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  • Logical Domain Modeling Made Simple

    - by Knut Vatsendvik
    How can logical domain modeling be made simple and collaborative? Many non-technical end-users, managers and business domain experts find it difficult to understand the visual models offered by many UML tools. This creates trouble in capturing and verifying the information that goes into a logical domain model. The tools are also too advanced and complex for a non-technical user to learn and use. We have therefore, in our current project, ended up with using Confluence as tool for designing the logical domain model with the help of a few very useful plugins. Big thanks to Ole Nymoen and Per Spilling for their expertise in this field that made this posting possible. Confluence Plugins Here is a list of Confluence plugins used in this solution. Install these before trying out the macros used below. Plugin Description Copy Space Allows a space administrator to copy a space, including the pages within the space Metadata Supports adding metadata to Wiki pages Label Manages labeling of pages Linking Contains macros for linking to templates, the dashboard and other Table Enhances the table capability in Confluence Creating a Confluence Space First we need to create a new confluence space for the domain model. Click the link Create a Space located below the list of spaces on the Dashboard. Please contact your Confluence administrator is you do not have permissions to do this.   For illustrative purpose all attributes and entities in this posting are based on my imaginary project manager domain model. When a logical domain model is good enough for being implemented, do a copy of the Confluence Space (see Copy Space plugin). In this way you create a stable version of the logical domain model while further design can continue with the new copied space. Typical will the implementation phase result in a database design and/or a XSD schema design. Add Space Templates Go to the Home page of your Confluence Space. Navigate to the Browse drop-down menu and click on Advanced. Then click the Templates option in the left navigation panel. Click Add New Space Template to add the following three templates. Name: attribute {metadata-list} || Name | | || Type | | || Format | | || Description | | {metadata-list} {add-label:attribute} Name: primary-type {metadata-list} || Name | || || Type | || || Format | || || Description | || {metadata-list} {add-label:primary-type} Name: complex-type {metadata-list} || Name | || || Description |  || {metadata-list} h3. Attributes || Name || Type || Format || Description || | [name] | {metadata-from:name|Type} | {metadata-from:name|Format} | {metadata-from:name|Description} | {add-label:complex-type,entity} The metadata-list macro (see Metadata plugin) will save a list of metadata values to the page. The add-label macro (see Label plugin) will automatically label the page. Primary Types Page Our first page to add will act as container for our primary types. Switch to Wiki markup when adding the following content to the page. | (+) {add-page:template=primary-type|parent=@self}Add new primary type{add-page} | {metadata-report:Name,Type,Format,Description|sort=Name|root=@self|pages=@descendents} Once the page is created, click the Add new primary type (create-page macro) to start creating a new pages. Here is an example of input to the LocalDate page. Embrace the LocalDate with square brackets [] to make the page linkable. Again switch to Wiki markup before editing. {metadata-list} || Name | [LocalDate] || || Type | Date || || Format | YYYY-MM-DD || || Description | Date in local time zone. YYYY = year, MM = month and DD = day || {metadata-list} {add-label:primary-type} The metadata-report macro will show a tabular report of all child pages.   Attributes Page The next page will act as container for all of our attributes. | (+) {add-page:template=attribute|parent=@self|title=attribute}Add new attribute{add-page} | {metadata-report:Name,Type,Format,Description|sort=Name|pages=@descendants} Here is an example of input to the startDate page. {metadata-list} || Name | [startDate] || || Type | [LocalDate] || || Format | {metadata-from:LocalDate|Format} || || Description | The projects start date || {metadata-list} {add-label:attribute} Using the metadata-from macro we fetch the text from the previously created LocalDate page. Complex Types Page The last page in this example shows how attributes can be combined together to form more complex types.   h3. Intro Overview of complex types in the domain model. | (+) {add-page:template=complex-type|parent=@self}Add a new complex type{add-page}\\ | {metadata-report:Name,Description|sort=Name|root=@self|pages=@descendents} Here is an example of input to the ProjectType page. {metadata-list} || Name | [ProjectType] || || Description | Represents a project || {metadata-list} h3. Attributes || Name || Type || Format || Description || | [projectId] | {metadata-from:projectId|Type} | {metadata-from:projectId|Format} | {metadata-from:projectId|Description} | | [name] | {metadata-from:name|Type} | {metadata-from:name|Format} | {metadata-from:name|Description} | | [description] | {metadata-from:description|Type} | {metadata-from:description|Format} | {metadata-from:description|Description} | | [startDate] | {metadata-from:startDate|Type} | {metadata-from:startDate|Format} | {metadata-from:startDate|Description} | {add-label:complex-type,entity} Gives us this Conclusion Using a web-based corporate Wiki like Confluence to create a logical domain model increases the collaboration between people with different roles in the enterprise. It’s my believe that this helps the domain model to be more accurate, and better documented. In our real project we have more pages than illustrated here to complete the documentation. We do also still use UML tools to create different types of diagrams that Confluence do not support. As a last tip, an ImageMap plugin can make those diagrams clickable when used in pages. Enjoy!

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  • Is this the right strategy to convert an in-level order binary tree to a doubly linked list?

    - by Ankit Soni
    So I recently came across this question - Make a function that converts a in-level-order binary tree into a doubly linked list. Apparently, it's a common interview question. This is the strategy I had - Simply do a pre-order traversal of the tree, and instead of returning a node, return a list of nodes, in the order in which you traverse them. i.e return a list, and append the current node to the list at each point. For the base case, return the node itself when you are at a leaf. so you would say left = recursive_function(node.left) right = recursive_function(node.right) return(left.append(node.data)).append(right);` Is this the right approach?

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  • adding nodes to a binary search tree randomly deletes nodes

    - by SDLFunTimes
    Hi, stack. I've got a binary tree of type TYPE (TYPE is a typedef of data*) that can add and remove elements. However for some reason certain values added will overwrite previous elements. Here's my code with examples of it inserting without overwriting elements and it not overwriting elements. the data I'm storing: struct data { int number; char *name; }; typedef struct data data; # ifndef TYPE # define TYPE data* # define TYPE_SIZE sizeof(data*) # endif The tree struct: struct Node { TYPE val; struct Node *left; struct Node *rght; }; struct BSTree { struct Node *root; int cnt; }; The comparator for the data. int compare(TYPE left, TYPE right) { int left_len; int right_len; int shortest_string; /* find longest string */ left_len = strlen(left->name); right_len = strlen(right->name); if(right_len < left_len) { shortest_string = right_len; } else { shortest_string = left_len; } /* compare strings */ if(strncmp(left->name, right->name, shortest_string) > 1) { return 1; } else if(strncmp(left->name, right->name, shortest_string) < 1) { return -1; } else { /* strings are equal */ if(left->number > right->number) { return 1; } else if(left->number < right->number) { return -1; } else { return 0; } } } And the add method struct Node* _addNode(struct Node* cur, TYPE val) { if(cur == NULL) { /* no root has been made */ cur = _createNode(val); return cur; } else { int cmp; cmp = compare(cur->val, val); if(cmp == -1) { /* go left */ if(cur->left == NULL) { printf("adding on left node val %d\n", cur->val->number); cur->left = _createNode(val); } else { return _addNode(cur->left, val); } } else if(cmp >= 0) { /* go right */ if(cur->rght == NULL) { printf("adding on right node val %d\n", cur->val->number); cur->rght = _createNode(val); } else { return _addNode(cur->rght, val); } } return cur; } } void addBSTree(struct BSTree *tree, TYPE val) { tree->root = _addNode(tree->root, val); tree->cnt++; } The function to print the tree: void printTree(struct Node *cur) { if (cur == 0) { printf("\n"); } else { printf("("); printTree(cur->left); printf(" %s, %d ", cur->val->name, cur->val->number); printTree(cur->rght); printf(")\n"); } } Here's an example of some data that will overwrite previous elements: struct BSTree myTree; struct data myData1, myData2, myData3; myData1.number = 5; myData1.name = "rooty"; myData2.number = 1; myData2.name = "lefty"; myData3.number = 10; myData3.name = "righty"; initBSTree(&myTree); addBSTree(&myTree, &myData1); addBSTree(&myTree, &myData2); addBSTree(&myTree, &myData3); printTree(myTree.root); Which will print: (( righty, 10 ) lefty, 1 ) Finally here's some test data that will go in the exact same spot as the previous data, but this time no data is overwritten: struct BSTree myTree; struct data myData1, myData2, myData3; myData1.number = 5; myData1.name = "i"; myData2.number = 5; myData2.name = "h"; myData3.number = 5; myData3.name = "j"; initBSTree(&myTree); addBSTree(&myTree, &myData1); addBSTree(&myTree, &myData2); addBSTree(&myTree, &myData3); printTree(myTree.root); Which prints: (( j, 5 ) i, 5 ( h, 5 ) ) Does anyone know what might be going wrong? Sorry if this post was kind of long.

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  • Tree Surgeon 2.0 - The future on the T4 Express

    - by Malcolm Anderson
    If you've never been a fan of TreeSurgeon (http://treesurgeon.codeplex.com/) then skip this post.However, if have been there have been some interesting developments over the last couple of years.The biggest one is T4Recently Bill Simser wrote a detailed post about the potential future of tree surgeon, called "Tree Surgeon - Alive and Kicking or Dead and Buried" He raised the question:Times have changed. Since that last release in 2008 so much has changed for .NET developers. The question is, today is the project still viable? Do we still need a tool to generate a project tree given that we have things like scaffolding systems, NuGet, and T4 templates. Or should we just give the project its rightful and respectful send off as its had a good life and has outlived its usefulness.For myself, the answer is, keep it.I've spent the last couple of years doing agile engineering coaching and architecture and from my experience, I can tell you, there are a lot of shops out there that would benefit from having Tree Surgeon as a viable product.  Many would benefit simply from having the software engineering information that is embedded in the tree surgeon site be floating around their conversation.Little things like, keep all of your software needed to run the build, with the build in the version control system.Have your developers and the build system using the same build.Have a one-touch buildSeparate your code from your interfacePut unit tests in first, not lastI've seen companies with great developers suffer from the problems that naturally come from builds taking 3 and 4 hours to run.  It takes work to get that build down to 10 minutes, but the benefits are always worth it.  Tree Surgeon gives you a leg up, by starting you off with a project that you can drop into your Continuous Integration system, right out of the box.Well, it used to be right out of the box.  Today, you have to play with the project to make it work for you, but even with the issues (it hasn't been updated since 2008) it still gives you a framework, with logical separations that you can build from.If you have used Tree Surgeon in the past, take a few minutes and drop a comment about what difference it made in your development style, and what you are doing differently today because of it.

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  • Worst Case number of rotations for BST to AVL algorithm?

    - by spacker_lechuck
    I have a basic algorithm below and I know that the worst case input BST is one that has degenerated to a linked list from inserts to only one side. How would I compute the worst case complexity in terms of number of rotations for this BST to AVL conversion algorithm? IF tree is right heavy { IF tree's right subtree is left heavy { Perform Double Left rotation } ELSE { Perform Single Left rotation } } ELSE IF tree is left heavy { IF tree's left subtree is right heavy { Perform Double Right rotation } ELSE { Perform Single Right rotation } }

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  • GWT: Change padding of tree rows?

    - by Epaga
    A GWT tree looks roughly like this: <div class="gwt-Tree"> <div style="padding-top: 3px; padding-right: 3px; padding-bottom: 3px; margin-left: 0px; padding-left: 23px;"> <div style="display:inline;" class="gwt-TreeItem"> <table> ... </table> </div> </div> <div ...> </div> ... </div> My question is: how should I change the padding of the individual tree rows? I suppose I could do something along the lines of setting CSS rules for .gwt-Tree > div but that seems hacky. Is there a more elegant way?

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  • red black tree balancing?

    - by Anirudh Kaki
    i am working to generate tango tree, where i need to check whether every sub tree in tango is balanced or not. if its not balanced i need to make it balance? I trying so hard to make entire RB-tree balance but i not getting any proper logic so can any one help me out?? here i am adding code to check how to find my tree is balanced are not but when its not balanced how can i make it balance. static boolean verifyProperty5(rbnode n) { int left = 0, right = 0; if (n != null) { bh++; left = blackHeight(n.left, 0); right = blackHeight(n.right, 0); } if (left == right) { System.out.println("black height is :: " + bh); return true; } else { System.out.println("in balance"); return false; } } public static int blackHeight(rbnode root, int len) { bh = 0; blackHeight(root, path1, len); return bh; } private static void blackHeight(rbnode root, int path1[], int len) { if (root == null) return; if (root.color == "black"){ root.black_count = root.parent.black_count+1; } else{ root.black_count = root.parent.black_count; } if ((root.left == null) && (root.right == null)) { bh = root.black_count; } blackHeight(root.left, path1, len); blackHeight(root.right, path1, len); }

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  • C++ find largest BST in a binary tree

    - by fonjibe
    what is your approach to have the largest BST in a binary tree? I refer to this post where a very good implementation for finding if a tree is BST or not is bool isBinarySearchTree(BinaryTree * n, int min=std::numeric_limits<int>::min(), int max=std::numeric_limits<int>::max()) { return !n || (min < n->value && n->value < max && isBinarySearchTree(n->l, min, n->value) && isBinarySearchTree(n->r, n->value, max)); } It is quite easy to implement a solution to find whether a tree contains a binary search tree. i think that the following method makes it: bool includeSomeBST(BinaryTree* n) { if(!isBinarySearchTree(n)) { if(!isBinarySearchTree(n->left)) return isBinarySearchTree(n->right); } else return true; else return true; } but what if i want the largest BST? this is my first idea, BinaryTree largestBST(BinaryTree* n) { if(isBinarySearchTree(n)) return n; if(!isBinarySearchTree(n->left)) { if(!isBinarySearchTree(n->right)) if(includeSomeBST(n->right)) return largestBST(n->right); else if(includeSomeBST(n->left)) return largestBST(n->left); else return NULL; else return n->right; } else return n->left; } but its not telling the largest actually. i struggle to make the comparison. how should it take place? thanks

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  • Why can't RB-Tree be a list?

    - by Alex
    Hey everyone. I have a problem with the rb-trees. according to wikipedia, rb-tree needs to follow the following: A node is either red or black. The root is black. (This rule is used in some definitions and not others. Since the root can always be changed from red to black but not necessarily vice-versa this rule has little effect on analysis.) All leaves are black. Both children of every red node are black. Every simple path from a given node to any of its descendant leaves contains the same number of black nodes. As we know, an rb-tree needs to be balanced and has the height of O(log(n)). But, if we insert an increasing series of numbers (1,2,3,4,5...) and theoretically we will get a tree that will look like a list and will have the height of O(n) with all its nodes black, which doesn't contradict the rb-tree properties mentioned above. So, where am I wrong?? thanks.

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  • navigate all items in a wpf tree view

    - by Brian Leahy
    I want to be able to traverse the visual ui tree looking for an element with an ID bound to the visual element's Tag property. I'm wondering how i do this. Controls don't have children to traverse. I started using LogicalTreeHelper.GetChildren, which seems to work as intended, up until i hit a TreeView control... then LogicalTreeHelper.GetChildren doesnt return any children. Note: the purpose is to find the visual UI element that corresponds to the data item. That is, given an ID of the item, Go find the UI element displaying it. Edit: I am apparently am not explaining this well enough. I am binding some data objects to a TreeView control and then wanting to select a specific item programaticly given that business object's ID. I dont see why it's so hard to travers the visual tree and find the element i want, as the data object's ID is in the Tag property of the appropriate visual element. I'm using Mole and I am able to find the UI element with the appropriate ID in it's Tag. I just cannot find the visual element in code. LogicalTreeHelper does not traverse any items in the tree. Neither does ItemContainerGenerator.ContainerFromItem retrieve anything for items in the tree view.

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  • Evaluate an expression tree

    - by Phronima
    Hi, This project that I'm working on requires that an expression tree be constructed from a string of single digit operands and operators both represented as type char. I did the implmentation and the program up to that point works fine. I'm able to print out the inorder, preorder and postorder traversals in the correct way. The last part calls for evaulating the expression tree. The parameters are an expression tree "t" and its root "root". The expression tree is ((3+2)+(6+2)) which is equal to 13. Instead I get 11 as the answer. Clearly I'm missing something here and I've done everything short of bashing my head against the desk. I would greatly appreciate it if someone can point me in the right direction. (Note that at this point I'm only testing addition and will add in the other operators when I get this method working.) public int evalExpression( LinkedBinaryTree t, BTNode root ) { if( t.isInternal( root ) ) { int x = 0, y = 0, value = 0; char operator = root.element(); if( root.getLeft() != null ) x = evalExpression(t, t.left( root ) ); if( root.getRight() != null ) y = evalExpression(t, t.right( root ) ); if( operator == '+' ) { value = value + Character.getNumericValue(x) + Character.getNumericValue(y); } return value; } else { return root.element(); } }

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  • Flex Tree Properties, Null Reference?

    - by mvrak
    I am pulling down a large XML file and I have no control over it's structure. I used a custom function to use the tag name to view the tree structure as a flex tree, but then it breaks. I am guessing it has something to do with my other function, one that calls attribute values from the selected node. See code. <mx:Tree x="254" y="21" width="498" height="579" id="xmllisttree" labelFunction="namer" dataProvider="{treeData}" showRoot="false" change="treeChanged(event)" /> //and the Cdata import mx.rpc.events.ResultEvent; [Bindable] private var fullXML:XMLList; private function contentHandler(evt:ResultEvent):void{ fullXML = evt.result.page; } [Bindable] public var selectedNode:Object; public function treeChanged(event:Event):void { selectedNode=Tree(event.target).selectedItem; } public function namer(item:Object):String { var node:XML = XML(item); var nodeName:QName = node.name(); var stringtest:String ="bunny"; return nodeName.localName; } The error is TypeError: Error #1009: Cannot access a property or method of a null object reference. Where is the null reference?

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  • Sorting by some field and fetching whole tree from DB

    - by Niaxon
    Hello everyone, I am trying to do file browser in a tree form and have a problem to sort it somehow. I use PHP and MySQL for that. I've created mixed (nested set + adjacency) table 'element' with the following fields: element_id, left_key, right_key, level, parent_id, element_name, element_type (enum: 'folder','file'), element_size. Let's not discuss right now that it is better to move information about element (name, type, size) into other table. Function to scan specified directory and fill table work correctly. Noteworthy, i am adding elements to tree in specific order: folders first and only files. After that i can easily fetch and display whole table on the page using simple query: SELECT * FROM element WHERE 1=1 ORDER BY left_key With the result of that query and another function i can generate correct html code (<ul><li>... and so on). Now back to the question (finally, huh?). I am struggling to add sorting functionality. For example i want to order my result by size. Here i need to keep in my mind whole hierarchy of tree and rule: folders first, files later. I believe i can do that by generating in PHP recursive query: SELECT * FROM element WHERE parent_id = {$parentId} ORDER BY element_type (so folders would be first), size (or name for example) After that for each result which is folder i will send another query to get it's content. Also it's possible to fetch whole tree by left_key and after that sort it in PHP as array but i guess that would be worse :) I wonder if there is better and more efficient way to do such thing?

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  • R Tree 50,000 foot overview?

    - by roufamatic
    I'm working on a school project that involves taking a lat/long point and finding the top five closest points in a known list of places. The list is to be stored in memory, with the caveat that we must choose an "appropriate data structure" -- that is, we cannot simply store all the places in an array and compare distances one-by-one in a linear fashion. The teacher suggested grouping the place data by US State to prevent calculating the distance for places that are obviously too far away. I think I can do better. From my research online it seems like an R-Tree or one of its variants might be a neat solution. Unfortunately, that sentence is as far as I've gotten with understanding the actual technique, as the literature is simply too dense for my non-academic head. Can somebody give me a really high overview of what the process is for populating an R-Tree with lat/long data, and then traversing the tree to find those 5 nearest neighbors of a given point? Additionally the project is in C, and I don't have to reinvent the wheel on this, so if you've used an existing open source C implementation of an R Tree I'd be interested in your experiences.

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  • python recursive iteration exceeding limit for tree implementation

    - by user3698027
    I'm implementing a tree dynamically in python. I have defined a class like this... class nodeobject(): def __init__(self,presentnode=None,parent=None): self.currentNode = presentnode self.parentNode = parent self.childs = [] I have a function which gets possible childs for every node from a pool def findchildren(node, childs): # No need to write the whole function on how it gets childs Now I have a recursive function that starts with the head node (no parent) and moves down the chain recursively for every node (base case being the last node having no children) def tree(dad,children): for child in children: childobject = nodeobject(child,dad) dad.childs.append(childobject) newchilds = findchildren(child, children) if len(newchilds) == 0: lastchild = nodeobject(newchilds,childobject) childobject.childs.append(lastchild) loopchild = copy.deepcopy(lastchild) while loopchild.parentNode != None: print "last child" else: tree(childobject,newchilds) The tree formation works for certain number of inputs only. Once the pool gets bigger, it results into "MAXIMUM RECURSION DEPTH EXCEEDED" I have tried setting the recursion limit with set.recursionlimit() and it doesn't work. THe program crashes. I want to implement a stack for recursion, can someone please help, I have gone no where even after trying for a long time ?? Also, is there any other way to fix this other than stack ?

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  • Simulation tree command in C

    - by Ecle
    I have to create the simulation of tree command in C, this is my current code: #include <stdio.h> #include <sys/types.h> #include <sys/stat.h> #include <dirent.h> #include <string.h> main(int argc, char *argv[]){ int i; if(argc < 2){ printf("\nError. Use: %s directory\n", argv[0]); system("exit"); } for(i=1;i<argc;i++) //if(argv[i][0] != '-') tree(argv[i]); } tree(char *ruta){ DIR *dirp; struct dirent *dp; static nivel = 0; struct stat buf; char fichero[256]; int i; if((dirp = opendir(path)) == NULL){ perror(path); return; } while((dp = readdir(dirp)) != NULL){ printf(fichero, "%s/%s", path, dp->d_name); if((buf.st_mode & S_IFMT) == S_IFDIR){ for(i=0;i<nivel;i++) printf("\t"); printf("%s\n", dp->d_name); ++nivel; tree(fichero); --nivel; } } } Apparently, it works! (due to it compiles correctly) But I don't why. I can't pass the correct arguments to execute this. Thank you so much, people.

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  • Need some help understanding IO Statistics

    - by Abe Miessler
    I have a query that has a very costly INDEX SEEK operation in the execution plan. In order to track down the cause i set IO STATISTICS on and ran it. In the problem section it gave the following statistics: Table '#TempStudents_Enrollment2_____________________________________000000004D5F'. Scan count 0, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table 'Worktable'. Scan count 0, logical reads 0, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#TempRace2______________________________________________000000004D58'. Scan count 1, logical reads 1, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table 'Worktable'. Scan count 0, logical reads 0, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table 'RefRace'. Scan count 120, logical reads 240, physical reads 1, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table 'RefFedEnctyRaceCatg'. Scan count 18, logical reads 36, physical reads 2, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#43B0BA0F'. Scan count 1, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#42BC95D6'. Scan count 1, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#41C8719D'. Scan count 1, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#40D44D64'. Scan count 1, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#LEA2_________________________________________________000000004D56'. Scan count 1, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#39332B9C'. Scan count 1, logical reads 60, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#School2________________________________________________000000004D57'. Scan count 1, logical reads 29164, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#GenderKey______________________________________________000000004D5A'. Scan count 1, logical reads 29164, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#LangAcqKey_____________________________________________000000004D5B'. Scan count 1, logical reads 29164, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#TransferCatKey___________________________________________000000004D5C'. Scan count 1, logical reads 29164, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#ResCatKey______________________________________________000000004D5D'. Scan count 1, logical reads 29164, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table 'RPT_SnapShot_1_4_StuPgm_Denorm'. Scan count 2344954, logical reads 4992518, physical reads 16, read-ahead reads 8, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#3FE0292B'. Scan count 1, logical reads 2344954, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table 'RPT_SnapShot_1_4_StuEnrlmt_Denorm'. Scan count 20, logical reads 87679, physical reads 0, read-ahead reads 87425, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table '#GradeKey_______________________________________________000000004D59'. Scan count 1, logical reads 1, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. What should I look for in here when i'm looking to improve the performance? The line with over 2 million for the Scan count looked suspicious to me but I really don't know. Does anyone see anything here that i should look into in more detail?

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  • Leveraging Logical Standby Databases in Oracle 11g Data Guard

    Oracle Data Guard still offers support for the venerable logical standby database in Oracle Database 11g. This article, investigates how data warehouse and data mart environments can effectively leverage logical standby database features, but simultaneously provide a final destination when failover from a primary database is mandated during disaster recovery.

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  • Mixed Solaris 10 and 11 versions in logical domains on the same server

    - by jsavit
    One question that comes up frequently is whether you can mix Solaris 10 and Solaris 11 in different logical domains under Oracle VM Server for SPARC. The answer is yes depending only on the system software requirements for the underlying hardware platform. Different versions of Solaris 10 and 11 can exist side-by-side on the same server and can act as control, service, I/O or guest domains subject only to the minimum software levels documented in the System Requirements section of the Oracle VM Server for SPARC Release Notes. Here's an example just taken from a running system. First, here's the control domain, which is running Solaris 10. I've highlighted a guest running Solaris 11. # uname -a SunOS atl-sewr-24 5.10 Generic_147440-01 sun4v sparc SUNW,SPARC-Enterprise-T5220 # ldm -V Logical Domains Manager (v 2.1) Hypervisor control protocol v 1.7 Using Hypervisor MD v 1.3 System PROM: Hypervisor v. 1.10.0 @(#)Hypervisor 1.10.0 2011/04/27 16:19\015 # ldm list NAME STATE FLAGS CONS VCPU MEMORY UTIL UPTIME primary active -n-cv- SP 16 4G 1.6% 120d 17h atl-sewr-pool-148 active -n---- 5001 8 2G 0.1% 119d 21h atl-sewr-pool-152 active -n---- 5000 8 4G 0.2% 112d 19h atl-sewr-pool-154 active -n---- 5002 8 2G 0.1% 120d 15h atl-sewr-pool-155 active -n---- 5003 16 2G 0.0% 26d 14h 30m This system is running Oracle VM Server 2.1 with a Solaris 10 control domain. Hmm, I should update this machine to 2.2 when I get a few free moments. Upgrading is very straightforward. Here's a display logging into the highlighted guest: Last login: Mon May 21 10:18:16 2012 from dhcp-adc-twvpn- Oracle Corporation SunOS 5.11 11.0 November 2011 sewr@atl-sewr-pool-152:~$ uname -a SunOS atl-sewr-pool-152 5.11 11.0 sun4v sparc SUNW,SPARC-Enterprise-T5220 sewr@atl-sewr-pool-152:~$ cat /etc/release Oracle Solaris 11 11/11 SPARC Copyright (c) 1983, 2011, Oracle and/or its affiliates. All rights reserved. Assembled 18 October 2011 sewr@atl-sewr-pool-152:~$ sudo virtinfo -ct Password: Domain role: LDoms guest Control domain: atl-sewr-24 sewr@atl-sewr-pool-152:~$ That's running the GA version of Solaris 11, so I probably should update that some time too. Note the use of the virtinfo -ct command that lets the guest get information about the hosting environment. Summary You can mix and match versions of Solaris in logical domains. All the different combinations work: Solaris 10 and/or Solaris 11 control and service domains with Solaris 10 and/or Solaris 11 guests. Mixing different guest OS levels on the same server is one of the traditional reasons for using virtual machines in the first place since virtual machines were invented some 40 years ago, used to run production and test systems in parallel while upgrading OS levels. This can easily be done with Oracle VM Server for SPARC (Logical Domains).

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  • Dynamic Tree control in Flex 3

    - by nimmyliji
    I am looking for sample code to create a dynamic Tree control in Flex using a Collection of objects obtained from the backend(Perl cgi). So, initially the Tree will display the root nodes. Clicking root node, will invoke the data for populating the child nodes (basically adding child nodes on demand). Clicking child nodes will pull another collection add child nodes of child node. So, lets assume Initially the Tree will display - Root1 Root2 Root3 Clicking Root1 will display something like this - Root1 Child 1 Child 2 Root2 Root3 And Clicking Child1 will display - Root1 Child 1 Child1 of Child 1 Child2 of Child 1 Child 2 Root2 Root3 Is it possible? Thanks in advance...

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  • Metalanguage like BNF or XML-Schema to validate a tree-instance against a tree-model

    - by Stefan
    Hi! I'm implementing a new machine learning algorithm in Java that extracts a prototype datastructure from a set of structured datasets (tree-structure). As im developing a generic library for that purpose, i kept my design independent from concrete data-representations like XML. My problem now is that I need a way to define a data model, which is basically a ruleset describing valid trees, against which a set of trees is being matched. I thought of using BNF or a similar dialect. Basically I need a way to iterate through the space of all valid TreeNodes defined by the ModelTree (Like a search through the search space for algorithms like A*) so that i can compare my set of concrete trees with the model. I know that I'll have to deal with infinite spaces there but first things first. I know, it's rather tricky (and my sentences are pretty bumpy) but I would appreciate any clues. Thanks in advance, Stefan

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  • how to handle duplicates in AVL tree

    - by Ahmed Kotb
    I want to make my avl tree support adding duplicates but there is a problem with the default behavior of the binary search tree with duplicates that the rotation could make nodes with equal key be on the left and the right of the parent for example adding A,A,A will cause the tree to do a rotation to be something like that A / \ A A so getting all the entries with that key will be a problem...and searching in both direction is not something good the solution that i have thought of is making each node stores an array of duplicates so when adding a new item that already exists will be just adding a new item to the array , removing item with key will delete the whole node while the find all items with key will return that array. are there are any other methods to handle duplicates ?

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  • Object tree navigation language in Java

    - by lewap
    In the system which I'm currently developing I often have to navigate an object tree and based on its state and values take actions. In normal Java this results in tedious for loops, if statements etc... Are there alternative ways to achieve tree navigation, similar to XPath for XML? I know there is JXPath and OGNL, but do you know any other libraries for such purpose? Do you know any libraries which generate bytecodes for specific tree navigation expressions to make the processing as fast as Java native fors and ifs?

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  • Manually Building a Tree in CakePHP

    - by mabwi
    I'm currently working on building an application in CakePHP. There's a quite extensive existing data set that's conceptually a tree, but wasn't previously stored as one. What I mean by that, is there's no real relationship defined in the data. The problem I'm having is getting it to work correctly with the CakePHP tree behaviour. Because I have to set all the values on existing data - as opposed to Cake setting up the structure as elements are inserted - I need to understand how the lft/rght values work. So, I guess the question is: How does the structure data work, specifically the lft/rght values? How do I set it up so that the data comes out rationally, without inserting them one at a time? It's a 2 level tree, with Sections and sub-sections. Thanks for the help

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  • Write a tree class in Java where each level has a unique object type

    - by user479576
    I need to write a tree class in Java where each level has a unique object type. The way it is written below does not take advantage of generics and causes alot of duplicate code. Is there a way to write this with Generics ? public class NodeB { private String nodeValue; //private List<NodeB> childNodes; // constructors // getters/setters } public class NodeA { private String value; private List<NodeB> childNodes; // constructors // getters/setters } public class Tree { private String value; private List<NodeA> childNodes; // constructors // tree methods }

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