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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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• Silverlight Tree View with Multiple Levels

  - by psheriff

  There are many examples of the Silverlight Tree View that you will find on the web, however, most of them only show you how to go to two levels. What if you have more than two levels? This is where understanding exactly how the Hierarchical Data Templates works is vital. In this blog post, I am going to break down how these templates work so you can really understand what is going on underneath the hood. To start, let’s look at the typical two-level Silverlight Tree View that has been hard coded with the values shown below: <sdk:TreeView>  <sdk:TreeViewItem Header="Managers">    <TextBlock Text="Michael" />    <TextBlock Text="Paul" />  </sdk:TreeViewItem>  <sdk:TreeViewItem Header="Supervisors">    <TextBlock Text="John" />    <TextBlock Text="Tim" />    <TextBlock Text="David" />  </sdk:TreeViewItem></sdk:TreeView> Figure 1 shows you how this tree view looks when you run the Silverlight application. Figure 1: A hard-coded, two level Tree View. Next, let’s create three classes to mimic the hard-coded Tree View shown above. First, you need an Employee class and an EmployeeType class. The Employee class simply has one property called Name. The constructor is created to accept a “name” argument that you can use to set the Name property when you create an Employee object. public class Employee{  public Employee(string name)  {    Name = name;  }   public string Name { get; set; }} Finally you create an EmployeeType class. This class has one property called EmpType and contains a generic List<> collection of Employee objects. The property that holds the collection is called Employees. public class EmployeeType{  public EmployeeType(string empType)  {    EmpType = empType;    Employees = new List<Employee>();  }   public string EmpType { get; set; }  public List<Employee> Employees { get; set; }} Finally we have a collection class called EmployeeTypes created using the generic List<> class. It is in the constructor for this class where you will build the collection of EmployeeTypes and fill it with Employee objects: public class EmployeeTypes : List<EmployeeType>{  public EmployeeTypes()  {    EmployeeType type;            type = new EmployeeType("Manager");    type.Employees.Add(new Employee("Michael"));    type.Employees.Add(new Employee("Paul"));    this.Add(type);     type = new EmployeeType("Project Managers");    type.Employees.Add(new Employee("Tim"));    type.Employees.Add(new Employee("John"));    type.Employees.Add(new Employee("David"));    this.Add(type);  }} You now have a data hierarchy in memory (Figure 2) which is what the Tree View control expects to receive as its data source. Figure 2: A hierachial data structure of Employee Types containing a collection of Employee objects. To connect up this hierarchy of data to your Tree View you create an instance of the EmployeeTypes class in XAML as shown in line 13 of Figure 3. The key assigned to this object is “empTypes”. This key is used as the source of data to the entire Tree View by setting the ItemsSource property as shown in Figure 3, Callout #1. Figure 3: You need to start from the bottom up when laying out your templates for a Tree View. The ItemsSource property of the Tree View control is used as the data source in the Hierarchical Data Template with the key of employeeTypeTemplate. In this case there is only one Hierarchical Data Template, so any data you wish to display within that template comes from the collection of Employee Types. The TextBlock control in line 20 uses the EmpType property of the EmployeeType class. You specify the name of the Hierarchical Data Template to use in the ItemTemplate property of the Tree View (Callout #2). For the second (and last) level of the Tree View control you use a normal <DataTemplate> with the name of employeeTemplate (line 14). The Hierarchical Data Template in lines 17-21 sets its ItemTemplate property to the key name of employeeTemplate (Line 19 connects to Line 14). The source of the data for the <DataTemplate> needs to be a property of the EmployeeTypes collection used in the Hierarchical Data Template. In this case that is the Employees property. In the Employees property there is a “Name” property of the Employee class that is used to display the employee name in the second level of the Tree View (Line 15). What is important here is that your lowest level in your Tree View is expressed in a <DataTemplate> and should be listed first in your Resources section. The next level up in your Tree View should be a <HierarchicalDataTemplate> which has its ItemTemplate property set to the key name of the <DataTemplate> and the ItemsSource property set to the data you wish to display in the <DataTemplate>. The Tree View control should have its ItemsSource property set to the data you wish to display in the <HierarchicalDataTemplate> and its ItemTemplate property set to the key name of the <HierarchicalDataTemplate> object. It is in this way that you get the Tree View to display all levels of your hierarchical data structure. Three Levels in a Tree View Now let’s expand upon this concept and use three levels in our Tree View (Figure 4). This Tree View shows that you now have EmployeeTypes at the top of the tree, followed by a small set of employees that themselves manage employees. This means that the EmployeeType class has a collection of Employee objects. Each Employee class has a collection of Employee objects as well. Figure 4: When using 3 levels in your TreeView you will have 2 Hierarchical Data Templates and 1 Data Template. The EmployeeType class has not changed at all from our previous example. However, the Employee class now has one additional property as shown below: public class Employee{  public Employee(string name)  {    Name = name;    ManagedEmployees = new List<Employee>();  }   public string Name { get; set; }  public List<Employee> ManagedEmployees { get; set; }} The next thing that changes in our code is the EmployeeTypes class. The constructor now needs additional code to create a list of managed employees. Below is the new code. public class EmployeeTypes : List<EmployeeType>{  public EmployeeTypes()  {    EmployeeType type;    Employee emp;    Employee managed;     type = new EmployeeType("Manager");    emp = new Employee("Michael");    managed = new Employee("John");    emp.ManagedEmployees.Add(managed);    managed = new Employee("Tim");    emp.ManagedEmployees.Add(managed);    type.Employees.Add(emp);     emp = new Employee("Paul");    managed = new Employee("Michael");    emp.ManagedEmployees.Add(managed);    managed = new Employee("Sara");    emp.ManagedEmployees.Add(managed);    type.Employees.Add(emp);    this.Add(type);     type = new EmployeeType("Project Managers");    type.Employees.Add(new Employee("Tim"));    type.Employees.Add(new Employee("John"));    type.Employees.Add(new Employee("David"));    this.Add(type);  }} Now that you have all of the data built in your classes, you are now ready to hook up this three-level structure to your Tree View. Figure 5 shows the complete XAML needed to hook up your three-level Tree View. You can see in the XAML that there are now two Hierarchical Data Templates and one Data Template. Again you list the Data Template first since that is the lowest level in your Tree View. The next Hierarchical Data Template listed is the next level up from the lowest level, and finally you have a Hierarchical Data Template for the first level in your tree. You need to work your way from the bottom up when creating your Tree View hierarchy. XAML is processed from the top down, so if you attempt to reference a XAML key name that is below where you are referencing it from, you will get a runtime error. Figure 5: For three levels in a Tree View you will need two Hierarchical Data Templates and one Data Template. Each Hierarchical Data Template uses the previous template as its ItemTemplate. The ItemsSource of each Hierarchical Data Template is used to feed the data to the previous template. This is probably the most confusing part about working with the Tree View control. You are expecting the content of the current Hierarchical Data Template to use the properties set in the ItemsSource property of that template. But you need to look to the template lower down in the XAML to see the source of the data as shown in Figure 6. Figure 6: The properties you use within the Content of a template come from the ItemsSource of the next template in the resources section. Summary Understanding how to put together your hierarchy in a Tree View is simple once you understand that you need to work from the bottom up. Start with the bottom node in your Tree View and determine what that will look like and where the data will come from. You then build the next Hierarchical Data Template to feed the data to the previous template you created. You keep doing this for each level in your Tree View until you get to the last level. The data for that last Hierarchical Data Template comes from the ItemsSource in the Tree View itself. NOTE: You can download the sample code for this article by visiting my website at http://www.pdsa.com/downloads. Select “Tips & Tricks”, then select “Silverlight TreeView with Multiple Levels” from the drop down list.

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• AVL tree vs. B-tree

  - by Phenom

  How is an AVL tree different from a B-tree?

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• Finding if a Binary Tree is a Binary Search Tree

  - by dharam

  Today I had an interview where I was asked to write a program which takes a Binary Tree and returns true if it is also a Binary Search Tree otherwise false. My Approach1: Perform an inroder traversal and store the elements in O(n) time. Now scan through the array/list of elements and check if element at ith index is greater than element at (i+1)th index. If such a condition is encountered, return false and break out of the loop. (This takes O(n) time). At the end return true. But this gentleman wanted me to provide an efficient solution. I tried but I was unsuccessfult, because to find if it is a BST I have to check each node. Moreover he was pointing me to think over recusrion. My Approach 2: A BT is a BST if for any node N N-left is < N and N-right N , and the INorder successor of left node of N is less than N and the inorder successor of right node of N is greater than N and the left and right subtrees are BSTs. But this is going to be complicated and running time doesn't seem to be good. Please help if you know any optimal solution. Thanks.

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• New to AVL tree implementation.

  - by nn

  I am writing a sliding window compression algorithm (LZ77) that searches for phrases in a "moving" dictionary. So far I have written a BST where each node is stored in an array and it's index in the array is also the value of the starting position in the window itself. I am now looking at transforming the BST to an AVL tree. I am a little confused at the sample implementations I have seen. Some only appear to store the balance factors whereas others store the height of each tree. Are there any performance advantage/disadvantages of storing the height and/or balance factor for each node? Apologies if this is a very simple question, but I'm still not visualizing how I want to restructure my BST to implement height balancing. Thanks.

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• Insert Record by Drag & Drop from ADF Tree to ADF Tree Table

  - by arul.wilson(at)oracle.com

  If you want to create record based on the values Dragged from ADF Tree and Dropped on a ADF Tree Table, then here you go.UseCase DescriptionUser Drags a tree node from ADF Tree and Drops it on a ADF Tree Table node. A new row gets added in the Tree Table based on the source tree node, subsequently a record gets added to the database table on which Tree table in based on.Following description helps to achieve this using ADF BC.Run the DragDropSchema.sql to create required tables.Create Business Components from tables (PRODUCTS, COMPONENTS, SUB_COMPONENTS, USERS, USER_COMPONENTS) created above.Add custom method to App Module Impl, this method will be used to insert record from view layer.   public String createUserComponents(String p_bugdbId, String p_productId, String p_componentId, String p_subComponentId){    Row newUserComponentsRow = this.getUserComponentsView1().createRow();    try {      newUserComponentsRow.setAttribute("Bugdbid", p_bugdbId);      newUserComponentsRow.setAttribute("ProductId", new oracle.jbo.domain.Number(p_productId));      newUserComponentsRow.setAttribute("Component1", p_componentId);      newUserComponentsRow.setAttribute("SubComponent", p_subComponentId);    } catch (Exception e) {        e.printStackTrace();        return "Failure";    }        return "Success";  }Expose this method to client interface.To display the root node we need a custom VO which can be achieved using below query. SELECT Users.ACTIVE, Users.BUGDB_ID, Users.EMAIL, Users.FIRSTNAME, Users.GLOBAL_ID, Users.LASTNAME, Users.MANAGER_ID, Users.MANAGER_PRIVILEGEFROM USERS UsersWHERE Users.MANAGER_ID is NULLCreate VL between UsersView and UsersRootNodeView VOs.Drop ProductsView from DC as ADF Tree to jspx page.Add Tree Level Rule based on ComponentsView and SubComponentsView.Drop UsersRootNodeView as ADF Tree TableAdd Tree Level Rules based on UserComponentsView and UsersView.Add DragSource to ADF Tree and CollectionDropTarget to ADF Tree Table respectively.Bind CollectionDropTarget's DropTarget to backing bean and implement method of signature DnDAction (DropEvent), this method gets invoked when Tree Table encounters a drop action, here details required for creating new record are captured from the drag source and passed to 'createUserComponents' method. public DnDAction onTreeDrop(DropEvent dropEvent) {      String newBugdbId = "";      String msgtxt="";            try {          // Getting the target node bugdb id          Object serverRowKey = dropEvent.getDropSite();          if (serverRowKey != null) {                  //Code for Tree Table as target              String dropcomponent = dropEvent.getDropComponent().toString();              dropcomponent = (String)dropcomponent.subSequence(0, dropcomponent.indexOf("["));              if (dropcomponent.equals("RichTreeTable")){                RichTreeTable richTreeTable = (RichTreeTable)dropEvent.getDropComponent();                richTreeTable.setRowKey(serverRowKey);                int rowIndexTreeTable = richTreeTable.getRowIndex();                //Drop Target Logic                if (((JUCtrlHierNodeBinding)richTreeTable.getRowData(rowIndexTreeTable)).getAttributeValue()==null) {                  //Get Parent                  newBugdbId = (String)((JUCtrlHierNodeBinding)richTreeTable.getRowData(rowIndexTreeTable)).getParent().getAttributeValue();                } else {                  if (isNum(((JUCtrlHierNodeBinding)richTreeTable.getRowData(rowIndexTreeTable)).getAttributeValue().toString())) {                    //Get Parent's parent                              newBugdbId = (String)((JUCtrlHierNodeBinding)richTreeTable.getRowData(rowIndexTreeTable)).getParent().getParent().getAttributeValue();                  } else{                      //Dropped on USER                                          newBugdbId = (String)((JUCtrlHierNodeBinding)richTreeTable.getRowData(rowIndexTreeTable)).getAttributeValue();                  }                  }              }           }                     DataFlavor<RowKeySet> df = DataFlavor.getDataFlavor(RowKeySet.class);          RowKeySet droppedValue = dropEvent.getTransferable().getData(df);            Object[] keys = droppedValue.toArray();          Key componentKey = null;          Key subComponentKey = null;           // binding for createUserComponents method defined in AppModuleImpl class  to insert record in database.                      operationBinding = bindings.getOperationBinding("createUserComponents");            // get the Product, Component, Subcomponent details and insert to UserComponents table.          // loop through the keys if more than one comp/subcomponent is select.                   for (int i = 0; i < keys.length; i++) {                  System.out.println("in for :"+i);              List list = (List)keys[i];                  System.out.println("list "+i+" : "+list);              System.out.println("list size "+list.size());              if (list.size() == 1) {                                // we cannot drag and drop  the highest node !                                msgtxt="You cannot drop Products, please drop Component or SubComponent from the Tree.";                  System.out.println(msgtxt);                                this.showInfoMessage(msgtxt);              } else {                  if (list.size() == 2) {                    // were doing the first branch, in this case all components.                    componentKey = (Key)list.get(1);                    Object[] droppedProdCompValues = componentKey.getAttributeValues();                    operationBinding.getParamsMap().put("p_bugdbId",newBugdbId);                    operationBinding.getParamsMap().put("p_productId",droppedProdCompValues[0]);                    operationBinding.getParamsMap().put("p_componentId",droppedProdCompValues[1]);                    operationBinding.getParamsMap().put("p_subComponentId","ALL");                    Object result = operationBinding.execute();              } else {                    subComponentKey = (Key)list.get(2);                    Object[] droppedProdCompSubCompValues = subComponentKey.getAttributeValues();                    operationBinding.getParamsMap().put("p_bugdbId",newBugdbId);                    operationBinding.getParamsMap().put("p_productId",droppedProdCompSubCompValues[0]);                    operationBinding.getParamsMap().put("p_componentId",droppedProdCompSubCompValues[1]);                    operationBinding.getParamsMap().put("p_subComponentId",droppedProdCompSubCompValues[2]);                    Object result = operationBinding.execute();                  }                   }            }                        /* this.getCil1().setDisabled(false);            this.getCil1().setPartialSubmit(true); */                      return DnDAction.MOVE;        } catch (Exception ex) {          System.out.println("drop failed with : " + ex.getMessage());          ex.printStackTrace();                  /* this.getCil1().setDisabled(true); */          return DnDAction.NONE;          }    } Run jspx page and drop a Component or Subcomponent from Products Tree to UserComponents Tree Table.

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• make tree in scheme

  - by ???

  (define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (define (mktree order items_list) (cond ((= (length items_list) 1) (make-tree (car items_list) '() '())) (else (insert2 order (car items_list) (mktree order (cdr items_list)))))) (define (insert2 order x t) (cond ((null? t) (make-tree x '() '())) ((order x (entry t)) (make-tree (entry t) (insert2 order x (left-branch t)) (right-branch t))) ((order (entry t) x ) (make-tree (entry t) (left-branch t) (insert2 order x (right-branch t)))) (else t))) The result is: (mktree (lambda (x y) (< x y)) (list 7 3 5 1 9 11)) (11 (9 (1 () (5 (3 () ()) (7 () ()))) ()) ()) But I'm trying to get: (7 (3 (1 () ()) (5 () ())) (9 () (11 () ()))) Where is the problem?

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• Can't find a Windows Explorer alternative that has FULL-TREE View

  - by samJL

  I cannot find a Windows Explorer alternative that has full-tree view functionality like Path Finder on the Mac This is the type of view I am looking for (screenshot of Path Finder on Mac): By full-tree I mean: A tree view that includes files in addition to folders, and can be operated as its own pane-- not as a shared pane or attached pane which is characteristic of Q-Dir. Q-Dir and most others simply stick a folder tree pane on to a file list pane, which is not as useful (think Ruby on Rails application or anything with MVC-- I want to be able to pop open folders in the tree and have them stay open as I work between them). I have tried xplorer2 XYplorer Nexus File Free Commander muCommander CubicExplorer Double Commander Q-Dir Explorer++ Unreal Commander wxCommander Power Desk Directory Opus Has anyone seen a Windows app that has a full-tree view? I can't believe such a simple feature is so hard to find Thanks

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• Figuring a max repetitive sub-tree in an object tree

  - by bonomo

  I am trying to solve a problem of finding a max repetitive sub-tree in an object tree. By the object tree I mean a tree where each leaf and node has a name. Each leaf has a type and a value of that type associated with that leaf. Each node has a set of leaves / nodes in certain order. Given an object tree that - we know - has a repetitive sub-tree in it. By repetitive I mean 2 or more sub-trees that are similar in everything (names/types/order of sub-elements) but the values of leaves. No nodes/leaves can be shared between sub-trees. Problem is to identify these sub-trees of the max height. I know that the exhaustive search can do the trick. I am rather looking for more efficient approach.

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• Unit testing to prove balanced tree

  - by Darrel Hoffman

  I've just built a self-balancing tree (red-black) in Java (language should be irrelevant for this question though), and I'm trying to come up with a good means of testing that it's properly balanced. I've tested all the basic tree operations, but I can't think of a way to test that it is indeed well and truly balanced. I've tried inserting a large dictionary of words, both pre-sorted and un-sorted. With a balanced tree, those should take roughly the same amount of time, but an unbalanced tree would take significantly longer on the already-sorted list. But I don't know how to go about testing for that in any reasonable, reproducible way. (I've tried doing millisecond tests on these, but there's no noticeable difference - probably because my source data is too small.) Is there a better way to be sure that the tree is really balanced? Say, by looking at the tree after it's created and seeing how deep it goes? (That is, without modifying the tree itself by adding a depth field to each node, which is just wasteful if you don't need it for anything other than testing.)

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• With ' N ' no of nodes, how many different Binary and Binary Search Trees possible ?

  - by siva

  For Binary tree: You no need to consider tree node values, I am interested about different tree topologies with 'N' nodes. For Binary Search Tree: We obviously consider tree node values.

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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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• Nicely printing/showing a binary tree in Haskell

  - by nicole

  I have a tree data type: data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a ...and I need to make it an instance of Show, without using deriving. I have found that nicely displaying a little branch with two leaves is easy: instance (Show a, Show b) => Show (Tree a b) where show (Leaf x) = show x show (Branch val l r) = " " ++ show val ++ "\n" ++ show l ++ " " ++ show r But how can I extend a nice structure to a tree of arbitrary size? It seems like determining the spacing would require me to know just how many leaves will be at the very bottom (or maybe just how many leaves there are in total) so that I can allocate all the space I need there and just work 'up.' I would probably need to call a size function. I can see this being workable, but is that making it harder than it is?

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• Max Value of a Multiway Tree in OCaml

  - by Trigork

  I'm an IT Student and kind of a newbie to OCaml Recently, studying for an exam I found this exercise. Given: type 'a tree = Tree of 'a * 'a tree list Define a function mtree : 'a tree -'a, that returns the greatest value of all the nodes in a multiway tree, following the usual order relation in OCaml (<=) I've come to do something like this below, which, of course, is not working. let rec mtree (Tree (r, sub)) = let max_node (Tree(a, l1)) (Tree(b, l2)) = if a >= b then (Tree(a, l1)) else (Tree(b, l2)) in List.fold_left max_node r sub;; After reading an answer to this I'm posting the fixed code. let rec mtree (Tree(r,sub)) = let max_node (Tree(a, l1)) (Tree(b, l2)) = if a >= b then a else b in List.fold_left (max_node) r (List.map mtree sub);; The idea is the same, fold the list comparing the nodes making use of my local function to do so and travel through the tree by calling the function itself over the nodes lists of the consecutive levels. Is still not working, though. Now complains about max_node in the fold_left part. Error: This expression has type 'a tree -> 'a tree -> 'a but an expression was expected of type 'a tree -> 'a tree -> 'a tree And here I'm definitely lost because I can't see why does it expects to return an 'a tree

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• Paged binary trees vs. AVL trees and B-trees

  - by Phenom

  How are paged binary trees different from AVL trees and B-trees?

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• HP to Cisco spanning tree root flapping

  - by Tim Brigham

  Per a recent question I recently configured both my HP (2x 2900) and Cisco (1x 3750) hardware to use MSTP for interoperability. I thought this was functional until I applied the change to the third device (HP switch 1 below) at which time the spanning tree root started flapping causing performance issues (5% packet loss) between my two HP switches. I'm not sure why. 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. I'd prefer the Cisco stack to act as the root. EDIT: There is one specific line - 'spanning-tree 1 path-cost 500000' in the HP switch 2 that I didn't add and was preexisting. I'm not sure if it could have the kind of impact that I'm describing. I'm more a security and monitoring guy then networking. EDIT 2: I'm starting to believe the problem lies in the fact that the value for my MST 0 instance on the Cisco is still at the default 32768. I worked up a diagram: This is based on every show command I could find for STP. I'll make this change after hours and see if it helps. Cisco 3750 Config: version 12.2 spanning-tree mode mst spanning-tree extend system-id spanning-tree mst configuration name mstp revision 1 instance 1 vlan 1, 40, 70, 100, 250 spanning-tree mst 1 priority 0 vlan internal allocation policy ascending interface TenGigabitEthernet1/1/1 switchport trunk encapsulation dot1q switchport mode trunk ! interface TenGigabitEthernet2/1/1 switchport trunk encapsulation dot1q switchport mode trunk ! interface Vlan1 no ip address ! interface Vlan100 ip address 192.168.100.253 255.255.255.0 ! Cisco 3750 show spanning tree: show spanning-tree MST0 Spanning tree enabled protocol mstp Root ID Priority 32768 Address 0004.ea84.5f80 Cost 200000 Port 53 (TenGigabitEthernet1/1/1) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32768 (priority 32768 sys-id-ext 0) Address a44c.11a6.7c80 Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Te1/1/1 Root FWD 2000 128.53 P2p MST1 Spanning tree enabled protocol mstp Root ID Priority 1 Address a44c.11a6.7c80 This bridge is the root Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 1 (priority 0 sys-id-ext 1) Address a44c.11a6.7c80 Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Te1/1/1 Desg FWD 2000 128.53 P2p Cisco 3750 show logging: %LINEPROTO-5-UPDOWN: Line protocol on Interface Vlan1, changed state to down %LINEPROTO-5-UPDOWN: Line protocol on Interface Vlan100, changed state to down %LINEPROTO-5-UPDOWN: Line protocol on Interface Vlan1, changed state to up %LINEPROTO-5-UPDOWN: Line protocol on Interface Vlan100, changed state to up %LINEPROTO-5-UPDOWN: Line protocol on Interface Vlan1, changed state to down %LINEPROTO-5-UPDOWN: Line protocol on Interface Vlan1, changed state to up HP Switch 1: ; J9049A Configuration Editor; Created on release #T.13.71 vlan 1 name "DEFAULT_VLAN" untagged 1-8,10,13-16,18-23,A1-A4 ip address 100.100.100.17 255.255.255.0 no untagged 9,11-12,17,24 exit vlan 100 name "192.168.100" untagged 9,11-12,17,24 tagged 1-8,10,13-16,18-23,A1-A4 no ip address exit vlan 21 name "Users_2" tagged 1,A1-A4 no ip address exit vlan 40 name "Cafe" tagged 1,4,7,A1-A4 no ip address exit vlan 250 name "Firewall" tagged 1,4,7,A1-A4 no ip address exit vlan 70 name "DMZ" tagged 1,4,7-8,13,A1-A4 no ip address exit spanning-tree spanning-tree config-name "mstp" spanning-tree config-revision 1 spanning-tree instance 1 vlan 1 40 70 100 250 password manager password operator HP Switch 1 show spanning tree: show spanning-tree Multiple Spanning Tree (MST) Information STP Enabled : Yes Force Version : MSTP-operation IST Mapped VLANs : 2-39,41-69,71-99,101-249,251-4094 Switch MAC Address : 0021f7-126580 Switch Priority : 32768 Max Age : 20 Max Hops : 20 Forward Delay : 15 Topology Change Count : 363,490 Time Since Last Change : 14 hours CST Root MAC Address : 0004ea-845f80 CST Root Priority : 32768 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 | Auto 128 Disabled | HP Switch 1 Logging: I removed the date / time fields since they are inaccurate (no NTP configured on these switches) 00839 stp: MSTI 1 Root changed from 0:a44c11-a67c80 to 32768:0021f7-126580 00839 stp: MSTI 1 Root changed from 32768:0021f7-126580 to 0:a44c11-a67c80 00842 stp: MSTI 1 starved for an MSTI Msg Rx on port A4 from 0:a44c11-a67c80 00839 stp: MSTI 1 Root changed from 0:a44c11-a67c80 to 32768:0021f7-126580 00839 stp: MSTI 1 Root changed from 32768:0021f7-126580 to 0:a44c11-a67c80 00839 stp: MSTI 1 Root changed from 0:a44c11-a67c80 to ... HP Switch 2 Configuration: ; J9146A Configuration Editor; Created on release #W.14.49 vlan 1 name "DEFAULT_VLAN" untagged 1,3-17,21-24,A1-A2,B2 ip address 100.100.100.36 255.255.255.0 no untagged 2,18-20,B1 exit vlan 100 name "192.168.100" untagged 2,18-20 tagged 1,3-17,21-24,A1-A2,B1-B2 no ip address exit vlan 21 name "Users_2" tagged 1,A1-A2,B2 no ip address exit vlan 40 name "Cafe" tagged 1,13-14,16,A1-A2,B2 no ip address exit vlan 250 name "Firewall" tagged 1,13-14,16,A1-A2,B2 no ip address exit vlan 70 name "DMZ" tagged 1,13-14,16,A1-A2,B2 no ip address exit logging 192.168.100.18 spanning-tree spanning-tree 1 path-cost 500000 spanning-tree config-name "mstp" spanning-tree config-revision 1 spanning-tree instance 1 vlan 1 40 70 100 250 HP Switch 2 Spanning Tree: show spanning-tree Multiple Spanning Tree (MST) Information STP Enabled : Yes Force Version : MSTP-operation IST Mapped VLANs : 2-39,41-69,71-99,101-249,251-4094 Switch MAC Address : 0024a8-cd6000 Switch Priority : 32768 Max Age : 20 Max Hops : 20 Forward Delay : 15 Topology Change Count : 21,793 Time Since Last Change : 14 hours CST Root MAC Address : 0004ea-845f80 CST Root Priority : 32768 CST Root Path Cost : 200000 CST Root Port : A1 IST Regional Root MAC Address : 0021f7-126580 IST Regional Root Priority : 32768 IST Regional Root Path Cost : 2000 IST Remaining Hops : 19 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 Forwarding | 0024a8-cd6000 2 Yes No B2 | Auto 128 Disabled | HP Switch 2 Logging: I removed the date / time fields since they are inaccurate (no NTP configured on these switches) 00839 stp: CST Root changed from 32768:0021f7-126580 to 32768:0004ea-845f80 00839 stp: IST Root changed from 32768:0021f7-126580 to 32768:0024a8-cd6000 00839 stp: CST Root changed from 32768:0004ea-845f80 to 32768:0024a8-cd6000 00839 stp: CST Root changed from 32768:0024a8-cd6000 to 32768:0004ea-845f80 00839 stp: CST Root changed from 32768:0004ea-845f80 to 32768:0024a8-cd6000 00435 ports: port B1 is Blocked by STP 00839 stp: CST Root changed from 32768:0024a8-cd6000 to 32768:0021f7-126580 00839 stp: IST Root changed from 32768:0024a8-cd6000 to 32768:0021f7-126580 00839 stp: CST Root changed from 32768:0021f7-126580 to 32768:0004ea-845f80

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• Binary to strings as binary? C #/.NET

  - by acidzombie24

  Redis keys are binary safe. I'd like to mess around and put binary into redis using C#. My client of choice doesn't support writing binary keys it uses keys and it make sense. However i am just fooling around so tell me how i can do this. How do i convert a raw byte[] into a string? At first i was thinking about converting a byte[] to a utf8 string however unicode has some checks to see if its valid or not. So raw binary should fail. Actually i tried it out. Instead of failing i got a strange result. My main question is how do i convert a raw byte[] to the equivalent string? My unimportant question is why did i get a 512 byte string instead of an exception saying this is not a valid UTF8 string? code var rainbow = new byte[256]; for (int i = 0; i < 256; i++) { rainbow[i] = (byte)i; } var sz = Encoding.UTF8.GetString(rainbow); var szarr = Encoding.UTF8.GetBytes(sz); Console.WriteLine("{0} {1} {2}", ByteArraysEqual(szarr, rainbow), szarr.Length, rainbow.Length); Output False 512 256

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• 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

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• drawing hierarchical tree with orthogonal lines ( HV-Drawing – Binary Tree)

  - by user267530

  Hi I need to work on drawing a hierarchical tree structure (HV-Drawing – Binary Tree) with orthogonal lines(straight rectangular connecting lines) between root and children ( like the following: http://lab.kapit.fr/display/visualizationlayouts/Hierarchical+Tree+layout ). I want to know if there are any open source examples of the algorithm of drawing trees like that so that I can implement the same algorithm in actionscript. Thanks Palash

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• Interval tree algorithm that supports merging of intervals with no overlap

  - by Dave Griffiths

  I'm looking for an interval tree algorithm similar to the red-black interval tree in CLR but that supports merging of intervals by default so that there are never any overlapping intervals. In other words if you had a tree containing two intervals [2,3] and [5,6] and you added the interval [4,4], the result would be a tree containing just one interval [2,6]. Thanks Update: the use case I'm considering is calculating transitive closure. Interval sets are used to store the successor sets because they have been found to be quite compact. But if you represent interval sets just as a linked list I have found that in some situations they can become quite large and hence so does the time required to find the insertion point. Hence my interest in interval trees. Also there may be quite a lot of merging one tree with another (i.e. a set OR operation) - if both trees are large then it may be better to create a new tree using inorder walks of both trees rather than repeated insertions of each interval.

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• Are there any B-tree programs or sites that show visually how a B-tree works

  - by Phenom

  I found this website that lets you insert and delete items from a B-tree and shows you visually what the B-tree looks like: java b-tree I'm looking for another website or program similar to this. This site does not allow you to specify a B-tree of order 4 (4 pointers and 3 elements), it only lets you specify B-trees with an even number of elements. Also, if possible, I'd like to be able to insert letters instead of numbers. I think I actually found a different site but that was a while ago and can't find it anymore.

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• Why we do not use PowerConnect to access PeopleSoft Tree

  - by dylan.wan

  1. It does not allow you to use parameters to the PeopleSoft connect. It may be changed later. However, it was a big issue when we try to address customer issues. 2. It requires EFFDT as an option. It expect that people change the EFFDT using Mapping Editor. How can a business user does that every month? 3. It asks for a Tree Name. Many PeopleSoft tree structure supports multiple trees. Tree is just a header of the hierarchy. Whenever you add a new Tree, you need to create a new mapping!! It does not make sense to use PowerConnect due to the customer demands. All requirements are from customers. We have no choice but stop using it.

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• Generated 3d tree meshes

  - by Jari Komppa

  I did not find a question on these lines yet, correct me if I'm wrong. Trees (and fauna in general) are common in games. Due to their nature, they are a good candidate for procedural generation. There's SpeedTree, of course, if you can afford it; as far as I can tell, it doesn't provide the possibility of generating your tree meshes at runtime. Then there's SnappyTree, an online webgl based tree generator based on the proctree.js which is some ~500 lines of javascript. One could use either of above (or some other tree generator I haven't stumbled upon) to create a few dozen tree meshes beforehand - or model them from scratch in a 3d modeller - and then randomly mirror/scale them for a few more variants.. But I'd rather have a free, linkable tree mesh generator. Possible solutions: Port proctree.js to c++ and deal with the open source license (doesn't seem to be gpl, so could be doable; the author may also be willing to co-operate to make the license even more free). Roll my own based on L-systems. Don't bother, just use offline generated trees. Use some other method I haven't found yet.

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• apt-get update error - binary-i386, binary-amd64 [duplicate]

  - by magamig

  This question already has an answer here: How can I fix a 404 Error when updating packages? 5 answers When I run: sudo apt-get update It shows me the following error: W: Failed to fetch http://ppa.launchpad.net/directhex/ppa/ubuntu/dists/trusty/main/binary-amd64/Packages 404 Not Found W: Failed to fetch http://ppa.launchpad.net/directhex/ppa/ubuntu/dists/trusty/main/binary-i386/Packages 404 Not Found E: Some index files failed to download. They have been ignored, or old ones used instead. I have googled for solutions, but none of what I found, have worked for me. Please give me your suggestions

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• Behavior tree implementation details

  - by angryInsomniac

  I have been looking around for implementation details of behavior trees, the best descriptions I found were by Alex Champarand and some of Damian Isla's talk about AI in Halo 2 (the video of which is locked up in the GDC vault sadly). However, both descriptions fall short of helping one actually create a BT, one particular question has been bugging me for a while. When is the tree in a behavior tree evaluated? Furthermore: If the tree is in the middle of executing a sequence of actions (patrolling waypoints) and a higher priority impulse comes in (distraction sound) , how to switch to that side of the tree seamlessly without resorting to a state machine like system and if it is decided that the impulse was irrelevant (the distraction is too far away to affect this guard), how to go back to the last thing that the guard was doing ? I have quite a few questions like this and I don't wish to flood the board with separate queries so if you know of any resource where questions like these can be answered I would be very grateful.

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