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  • Retaining expanded/contracted nodes in Word 2010's navigation pane

    - by msorens
    I have a lengthy document in Word 2010 which makes copious use of standard header styles. These display hierarchically (and very usefully!) in the navigation pane. But the list of these sections and subsections in the navigation pane is somewhat lengthy, with all nodes being expanded by default. I am now working near the end of the document so every time I open the document I contract the first so many header 1 nodes to compress the list to fit on one screen. I then have direct access to the later sections while still letting me access the earlier sections if needed. The problem is that as soon as I close the document all the customization I did in the navigation pane is lost. Is there any option or setting somewhere that will retain the expanded or contracted state of nodes in the navigation pane?

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  • Problem with building tree bottom up

    - by Esmond
    Hi, I have problems building a binary tree from the bottom up. THe input of the tree would be internal nodes of the trees with the children of this node being the leaves of the eventual tree. So initially if the tree is empty the root would be the first internal node. Afterwards, The next internal node to be added would be the new root(NR), with the old root(OR) being one of the child of NR. And so on. The problem i have is that whenever i add a NR, the children of the OR seems to be lost when i do a inOrder traversal. This is proven to be the case when i do a getSize() call which returns the same number of nodes before and after addNode(Tree,Node) Any help with resolving this problem is appreciated edited with the inclusion of node class code. both tree and node classes have the addChild methods because i'm not very sure where to put them for it to be appropriated. any comments on this would be appreciated too. The code is as follows: import java.util.*; public class Tree { Node root; int size; public Tree() { root = null; } public Tree(Node root) { this.root = root; } public static void setChild(Node parent, Node child, double weight) throws ItemNotFoundException { if (parent.child1 != null && parent.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (parent.child1 != null) { parent.child2 = child; child.parent = parent; parent.c2Weight = weight; } else { parent.child1 = child; child.parent = parent; parent.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public static Tree addNode(Tree tree, Node node) throws ItemNotFoundException { Tree tree1; if (tree.root == null) { tree.root = node; } else if (tree.root.getSeq().equals(node.getChild1().getSeq()) || tree.root.getSeq().equals(node.getChild2().getSeq())) { Node oldRoot = tree.root; oldRoot.setParent(node); tree.root = node; } else { //form a disjoint tree and merge the 2 trees tree1 = new Tree(node); tree = mergeTree(tree, tree1); } System.out.print("addNode2 = "); if(tree.root != null ) { Tree.inOrder(tree.root); } System.out.println(); return tree; } public static Tree mergeTree(Tree tree, Tree tree1) { String root = "root"; Node node = new Node(root); tree.root.setParent(node); tree1.root.setParent(node); tree.root = node; return tree; } public static int getSize(Node root) { if (root != null) { return 1 + getSize(root.child1) + getSize(root.child2); } else { return 0; } } public static boolean isEmpty(Tree Tree) { return Tree.root == null; } public static void inOrder(Node root) { if (root != null) { inOrder(root.child1); System.out.print(root.sequence + " "); inOrder(root.child2); } } } public class Node { Node child1; Node child2; Node parent; double c1Weight; double c2Weight; String sequence; boolean isInternal; public Node(String seq) { sequence = seq; child1 = null; c1Weight = 0; child2 = null; c2Weight = 0; parent = null; isInternal = false; } public boolean hasChild() { if (this.child1 == null && this.child2 == null) { this.isInternal = false; return isInternal; } else { this.isInternal = true; return isInternal; } } public String getSeq() throws ItemNotFoundException { if (this.sequence == null) { throw new ItemNotFoundException("No such node"); } else { return this.sequence; } } public void setChild(Node child, double weight) throws ItemNotFoundException { if (this.child1 != null && this.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (this.child1 != null) { this.child2 = child; this.c2Weight = weight; } else { this.child1 = child; this.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public void setParent(Node parent){ this.parent = parent; } public Node getParent() throws ItemNotFoundException { if (this.parent == null) { throw new ItemNotFoundException("This Node has no parent"); } else { return this.parent; } } public Node getChild1() throws ItemNotFoundException { if (this.child1 == null) { throw new ItemNotFoundException("There is no child1"); } else { return this.child1; } } public Node getChild2() throws ItemNotFoundException { if (this.child2 == null) { throw new ItemNotFoundException("There is no child2"); } else { return this.child2; } } }

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  • Behaviour tree code example?

    - by jokoon
    http://altdevblogaday.org/2011/02/24/introduction-to-behavior-trees/ Obviously the most interesting article I found on this website. What do you think about it ? It lacks some code example, don't you know any ? I also read that state machines are not very flexible compared to behaviour trees... On top of that I'm not sure if there is a true link between state machines and the state pattern... is there ?

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  • Navigation (controller) in Tabbar (c) in Navigation (c) (iphone)

    - by Jonathan
    I want to have a TabBar controller inside a navigation controller. So that when an item is selected on the first Navigation Controller it pushes the TabBar into view. Inside this tabbar on the first tab is another navigation controller. However I only want one navigation bar. I've come up with 2 ways but not sure which way is better (Ie more acceptable etc)? 1) The first navigation controller isn't actually a navigation controller but to the user it looks like one. So when a cell is selected on it's table view the first navC's view is removed from the superview and the TabBarC's view is added, animation would have to be done manually. 2)The first NavC is actually a NavC and when an item is selected and the TabBar is pushed on to the screen the first NavC's navigationbar is hidden so that the first tab's navigationBar is the only nav bar on screen.

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  • How to procedurally (create) grow an artistic (2D) tree in real-time (L-System?).

    - by lalan
    Recently I programmed an L-system module, It got me interested further. I am a Plants vs Zombies junkie as well, really liked the concept of Tree of Wisdom. Would love to create similar procedural art just for fun and learn more. Question: How should I approach the process of creating an artistic tree (2d perhaps with fixed camera/perspective) dynamically? Ideally I would like to start with a plant (only a stem with a leaf) and grow it dynamically using some influence (input/user action) over its structure. These influences may result in different type of branching, curves in branches, its spread, location of fruits, color of flowers, etc. Want it to be really full of life/spirit. :) Plants vs Zombies: Tree of wisdom It would be great to dynamically grow a similar tree, but with lot more variation and animations happening. My Background: Student / Programmer, have used few game engines (Ogre3d, cocos2d, unity). Haven't really programmed directly using openGL, trying to fix that :). I am ready to spend considerable time, Please let me know about the APIs? and how would an expert like you would take on this problem? Why 2D? I think it's easier to solve the problem only considering 2 dimensions. Artistic inspirations: Only the tree, with fruits and leaves, without the shrubs at the bottom The large tree (visible branches, green leaves, flowers, fruits, etc) on the left, behind monkey. PixelJunk's Eden (Art style inspiration). Procedurally Generated Apple Tree using Fractals Please let me know if it was easy for you to understand the question, I may elaborate further. I hope a discussion of various approach would be helpful for everyone. You guys are awesome.

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  • Upgrade tree to 1.6?

    - by Pureferret
    I'm trying to upgrade my version of tree to 1.6 on ubuntu 12.04. I've d'loaded, ran make and make install in the terminal using the sudo command. ~/tree-1.6.0$ sudo make make: Nothing to be done for `all'. I've already run sudo make here ~/tree-1.6.0$ sudo make install install -d /usr/bin install -d /usr/man/man1 if [ -e tree ]; then \ install -s tree /usr/bin/tree; \ fi install doc/tree.1 /usr/man/man1/tree.1 What's this output though? It's not updated. I've checked the man page, and -du doesn't work. How am I supposed to update tree if not via the terminal?

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  • Configure spanning tree from HP to Cisco hardware

    - by Tim Brigham
    I have three switches I'd like to configure in a loop - a Cisco stack (3750s) and two HP 2900 series. Each is connected to the next with a 10 gig backplane of one form or another. How do I configure the spanning tree on these systems to make this function correctly? From the documents I've looked at it looks like I need to set both sets of hardware to use MST mode but I'm not sure past that point. The trunking, etc is all set up as needed. HP Switch 1 A4 connected to Cisco 1/0/1. HP Switch 2 B2 connected to Cisco 2/0/1. HP Switch 1 A2 connected to HP Switch 2 A1. HP Switch 1 show spanning-tree Multiple Spanning Tree (MST) Information STP Enabled : Yes Force Version : MSTP-operation IST Mapped VLANs : 1-4094 Switch MAC Address : 0021f7-126580 Switch Priority : 32768 Max Age : 20 Max Hops : 20 Forward Delay : 15 Topology Change Count : 352,485 Time Since Last Change : 2 secs CST Root MAC Address : 0018ba-c74268 CST Root Priority : 1 CST Root Path Cost : 200000 CST Root Port : 1 IST Regional Root MAC Address : 0021f7-126580 IST Regional Root Priority : 32768 IST Regional Root Path Cost : 0 IST Remaining Hops : 20 Root Guard Ports : TCN Guard Ports : BPDU Protected Ports : BPDU Filtered Ports : PVST Protected Ports : PVST Filtered Ports : | Prio | Designated Hello Port Type | Cost rity State | Bridge Time PtP Edge ----- --------- + --------- ---- ---------- + ------------- ---- --- ---- ... A1 | Auto 128 Disabled | A2 10GbE-CX4 | 2000 128 Forwarding | 0021f7-126580 2 Yes No A3 10GbE-CX4 | Auto 128 Disabled | A4 10GbE-SR | 2000 128 Forwarding | 0021f7-126580 2 Yes No HP Switch 2 show spanning-tree Multiple Spanning Tree (MST) Information STP Enabled : Yes Force Version : MSTP-operation IST Mapped VLANs : 1-4094 Switch MAC Address : 0024a8-cd6000 Switch Priority : 32768 Max Age : 20 Max Hops : 20 Forward Delay : 15 Topology Change Count : 19,623 Time Since Last Change : 32 secs CST Root MAC Address : 0018ba-c74268 CST Root Priority : 1 CST Root Path Cost : 202000 CST Root Port : A1 IST Regional Root MAC Address : 0024a8-cd6000 IST Regional Root Priority : 32768 IST Regional Root Path Cost : 0 IST Remaining Hops : 20 Root Guard Ports : TCN Guard Ports : BPDU Protected Ports : BPDU Filtered Ports : PVST Protected Ports : PVST Filtered Ports : | Prio | Designated Hello Port Type | Cost rity State | Bridge Time PtP Edge ----- --------- + --------- ---- ---------- + ------------- ---- --- ---- ... A1 10GbE-CX4 | 2000 128 Forwarding | 0021f7-126580 2 Yes No A2 10GbE-CX4 | Auto 128 Disabled | B1 SFP+SR | 2000 128 Blocking | a44c11-a67c80 2 Yes No B2 | Auto 128 Disabled | Cisco Stack 1 show spanning-tree ... (additional VLANs) VLAN0100 Spanning tree enabled protocol ieee Root ID Priority 1 Address 0018.bac7.426e Cost 2 Port 107 (TenGigabitEthernet2/1/1) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32868 (priority 32768 sys-id-ext 100) Address a44c.11a6.7c80 Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Te1/1/1 Desg FWD 2 128.53 P2p Te2/1/1 Root FWD 2 128.107 P2p

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  • Gson Deserialize to Java Tree

    - by MountainX
    I need to deserialize some JSON to a Java tree structure that contains TreeNodes and NodeData. TreeNodes are thin wrappers around NodeData. I'll provide the JSON and the classes below. I have looked at the usual Gson help sources, including here, but I can't seem to come up with the solution. Serialization works fine with Gson. The JSON below was produced by Gson. But deserialization is the problem I need help with. Can someone show me how to write the deserializer (or suggest an alternative approach using Gson best practices)? Here is my JSON. The "data" element corresponds to class NodeData, and the "subList" JSON element corresponds to Java class TreeNode. { "data": { "version": "032", "name": "root", "path": "/", "id": "1", "parentId": "0", "toolTipText": "rootNode" }, "subList": [ { "data": { "version": "032", "name": "level1", "labelText": "Some Label Text at Level1", "path": "/root", "id": "2", "parentId": "1", "toolTipText": "a tool tip for level1" }, "subList": [ { "data": { "version": "032", "name": "level1_1", "labelText": "Label level1_1", "path": "/root/level1", "id": "3", "parentId": "2", "toolTipText": "ToolTipText for level1_1" } }, { "data": { "version": "032", "name": "level1_2", "labelText": "Label level1_2", "path": "/root/level1", "id": "4", "parentId": "2", "toolTipText": "ToolTipText for level1_2" } } ] }, { "data": { "version": "032", "name": "level2", "path": "/root", "id": "5", "parentId": "1", "toolTipText": "ToolTipText for level2" }, "subList": [ { "data": { "version": "032", "name": "level2_1", "labelText": "Label level2_1", "path": "/root/level2", "id": "6", "parentId": "5", "toolTipText": "ToolTipText for level2_1" }, "subList": [ { "data": { "version": "032", "name": "level2_1_1", "labelText": "Label level2_1_1", "path": "/root/level2/level2_1", "id": "7", "parentId": "6", "toolTipText": "ToolTipText for level2_1_1" } } ] } ] } ] } Here are the Java classes: public class Tree { private TreeNode rootElement; private HashMap<String, TreeNode> indexById; private HashMap<String, TreeNode> indexByKey; private long nextAvailableID = 0; public Tree() { indexById = new HashMap<String, TreeNode>(); indexByKey = new HashMap<String, TreeNode>(); } public long getNextAvailableID() { return this.nextAvailableID; } ... [snip] ... } public class TreeNode { private Tree tree; private NodeData data; public List<TreeNode> subList; private HashMap<String, TreeNode> indexById; private HashMap<String, TreeNode> indexByKey; //this default ctor is used only for Gson deserialization public TreeNode() { this.tree = new Tree(); indexById = tree.getIdIndex(); indexByKey = tree.getKeyIndex(); this.makeRoot(); tree.setRootElement(this); } //makes this node the root node. Calling this obviously has side effects. public NodeData makeRoot() { NodeData rootProp = new NodeData(TreeFactory.version, "example", "rootNode"); String nextAvailableID = getNextAvailableID(); if (!nextAvailableID.equals("1")) { throw new IllegalStateException(); } rootProp.setId(nextAvailableID); rootProp.setParentId("0"); rootProp.setKeyPathOnly("/"); rootProp.setSchema(tree); this.data = rootProp; rootProp.setNode(this); indexById.put(rootProp.getId(), this); indexByKey.put(rootProp.getKeyFullName(), this); return rootProp; } ... [snip] ... } public class NodeData { protected static Tree tree; private LinkedHashMap<String, String> keyValMap; protected String version; protected String name; protected String labelText; protected String path; protected String id; protected String parentId; protected TreeNode node; protected String toolTipText;//tool tip or help string protected String imagePath;//for things like images; not persisted to properties protected static final String delimiter = "/"; //this default ctor is used only for Gson deserialization public NodeData() { this("NOT_SET", "NOT_SET", "NOT_SET"); } ... [snip] ... } Side note: The tree data structure is a bit strange, as it includes indexes. Obviously, this isn't a typical search tree. In fact, the tree is used mainly to create a hierarchical path element (String) in each NodeData element. (Example: "path": "/root/level2/level2_1".) The indexes are actually used for NodeData retrieval.

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  • How does a BSP tree work for Z sorting?

    - by Jenko
    I'm developing a 3D engine in software, and so I must compute Z sorting manually. I'm currently using the painters algorithm to sort triangles and then drawing them back-to-front. This causes artifacts that I'm trying to correct. Would using a dynamic BSP-tree ensure "correct Z sorting" of triangles? Why? Because the bounding volumes of triangles would be similar? Since I would have a single "world" BSP tree, would I have to remove and re-add any moved/scaled/rotated object into the tree? Is it possible to add triangles into a BSP tree without the expensive cutting process? Why do you need to cut triangles on the axis planes anyway? Is it faster to traverse a BSP tree from any angle, than to sort all tris each draw like the painters algorithm?

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  • Implement Tree/Details With Taskflow Regions Using EJB

    - by Deepak Siddappa
    This article describes on Display Tree/Details using taskflow regions.Use Case DescriptionLet us take scenario where we need to display Tree/Details, left region contains category hierarchy with items listed in a tree structure (ex:- Region-Countries-Locations-Departments in tree format) and right region contains the Employees list.In detail, Here User may drills down through categories using a tree until Employees are listed. Clicking the tree node name displays Employee list in the adjacent pane related to particular tree node. Implementation StepsThe script for creating the tables and inserting the data required for this application CreateSchema.sql Lets create a Java EE Web Application with Entities based on Regions, Countries, Locations, Departments and Employees table. Create a Stateless Session Bean and data control for the Stateless Session Bean. Add the below code to the session bean and expose the method in local/remote interface and generate a data control for that.Note:- Here in the below code "em" is a EntityManager. public List<Employees> empFilteredByTreeNode(String treeNodeType, String paramValue) { String queryString = null; try { if (treeNodeType == "null") { queryString = "select * from Employees emp ORDER BY emp.employee_id ASC"; } else if (Pattern.matches("[a-zA-Z]+[_]+[a-zA-Z]+[_]+[[0-9]+]+", treeNodeType)) { queryString = "select * from employees emp INNER JOIN departments dept\n" + "ON emp.department_id = dept.department_id JOIN locations loc\n" + "ON dept.location_id = loc.location_id JOIN countries cont\n" + "ON loc.country_id = cont.country_id JOIN regions reg\n" + "ON cont.region_id = reg.region_id and reg.region_name = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.contains("regionsFindAll_bc_countriesList_1")) { queryString = "select * from employees emp INNER JOIN departments dept \n" + "ON emp.department_id = dept.department_id JOIN locations loc \n" + "ON dept.location_id = loc.location_id JOIN countries cont \n" + "ON loc.country_id = cont.country_id and cont.country_name = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.contains("regionsFindAll_bc_locationsList_1")) { queryString = "select * from employees emp INNER JOIN departments dept ON emp.department_id = dept.department_id JOIN locations loc ON dept.location_id = loc.location_id and loc.city = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.trim().contains("regionsFindAll_bc_departmentsList_1")) { queryString = "select * from Employees emp INNER JOIN Departments dept ON emp.DEPARTMENT_ID = dept.DEPARTMENT_ID and dept.DEPARTMENT_NAME = '" + paramValue + "'"; } } catch (NullPointerException e) { System.out.println(e.getMessage()); } return em.createNativeQuery(queryString, Employees.class).getResultList(); } In the ViewController project, create two ADF taskflow with page Fragments and name them as FirstTaskflow and SecondTaskflow respectively. Open FirstTaskflow,from component palette drop view(Page Fragment) name it as TreeList.jsff. Open SeconfTaskflow, from component palette drop view(Page Fragment) name it as EmpList.jsff and create two paramters in its overview parameters tab as shown in below image. Open TreeList.jsff , from data control palette drop regionsFindAll->Tree as ADF Tree. In Edit Tree Binding dialog, for Tree Level Rules select the display attributes as follows:-model.Regions - regionNamemodel.Countries - countryNamemodel.Locations - citymodel.Departments - departmentName In structure panel, click on af:Tree - t1 and select selectionListener with edit property. Create a "TreeBean" managed bean with scope as "session" as shown in below Image. Create new method as getTreeNodeSelectedValue and click ok. Open TreeBean managed bean and add the below code: private String treeNodeType; private String paramValue; public void getTreeNodeSelectedValue(SelectionEvent selectionEvent) { RichTree tree = (RichTree)selectionEvent.getSource(); RowKeySet addedSet = selectionEvent.getAddedSet(); Iterator i = addedSet.iterator(); TreeModel model = (TreeModel)tree.getValue(); model.setRowKey(i.next()); JUCtrlHierNodeBinding node = (JUCtrlHierNodeBinding)tree.getRowData(); //oracle.jbo.Row Row rw = node.getRow(); Object selectedTreeNode = node.getAttribute(0); Object treeListType = node.getBindings(); String treeNodeType = treeListType.toString(); this.setParamValue(selectedTreeNode.toString()); this.setTreeNodeType(treeNodeType); } public void setTreeNodeType(String treeNodeType) { this.treeNodeType = treeNodeType; } public String getTreeNodeType() { return treeNodeType; } public void setParamValue(String paramValue) { this.paramValue = paramValue; } public String getParamValue() { return paramValue; }<br /> Open EmpList.jsff , from data control palette drop empFilteredByTreeNode->Employees->Table as ADF Read-only Table. After selecting the  Employees result set, in Edit Action Binding dialog window pass the pageFlowScope parameters as shown in below Image. In empList.jsff page, click Binding tab and click on Create Executable binding and select Invoke action and follow as shown in below image. Edit executeEmpFiltered invoke action properties and set the Refresh to ifNeeded, So when ever the page needs the method will be executed. Create Main.jspx page with page template as Oracle Three Column Layout. Drop FirstTaskflow as Region in start facet and drop SecondTaskflow as Region in center facet, Edit task Flow Binding dialog window pass the Input Paramters as shown in below Image. Run the Main.jspx, tree will be displayed in left region and emp details will displyaed on the right region. Click on the Americas in tree node, all emp related to the Americas related will be displayed. Click on Americas->United States of America->South San Francisco->Accounting, only employee belongs to the Accounting department will be displayed.

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  • public family tree

    - by Remus Rigo
    Hi all Does anyone know a ancestry site that allows you to create a public profile or tree, so that other visitors can see your family tree. In all sites that I have found (dynastree.com, familylink.com, ancestry.com, genebase.com), if someone wants to see your family tree, they must be members or register. thanks

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  • Visualizing Parallel Branch and Bound Tree Exploration

    - by Akhil
    I have written a parallel program that does a depth first branch and bound exploration of a tree. I can dump the id's (id's are like this 0, 00, 01, 0000, 0001, etc.) of the nodes at frequent intervals to know the frontier of the tree that is being explored at that instant in the tree. The challenge is to visualize the tree exploration with time. Any ideas? e.g. I can draw trees(e.g. using graphViz) at different times and create a movie out of it. Looking for ideas to facilitate this visualization - some better ways to do so or easy tools that can help me make the visualization

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  • 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.

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  • 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?

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  • 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?

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  • Dojo JsonRest store and dijit.Tree

    - by user1427712
    I'm having a some problem making JSonRest store and dijit.Tree with ForestModel. I've tried some combination of JsonRestStore and json data format following many tips on the web, with no success. At the end, taking example form here http://blog.respondify.se/2011/09/using-dijit-tree-with-the-new-dojo-object-store/ I've made up this simple page (I'm using dojotolkit 1.7.2) <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title>Tree Model Explorer</title> <script type="text/javascript"> djConfig = { parseOnLoad : true, isDebug : true, } </script> <script type="text/javascript" djConfig="parseOnLoad: true" src="lib/dojo/dojo.js"></script> <script type="text/javascript"> dojo.require("dojo.parser"); dojo.require("dijit.Tree"); dojo.require("dojo.store.JsonRest"); dojo.require("dojo.data.ObjectStore"); dojo.require("dijit.tree.ForestStoreModel"); dojo.addOnLoad(function() { var objectStore = new dojo.store.JsonRest({ target : "test.json", labelAttribute : "name", idAttribute: "id" }); var dataStore = new dojo.data.ObjectStore({ objectStore : objectStore }); var treeModel = new dijit.tree.ForestStoreModel({ store : dataStore, deferItemLoadingUntilExpand : true, rootLabel : "Subjects", query : { "id" : "*" }, childrenAttrs : [ "children" ] }); var tree = new dijit.Tree({ model : treeModel }, 'treeNode'); tree.startup(); }); </script> </head> <body> <div id="treeNode"></div> </body> </html> My rest service responds the following json { data: [ { "id": "PippoId", "name": "Pippo", "children": [] }, { "id": "PlutoId", "name": "Pluto", "children": [] }, { "id": "PaperinoId", "name": "Paperino", "children": [] } ]} I've tried also with the following response (actually my final intention n is to use lazy loading for the tree) { data: [ { "id": "PippoId", "name": "Pippo", "$ref": "author0", "children": true }, { "id": "PlutoId", "name": "Pluto", "$ref": "author1", "children": true }, { "id": "PaperinoId", "name": "Paperino", "$ref": "author2", "children": true } ]} Neither of the two works. I see no error message in firebug. I simply see the root "Subject" on the page. Thanks to anybody could help in some way.

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  • 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,[])]

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  • Zend Navigation add extra id

    - by lander
    I made a menu with zend_navigation, using an ini file. That work well. If I look at the code, i see that zend has added a class, to the ul element. With a name that i use in my ini file. But now I want to add also a id to the ul element. How can I do this? protected function _initNavigation() { $this->bootstrap('layout'); $layout = $this->getResource('layout'); $view = $layout->getView(); $iniOptions = array('allowModifications' => true); $config = new Zend_Config_Ini(APPLICATION_PATH . '/configs/navigation.ini', 'nav', $iniOptions); $config->id = 1; $navigation = new Zend_Navigation($config->navigation); $view->navigation($navigation); }

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  • Android google.navigation Intent Modes?

    - by Patrick Mahoney
    I'm currently developing an application that will launch a navigation intent. I know that this isn't an official API, but it works perfectly the way I want it to. I allow the user to select driving, walking, and bus navigation to a location. The intent to launch directly into Google Maps Navigation looks like this: google.navigation:ll= + a latitude and longitude, then + &mode= then your mode of transportation. For example, to navigate using walking directions to a certain area: google.navigation:ll=blah,blah&mode=w Driving is default, or &mode=d, and biking is &mode=b, but I can't figure out bus (public transit). Has anyone done this before? Thanks! Edit: So far, I've found that mode=public gives bike directions, mode=transit gives driving, frustratingly, mode=bus also returns bike directions.

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  • 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.

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  • Implementing rounded corners on slide down navigation menu

    - by Nick
    I am working on the slide down menu you can see here. I have rounded corners on both ul#navigation and ul.subnavigation. When the submenu slides down it is possible to see the border at the bottom of ul.subnavigation overlap with the content of ul#navigation, when I would like it to slide down smoothly, without the 'flicker'. I am aware that this issue is caused by the rounded corners. I need ul.subnavigation to cover the rounded corners at the bottom of ul#navigation when the menu drops down, without seeing the double border-bottom issue. I hope this is clear! Code is below. Thanks, Nick HTML <ul id="navigation"> <li class="dropdown"><a href="#">menu</a> <ul class="sub_navigation"> <li><a href="#">home</a></li> <li><a href="#">help</a></li> <li><a href="#">disable tips</a></li> </ul> </li> </ul> JQUERY $('.dropdown').hover(function() { $(this).find('.sub_navigation').slideToggle(); });? CSS ul#navigation, ul.sub_navigation { margin:0; padding:0; list-style-type:none; min-width:100px; background-color: white; font-size:15px; font-family: Trebuchet MS; text-align: center; -khtml-border-radius: 0 0 5px 5px; -moz-border-radius: 0 0 5px 5px; -webkit-border-radius: 0 0 5px 5px; border-radius: 0 0 5px 5px; border:1px black solid; border-top:none; } ul.sub_navigation { margin-left:-1px; position: absolute; top:28px; } ul#navigation { float:left; position:absolute; top:0; } ul#navigation li { float:left; min-width:100px; } ul.sub_navigation { position:absolute; display:none; } ul.sub_navigation li { clear:both; } a, a:active, a:visited { display:block; padding:7px; }

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  • 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.

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  • 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)

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  • 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/

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  • 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)?

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