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  • WPF, notify a child in the element tree about an event in a parent

    - by jester
    I am developing a WPF app and I want an event in a parent to be notified to several of its children in the element tree, so that each of them can take an action accordingly. I know that a custom RoutedEvent can be used to signal in the other direction from a child to one of its ancestors by bubbling the event upwards, so that any of the ancestor elements can handle the event. What I want is the children to be notified about an event in the parent and they handle them appropriately. What is the best strategy to achieve this? EDIT: Clarifying the comments : Say I have a parent UserControl. It has a TabControl and its contents are several nested child UserControls. Now consider a scenario where I want the TabControl.SelectionChanged() event to cause some changes in each of the child UserControl. How to achieve this? (The contents of each tab is a UserControl which themselves may contain another few levels of children UserControls. I want the UserControl in the bottom level to know about the SelectionChanged() event and respond accordingly).

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  • Using recursion to to trim a binary tree based on a given min and max value

    - by Justin
    As the title says, I have to trim a binary tree based on a given min and max value. Each node stores a value, and a left/right node. I may define private helper methods to solve this problem, but otherwise I may not call any other methods of the class nor create any data structures such as arrays, lists, etc. An example would look like this: overallRoot _____[50]____________________ / \ __________[38] _______________[90] / \ / _[14] [42] [54]_____ / \ \ [8] [20] [72] \ / \ [26] [61] [83] trim(52, 65); should return: overallRoot [54] \ [61] My attempted solution has three methods: public void trim(int min, int max) { rootFinder(overallRoot, min, max); } First recursive method finds the new root perfectly. private void rootFinder(IntTreeNode node, int min, int max) { if (node == null) return; if (overallRoot.data < min) { node = overallRoot = node.right; rootFinder(node, min, max); } else if (overallRoot.data > max) { node = overallRoot = node.left; rootFinder(node, min, max); } else cutter(overallRoot, min, max); } This second method should eliminate any further nodes not within the min/max, but it doesn't work as I would hope. private void cutter(IntTreeNode node, int min, int max) { if (node == null) return; if (node.data <= min) { node.left = null; } if (node.data >= max) { node.right = null; } if (node.data < min) { node = node.right; } if (node.data > max) { node = node.left; } cutter(node.left, min, max); cutter(node.right, min, max); } This returns: overallRoot [54]_____ \ [72] / [61] Any help is appreciated. Feel free to ask for further explanation as needed.

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  • How can I recurse up a DOM tree?

    - by smartdirt
    So I have a series of nested ul elements as part of a tree like below: <ul> <li> <ul> <li>1.1</li> <li>1.2</li> </ul> <ul> <li>2.1</li> <li> <ul> <li>2.2</li> </ul> </li> </ul> <ul> <li>3.1</li> <li>3.2</li> </ul> </li> </ul> Let's say when 3.1 is the selected node and when the user clicks previous the selected node should then be 2.2. The bad news is that there could be any number of levels deep. How can I find the previous node (li) in relationship to the currently selected node using jquery?

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  • Populate a tree from Hierarchical data using 1 LINQ statement

    - by Midhat
    Hi. I have set up this programming exercise. using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication2 { class DataObject { public int ID { get; set; } public int ParentID { get; set; } public string Data { get; set; } public DataObject(int id, int pid, string data) { this.ID = id; this.ParentID = pid; this.Data = data; } } class TreeNode { public DataObject Data {get;set;} public List<DataObject> Children { get; set; } } class Program { static void Main(string[] args) { List<DataObject> data = new List<DataObject>(); data.Add(new DataObject(1, 0, "Item 1")); data.Add(new DataObject(2, 0, "Item 2")); data.Add(new DataObject(21, 2, "Item 2.1")); data.Add(new DataObject(22, 2, "Item 2.2")); data.Add(new DataObject(221, 22, "Item 2.2.1")); data.Add(new DataObject(3, 0, "Item 3")); } } } The desired output is a List of 3 treenodes, having items 1, 2 and 3. Item 2 will have a list of 2 dataobjects as its children member and so on. I have been trying to populate this tree (or rather a forest) using just 1 SLOC using LINQ. A simple group by gives me the desired data but the challenge is to organize it in TreeNode objects. Can someone give a hint or an impossibility result for this?

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  • Recursion through a directory tree in PHP

    - by phphelpplz
    I have a set of folders that has a depth of at least 4 or 5 levels. I'm looking to recurse through the directory tree as deep as it goes, and iterate over every file. I've gotten the code to go down into the first sets of subdirectories, but no deeper, and I'm not sure why. Any ideas? $count = 0; $dir = "/Applications/MAMP/htdocs/idahohotsprings.com"; function recurseDirs($main, $count){ $dir = "/Applications/MAMP/htdocs/idahohotsprings.com"; $dirHandle = opendir($main); echo "here"; while($file = readdir($dirHandle)){ if(is_dir($file) && $file != '.' && $file != '..'){ echo "isdir"; recurseDirs($file); } else{ $count++; echo "$count: filename: $file in $dir/$main \n<br />"; } } } recurseDirs($dir, $count); ?

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  • Tree-like queues

    - by Rehno Lindeque
    I'm implementing a interpreter-like project for which I need a strange little scheduling queue. Since I'd like to try and avoid wheel-reinvention I was hoping someone could give me references to a similar structure or existing work. I know I can simply instantiate multiple queues as I go along, I'm just looking for some perspective by other people who might have better ideas than me ;) I envision that it might work something like this: The structure is a tree with a single root. You get a kind of "insert_iterator" to the root and then push elements onto it (e.g. a and b in the example below). However, at any point you can also split the iterator into multiple iterators, effectively creating branches. The branches cannot merge into a single queue again, but you can start popping elements from the front of the queue (again, using a kind of "visitor_iterator") until empty branches can be discarded (at your discretion). x -> y -> z a -> b -> { g -> h -> i -> j } f -> b Any ideas? Seems like a relatively simple structure to implement myself using a pool of circular buffers but I'm following the "think first, code later" strategy :) Thanks

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  • breadth-first traversal of directory tree is not lazy

    - by user855443
    I try to traverse the diretory tree. A naive depth-first traversal seems not to produce the data in a lazy fashion and runs out of memory. I next tried a breadth first approach, which shows the same problem - it uses all the memory available and then crashes. the code i have is: getFilePathBreadtFirst :: FilePath -> IO [FilePath] getFilePathBreadtFirst fp = do fileinfo <- getInfo fp res :: [FilePath] <- if isReadableDirectory fileinfo then do children <- getChildren fp lower <- mapM getFilePathBreadtFirst children return (children ++ concat lower) return (children ++ concat () else return [fp] -- should only return the files? return res getChildren :: FilePath -> IO [FilePath] getChildren path = do names <- getUsefulContents path let namesfull = map (path </>) names return namesfull testBF fn = do -- crashes for /home/frank, does not go to swap fps <- getFilePathBreadtFirst fn putStrLn $ unlines fps I think all the code is either linear or tail recursive, and I would expect that the listing of filenames starts immediately, but in fact it does not. Where is the error in my code and my thinking? where have I lost lazy evaluation?

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  • Optimizing C++ Tree Generation

    - by cam
    Hi, I'm generating a Tic-Tac-Toe game tree (9 seconds after the first move), and I'm told it should take only a few milliseconds. So I'm trying to optimize it, I ran it through CodeAnalyst and these are the top 5 calls being made (I used bitsets to represent the Tic-Tac-Toe board): std::_Iterator_base::_Orphan_me std::bitset<9::test std::_Iterator_base::_Adopt std::bitset<9::reference::operator bool std::_Iterator_base::~_Iterator_base void BuildTreeToDepth(Node &nNode, const int& nextPlayer, int depth) { if (depth > 0) { //Calculate gameboard states int evalBoard = nNode.m_board.CalculateBoardState(); bool isFinished = nNode.m_board.isFinished(); if (isFinished || (nNode.m_board.isWinner() > 0)) { nNode.m_winCount = evalBoard; } else { Ticboard tBoard = nNode.m_board; do { int validMove = tBoard.FirstValidMove(); if (validMove != -1) { Node f; Ticboard tempBoard = nNode.m_board; tempBoard.Move(validMove, nextPlayer); tBoard.Move(validMove, nextPlayer); f.m_board = tempBoard; f.m_winCount = 0; f.m_Move = validMove; int currPlay = (nextPlayer == 1 ? 2 : 1); BuildTreeToDepth(f,currPlay, depth - 1); nNode.m_winCount += f.m_board.CalculateBoardState(); nNode.m_branches.push_back(f); } else { break; } }while(true); } } } Where should I be looking to optimize it? How should I optimize these 5 calls (I don't recognize them=.

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  • Any way to recover ext4 filesystems from a deleted LVM logical volume?

    - by Vegar Nilsen
    The other day I had a proper brain fart moment while expanding a disk on a Linux guest under Vmware. I stretched the Vmware disk file to the desired size and then I did what I usually do on Linux guests without LVM: I deleted the LVM partition and recreated it, starting in the same spot as the old one, but extended to the new size of the disk. (Which will be followed by fsck and resize2fs.) And then I realized that LVM doesn't behave the same way as ext2/3/4 on raw partitions... After restoring the Linux guest from the most recent backup (taken only five hours earlier, luckily) I'm now curious on how I could have recovered from the following scenario. It's after all virtually guaranteed that I'll be a dumb ass in the future as well. Virtual Linux guest with one disk, partitioned into one /boot (primary) partition (/dev/sda1) of 256MB, and the rest in a logical, extended partition (/dev/sda5). /dev/sda5 is then setup as a physical volume with pvcreate, and one volume group (vgroup00) created on top of it with the usual vgcreate command. vgroup00 is then split into two logical volumes root and swap, which are used for / and swap, logically. / is an ext4 file system. Since I had backups of the broken guest I was able to recreate the volume group with vgcfgrestore from the backup LVM setup found under /etc/lvm/backup, with the same UUID for the physical volume and all that. After running this I had two logical volumes with the same size as earlier, with 4GB free space where I had stretched the disk. However, when I tried to run "fsck /dev/mapper/vgroup00-root" it complained about a broken superblock. I tried to locate backup superblocks by running "mke2fs -n /dev/mapper/vgroup00-root" but none of those worked either. Then I tried to run TestDisk but when I asked it to find superblocks it only gave an error about not being able to open the file system due to a broken file system. So, with the default allocation policy for LVM2 in Ubuntu Server 10.04 64-bit, is it possible that the logical volumes are allocated from the end of the volume group? That would definitely explain why the restored logical volumes didn't contain the expected data. Could I have recovered by recreating /dev/sda5 with exactly the same size and disk position as earlier? Are there any other tools I could have used to find and recover the file system? (And clearly, the question is not whether or not I should have done this in a different way from the start, I know that. This is a question about what to do when shit has already hit the fan.)

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  • How to resize / enlarge / grow a non-LVM ext4 partition

    - by Mischa
    I have already searched the forums, but couldnt find a good suitable answer: I have an Ubuntu Server 10.04 as KVM Host and a guest system, that also runs 10.04. The host system uses LVM and there are three logical volumes, which are provided to the guest as virtual block devices - one for /, one for /home and one for swap. The guest had been partitioned without LVM. I have already enlarged the logical volume in the host system - the guest successfully sees the bigger virtual disk. However, this virtual disk contains one "good old" partition, which still has the old small size. The output of fdisk -l is me@produktion:/$ LC_ALL=en_US sudo fdisk -l Disk /dev/vda: 32.2 GB, 32212254720 bytes 255 heads, 63 sectors/track, 3916 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000c8ce7 Device Boot Start End Blocks Id System /dev/vda1 * 1 3917 31455232 83 Linux Disk /dev/vdb: 2147 MB, 2147483648 bytes 244 heads, 47 sectors/track, 365 cylinders Units = cylinders of 11468 * 512 = 5871616 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000f2bf7 Device Boot Start End Blocks Id System /dev/vdb1 1 366 2095104 82 Linux swap / Solaris Partition 1 has different physical/logical beginnings (non-Linux?): phys=(0, 32, 33) logical=(0, 43, 28) Partition 1 has different physical/logical endings: phys=(260, 243, 47) logical=(365, 136, 44) Disk /dev/vdc: 225.5 GB, 225485783040 bytes 255 heads, 63 sectors/track, 27413 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00027f25 Device Boot Start End Blocks Id System /dev/vdc1 1 9138 73398272 83 Linux The output of parted print all is Model: Virtio Block Device (virtblk) Disk /dev/vda: 32.2GB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 1049kB 32.2GB 32.2GB primary ext4 boot Model: Virtio Block Device (virtblk) Disk /dev/vdb: 2147MB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 1049kB 2146MB 2145MB primary linux-swap(v1) Model: Virtio Block Device (virtblk) Disk /dev/vdc: 225GB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 1049kB 75.2GB 75.2GB primary ext4 What I want to achieve is to simply grow or resize the partition /dev/vdc1 so that it uses the whole space provided by the virtual block device /dev/vdc. The problem is, that when I try to do that with parted, it complains: (parted) select /dev/vdc Using /dev/vdc (parted) print Model: Virtio Block Device (virtblk) Disk /dev/vdc: 225GB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 1049kB 75.2GB 75.2GB primary ext4 (parted) resize 1 WARNING: you are attempting to use parted to operate on (resize) a file system. parted's file system manipulation code is not as robust as what you'll find in dedicated, file-system-specific packages like e2fsprogs. We recommend you use parted only to manipulate partition tables, whenever possible. Support for performing most operations on most types of file systems will be removed in an upcoming release. Start? [1049kB]? End? [75.2GB]? 224GB Error: File system has an incompatible feature enabled. Compatible features are has_journal, dir_index, filetype, sparse_super and large_file. Use tune2fs or debugfs to remove features. So what can I do? This is a headless production system. What is a safe way to grow this partition? I CAN unmount it, though - so this is not the problem.

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  • B-trees that use redistribution on insertion

    - by Phenom
    If I insert the following keys into a B-tree of order 4 (meaning 4 pointers and 3 elements in each node), I get the following B-tree. G / \ A IY Would it look any different if redistribution on insertion were used? How does redistribution on insertion work?

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  • C++ vector and segmentation faults

    - by Headspin
    I am working on a simple mathematical parser. Something that just reads number = 1 + 2; I have a vector containing these tokens. They store a type and string value of the character. I am trying to step through the vector to build an AST of these tokens, and I keep getting segmentation faults, even when I am under the impression my code should prevent this from happening. Here is the bit of code that builds the AST: struct ASTGen { const vector<Token> &Tokens; unsigned int size, pointer; ASTGen(const vector<Token> &t) : Tokens(t), pointer(0) { size = Tokens.size() - 1; } unsigned int next() { return pointer + 1; } Node* Statement() { if(next() <= size) { switch(Tokens[next()].type) { case EQUALS : Node* n = Assignment_Expr(); return n; } } advance(); } void advance() { if(next() <= size) ++pointer; } Node* Assignment_Expr() { Node* lnode = new Node(Tokens[pointer], NULL, NULL); advance(); Node* n = new Node(Tokens[pointer], lnode, Expression()); return n; } Node* Expression() { if(next() <= size) { advance(); if(Tokens[next()].type == SEMICOLON) { Node* n = new Node(Tokens[pointer], NULL, NULL); return n; } if(Tokens[next()].type == PLUS) { Node* lnode = new Node(Tokens[pointer], NULL, NULL); advance(); Node* n = new Node(Tokens[pointer], lnode, Expression()); return n; } } } }; ... ASTGen AST(Tokens); Node* Tree = AST.Statement(); cout << Tree->Right->Data.svalue << endl; I can access Tree->Data.svalue and get the = Node's token info, so I know that node is getting spawned, and I can also get Tree->Left->Data.svalue and get the variable to the left of the = I have re-written it many times trying out different methods for stepping through the vector, but I always get a segmentation fault when I try to access the = right node (which should be the + node) Any help would be greatly appreciated.

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  • Binary Trees in Scheme

    - by Javier
    Consider the following BNF defining trees of numbers. Notice that a tree can either be a leaf, a node-1 with one subtrees, or a node-2 with two subtrees. tree ::= (’leaf number) | (’node-1 tree) | (’node-2 tree tree) a. Write a template for recursive procedures on these trees. b. Define the procedure (leaf-count t) that returns the number of leaves in t > (leaf-count ’(leaf 5)) 1 > (leaf-count ’(node-2 (leaf 25) (leaf 17))) 2 > (leaf-count ’(node-1 (node-2 (leaf 4) (node-2 (leaf 2) (leaf 3))))) 3 Here's what I have so far: ;define what a leaf, node-1, and node-2 is (define leaf list) (define node-1 list) (define node-2 list) ;procedure to decide if a list is a leaf or a node (define (leaf? tree) (number? (car tree))) (define (node? tree) (pair? (car tree))) (define (leaf-count tree) (cond ((null? tree) 0) ((number? tree) 0) ((leaf? tree) 1) (else (+ (leaf-count (car tree)) (leaf-count (cdr tree)))))) It looks like it should run just fine, but when I try to run it using a simple test case like (leaf-count '(leaf 5)) I get the following error message: car: expects argument of type pair; given leaf What does this error message mean? I am defining a leaf as a list. But for some reason, it's not seeing that and gives me that error message.

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  • How to make a tree view from MySQL and PHP and jquery

    - by Mac Taylor
    hey guys i need to show a treeview of my categories , saved in my mysql database . Database table : table : cats : columns: id,name,parent Here is a sample of what I want the markup to be like: <ul id="browser" class="filetree"> <li><span class="folder">Folder 1</span> <ul> <li><span class="file">Item 1.1</span></li> </ul> </li> <li><span class="folder">Folder 2</span> <ul> <li><span class="folder">Subfolder 2.1</span> <ul id="folder21"> <li><span class="file">File 2.1.1</span></li> <li><span class="file">File 2.1.2</span></li> </ul> </li> <li><span class="file">File 2.2</span></li> </ul> </li> <li><span class="file">File 4</span></li> </ul> i used this script to show treeview : http://www.dynamicdrive.com/dynamicindex1/treeview now problem is in php part : //function to build tree menu from db table test1 function tree_set($index) { global $menu; $q=mysql_query("select * from cats where parent='$index'"); if(!mysql_num_rows($q)) return; $menu .= '<ul>'."\n"; while($arr=mysql_fetch_assoc($q)) { $menu .= '<li>'; $menu .= '<span class="file">'.$arr['name'].'</span>';//you can add another output there $menu .=tree_set("".$arr['id'].""); $menu .= '</li>'."\n"; } $menu.= '</ul>'."\n"; return $menu; } //variable $menu must be defined before the function call $menu = ' <link rel="stylesheet" href="modules/Topics/includes/jquery.treeview.css" /> <script src="modules/Topics/includes/lib/jquery.cookie.js" type="text/javascript"></script> <script src="modules/Topics/includes/jquery.treeview.js" type="text/javascript"></script> <script type="text/javascript" src="modules/Topics/includes/demo/demo.js"></script> <ul id="browser" class="filetree">'."\n"; $menu .= tree_set(0); $menu .= '</ul>'; echo $menu; i even asked in this forum : http://forums.tizag.com/showthread.php?p=60649 problem is in php part of my codes that i mentioned . i cant show sub menus , i mean , really i dont know how to show sub menus is there any chance of a pro php coder helping me here ?

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  • Demantra 7.3.1.3 Controlling MDP_MATRIX Combinations Assigned to Forecasting Tasks Using TargetTaskSize

    - by user702295
    New 7.3.1.3 parameter: TargetTaskSize Old parameter: BranchID  Multiple, deprecated  7.3.1.3 onwards Parameter Location: Parameters > System Parameters > Engine > Proport   Default: 0   Engine Mode: Both   Details: Specifies how many MDP_MATRIX combinations the analytical engine attempts to assign to each forecasting task.  Allocation will be affected by forecsat tree branch size.  TaskTargetSize is automcatically calculated.  It holds the perferred branch size, in number of combinations in the lowest level. This parameter is adjusted to a lower value for smaller schemas, depending on the number of available engines.   - As the forecast is generated the engine goes up the tree using max_fore_level and not top_level -1.  Max_fore_level has     to be less than or equal to top_level -1.  Due to this requirement, combinations falling under the same top level -1     member must be in the same task.  A member of the top level -1 of the forecast tree is known as a branch.  An engine     task is therefore comprised of one or more branches.     - Reveal current task size       go to Engine Administrator --> View --> Branch Information and run the application on your Demantra schema.  This will be deprecated in 7.3.1.3 since there is no longer a means of adjusting the brach size directly.  The focus is now on proper hierarchy / forecast design.     - Control of tasks       The number of tasks created is the lowest of number of branches, as defined by top level -1 members in forecast       tree, and engine sessions and the value of TargetTaskSize.  You are used to using the branch multiplier in this       calculation.  As of 7.3.1.3, the branch ID multiple is deprecated.     - Discovery of current branch size       To resolve this you must review the 2nd highest level in the forecast tree (below highest/highest) as this is the       level which determines the size of the branches.  If a few resulting tasks are too large it is recommended that       the forecast tree level driving branches be revised or at times completely removed from the forecast tree.     - Control of foreacast tree branch size         - Run the following sql to determine how even the branches are being split by the engine:             select count(*),branch_id from mdp_matrix where prediction_status = 1 and do_fore = 1 group by branch_id;             This will give you an understanding if some of the individual branches have an unusually large number of           rows and thus might indicate that the engine is not efficiently dividing up the parallel tasks.         - Based on the results of this sql, we may want to adjust the branch id multiplier and/or the number of engines           (both of these settings are found in the Engine Administrator)           select count(*), level_id from mdp_matrix where prediction_status = 1 and do_fore = 1 group by level_id;           This will give us an understanding at which level of the Forecast tree where the forecast is being generated.            Having a majority of combinations higher on the forecast tree might indicate either a poorly designed forecast           tree and/or engine parameters that are too strict           Based on the results of this we would adjust the Forecast Tree to see if choosing a different hierarchy might           produce a forecast, with more combinations, at a lower level.           For example:             - Review the 2nd highest level in the forecast tree, below highest/highest, as this is the level which               determines the size of the branches.             - If a few resulting tasks are too large it is recommended that the forecast tree level driving branches               be revised or at times completely removed from the forecast tree.               - For example, if the highest level of the forecast tree is set to Brand/All Locations.             - You have 10 brands but 2 of the brands account for 67% and 29% of all combinations.             - There is a distinct possibility that the tasks resulting from these 2 branches will be too large for               a single engine to process.  Some possible solutions could be to remove the Brand level and instead               use a different product grouping which has a more even distribution, possibly Product Group.               - It is also possible to add a location dimension to this forecast tree level, for example Customer.                This will also reduce forecast tree branch size and will deliver a balanced task allocation.             - A correctly configured Forecast Tree is something that is done by the Implementation team and is               not the responsibility of Oracle Support.  Allocation will be affected by forecast tree branch size.  When TargetTaskSize is set to 0, the default value, the system automatically calculates a value for 'TargetTaskSize' depending on the number of engines.   - QUESTION:  Does this mean that if TargetTaskSize is 1, we use tree branch size to allocate branches to tasks instead                of automatically calculating the size?     ANSWER: DEV Strongly recommends that the setting of TargetTaskSize remain at the DEFAULT of ZERO (0).   - How to control the number of engines?     Determine how many CPUs are on the machine(s) that is (are) running the engine.  As mentioned earlier, the general     rule is that you should designate 2 engines per each CPU that is available.  So for example, if you are running the     engine on a machine that has 4 CPU then you can have up to 8 engines designated in the Engine Administrator.  In this     type of architecture then instead of having one 'localhost' in your Engine Settings Screen, you would have 'localhost'     repeated eight times in this field.     Where do I set the number of engines?                 To add multiples computers where engine will run, please do a back-up of Settings.xml file under         Analytical Engines\bin\ folder, then edit it and add there the selected machines.                 Example, this will allow 3 engines to start:         - <Entry>           <Key argument="ComputerNames" />           <Value type="string" argument="localhost,localhost,localhost" />           </Entry Otherwise, if there are no additional engines defined, the calculated value of 'TargetTaskSize' is used. (Oracle does not recommend changing the default value.) The TargetTaskSize holds the engines prefered branch size, in number of level 1 combinations.   - Level 1 combinations, known as group size The engine manager will use this parameter to attempt creating branches with similar size.   * The engine manager will not create engines that do not have a branch. The engine divider algorithm uses the value of 'TargetTaskSize' as a system-preferred branch size to create branches that are more equal in size which improves engine performance.  The engine divider will try to add as many tasks as possible to an existing branch, up to the limit of 'TargetTaskSize' level 1 combinations, before adding new branches. Coming up next: - The engine divider - Group size - Level 1 combinations - MAX_FORE_LEVEL - Engine Parameters  

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  • NEC Corporation uPD720200 USB 3.0 controller doesn't run at full speed

    - by Radek Zyskowski
    I have fresh install of Ubuntu 10.10. I have external HD on USB 3.0. Trying to connect this via PCI Express NEC controller. dmesg: [ 8966.820078] usb 6-3: new high speed USB device using xhci_hcd and address 0 [ 8966.839831] xhci_hcd 0000:02:00.0: WARN: short transfer on control ep [ 8966.840580] xhci_hcd 0000:02:00.0: WARN: short transfer on control ep [ 8966.841329] xhci_hcd 0000:02:00.0: WARN: short transfer on control ep [ 8966.842079] xhci_hcd 0000:02:00.0: WARN: short transfer on control ep [ 8966.843343] scsi8 : usb-storage 6-3:1.0 [ 8967.847144] scsi 8:0:0:0: Direct-Access SAMSUNG HD204UI 1AQ1 PQ: 0 ANSI: 5 [ 8967.847589] sd 8:0:0:0: Attached scsi generic sg2 type 0 [ 8967.847923] sd 8:0:0:0: [sdb] 3907029168 512-byte logical blocks: (2.00 TB/1.81 TiB) [ 8967.848341] xhci_hcd 0000:02:00.0: WARN: Stalled endpoint [ 8967.850959] sd 8:0:0:0: [sdb] Write Protect is off [ 8967.850963] sd 8:0:0:0: [sdb] Mode Sense: 23 00 00 00 [ 8967.850966] sd 8:0:0:0: [sdb] Assuming drive cache: write through [ 8967.851818] xhci_hcd 0000:02:00.0: WARN: Stalled endpoint [ 8967.852365] sd 8:0:0:0: [sdb] Assuming drive cache: write through [ 8967.852370] sdb: sdb1 [ 8967.871315] xhci_hcd 0000:02:00.0: WARN: Stalled endpoint [ 8967.871853] sd 8:0:0:0: [sdb] Assuming drive cache: write through [ 8967.871856] sd 8:0:0:0: [sdb] Attached SCSI disk [ 8967.950728] xhci_hcd 0000:02:00.0: WARN: Stalled endpoint [ 8967.951355] sd 8:0:0:0: [sdb] Sense Key : Recovered Error [current] [descriptor] [ 8967.951361] Descriptor sense data with sense descriptors (in hex): [ 8967.951363] 72 01 04 1d 00 00 00 0e 09 0c 00 00 00 00 00 00 [ 8967.951375] 00 00 00 00 00 50 [ 8967.951380] sd 8:0:0:0: [sdb] ASC=0x4 ASCQ=0x1d [ 8968.790076] xhci_hcd 0000:02:00.0: HC died; cleaning up [ 8968.790076] usb 6-3: USB disconnect, address 2 [ 8999.008554] scsi 8:0:0:0: [sdb] Unhandled error code [ 8999.008558] scsi 8:0:0:0: [sdb] Result: hostbyte=DID_TIME_OUT driverbyte=DRIVER_OK [ 8999.008562] scsi 8:0:0:0: [sdb] CDB: Read(10): 28 00 74 70 97 39 00 00 3e 00 [ 8999.008573] end_request: I/O error, dev sdb, sector 1953535801 [ 8999.008578] Buffer I/O error on device sdb1, logical block 1953535738 [ 8999.008582] Buffer I/O error on device sdb1, logical block 1953535739 [ 8999.008585] Buffer I/O error on device sdb1, logical block 1953535740 [ 8999.008589] Buffer I/O error on device sdb1, logical block 1953535741 [ 8999.008592] Buffer I/O error on device sdb1, logical block 1953535742 [ 8999.008595] Buffer I/O error on device sdb1, logical block 1953535743 [ 8999.008600] Buffer I/O error on device sdb1, logical block 1953535744 [ 8999.008603] Buffer I/O error on device sdb1, logical block 1953535745 [ 8999.008606] Buffer I/O error on device sdb1, logical block 1953535746 [ 8999.008609] Buffer I/O error on device sdb1, logical block 1953535747 [ 8999.008642] scsi 8:0:0:0: rejecting I/O to offline device [ 8999.008747] scsi 8:0:0:0: [sdb] Unhandled error code [ 8999.008749] scsi 8:0:0:0: [sdb] Result: hostbyte=DID_NO_CONNECT driverbyte=DRIVER_OK [ 8999.008752] scsi 8:0:0:0: [sdb] CDB: Read(10): 28 00 74 70 97 77 00 00 3e 00 [ 8999.008760] end_request: I/O error, dev sdb, sector 1953535863 sudo lspci -v 2:00.0 USB Controller: NEC Corporation uPD720200 USB 3.0 Host Controller (rev 03) (prog-if 30) Physical Slot: 32 Flags: bus master, fast devsel, latency 0, IRQ 16 Memory at fe9fe000 (64-bit, non-prefetchable) [size=8K] Capabilities: [50] Power Management version 3 Capabilities: [70] MSI: Enable- Count=1/8 Maskable- 64bit+ Capabilities: [90] MSI-X: Enable- Count=8 Masked- Capabilities: [a0] Express Endpoint, MSI 00 Capabilities: [100] Advanced Error Reporting Capabilities: [140] Device Serial Number ff-ff-ff-ff-ff-ff-ff-ff Capabilities: [150] #18 Kernel driver in use: xhci_hcd Kernel modules: xhci-hcd If I try to put into this controller any USB 2.0, it works fine. But USB 3.0 nope. Any idea?

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  • New record may be written twice in clusterd index structure

    - by Cupidvogel
    As per the article at Microsoft, under the Test 1: INSERT Performance section, it is written that For the table with the clustered index, only a single write operation is required since the leaf nodes of the clustered index are data pages (as explained in the section Clustered Indexes and Heaps), whereas for the table with the nonclustered index, two write operations are required—one for the entry into the index B-tree and another for the insert of the data itself. I don't think that is necessarily true. Clustered Indexes are implemented through B+ tree structures, right? If you look at at this article, which gives a simple example of inserting into a B+ tree, we can see that when 8 is initially inserted, it is written only once, but then when 5 comes in, it is written to the root node as well (thus written twice, albeit not initially at the time of insertion). Also when 8 comes in next, it is written twice, once at the root and then at the leaf. So won't it be correct to say, that the number of rewrites in case of a clustered index is much less compared to a NIC structure (where it must occur every time), instead of saying that rewrite doesn't occur in CI at all?

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  • Cannot view, use, or open CDs or DVDs in Ubuntu 12.04

    - by user67592
    I am fairly new to Ubuntu 12.04 and I have encountered a rather irritating problem. Whenever I insert a CD or DVD (whether it have data, music, movies, or nothing at all), nothing pops up saying "you have inserted a CD", "play with Rhythmbox?" etc. It doesn't show the CD in the launcher/dock or anything of the sort. This is especially peculiar because not only do I have a standard IDE built-in optical drive, but I have an external USB optical drive. Neither work. In addition, whenever I go to "Computer///" and I click (double click, right click, or even left click) on "CD/DVD Drive" nothing happens, when I right click and select "Open" nothing happens either [for either of the two drives (both are listed in Computer///)] And if I insert a blank disk and go to a disk burning program such as Brasero, and try to burn to the drive it detects no CDs or DVDs of any kind. I'm rather stumped and can't seem to find a question similar to this. :( Thanks for all your help in advance!! :) ~Preston Output of sudo lshw *-cdrom description: DVD-RAM writer product: CD/DVDW TS-H652M vendor: TSSTcorp physical id: 0.0.0 bus info: scsi@5:0.0.0 logical name: /dev/cdrom logical name: /dev/cdrw logical name: /dev/dvd logical name: /dev/dvdrw logical name: /dev/sr0 version: 0414 capabilities: removable audio cd-r cd-rw dvd dvd-r dvd-ram configuration: ansiversion=5 status=nodisc *-cdrom description: DVD reader product: DVD Writer 300n vendor: HP physical id: 0.0.0 bus info: scsi@4:0.0.0 logical name: /dev/cdrom2 logical name: /dev/cdrw2 logical name: /dev/dvd2 logical name: /dev/sr1 version: 1.25 serial: [ capabilities: removable audio cd-r cd-rw dvd configuration: status=nodisc

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  • More localized, efficient Lowest Common Ancestor algorithm given multiple binary trees?

    - by mstksg
    I have multiple binary trees stored as an array. In each slot is either nil (or null; pick your language) or a fixed tuple storing two numbers: the indices of the two "children". No node will have only one child -- it's either none or two. Think of each slot as a binary node that only stores pointers to its children, and no inherent value. Take this system of binary trees: 0 1 / \ / \ 2 3 4 5 / \ / \ 6 7 8 9 / \ 10 11 The associated array would be: 0 1 2 3 4 5 6 7 8 9 10 11 [ [2,3] , [4,5] , [6,7] , nil , nil , [8,9] , nil , [10,11] , nil , nil , nil , nil ] I've already written simple functions to find direct parents of nodes (simply by searching from the front until there is a node that contains the child) Furthermore, let us say that at relevant times, both all trees are anywhere between a few to a few thousand levels deep. I'd like to find a function P(m,n) to find the lowest common ancestor of m and n -- to put more formally, the LCA is defined as the "lowest", or deepest node in which have m and n as descendants (children, or children of children, etc.). If there is none, a nil would be a valid return. Some examples, given our given tree: P( 6,11) # => 2 P( 3,10) # => 0 P( 8, 6) # => nil P( 2,11) # => 2 The main method I've been able to find is one that uses an Euler trace, which turns the given tree, with a node A to be the invisible parent of 0 and 1 with a depth of -1, into: A-0-2-6-2-7-10-7-11-7-2-0-3-0-A-1-4-1-5-8-5-9-5-1-A And from that, simply find the node between your given m and n that has the lowest number; For example, to find P(6,11), look for a 6 and an 11 on the trace. The number between them that is the lowest is 2, and that's your answer. If A is in between them, return nil. -- Calculating P(6,11) -- A-0-2-6-2-7-10-7-11-7-2-0-3-0-A-1-4-1-5-8-5-9-5-1-A ^ ^ ^ | | | m lowest n Unfortunately, I do believe that finding the Euler trace of a tree that can be several thousands of levels deep is a bit machine-taxing...and because my tree is constantly being changed throughout the course of the programming, every time I wanted to find the LCA, I'd have to re-calculate the Euler trace and hold it in memory every time. Is there a more memory efficient way, given the framework I'm using? One that maybe iterates upwards? One way I could think of would be the "count" the generation/depth of both nodes, and climb the lowest node until it matched the depth of the highest, and increment both until they find someone similar. But that'd involve climbing up from level, say, 3025, back to 0, twice, to count the generation, and using a terribly inefficient climbing-up algorithm in the first place, and then re-climbing back up. Are there any other better ways?

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  • Right rotate of tree in Haskell: how is it work?

    - by Roman
    I don't know haskell syntax, but I know some FP concepts (like algebraic data types, pattern matching, higher-order functions ect). Can someone explain please, what does this code mean: data Tree ? = Leaf ? | Fork ? (Tree ?) (Tree ?) rotateR tree = case tree of Fork q (Fork p a b) c -> Fork p a (Fork q b c) As I understand, first line is something like Tree-type declaration (but I don't understand it exactly). Second line includes pattern matching (I don't understand as well why do we need to use pattern matching here). And third line does something absolutely unreadable for non-haskell developer. I've found definition of Fork as fork (f,g) x = (f x, g x) but I can't move further anymore.

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  • How to log the output from cmd tree command using Apache Ant exec task?

    - by S.N
    Hi, I am trying to log the output from cmd tree command using ant with the following: <exec dir="${basedir}" executable="cmd" output="output.txt"> <arg value="tree" /> </exec> However, I am seeing the following in the "output.txt": Microsoft Windows XP [Version 5.1.2600] (C) Copyright 1985-2001 Microsoft Corp. When I run the command in the windows cmd: C:\tree>tree I get something like: C:\tree +---test +---test Can anyone tell me how to write a Ant script to print the tree structure in to a file?

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  • Prolog Family tree

    - by Tania
    Hi I have a Question in prolog , I did it but its not showing answers When i ask about the brothers,sisters,uncles,aunts This is what I wrote, what's wrong ? /*uncle(X, Y) :– male(X), sibling(X, Z), parent(Z, Y).*/ /*uncle(X, Y) :– male(X), spouse(X, W), sibling(W, Z), parent(Z, Y).*/ uncle(X,Y) :- parent(Z,Y), brother(X,Z). aunt(X,Y) :- parent(Z,Y), sister(X,Z). sibling(X, Y) :- parent(Z, X), parent(Z, Y), X \= Y. sister(X, Y) :- sibling(X, Y), female(X). brother(X, Y) :- sibling(X, Y), male(X). parent(Z,Y) :- father(Z,Y). parent(Z,Y) :- mother(Z,Y). grandparent(C,D) :- parent(C,E), parent(E,D). aunt(X, Y) :– female(X), sibling(X, Z), parent(Z, Y). aunt(X, Y) :– female(X), spouse(X, W), sibling(W, Z), parent(Z, Y). male(john). male(bob). male(bill). male(ron). male(jeff). female(mary). female(sue). female(nancy). mother(mary, sue). mother(mary, bill). mother(sue, nancy). mother(sue, jeff). mother(jane, ron). father(john, sue). father(john, bill). father(bob, nancy). father(bob, jeff). father(bill, ron). sibling(bob,bill). sibling(sue,bill). sibling(nancy,jeff). sibling(nancy,ron). sibling(jell,ron). And one more thing, how do I optimize the rule of the brother so that X is not brother to itself.

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  • WPF Tree doesn't work

    - by phenevo
    Could you tell me why I can't see subItems? I've got winforms apps and I added my wpfusercontrol:ObjectsAndZonesTree ServiceProvider is my webservice. Adn method to get listofcountires with subitems works properly (i get countires, regions from this countires, provinces etc...) ElementHost elementHost = new ElementHost { Width = 150, Height = 50, Dock = DockStyle.Fill, Child = new ObjectsAndZonesTree() }; this.splitContainer3.Panel1.Controls.Add(elementHost); XAML: <TreeView Name="GroupView" Grid.Row="0" Grid.Column="0" ItemsSource="{Binding}"> <TreeView.Resources> <HierarchicalDataTemplate DataType="{x:Type ServiceProvider:Country }" ItemsSource="{Binding Items}"> <TextBlock Text="{Binding Path=Name}" /> </HierarchicalDataTemplate> <DataTemplate DataType="{x:Type ServiceProvider:Region}" > <TextBlock Text="{Binding Path=Name}" /> </DataTemplate> <DataTemplate DataType="{x:Type ServiceProvider:Province}" > <TextBlock Text="{Binding Path=Name}" /> </DataTemplate> </TreeView.Resources> </TreeView> XAML.CS public ObjectsAndZonesTree() { InitializeComponent(); LoadView(); } private void LoadView() { GroupView.ItemsSource = new ServiceProvider().GetListOfObjectsAndZones(); } class Country: public class Country { string _name; [XmlAttribute] public string Name { get { return _name; } set { _name = value; } } string _code; [XmlAttribute] public string Code { get { return _code; } set { _code = value; } } string _continentCode; [XmlAttribute] public string ContinentCode { get { return _continentCode; } set { _continentCode = value; } } public Region[] ListOfRegions { get { return _listOfRegions; } set { _listOfRegions = value; } } private Region[] _listOfRegions; public IList<object> Items { get { IList<object> childNodes = new List<object>(); foreach (var group in this.ListOfRegions) childNodes.Add(group); return childNodes; } } } Class Region: public class Region { private Province[] _listOfProvinces; private string _name; private string _code; public Province[] ListOfProvinces { get { return _listOfProvinces; } set { _listOfProvinces = value; } } public string Name { get { return _name; } set { _name = value; } } public string Code { get { return _code; } set { _code = value; } } public string CountryCode { get { return _countryCode; } set { _countryCode = value; } } private string _countryCode; public IList<object> Items { get { IList<object> childNodes = new List<object>(); foreach (var group in this.ListOfProvinces) childNodes.Add(group); return childNodes; } } } It displays me only list of countires.

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