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  • Oracle tuning optimizer index cost adj and optimizer index caching

    - by Darryl Braaten
    What is the correct way to set the optimizer index cost adj parameter for Oracle. As a developer I have observed huge performance improvements as this parameter is lowered. Common queries are reduced from 2 seconds to 200ms. There are lots of warnings on the net that lowering this value will cause dire issues with the database, but no detail is given on what will start going wrong. I am currently only seeing only an upside, much improved application performance and no downside. I need to better understand the possible negative repercussions of adjusting these parameters.

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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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  • Multiple columns in a single index versus multiple indexes

    - by Tim Coker
    The short version of my question is what's the difference between three indexes each indexing a single column and one index indexing three columns. Background follows. I'm primarily a programmer but have to do DBA work because we don't have a DBA. I'm evaluating our indexes versus the queries run against a particular table. The table as 3 columns that I'm often filtering against or getting the max value of. Most of the time the queries look like select max(col_a) from table where col_b = 'avalue' or select col_c from table where col_b = 'avalue' and col_a = 'anothervalue' All columns are independently indexed. My question is would I see any difference if I had an index that indexed col_b and col_a together since they can appear in a where clause together?

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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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  • SQL SERVER – Shrinking Database is Bad – Increases Fragmentation – Reduces Performance

    - by pinaldave
    Earlier, I had written two articles related to Shrinking Database. I wrote about why Shrinking Database is not good. SQL SERVER – SHRINKDATABASE For Every Database in the SQL Server SQL SERVER – What the Business Says Is Not What the Business Wants I received many comments on Why Database Shrinking is bad. Today we will go over a very interesting example that I have created for the same. Here are the quick steps of the example. Create a test database Create two tables and populate with data Check the size of both the tables Size of database is very low Check the Fragmentation of one table Fragmentation will be very low Truncate another table Check the size of the table Check the fragmentation of the one table Fragmentation will be very low SHRINK Database Check the size of the table Check the fragmentation of the one table Fragmentation will be very HIGH REBUILD index on one table Check the size of the table Size of database is very HIGH Check the fragmentation of the one table Fragmentation will be very low Here is the script for the same. USE MASTER GO CREATE DATABASE ShrinkIsBed GO USE ShrinkIsBed GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Create FirstTable CREATE TABLE FirstTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_FirstTable_ID] ON FirstTable ( [ID] ASC ) ON [PRIMARY] GO -- Create SecondTable CREATE TABLE SecondTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_SecondTable_ID] ON SecondTable ( [ID] ASC ) ON [PRIMARY] GO -- Insert One Hundred Thousand Records INSERT INTO FirstTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Insert One Hundred Thousand Records INSERT INTO SecondTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO Let us check the table size and fragmentation. Now let us TRUNCATE the table and check the size and Fragmentation. USE MASTER GO CREATE DATABASE ShrinkIsBed GO USE ShrinkIsBed GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Create FirstTable CREATE TABLE FirstTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_FirstTable_ID] ON FirstTable ( [ID] ASC ) ON [PRIMARY] GO -- Create SecondTable CREATE TABLE SecondTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_SecondTable_ID] ON SecondTable ( [ID] ASC ) ON [PRIMARY] GO -- Insert One Hundred Thousand Records INSERT INTO FirstTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Insert One Hundred Thousand Records INSERT INTO SecondTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO You can clearly see that after TRUNCATE, the size of the database is not reduced and it is still the same as before TRUNCATE operation. After the Shrinking database operation, we were able to reduce the size of the database. If you notice the fragmentation, it is considerably high. The major problem with the Shrink operation is that it increases fragmentation of the database to very high value. Higher fragmentation reduces the performance of the database as reading from that particular table becomes very expensive. One of the ways to reduce the fragmentation is to rebuild index on the database. Let us rebuild the index and observe fragmentation and database size. -- Rebuild Index on FirstTable ALTER INDEX IX_SecondTable_ID ON SecondTable REBUILD GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO You can notice that after rebuilding, Fragmentation reduces to a very low value (almost same to original value); however the database size increases way higher than the original. Before rebuilding, the size of the database was 5 MB, and after rebuilding, it is around 20 MB. Regular rebuilding the index is rebuild in the same user database where the index is placed. This usually increases the size of the database. Look at irony of the Shrinking database. One person shrinks the database to gain space (thinking it will help performance), which leads to increase in fragmentation (reducing performance). To reduce the fragmentation, one rebuilds index, which leads to size of the database to increase way more than the original size of the database (before shrinking). Well, by Shrinking, one did not gain what he was looking for usually. Rebuild indexing is not the best suggestion as that will create database grow again. I have always remembered the excellent post from Paul Randal regarding Shrinking the database is bad. I suggest every one to read that for accuracy and interesting conversation. Let us run following script where we Shrink the database and REORGANIZE. -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO -- Shrink the Database DBCC SHRINKDATABASE (ShrinkIsBed); GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO -- Rebuild Index on FirstTable ALTER INDEX IX_SecondTable_ID ON SecondTable REORGANIZE GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO You can see that REORGANIZE does not increase the size of the database or remove the fragmentation. Again, I no way suggest that REORGANIZE is the solution over here. This is purely observation using demo. Read the blog post of Paul Randal. Following script will clean up the database -- Clean up USE MASTER GO ALTER DATABASE ShrinkIsBed SET SINGLE_USER WITH ROLLBACK IMMEDIATE GO DROP DATABASE ShrinkIsBed GO There are few valid cases of the Shrinking database as well, but that is not covered in this blog post. We will cover that area some other time in future. Additionally, one can rebuild index in the tempdb as well, and we will also talk about the same in future. Brent has written a good summary blog post as well. Are you Shrinking your database? Well, when are you going to stop Shrinking it? Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Index, SQL Performance, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, SQLServer, T SQL, Technology

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  • lucene index missing files

    - by Akhil
    I have _0.cfs file of a lucene index directory but segments.gen and segments_2 are missing. Can I generate the segments.gen and segments_2 files without having to regenerate the _0.cfs file. Does these "segments" files contain any index specific data, which will thus force me to regnerate the entire index again. Or can I just generate the two "segments" file by copying these from another lucen index directory gnerated with the same lucene version.

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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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  • Index of elements, jQuery or Javascript

    - by ozsenegal
    I've a table that contains 3 columns. I need to bind an event that fires off whenever one of those columns is clicked using jQuery. However, I need to know the index of the column clicked. i.e: First column (index 0), Second column (index 1), Third column (index 2), and so on... How can I do that?

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Beware Sneaky Reads with Unique Indexes

    - by Paul White NZ
    A few days ago, Sandra Mueller (twitter | blog) asked a question using twitter’s #sqlhelp hash tag: “Might SQL Server retrieve (out-of-row) LOB data from a table, even if the column isn’t referenced in the query?” Leaving aside trivial cases (like selecting a computed column that does reference the LOB data), one might be tempted to say that no, SQL Server does not read data you haven’t asked for.  In general, that’s quite correct; however there are cases where SQL Server might sneakily retrieve a LOB column… Example Table Here’s a T-SQL script to create that table and populate it with 1,000 rows: CREATE TABLE dbo.LOBtest ( pk INTEGER IDENTITY NOT NULL, some_value INTEGER NULL, lob_data VARCHAR(MAX) NULL, another_column CHAR(5) NULL, CONSTRAINT [PK dbo.LOBtest pk] PRIMARY KEY CLUSTERED (pk ASC) ); GO DECLARE @Data VARCHAR(MAX); SET @Data = REPLICATE(CONVERT(VARCHAR(MAX), 'x'), 65540);   WITH Numbers (n) AS ( SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 0)) FROM master.sys.columns C1, master.sys.columns C2 ) INSERT LOBtest WITH (TABLOCKX) ( some_value, lob_data ) SELECT TOP (1000) N.n, @Data FROM Numbers N WHERE N.n <= 1000; Test 1: A Simple Update Let’s run a query to subtract one from every value in the some_value column: UPDATE dbo.LOBtest WITH (TABLOCKX) SET some_value = some_value - 1; As you might expect, modifying this integer column in 1,000 rows doesn’t take very long, or use many resources.  The STATITICS IO and TIME output shows a total of 9 logical reads, and 25ms elapsed time.  The query plan is also very simple: Looking at the Clustered Index Scan, we can see that SQL Server only retrieves the pk and some_value columns during the scan: The pk column is needed by the Clustered Index Update operator to uniquely identify the row that is being changed.  The some_value column is used by the Compute Scalar to calculate the new value.  (In case you are wondering what the Top operator is for, it is used to enforce SET ROWCOUNT). Test 2: Simple Update with an Index Now let’s create a nonclustered index keyed on the some_value column, with lob_data as an included column: CREATE NONCLUSTERED INDEX [IX dbo.LOBtest some_value (lob_data)] ON dbo.LOBtest (some_value) INCLUDE ( lob_data ) WITH ( FILLFACTOR = 100, MAXDOP = 1, SORT_IN_TEMPDB = ON ); This is not a useful index for our simple update query; imagine that someone else created it for a different purpose.  Let’s run our update query again: UPDATE dbo.LOBtest WITH (TABLOCKX) SET some_value = some_value - 1; We find that it now requires 4,014 logical reads and the elapsed query time has increased to around 100ms.  The extra logical reads (4 per row) are an expected consequence of maintaining the nonclustered index. The query plan is very similar to before (click to enlarge): The Clustered Index Update operator picks up the extra work of maintaining the nonclustered index. The new Compute Scalar operators detect whether the value in the some_value column has actually been changed by the update.  SQL Server may be able to skip maintaining the nonclustered index if the value hasn’t changed (see my previous post on non-updating updates for details).  Our simple query does change the value of some_data in every row, so this optimization doesn’t add any value in this specific case. The output list of columns from the Clustered Index Scan hasn’t changed from the one shown previously: SQL Server still just reads the pk and some_data columns.  Cool. Overall then, adding the nonclustered index hasn’t had any startling effects, and the LOB column data still isn’t being read from the table.  Let’s see what happens if we make the nonclustered index unique. Test 3: Simple Update with a Unique Index Here’s the script to create a new unique index, and drop the old one: CREATE UNIQUE NONCLUSTERED INDEX [UQ dbo.LOBtest some_value (lob_data)] ON dbo.LOBtest (some_value) INCLUDE ( lob_data ) WITH ( FILLFACTOR = 100, MAXDOP = 1, SORT_IN_TEMPDB = ON ); GO DROP INDEX [IX dbo.LOBtest some_value (lob_data)] ON dbo.LOBtest; Remember that SQL Server only enforces uniqueness on index keys (the some_data column).  The lob_data column is simply stored at the leaf-level of the non-clustered index.  With that in mind, we might expect this change to make very little difference.  Let’s see: UPDATE dbo.LOBtest WITH (TABLOCKX) SET some_value = some_value - 1; Whoa!  Now look at the elapsed time and logical reads: Scan count 1, logical reads 2016, physical reads 0, read-ahead reads 0, lob logical reads 36015, lob physical reads 0, lob read-ahead reads 15992.   CPU time = 172 ms, elapsed time = 16172 ms. Even with all the data and index pages in memory, the query took over 16 seconds to update just 1,000 rows, performing over 52,000 LOB logical reads (nearly 16,000 of those using read-ahead). Why on earth is SQL Server reading LOB data in a query that only updates a single integer column? The Query Plan The query plan for test 3 looks a bit more complex than before: In fact, the bottom level is exactly the same as we saw with the non-unique index.  The top level has heaps of new stuff though, which I’ll come to in a moment. You might be expecting to find that the Clustered Index Scan is now reading the lob_data column (for some reason).  After all, we need to explain where all the LOB logical reads are coming from.  Sadly, when we look at the properties of the Clustered Index Scan, we see exactly the same as before: SQL Server is still only reading the pk and some_value columns – so what’s doing the LOB reads? Updates that Sneakily Read Data We have to go as far as the Clustered Index Update operator before we see LOB data in the output list: [Expr1020] is a bit flag added by an earlier Compute Scalar.  It is set true if the some_value column has not been changed (part of the non-updating updates optimization I mentioned earlier). The Clustered Index Update operator adds two new columns: the lob_data column, and some_value_OLD.  The some_value_OLD column, as the name suggests, is the pre-update value of the some_value column.  At this point, the clustered index has already been updated with the new value, but we haven’t touched the nonclustered index yet. An interesting observation here is that the Clustered Index Update operator can read a column into the data flow as part of its update operation.  SQL Server could have read the LOB data as part of the initial Clustered Index Scan, but that would mean carrying the data through all the operations that occur prior to the Clustered Index Update.  The server knows it will have to go back to the clustered index row to update it, so it delays reading the LOB data until then.  Sneaky! Why the LOB Data Is Needed This is all very interesting (I hope), but why is SQL Server reading the LOB data?  For that matter, why does it need to pass the pre-update value of the some_value column out of the Clustered Index Update? The answer relates to the top row of the query plan for test 3.  I’ll reproduce it here for convenience: Notice that this is a wide (per-index) update plan.  SQL Server used a narrow (per-row) update plan in test 2, where the Clustered Index Update took care of maintaining the nonclustered index too.  I’ll talk more about this difference shortly. The Split/Sort/Collapse combination is an optimization, which aims to make per-index update plans more efficient.  It does this by breaking each update into a delete/insert pair, reordering the operations, removing any redundant operations, and finally applying the net effect of all the changes to the nonclustered index. Imagine we had a unique index which currently holds three rows with the values 1, 2, and 3.  If we run a query that adds 1 to each row value, we would end up with values 2, 3, and 4.  The net effect of all the changes is the same as if we simply deleted the value 1, and added a new value 4. By applying net changes, SQL Server can also avoid false unique-key violations.  If we tried to immediately update the value 1 to a 2, it would conflict with the existing value 2 (which would soon be updated to 3 of course) and the query would fail.  You might argue that SQL Server could avoid the uniqueness violation by starting with the highest value (3) and working down.  That’s fine, but it’s not possible to generalize this logic to work with every possible update query. SQL Server has to use a wide update plan if it sees any risk of false uniqueness violations.  It’s worth noting that the logic SQL Server uses to detect whether these violations are possible has definite limits.  As a result, you will often receive a wide update plan, even when you can see that no violations are possible. Another benefit of this optimization is that it includes a sort on the index key as part of its work.  Processing the index changes in index key order promotes sequential I/O against the nonclustered index. A side-effect of all this is that the net changes might include one or more inserts.  In order to insert a new row in the index, SQL Server obviously needs all the columns – the key column and the included LOB column.  This is the reason SQL Server reads the LOB data as part of the Clustered Index Update. In addition, the some_value_OLD column is required by the Split operator (it turns updates into delete/insert pairs).  In order to generate the correct index key delete operation, it needs the old key value. The irony is that in this case the Split/Sort/Collapse optimization is anything but.  Reading all that LOB data is extremely expensive, so it is sad that the current version of SQL Server has no way to avoid it. Finally, for completeness, I should mention that the Filter operator is there to filter out the non-updating updates. Beating the Set-Based Update with a Cursor One situation where SQL Server can see that false unique-key violations aren’t possible is where it can guarantee that only one row is being updated.  Armed with this knowledge, we can write a cursor (or the WHILE-loop equivalent) that updates one row at a time, and so avoids reading the LOB data: SET NOCOUNT ON; SET STATISTICS XML, IO, TIME OFF;   DECLARE @PK INTEGER, @StartTime DATETIME; SET @StartTime = GETUTCDATE();   DECLARE curUpdate CURSOR LOCAL FORWARD_ONLY KEYSET SCROLL_LOCKS FOR SELECT L.pk FROM LOBtest L ORDER BY L.pk ASC;   OPEN curUpdate;   WHILE (1 = 1) BEGIN FETCH NEXT FROM curUpdate INTO @PK;   IF @@FETCH_STATUS = -1 BREAK; IF @@FETCH_STATUS = -2 CONTINUE;   UPDATE dbo.LOBtest SET some_value = some_value - 1 WHERE CURRENT OF curUpdate; END;   CLOSE curUpdate; DEALLOCATE curUpdate;   SELECT DATEDIFF(MILLISECOND, @StartTime, GETUTCDATE()); That completes the update in 1280 milliseconds (remember test 3 took over 16 seconds!) I used the WHERE CURRENT OF syntax there and a KEYSET cursor, just for the fun of it.  One could just as well use a WHERE clause that specified the primary key value instead. Clustered Indexes A clustered index is the ultimate index with included columns: all non-key columns are included columns in a clustered index.  Let’s re-create the test table and data with an updatable primary key, and without any non-clustered indexes: IF OBJECT_ID(N'dbo.LOBtest', N'U') IS NOT NULL DROP TABLE dbo.LOBtest; GO CREATE TABLE dbo.LOBtest ( pk INTEGER NOT NULL, some_value INTEGER NULL, lob_data VARCHAR(MAX) NULL, another_column CHAR(5) NULL, CONSTRAINT [PK dbo.LOBtest pk] PRIMARY KEY CLUSTERED (pk ASC) ); GO DECLARE @Data VARCHAR(MAX); SET @Data = REPLICATE(CONVERT(VARCHAR(MAX), 'x'), 65540);   WITH Numbers (n) AS ( SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 0)) FROM master.sys.columns C1, master.sys.columns C2 ) INSERT LOBtest WITH (TABLOCKX) ( pk, some_value, lob_data ) SELECT TOP (1000) N.n, N.n, @Data FROM Numbers N WHERE N.n <= 1000; Now here’s a query to modify the cluster keys: UPDATE dbo.LOBtest SET pk = pk + 1; The query plan is: As you can see, the Split/Sort/Collapse optimization is present, and we also gain an Eager Table Spool, for Halloween protection.  In addition, SQL Server now has no choice but to read the LOB data in the Clustered Index Scan: The performance is not great, as you might expect (even though there is no non-clustered index to maintain): Table 'LOBtest'. Scan count 1, logical reads 2011, physical reads 0, read-ahead reads 0, lob logical reads 36015, lob physical reads 0, lob read-ahead reads 15992.   Table 'Worktable'. Scan count 1, logical reads 2040, physical reads 0, read-ahead reads 0, lob logical reads 34000, lob physical reads 0, lob read-ahead reads 8000.   SQL Server Execution Times: CPU time = 483 ms, elapsed time = 17884 ms. Notice how the LOB data is read twice: once from the Clustered Index Scan, and again from the work table in tempdb used by the Eager Spool. If you try the same test with a non-unique clustered index (rather than a primary key), you’ll get a much more efficient plan that just passes the cluster key (including uniqueifier) around (no LOB data or other non-key columns): A unique non-clustered index (on a heap) works well too: Both those queries complete in a few tens of milliseconds, with no LOB reads, and just a few thousand logical reads.  (In fact the heap is rather more efficient). There are lots more fun combinations to try that I don’t have space for here. Final Thoughts The behaviour shown in this post is not limited to LOB data by any means.  If the conditions are met, any unique index that has included columns can produce similar behaviour – something to bear in mind when adding large INCLUDE columns to achieve covering queries, perhaps. Paul White Email: [email protected] Twitter: @PaulWhiteNZ

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  • Reading from compressed lucene index

    - by Akhil
    I created a lucene index and compressed the index directory with bz2 or zip. I donot want to uncompress it. Is there any API call that can read the index from this zipped directory and thus allow searching and other functionalities. That is, can lucence IndexReader read the index from a compressed file. I saw that Lucnene IndexReader does not support "Reader" to open the index, otherwise I would have created a Reader class that uncompresses the file and streams the uncompressed version. Any alternatives to this are welcome. Thanks, Akhil

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  • how to effectively modify index

    - by daedlus
    Hej everyone, problem : I am looking for right way to convert an index from clustered to non-clustered Description : I have a table as below in sybase db: dbo.UserLog Id | UserId |time | .... This is hash partitioned using UserId. Currently it has 2 indexes UserId : non-clustered time: clustered This table has about 20 million records. I now want to make UserId as clustered index and time as non-clustered index. is it correct to user alter index to change from clustered to non-clustered or do i drop index and recreate. does the fact that userId is used in hash partitioning have any implications to this? To me alter seems way to go but I have not yet tried this.

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  • Use a vector to index a matrix without linear index

    - by David_G
    G'day, I'm trying to find a way to use a vector of [x,y] points to index from a large matrix in MATLAB. Usually, I would convert the subscript points to the linear index of the matrix.(for eg. Use a vector as an index to a matrix in MATLab) However, the matrix is 4-dimensional, and I want to take all of the elements of the 3rd and 4th dimensions that have the same 1st and 2nd dimension. Let me hopefully demonstrate with an example: Matrix = nan(4,4,2,2); % where the dimensions are (x,y,depth,time) Matrix(1,2,:,:) = 999; % note that this value could change in depth (3rd dim) and time (4th time) Matrix(3,4,:,:) = 888; % note that this value could change in depth (3rd dim) and time (4th time) Matrix(4,4,:,:) = 124; Now, I want to be able to index with the subscripts (1,2) and (3,4), etc and return not only the 999 and 888 which exist in Matrix(:,:,1,1) but the contents which exist at Matrix(:,:,1,2),Matrix(:,:,2,1) and Matrix(:,:,2,2), and so on (IRL, the dimensions of Matrix might be more like size(Matrix) = (300 250 30 200) I don't want to use linear indices because I would like the results to be in a similar vector fashion. For example, I would like a result which is something like: ans(time=1) 999 888 124 999 888 124 ans(time=2) etc etc etc etc etc etc I'd also like to add that due to the size of the matrix I'm dealing with, speed is an issue here - thus why I'd like to use subscript indices to index to the data. I should also mention that (unlike this question: Accessing values using subscripts without using sub2ind) since I want all the information stored in the extra dimensions, 3 and 4, of the i and jth indices, I don't think that a slightly faster version of sub2ind still would not cut it..

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