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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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  • 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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  • LDAP object class violation: attribute ou not allowed in suffix?

    - by Paramaeleon
    I am about to set up a LDAP directory. It is used as a tool to communicate user permissions from a web application to WebDav file system access, e.g. adding a user to the web platform shall allow login to the file system with the same credentials. There are no other usages intended. Following this German tutorial which encourages the use of the attributes c, o, ou etc. over dc, I configured the following suffix and root: suffix "ou=webtool,o=myOrg,c=de" rootdn "cn=ldapadmin,ou=webtool,o=myOrg,c=de" Server starts and I can connect to it by LDAP Admin, which reports “LDAP error: Object lacks”. Well, there aren’t any objects yet. I now want to create the root and admin elements from shell. I created an init.ldif file: dn: ou=webtool,o=myOrg,c=de objectclass: dcObject objectclass: organization dc: webtool o: webtool dn: cn=ldapadmin,ou=webtool,o=myOrg,c=de objectclass: organizationalRole cn: ldapadmin Trying to load the file runs into an error, telling me that ou is not allowed: server:~ # ldapadd -x -D "cn=ldapadmin,ou=webtool,o=myOrg,c=de" -W -f init.ldif Enter LDAP Password: adding new entry "ou=webtool,o=myOrg,c=de" ldap_add: Object class violation (65) additional info: attribute 'ou' not allowed I am not using ou anywhere except in the suffix, so the question: Isn’t it allowed here? What is allowed here? Here is my answer. I am not allowed to post it as answer for 8 hours, so don’t mind that it is part of the question by now. I will move it outside some day, if I don’t forget to do so. There are numberous dependencies for the creation of elements, and error messages are rather confusing if you don’t know of the concept. The objectclass isn’t necessarily dcObject for the databases’ root node, as it is likely to guess when you read several tutoriales. Instead, it must correspond to the object’s type: Here, for a name starting with ou=, it must be organizationalUnit. I found this piece of information in these tables [Link removed due to restriction: Oops! Your edit couldn't be submitted because: We're sorry, but as a spam prevention mechanism, new users can only post a maximum of two hyperlinks. Earn more than 10 reputation to post more hyperlinks. Link is below]. Further on, the object class dictates which properties must and can be added in the record. Here, organizationalUnit must have an ou: entry and must not have neither dc: nor o: entry. The healthy init.ldif file looks like that: dn: ou=webtool,o=myOrg,c=de objectclass: organizationalUnit ou: LDAP server for my webtool dn: cn=ldapadmin,ou=webtool,o=myOrg,c=de objectclass: organizationalRole cn: ldapadmin Note: The page also states: “While many objectClasses show no MUST attributes you must (ouch) follow any hierarchy […] to determine if this is the really case.” I thought that would mean my root record would have to provide the must fields for c= and o= (c: and o:, respectively) but this isn’t the case. Link in answer is (1): http :// www (dot) zytrax (dot) com/books/ldap/ape/ "Appendix E: LDAP - Object Classes and Attributes"

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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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  • How to eliminate the domain suffix off my user profile folder when migrating to a new domain?

    - by Jerry Dodge
    We have just upgraded a decade old SBS 2003 server to a brand new SBS 2011 machine. During the process, over 30 other client/server machines on that domain also needed to be dis-joined and re-joined from the old domain to the new one. These domains have different names and is not migrated in any way. It's built from scratch. Since each client machine had very unique user profiles under this domain, we needed to make sure these were all backed up and migrated over to the new domain. For the most part, profiles were migrated with no hassle, just by renaming the user profile folder names. However, in one case, when I log in to my domain account, it creates a profile folder with a suffix of the new domain name. I have replaced all the files in the profile's root which begin with "ntuser" with the files of the new profile. The only problem is half the applications can't find their data, because the folder name is different. How can I change this folder name and maintain this profile on the new domain? I have deleted every user account (except admin), deleted their profiles/folders, removed them from the registry, and made sure every trace of this account was gone. The computer was basically a dummy with only an admin account. Then, I log into the machine under my new domain user account (same username as the old domain). It creates a profile folder with my username plus a suffix extension of the new domain name. The client machine is Windows 7 Ultimate, the old server was SBS 2003, and the new server is SBS 2011.

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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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  • Storing hierarchical (parent/child) data in Python/Django: MPTT alternative?

    - by Parand
    I'm looking for a good way to store and use hierarchical (parent/child) data in Django. I've been using django-mptt, but it seems entirely incompatible with my brain - I end up with non-obvious bugs in non-obvious places, mostly when moving things around in the tree: I end up with inconsistent state, where a node and its parent will disagree on their relationship. My needs are simple: Given a node: find its root find its ancestors find its descendants With a tree: easily move nodes (ie. change parent) My trees will be smallish (at most 10k nodes over 20 levels, generally much much smaller, say 10 nodes with 1 or 2 levels). I have to think there has to be an easier way to do trees in python/django. Are there other approaches that do a better job of maintaining consistency?

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  • org-mode agenda tree view

    - by Ray
    In org-mode's daily/weekly agenda view, is there a way to display the full context of the entries? My reading of the code is that it finds the first heading above the timestamp and displays that. However, in my case, that heading is often 3-4 levels deep and doesn't make sense without the bullets above it. It also doesn't seem like there are hooks to easily change that. Filtering is trivial, but not changing the fundamental presentation format.

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  • Red-Black trees - Erasing a node with two non-leaf children

    - by SalamiArmi
    Hi all, I've been implementing my own version of a red-black tree, mostly basing my algorithms from Wikipedia (http://en.wikipedia.org/wiki/Red-black_tree). Its fairly concise for the most part, but there's one part that I would like clarification on. When erasing a node from the tree that has 2 non-leaf (non-NULL) children, it says to move either side's children into the deletable node, and remove that child. I'm a little confused as to which side to remove from, based on that. Do I pick the side randomly, do I alternate betweek sides, or do I stick to the same side for every future deletion?

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  • java Process stop entire process tree

    - by ages04
    I am using Java Runtime to run commands, including certain CVS commands. I use: process = runtime.exec ("cmd /C cvs..."); format for running the Process in Java I need to have the option of stopping it. For this I use the Java Process destroy method process.destroy(); However only the cmd is stopped not the cvs process. It continues to run as a separate process without the cmd process as the parent. There are many references to this on the internet, but I haven't found any satisfactory solution. Thanks

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