Search Results

Search found 4173 results on 167 pages for 'dijit tree'.

Page 4/167 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Windows 7 Windows Explorer jumpy tree view

    - by P a u l
    Is there any way to get Windows Explorer tree view in Windows 7 to stop jumping? I think they really messed up this design. Click a node to expand a deeper level and it instantly scrolls the tree vertically to a new location. This is not a good feature since my eye completely loses the node it was focused on and I have to hunt for where I was. I want the tree view to remain fixed where it is unless I scroll it myself.

    Read the article

  • 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

    Read the article

  • Dijit.Dialog irregular behaviour of scroll bar making dialog box unusable

    - by arachnica
    I use a dojo dialog box to display a page pulled from another part of the site. The page being pulled is long - so I use the css attribute: `max-height: 900px; overflow:auto; To make sure it displays properly. For a long page, it displays a 900px high dialog box with a scrollbar down the right hand side side. However the scrollbar is going up to the dialog box title and shifting the close box icon to the left. In internet explorer, when a user moves his mouse of the close box icon, it moves to the right. If you move your mouse to the right it moves left - so you can never actually close the box with the icon. I have tried clearing the content in the box and just making it very long, and the same thing still happens. Oddly enough, if I close the window using esc and click pull up the dialog box again, it works fine. Any way I can make it display properly at all times. I am using Dojo version 1.3. Thanks

    Read the article

  • check if a tree is complete standard ml

    - by aizen92
    I want to make a function in standard ml that checks if a tree is complete or not, the function somehow works, but its giving me the wrong type and a warning of non-exhaustive cases The tree code: datatype 'data tree = EMPTY | NODE of 'data tree * 'data * 'data tree; fun isComplete EMPTY = true | isComplete (NODE(x, y, z)) = if (x = EMPTY andalso z <> EMPTY) orelse (x <> EMPTY andalso z = EMPTY) then false else true; Now the above function's type is: ''a tree -> bool but the required type is 'a tree -> bool The warning I'm having is: stdIn:169.8 Warning: calling polyEqual stdIn:169.26 Warning: calling polyEqual stdIn:169.45-169.47 Warning: calling polyEqual stdIn:169.64-169.66 Warning: calling polyEqual stdIn:124.1-169.94 Warning: match nonexhaustive NODE (x,y,z) => ... What is the problem I'm having?

    Read the article

  • How to find largest common sub-tree in the given two binary search trees?

    - by Bhushan
    Two BSTs (Binary Search Trees) are given. How to find largest common sub-tree in the given two binary trees? EDIT 1: Here is what I have thought: Let, r1 = current node of 1st tree r2 = current node of 2nd tree There are some of the cases I think we need to consider: Case 1 : r1.data < r2.data 2 subproblems to solve: first, check r1 and r2.left second, check r1.right and r2 Case 2 : r1.data > r2.data 2 subproblems to solve: - first, check r1.left and r2 - second, check r1 and r2.right Case 3 : r1.data == r2.data Again, 2 cases to consider here: (a) current node is part of largest common BST compute common subtree size rooted at r1 and r2 (b)current node is NOT part of largest common BST 2 subproblems to solve: first, solve r1.left and r2.left second, solve r1.right and r2.right I can think of the cases we need to check, but I am not able to code it, as of now. And it is NOT a homework problem. Does it look like?

    Read the article

  • F#: Recursive collect and filter over N-ary Tree

    - by RodYan
    This is hurting my brain! I want to recurse over a tree structure and collect all instances that match some filter into one list. Here's a sample tree structure type Tree = | Node of int * Tree list Here's a test sample tree: let test = Node((1, [Node(2, [Node(3,[]); Node(3,[])]); Node(3,[])])) Collecting and filtering over nodes with and int value of 3 should give you output like this: [Node(3,[]);Node(3,[]);Node(3,[])]

    Read the article

  • In-order tree traversal

    - by Chris S
    I have the following text from an academic course I took a while ago about in-order traversal (they also call it pancaking) of a binary tree (not BST): In-order tree traversal Draw a line around the outside of the tree. Start to the left of the root, and go around the outside of the tree, to end up to the right of the root. Stay as close to the tree as possible, but do not cross the tree. (Think of the tree — its branches and nodes — as a solid barrier.) The order of the nodes is the order in which this line passes underneath them. If you are unsure as to when you go “underneath” a node, remember that a node “to the left” always comes first. Here's the example used (slightly different tree from below) However when I do a search on google, I get a conflicting definition. For example the wikipedia example: Inorder traversal sequence: A, B, C, D, E, F, G, H, I (leftchild,rootnode,right node) But according to (my understanding of) definition #1, this should be A, B, D, C, E, F, G, I, H Can anyone clarify which definition is correct? They might be both describing different traversal methods, but happen to be using the same name. I'm having trouble believing the peer-reviewed academic text is wrong, but can't be certain.

    Read the article

  • Scene graphs and spatial partitioning structures: What do you really need?

    - by tapirath
    I've been fiddling with 2D games for awhile and I'm trying to go into 3D game development. I thought I should get my basics right first. From what I read scene graphs hold your game objects/entities and their relation to each other like 'a tire' would be the child of 'a vehicle'. It's mainly used for frustum/occlusion culling and minimizing the collision checks between the objects. Spatial partitioning structures on the other hand are used to divide a big game object (like the map) to smaller parts so that you can gain performance by only drawing the relevant polygons and again minimizing the collision checks to those polygons only. Also a spatial partitioning data structure can be used as a node in a scene graph. But... I've been reading about both subjects and I've seen a lot of "scene graphs are useless" and "BSP performance gain is irrelevant with modern hardware" kind of articles. Also some of the game engines I've checked like gameplay3d and jmonkeyengine are only using a scene graph (That also may be because they don't want to limit the developers). Whereas games like Quake and Half-Life only use spatial partitioning. I'm aware that the usage of these structures very much depend on the type of the game you're developing so for the sake of clarity let's assume the game is a FPS like Counter-Strike with some better outdoor environment capabilities (like a terrain). The obvious question is which one is needed and why (considering the modern hardware capabilities). Thank you.

    Read the article

  • On Windows 7, dir or tree can't show unicode characters, even starting cmd with cmd /U

    - by Jian Lin
    On Windows 7, dir or tree can't show unicode characters, even starting cmd with cmd /U So I would press Window Key + R to run something, and type in cmd /U so that the content might handle Unicode. And then using dir or tree /F, the content in Unicode won't show as Unicode. (in Window Explorer (file manager), the Unicode will show) Is there a way to handle it? To get Unicode characters to test your filenames, you can go to http://news.google.com/news?edchanged=1&ned=tw and you will be able to get many Unicode characters there (UTF-8)

    Read the article

  • How to Serialize Binary Tree

    - by Veljko Skarich
    I went to an interview today where I was asked to serialize a binary tree. I implemented an array-based approach where the children of node i (numbering in level-order traversal) were at the 2*i index for the left child and 2*i + 1 for the right child. The interviewer seemed more or less pleased, but I'm wondering what serialize means exactly? Does it specifically pertain to flattening the tree for writing to disk, or would serializing a tree also include just turning the tree into a linked list, say. Also, how would we go about flattening the tree into a (doubly) linked list, and then reconstructing it? Can you recreate the exact structure of the tree from the linked list? Thank you/

    Read the article

  • recursion tree and binary tree cost calculation

    - by Tony
    Hi all, I've got the following recursion: T(n) = T(n/3) + T(2n/3) + O(n) The height of the tree would be log3/2 of 2. Now the recursion tree for this recurrence is not a complete binary tree. It has missing nodes lower down. This makes sense to me, however I don't understand how the following small omega notation relates to the cost of all leaves in the tree. "... the total cost of all leaves would then be Theta (n^log3/2 of 2) which, since log3/2 of 2 is a constant strictly greater then 1, is small omega(n lg n)." Can someone please help me understand how the Theta(n^log3/2 of 2) becomes small omega(n lg n)?

    Read the article

  • SQL server - climb up in the tree structure

    - by Vytas999
    Hello. I have some sql table, named Object, which saves tree data in fields ObjectID, ParentID, and others. I have implemented recurse procedure, which select everything down by objectID from tree, like this: 1. 1.1. 1.2. 1.2.1. ... Now o need to "Climb up" - by some ObjectID i need to select everything Up, like this: 1.2.1. 1.2. 1. How i can do that? In example, my "down" procedure looks like: ALTER PROCEDURE [dbo].[Object_SelectDownByRoot_Simple] @ObjectID int AS WITH tree (ObjectID, ParentID, ObjectName, ObjectCode) AS ( SELECT ObjectID, ParentID, ObjectName, ObjectCode FROM dbo.[ObjectQ] ofs WHERE( ObjectID = @ObjectID ) UNION ALL SELECT ofs.ObjectID, ofs.ParentID, ofs.ObjectName, ofs.ObjectCode FROM dbo.[ObjectQ] ofs JOIN tree ON tree.ObjectID = ofs.ParentID ) SELECT ObjectID, ParentID, ObjectName, ObjectCode FROM tree

    Read the article

  • how do i supply data to my gwt tree

    - by molleman
    Hello, So i need to create a tree with tree items for my gwt project. i am using the composite pattern to store all the information i need to be placed within a tree. A User has a root Folder that extends Hierarchy, this root Folder then has a list of Hierarchy objects, that can be FileLocations or Folders. Trouble i am having is building my tree based on this pattern. this data is all stored using hibernate in a mysql database How would i be able to implement this as a tree in gwt. Also the tree item that i create would have to reference back to the object so i can rename or move it.

    Read the article

  • On Windows 7, dir or tree can't show unicode characters, even starting cmd with cmd /U

    - by ????
    On Windows 7, dir or tree can't show unicode characters, even starting cmd with cmd /U So I would press Window Key + R to run something, and type in cmd /U so that the content might handle Unicode. And then using dir or tree /F, the content in Unicode won't show as Unicode. (in Window Explorer (file manager), the Unicode will show) Is there a way to handle it? To get Unicode characters to test your filenames, you can go to http://news.google.com/news?edchanged=1&ned=tw and you will be able to get many Unicode characters there (UTF-8)

    Read the article

  • Pure functional bottom up tree algorithm

    - by Axel Gneiting
    Say I wanted to write an algorithm working on an immutable tree data structure that has a list of leaves as its input. It needs to return a new tree with changes made to the old tree going upwards from those leaves. My problem is that there seems to be no way to do this purely functional without reconstructing the entire tree checking at leaves if they are in the list, because you always need to return a complete new tree as the result of an operation and you can't mutate the existing tree. Is this a basic problem in functional programming that only can be avoided by using a better suited algorithm or am I missing something?

    Read the article

  • Correcting tree from messed up file tree in NTFS partition

    - by Fullmooninu
    It's a real messed situation, but I'm quite at the end of my options. It's my personal hardrive, so it's very important for me, and yes, I have no backup =( The short story: 1) I have two discs. One with Windows, and another where I had a bit of empty space at the front of the disk, so i could install Linux. The rest was occupied by a 1.8TB NTFS partition filled with data. 2) I installed Linux, and after a while realized there was not enough space for everything, so I tried using Gparted, and told it to re-size the NTFS partition, to a lesser size. 3) The system jammed. I had to reboot and broke the Resizing operation. Here's what I did to fix it: a) Rebooted into Linux Live, and used Testdisk,to deep analyze the disk, and recover the possible partitions. It found several versions of the NTFS partitions, probably made during the resizing. I told Testdisk to open every one of them, and only one could list its files. When trying to open the other options on Testdisk, it showed an error message. I assumed the one without errors, to be the correct one, and I told Testdisk to recover the partition, and write a new MBR. b) The partition had errors, and Linux has a NTFS fixing tool, used it, but the system still had errors. c) So I booted into windows and use chkdsk to correct all errors in the partition. d) Everything seems fine, but now, back in Windows, when I open one file, it opens another file, or part of another file. As in, some files took up the position of other files. What I think happened is that I recovered an old tree, and not the most current one. And that one just happened to be intact, while the most recent one was damaged. As such, the files that were moved during the failed resizing, were now, during the automatic correction, assumed wrongly to be in their correct places. So when I open a file, it tries to open another one. Radiohead - Creep.mp3 will open and it will actually be a bit from another song, or even code from a jpg. Some files seem to be all right, but others have seemed to have had their position taken by others. Anyone knows of something really powerful that can help me solve this?

    Read the article

  • Array "tree" creation from db table

    - by Tural Teyyuboglu
    Trying to create array tree for db driven navigation. Getting following errur: array_key_exists() expects exactly 2 parameters, 1 given on line if (!array_key_exists($tree[$parent]['children'][$id])) Function looks like that $tree = array(); $sql = "SELECT id, parent, name FROM menu WHERE parent ... etc.... "; $results = mysql_query($sql) or die(mysql_error()); while(list($id, $parent, $name) = mysql_fetch_assoc($results)) { $tree[$id] = array('name' => $name, 'children' => array(), 'parent' => $parent); if (!array_key_exists($tree[$parent]['children'][$id])) { $tree[$parent]['children'][$id] = $id; } } Db structure How can I fix that? Whats wrong in this function?

    Read the article

  • Most efficient Implementation a Tree in C++

    - by Topo
    I need to write a tree where each element may have any number of child elements, and because of this each branch of the tree may have any length. The tree is only going to receive elements at first and then it is going to use exclusively for iterating though it's branches in no specific order. The tree will have several million elements and must be fast but also memory efficient. My plan makes a node class to store the elements and the pointers to its children. When the tree is fully constructed, it would be transformed it to an array or something faster and if possible, loaded to the processor's cache. Construction and the search on the tree are two different problems. Can I focus on how to solve each problem on the best way individually? The construction of has to be as fast as possible but it can use memory as it pleases. Then the transformation into a format that give us speed when iterating the tree's branches. This should preferably be an array to avoid going back and forth from RAM to cache in each element of the tree. So the real question is which is the structure to implement a tree to maximize insert speed, how can I transform it to a structure that gives me the best speed and memory?

    Read the article

  • Counting leaf nodes in hierarchical tree

    - by timn
    This code fills a tree with values based their depths. But when traversing the tree, I cannot manage to determine the actual number of children. node-cnt is always 0. I've already tried node-parent-cnt but that gives me lots of warnings in Valgrind. Anyway, is the tree type I've chosen even appropriate for my purpose? #include <string.h> #include <stdio.h> #include <stdlib.h> #ifndef NULL #define NULL ((void *) 0) #endif // ---- typedef struct _Tree_Node { // data ptr void *p; // number of nodes int cnt; struct _Tree_Node *nodes; // parent nodes struct _Tree_Node *parent; } Tree_Node; typedef struct { Tree_Node root; } Tree; void Tree_Init(Tree *this) { this->root.p = NULL; this->root.cnt = 0; this->root.nodes = NULL; this->root.parent = NULL; } Tree_Node* Tree_AddNode(Tree_Node *node) { if (node->cnt == 0) { node->nodes = malloc(sizeof(Tree_Node)); } else { node->nodes = realloc( node->nodes, (node->cnt + 1) * sizeof(Tree_Node) ); } Tree_Node *res = &node->nodes[node->cnt]; res->p = NULL; res->cnt = 0; res->nodes = NULL; res->parent = node; node->cnt++; return res; } // ---- void handleNode(Tree_Node *node, int depth) { int j = depth; printf("\n"); while (j--) { printf(" "); } printf("depth=%d ", depth); if (node->p == NULL) { goto out; } printf("value=%s cnt=%d", node->p, node->cnt); out: for (int i = 0; i < node->cnt; i++) { handleNode(&node->nodes[i], depth + 1); } } Tree tree; int curdepth; Tree_Node *curnode; void add(int depth, char *s) { printf("%s: depth (%d) > curdepth (%d): %d\n", s, depth, curdepth, depth > curdepth); if (depth > curdepth) { curnode = Tree_AddNode(curnode); Tree_Node *node = Tree_AddNode(curnode); node->p = malloc(strlen(s)); memcpy(node->p, s, strlen(s)); curdepth++; } else { while (curdepth - depth > 0) { if (curnode->parent == NULL) { printf("Illegal nesting\n"); return; } curnode = curnode->parent; curdepth--; } Tree_Node *node = Tree_AddNode(curnode); node->p = malloc(strlen(s)); memcpy(node->p, s, strlen(s)); } } void main(void) { Tree_Init(&tree); curnode = &tree.root; curdepth = 0; add(0, "1"); add(1, "1.1"); add(2, "1.1.1"); add(3, "1.1.1.1"); add(4, "1.1.1.1.1"); add(2, "1.1.2"); add(0, "2"); handleNode(&tree.root, 0); }

    Read the article

  • Updating a Minimum spanning tree when a new edge is inserted

    - by Lynette
    Hello, I've been presented the following problem in University: Let G = (V, E) be an (undirected) graph with costs ce = 0 on the edges e € E. Assume you are given a minimum-cost spanning tree T in G. Now assume that a new edge is added to G, connecting two nodes v, tv € V with cost c. a) Give an efficient algorithm to test if T remains the minimum-cost spanning tree with the new edge added to G (but not to the tree T). Make your algorithm run in time O(|E|). Can you do it in O(|V|) time? Please note any assumptions you make about what data structure is used to represent the tree T and the graph G. b)Suppose T is no longer the minimum-cost spanning tree. Give a linear-time algorithm (time O(|E|)) to update the tree T to the new minimum-cost spanning tree. This is the solution I found: Let e1=(a,b) the new edge added Find in T the shortest path from a to b (BFS) if e1 is the most expensive edge in the cycle then T remains the MST else T is not the MST It seems to work but i can easily make this run in O(|V|) time, while the problem asks O(|E|) time. Am i missing something? By the way we are authorized to ask for help from anyone so I'm not cheating :D Thanks in advance

    Read the article

  • Reduced Tree View in NetBeans IDE 7.2

    - by Geertjan
    Right-click within the Projects window in NetBeans IDE 7.2 and from the "View Java Packages As" menu, you can now choose "Reduced Tree".I never really understood the difference between "Reduced Tree" and the already existing "Tree". But it makes sense when you see it. Here's Reduced Tree view: And here's Tree view, where you can see that the "actions" and "nodes" packages above each have their own top level package nodes, which takes up more space than the above: What's cool is that your selected package view is persisted across restarts of the IDE. To be complete, here's the List view, which is the third option you have in the "View Java Packages As" menu: Seems to me like the new Reduced Tree view combines the best of the Tree view with the best of the List view! Related issue: http://netbeans.org/bugzilla/show_bug.cgi?id=53192

    Read the article

  • check if a tree is a binary search tree

    - by TimeToCodeTheRoad
    I have written the following code to check if a tree is a Binary search tree. Please help me check the code: Pair p{ boolean isTrue; int min; int max; } public boo lean isBst(BNode v){ return isBST1(v).isTrue; } public Pair isBST1(BNode v){ if(v==null) return new Pair(true, INTEGER.MIN,INTEGER.MAX); if(v.left==null && v.right==null) return new Pair(true, v.data, v.data); Pair pLeft=isBST1(v.left); Pair pRight=isBST1(v.right); boolean check=pLeft.max<v.data && v.data<= pRight.min; Pair p=new Pair(); p.isTrue=check&&pLeft.isTrue&&pRight.isTrue; p.min=pLeft.min; p.max=pRight.max; return p; } Note: This function checks if a tree is a binary search tree

    Read the article

  • how does NTFS actually work with B-tree ?

    - by bakra
    To improve performance, NTFS directories use a special data management structure called a B-tree. "B-tree" concept here refers to a "tree of storage units" that hold the contents of an individual directory. What I don't understand is where on the disk is this tree stored? Its surely not created every-time we reboot...that would take lots of time. and since its a tree(dynamic Data structure) unlike arrays it will grow. so space needs to be allocated every-time it grows. so how is this "dynamic meta-data" stored ?

    Read the article

  • Accessing selected node of richfaces tree from Javascript

    - by kazanaki
    Hello This should be a very simple question. I have a richfaces tree that is rendered using JSF. When the user clicks on a node I want a javascript function to run. Nothing more nothing less. No redirects, no re-submit, no-rerender, no Ajax. Just plain old Javascript. I have seen the onselected attribute of the tree and it indeed fires a Javascript method. But of course I want to know which node was clicked. Here is what I have so far <head> <script type="text/javascript"> function documentClicked(nodeRef) { alert("Node is "+nodeRef); } </script> </head> <rich:tree switchType="client" value="#{ajaxDocumentTree.rootNode}" var="document" onselected="documentClicked()" > <rich:treeNode iconLeaf="../images/tree/doc.gif" icon="../images/tree/doc.gif"> <h:outputText value="#{doc.friendlyName}" /> </rich:treeNode> But this does not work because nodeRef is undefined. I expected that the first argument of the callback would be the selected node but this is not the case. So the question is this: How do I fire a Javascript function with the selected node from a richfaces tree?

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >