can convert the XPath expression into custom expression which is close to english and vice versa. If no API is available, what would be the best approach? Thanks in advance.

    Read the article

  • transform file/directory structure into 'tree' in javascript

    - by dave
    I have an array of objects that looks like this: [{ name: 'test', size: 0, type: 'directory', path: '/storage/test' }, { name: 'asdf', size: 170, type: 'directory', path: '/storage/test/asdf' }, { name: '2.txt', size: 0, type: 'file', path: '/storage/test/asdf/2.txt' }] There could be any number of arbitrary path's, this is the result of iterating through files and folders within a directory. What I'm trying to do is determine the 'root' node of these. Ultimately, this will be stored in mongodb and use materialized path to determine it's relationships. In this example, /storage/test is a root with no parent. /storage/test/asdf has the parent of /storage/test which is the parent to /storage/test/asdf/2.txt. My question is, how would you go about iterating through this array, to determine the parent's and associated children? Any help in the right direction would be great! Thank you

    Read the article

  • Javascript / Jquery Tree Travesal question

    - by Copper
    Suppose I have the following <ul> <li>Item 1</li> <li>Item 2 <ul> <li>Sub Item</li> </ul> </li> <li>Item 3</li> </ul> This list is auto-generated by some other code (so adding exclusive id's/class' is out of the question. Suppose I have some jquery code that states that if I mouseover an li, it gets a background color. However, if I mouseover the "Sub Item" list item, "Item 2" will be highlighted as well. How can I make it so that if the user mouses over "Sub Item" it only puts a background color on that and not on "Item 2" as well?

    Read the article

  • How to find Sub-trees in non-binary tree

    - by kenny
    I have a non-binary tree. I want to find all "sub-trees" that are connected to root. Sub-tree is a a link group of tree nodes. every group is colored in it's own color. What would be be the best approach? Run recursion down and up for every node? The data structure of every treenode is a list of children, list of parents. (the type of children and parents are treenodes) Clarification: Group defined if there is a kind of "closure" between nodes where root itself is not part of the closure. As you can see from the graph you can't travel from pink to other nodes (you CAN NOT use root). From brown node you can travel to it's child so this form another group. Finally you can travel from any cyan node to other cyan nodes so the form another group

    Read the article

  • GUI question : representing large tree

    - by Peter
    I have a tree-like datastructure of some six levels deep, that I would like to represent on a single webpage (can be tabs, trees; ....) In each level both childnodes and content are possible. Presenting it like a real tree would be not very usable (too big). I was thinking in the lines of hiding parts of the tree when you drill down and presenting a breadcrumbs or the like to keep you informed as to where you are... I guess my question boils down to : any ideas / examples ? Tx!

    Read the article

  • How to find siblings of a tree?

    - by smallB
    On my interview for an internship, I was asked following question: On a whiteboard write the simplest algorithm with use of recursion which would take a root of a so called binary tree (so called because it is not strictly speaking binary tree) and make every child in this tree connected with its sibling. So if I have: 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 then the sibling to 2 would be 3, to four five, to five six and to seven eight. I didn't do this, although I was heading in the right direction. Later (next day) at home I did it, but with the use of a debugger. It took me better part of two hours and 50 lines of code. I personally think that this was very difficult question, almost impossible to do correctly on a whiteboard. How would you solve it on a whiteboard? How to apprehend this question without using a debugger?

    Read the article

  • Collaborative Whiteboard using WebSocket in GlassFish 4 - Text/JSON and Binary/ArrayBuffer Data Transfer (TOTD #189)

    - by arungupta
    This blog has published a few blogs on using JSR 356 Reference Implementation (Tyrus) as its integrated in GlassFish 4 promoted builds. TOTD #183: Getting Started with WebSocket in GlassFish TOTD #184: Logging WebSocket Frames using Chrome Developer Tools, Net-internals and Wireshark TOTD #185: Processing Text and Binary (Blob, ArrayBuffer, ArrayBufferView) Payload in WebSocket TOTD #186: Custom Text and Binary Payloads using WebSocket One of the typical usecase for WebSocket is online collaborative games. This Tip Of The Day (TOTD) explains a sample that can be used to build such games easily. The application is a collaborative whiteboard where different shapes can be drawn in multiple colors. The shapes drawn on one browser are automatically drawn on all other peer browsers that are connected to the same endpoint. The shape, color, and coordinates of the image are transfered using a JSON structure. A browser may opt-out of sharing the figures. Alternatively any browser can send a snapshot of their existing whiteboard to all other browsers. Take a look at this video to understand how the application work and the underlying code. The complete sample code can be downloaded here. The code behind the application is also explained below. The web page (index.jsp) has a HTML5 Canvas as shown: <canvas id="myCanvas" width="150" height="150" style="border:1px solid #000000;"></canvas> And some radio buttons to choose the color and shape. By default, the shape, color, and coordinates of any figure drawn on the canvas are put in a JSON structure and sent as a message to the WebSocket endpoint. The JSON structure looks like: { "shape": "square", "color": "#FF0000", "coords": { "x": 31.59999942779541, "y": 49.91999053955078 }} The endpoint definition looks like: @WebSocketEndpoint(value = "websocket",encoders = {FigureDecoderEncoder.class},decoders = {FigureDecoderEncoder.class})public class Whiteboard { As you can see, the endpoint has decoder and encoder registered that decodes JSON to a Figure (a POJO class) and vice versa respectively. The decode method looks like: public Figure decode(String string) throws DecodeException { try { JSONObject jsonObject = new JSONObject(string); return new Figure(jsonObject); } catch (JSONException ex) { throw new DecodeException("Error parsing JSON", ex.getMessage(), ex.fillInStackTrace()); }} And the encode method looks like: public String encode(Figure figure) throws EncodeException { return figure.getJson().toString();} FigureDecoderEncoder implements both decoder and encoder functionality but thats purely for convenience. But the recommended design pattern is to keep them in separate classes. In certain cases, you may even need only one of them. On the client-side, the Canvas is initialized as: var canvas = document.getElementById("myCanvas");var context = canvas.getContext("2d");canvas.addEventListener("click", defineImage, false); The defineImage method constructs the JSON structure as shown above and sends it to the endpoint using websocket.send(). An instant snapshot of the canvas is sent using binary transfer with WebSocket. The WebSocket is initialized as: var wsUri = "ws://localhost:8080/whiteboard/websocket";var websocket = new WebSocket(wsUri);websocket.binaryType = "arraybuffer"; The important part is to set the binaryType property of WebSocket to arraybuffer. This ensures that any binary transfers using WebSocket are done using ArrayBuffer as the default type seem to be blob. The actual binary data transfer is done using the following: var image = context.getImageData(0, 0, canvas.width, canvas.height);var buffer = new ArrayBuffer(image.data.length);var bytes = new Uint8Array(buffer);for (var i=0; i<bytes.length; i++) { bytes[i] = image.data[i];}websocket.send(bytes); This comprehensive sample shows the following features of JSR 356 API: Annotation-driven endpoints Send/receive text and binary payload in WebSocket Encoders/decoders for custom text payload In addition, it also shows how images can be captured and drawn using HTML5 Canvas in a JSP. How could this be turned in to an online game ? Imagine drawing a Tic-tac-toe board on the canvas with two players playing and others watching. Then you can build access rights and controls within the application itself. Instead of sending a snapshot of the canvas on demand, a new peer joining the game could be automatically transferred the current state as well. Do you want to build this game ? I built a similar game a few years ago. Do somebody want to rewrite the game using WebSocket APIs ? :-) Many thanks to Jitu and Akshay for helping through the WebSocket internals! Here are some references for you: JSR 356: Java API for WebSocket - Specification (Early Draft) and Implementation (already integrated in GlassFish 4 promoted builds) Subsequent blogs will discuss the following topics (not necessary in that order) ... Error handling Interface-driven WebSocket endpoint Java client API Client and Server configuration Security Subprotocols Extensions Other topics from the API

    Read the article

  • [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.

    Read the article

  • Quick Outline: Navigating Your PL/SQL Packages in Oracle SQL Developer

    - by thatjeffsmith
    If you’re browsing your packages using the Connections panel, you have a nice tree navigator to click around your packages and your variable, procedure, and functions. Click, click, click all day long, click, click, click while I sing this song… But What if you drill into your PL/SQL source from the worksheet and don’t have the Tree expanded? Let’s say you’re working on your script, something like - Hmm, what goes next again? So I need to reacquaint myself with just what my beer package requires, so I’m going to drill into it by doing a DESCRIBE (via SHIFT+F4), and now I have the package open. The package is open but the tree hasn’t auto-expanded. Please don’t tell me I have to do the click-click-click thing in the tree!?! Just Open the Quick Outline Panel Do you see it? Just right click in the procedure editor – select the ‘Quick Outline’ in the context menu, and voila! The navigational power of the tree, without needing to drill down the tree itself. If I want to drill into my procedure declaration, just click on said procedure name in the Quick Outline panel. This works for both package specs and bodies. Technically you can use this for stand alone procedures and functions, but the real power is demonstrated for packages.

    Read the article

  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

    Read the article

  • Binary Search Tree, cannot do traversal

    - by ihm
    Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

    Read the article

  • New binary analysis tool finds FOSS in device firmware

    <b>ars Technica:</b> "Software development company Loohuis Consulting and process management consultancy OpenDawn have released a new binary analysis tool that is designed to detect Linux and BusyBox in binary firmware. The program, which is freely available for download, is intended to aid open source license compliance efforts."

    Read the article

  • QotD: Maurizio Cimadamore on Project Lambda Binary Snapshots

    - by $utils.escapeXML($entry.author)
    I'm glad to announce that the first binary snapshots of the lambda repository are available at the following URL:http://jdk8.java.net/lambda/As you can imagine, as the implementation of the compiler/libraries is still under heavy development, there are still many rough corners that need to be polished. I'd like to thank you all for all the patience and the valuable feedback provided so far - please keep it coming!Maurizio Cimadamore announcing the Project Lambda binary snapshots on the lambda-dev OpenJDK mailing list.

    Read the article

  • How to cleanly add after-the-fact commits from the same feature into git tree

    - by Dennis
    I am one of two developers on a system. I make most of the commits at this time period. My current git workflow is as such: there is master branch only (no develop/release) I make a new branch when I want to do a feature, do lots of commits, and then when I'm done, I merge that branch back into master, and usually push it to remote. ...except, I am usually not done. I often come back to alter one thing or another and every time I think it is done, but it can be 3-4 commits before I am really done and move onto something else. Problem The problem I have now is that .. my feature branch tree is merged and pushed into master and remote master, and then I realize that I am not really done with that feature, as in I have finishing touches I want to add, where finishing touches may be cosmetic only, or may be significant, but they still belong to that one feature I just worked on. What I do now Currently, when I have extra after-the-fact commits like this, I solve this problem by rolling back my merge, and re-merging my feature branch into master with my new commits, and I do that so that git tree looks clean. One clean feature branch branched out of master and merged back into it. I then push --force my changes to origin, since my origin doesn't see much traffic at the moment, so I can almost count that things will be safe, or I can even talk to other dev if I have to coordinate. But I know it is not a good way to do this in general, as it rewrites what others may have already pulled, causing potential issues. And it did happen even with my dev, where git had to do an extra weird merge when our trees diverged. Other ways to solve this which I deem to be not so great Next best way is to just make those extra commits to the master branch directly, be it fast-forward merge, or not. It doesn't make the tree look as pretty as in my current way I'm solving this, but then it's not rewriting history. Yet another way is to wait. Maybe wait 24 hours and not push things to origin. That way I can rewrite things as I see fit. The con of this approach is time wasted waiting, when people may be waiting for a fix now. Yet another way is to make a "new" feature branch every time I realize I need to fix something extra. I may end up with things like feature-branch feature-branch-html-fix, feature-branch-checkbox-fix, and so on, kind of polluting the git tree somewhat. Is there a way to manage what I am trying to do without the drawbacks I described? I'm going for clean-looking history here, but maybe I need to drop this goal, if technically it is not a possibility.

    Read the article

  • How to perform a binary search on IList<T>?

    - by Daniel Brückner
    Simple question - given an IList<T> how do you perform a binary search without writing the method yourself and without copying the data to a type with build-in binary search support. My current status is the following. List<T>.BinarySearch() is not a member of IList<T> There is no equivalent of the ArrayList.Adapter() method for List<T> IList<T> does not inherit from IList, hence using ArrayList.Adapter() is not possible I tend to believe that is not possible with build-in methods, but I cannot believe that such a basic method is missing from the BCL/FCL. If it is not possible, who can give the shortest, fastest, smartest, or most beatiful binary search implementation for IList<T>? UPDATE We all know that a list must be sorted before using binary search, hence you can assume that it is. But I assume (but did not verify) it is the same problem with sort - how do you sort IList<T>? CONCLUSION There seems to be no build-in binary search for IList<T>. One can use First() and OrderBy() LINQ methods to search and sort, but it will likly have a performance hit. Implementing it yourself (as an extension method) seems the best you can do.

    Read the article

  • How to tell Subversion to display binary files using an external program?

    - by lamcro
    I have some code which, like java, is stored in a binary format, and I have the applications to display and modify this code setup in the Subversion's config file. But when I run svn diff for these file, Subversion prevents me =================================================================== Cannot display: file marked as a binary type. svn:mime-type = application/octet-stream I can still view them, but only with the --force argument Since all the files in the repository are of this binary code, how can I permanently force subversion to open the files for diff or edit mode?

    Read the article

  • How to create a complete binary tree of height 'h' using Python?

    - by Jack
    Here is the node structure class Node: def __init__(self, data): # initializes the data members self.left = None self.right = None self.parent = None self.data = data complete binary tree Definition: A binary tree in which every level, except possibly the deepest, is completely filled. At depth n, the height of the tree, all nodes must be as far left as possible. -- http://www.itl.nist.gov/div897/sqg/dads/HTML/completeBinaryTree.html I am looking for an efficient algorithm.

    Read the article

  • Transferring binary data through a SOAP webservice? C# / .NET

    - by Jason
    I have a webservice that returns the binary array of an object. Is there an easier way to transfer this with SOAP or does it need to be contained in XML? It's working, but I had to increase the send and receive buffer to a large value. How much is too much? Transferring binary in XML as an array seems really inefficient, but I can't see any way to add a binary attachment using .NET.

    Read the article

  • How can I find the common ancestor of two nodes in a binary tree?

    - by Siddhant
    The Binary Tree here is not a Binary Search Tree. Its just a Binary Tree. The structure could be taken as - struct node { int data; struct node *left; struct node *right; }; The maximum solution I could work out with a friend was something of this sort - Consider this binary tree (from http://lcm.csa.iisc.ernet.in/dsa/node87.html) : The inorder traversal yields - 8, 4, 9, 2, 5, 1, 6, 3, 7 And the postorder traversal yields - 8, 9, 4, 5, 2, 6, 7, 3, 1 So for instance, if we want to find the common ancestor of nodes 8 and 5, then we make a list of all the nodes which are between 8 and 5 in the inorder tree traversal, which in this case happens to be [4, 9, 2]. Then we check which node in this list appears last in the postorder traversal, which is 2. Hence the common ancestor for 8 and 5 is 2. The complexity for this algorithm, I believe is O(n) (O(n) for inorder/postorder traversals, the rest of the steps again being O(n) since they are nothing more than simple iterations in arrays). But there is a strong chance that this is wrong. :-) But this is a very crude approach, and I'm not sure if it breaks down for some case. Is there any other (possibly more optimal) solution to this problem?

    Read the article

  • Serialize a C# class to binary be used by C++. How to handle alignment?

    - by glenn.danthi
    I am currently serializing a C# class into a binary stream using BinaryWriter. I take each element of the class and write it out using BinaryWriter. This worked fine as the C++ application reading this binary file supported packed structs and hence the binary file could be loaded directly. Now I have got a request to handle alignment as a new application has popped up which cannot support packed structs. What's the best way to convert the C# class and exporting it out as a binary keeping both 2 byte as well as 4 byte alignment in mind? The user can choose the alignment.

    Read the article

  • Detecting if a file is binary or plain text?

    - by dr. evil
    How can I detect if a file is binary or a plain text? Basically my .NET app is processing batch files and extracting data however I don't want to process binary files. As a solution I'm thinking about analysing first X bytes of the file and if there are more unprintable characters than printable characters it should be binary. Is this the right way to do it? Is there nay better implementation for this task?

    Read the article

  • How can I implement a splay tree that performs the zig operation last, not first?

    - by Jakob
    For my Algorithms & Data Structures class, I've been tasked with implementing a splay tree in Haskell. My algorithm for the splay operation is as follows: If the node to be splayed is the root, the unaltered tree is returned. If the node to be splayed is one level from the root, a zig operation is performed and the resulting tree is returned. If the node to be splayed is two or more levels from the root, a zig-zig or zig-zag operation is performed on the result of splaying the subtree starting at that node, and the resulting tree is returned. This is valid according to my teacher. However, the Wikipedia description of a splay tree says the zig step "will be done only as the last step in a splay operation" whereas in my algorithm it is the first step in a splay operation. I want to implement a splay tree that performs the zig operation last instead of first, but I'm not sure how it would best be done. It seems to me that such an algorithm would become more complex, seeing as how one needs to find the node to be splayed before it can be determined whether a zig operation should be performed or not. How can I implement this in Haskell (or some other functional language)?

    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 | 15 16 17 18 19 20 21 22 23 24 25 26  | Next Page >