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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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  • How to add markers on Google Maps polylines based on distance along the line?

    - by mikl
    I am trying to create a Google Map where the user can plot the route he walked/ran/bicycled and see how long he ran. The GPolyline class with it’s getLength() method is very helpful in this regard (at least for Google Maps API V2), but I wanted to add markers based on distance, for example a marker for 1 km, 5 km, 10 km, etc., but it seems that there is no obvious way to find a point on a polyline based on how far along the line it is. Any suggestions?

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  • Update count every second causing massive memory problems

    - by Josh
    Just on my local machine, trying the run the following script causes my computer to crash... What am I doing wrong? (function($) { var count = '6824756980'; while (count > 0) { setInterval(function() { $('#showcount').html(Math.floor(count-1)); count--; }, 1000 ); } })(jQuery); All I need to do is subtract one from the var "count" and update/display it's value every second.

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  • Calculating confidence intervals for a non-normal distribution

    - by Josiah
    Hi all, First, I should specify that my knowledge of statistics is fairly limited, so please forgive me if my question seems trivial or perhaps doesn't even make sense. I have data that doesn't appear to be normally distributed. Typically, when I plot confidence intervals, I would use the mean +- 2 standard deviations, but I don't think that is acceptible for a non-uniform distribution. My sample size is currently set to 1000 samples, which would seem like enough to determine if it was a normal distribution or not. I use Matlab for all my processing, so are there any functions in Matlab that would make it easy to calculate the confidence intervals (say 95%)? I know there are the 'quantile' and 'prctile' functions, but I'm not sure if that's what I need to use. The function 'mle' also returns confidence intervals for normally distributed data, although you can also supply your own pdf. Could I use ksdensity to create a pdf for my data, then feed that pdf into the mle function to give me confidence intervals? Also, how would I go about determining if my data is normally distributed. I mean I can currently tell just by looking at the histogram or pdf from ksdensity, but is there a way to quantitatively measure it? Thanks!

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  • JavaScript apparently waits for each AJAX call before sending another in a loop

    - by itako
    Hello. Straight to the point: I have this javascript: for(item=1;item<5;item++) { xmlhttp=new XMLHttpRequest(); xmlhttp.open("GET",'zzz.php', true); xmlhttp.send(); } And in PHP file something like this: usleep(5);die('ok'); Now the problem is javascript seems to be waiting for each ajax call to be completed before sending another one. So the first response gets back after approx. 5 seconds, next after 10 seconds and so on. That's a very simplified version of what I do, since the real script involves using cURL in PHP and jQuery as JS lib. But the problem remains the same. Why do responses come back in 5 second intervals?

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  • (iphone) can i give different intervals between images when animating?

    - by Eugene
    Hi, I'm animating several image as follows. UIImageView* animationView = [[UIImageView alloc] initWithFrame: self.animationViewContainer.bounds]; animationView.animationImages = animationArray; animationView.animationDuration = 0.5; animationView.animationRepeatCount = 5; [animationView startAnimating]; What I'd like to do is, controlling duration between animationImages. For instance, show image1 for 0.3 sec image2 for 0.5 sec.. There must be some way to do this, but hard to find an answer. I've asked the same question here before, but wording of the question wasn't so clear. Thank you

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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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  • How To Run A Shell Script Again And Again Having X Interval Of Time?

    - by Muhammad Hassan
    I have a shell script in my Ubuntu Server 14.04 LTS at ./ShellScript.sh. I setup /etc/rc.local to run the shell script after boot but before login using below code. Run this: sudo nano /etc/rc.local then add following and save. #!/bin/sh -e # # rc.local # # This script is executed at the end of each multiuser runlevel. # Make sure that the script will "exit 0" on success or any other # value on error. # # In order to enable or disable this script just change the execution # bits. # # By default this script does nothing. #!/bin/bash ./ShellScript.sh exit 0 Now I want to run/execute this shell script again and again having 15min of time interval between every run after boot but before login. So Can I do it? Update 1:) When I run crontab -e then I got the following. Now What to do? no crontab for root - using an empty one Select an editor. To change later, run 'select-editor'. 1. /bin/ed 2. /bin/nano <---- easiest 3. /usr/bin/vim.basic 4. /usr/bin/vim.tiny Choose 1-4 [2]: After selecting 2, I got crontab: "/usr/bin/sensible-editor" exited with status 2 UPDATE 2:) Update ShellScript.sh like below... #!/bin/bash # Testing ShellScript... while true do echo "ShellScript Start Running..." ********************************** All My Shell Script Codes/Script/Commands ********************************** echo "ShellScript End Running..." exit 0 sleep 900 done Then Run this: sudo nano /etc/rc.local then add following and save. #!/bin/sh -e # # rc.local # # This script is executed at the end of each multiuser runlevel. # Make sure that the script will "exit 0" on success or any other # value on error. # # In order to enable or disable this script just change the execution # bits. # # By default this script does nothing. sh ./ShellScript.sh & exit 0

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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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  • Corosync :: Restarting some resources after Lan connectivity issue

    - by moebius_eye
    I am currently looking into corosync to build a two-node cluster. So, I've got it working fine, and it does what I want to do, which is: Lost connectivity between the two nodes gives the first node '10node' both Failover Wan IPs. (aka resources WanCluster100 and WanCluster101 ) '11node' does nothing. He "thinks" he still has his Failover Wan IP. (aka WanCluster101) But it doesn't do this: '11node' should restart the WanCluster101 resource when the connectivity with the other node is back. This is to prevent a condition where node10 simply dies (and thus does not get 11node's Failover Wan IP), resulting in a situation where none of the nodes have 10node's failover IP because 10node is down 11node has "given back" his failover Wan IP. Here's the current configuration I'm working on. node 10sch \ attributes standby="off" node 11sch \ attributes standby="off" primitive LanCluster100 ocf:heartbeat:IPaddr2 \ params ip="172.25.0.100" cidr_netmask="32" nic="eth3" \ op monitor interval="10s" \ meta is-managed="true" target-role="Started" primitive LanCluster101 ocf:heartbeat:IPaddr2 \ params ip="172.25.0.101" cidr_netmask="32" nic="eth3" \ op monitor interval="10s" \ meta is-managed="true" target-role="Started" primitive Ping100 ocf:pacemaker:ping \ params host_list="192.0.2.1" multiplier="500" dampen="15s" \ op monitor interval="5s" \ meta target-role="Started" primitive Ping101 ocf:pacemaker:ping \ params host_list="192.0.2.1" multiplier="500" dampen="15s" \ op monitor interval="5s" \ meta target-role="Started" primitive WanCluster100 ocf:heartbeat:IPaddr2 \ params ip="192.0.2.100" cidr_netmask="32" nic="eth2" \ op monitor interval="10s" \ meta target-role="Started" primitive WanCluster101 ocf:heartbeat:IPaddr2 \ params ip="192.0.2.101" cidr_netmask="32" nic="eth2" \ op monitor interval="10s" \ meta target-role="Started" primitive Website0 ocf:heartbeat:apache \ params configfile="/etc/apache2/apache2.conf" options="-DSSL" \ operations $id="Website-one" \ op start interval="0" timeout="40" \ op stop interval="0" timeout="60" \ op monitor interval="10" timeout="120" start-delay="0" statusurl="http://127.0.0.1/server-status/" \ meta target-role="Started" primitive Website1 ocf:heartbeat:apache \ params configfile="/etc/apache2/apache2.conf.1" options="-DSSL" \ operations $id="Website-two" \ op start interval="0" timeout="40" \ op stop interval="0" timeout="60" \ op monitor interval="10" timeout="120" start-delay="0" statusurl="http://127.0.0.1/server-status/" \ meta target-role="Started" group All100 WanCluster100 LanCluster100 group All101 WanCluster101 LanCluster101 location AlwaysPing100WithNode10 Ping100 \ rule $id="AlWaysPing100WithNode10-rule" inf: #uname eq 10sch location AlwaysPing101WithNode11 Ping101 \ rule $id="AlWaysPing101WithNode11-rule" inf: #uname eq 11sch location NeverLan100WithNode11 LanCluster100 \ rule $id="RAND1083308" -inf: #uname eq 11sch location NeverPing100WithNode11 Ping100 \ rule $id="NeverPing100WithNode11-rule" -inf: #uname eq 11sch location NeverPing101WithNode10 Ping101 \ rule $id="NeverPing101WithNode10-rule" -inf: #uname eq 10sch location Website0NeedsConnectivity Website0 \ rule $id="Website0NeedsConnectivity-rule" -inf: not_defined pingd or pingd lte 0 location Website1NeedsConnectivity Website1 \ rule $id="Website1NeedsConnectivity-rule" -inf: not_defined pingd or pingd lte 0 colocation Never -inf: LanCluster101 LanCluster100 colocation Never2 -inf: WanCluster100 LanCluster101 colocation NeverBothWebsitesTogether -inf: Website0 Website1 property $id="cib-bootstrap-options" \ dc-version="1.1.7-ee0730e13d124c3d58f00016c3376a1de5323cff" \ cluster-infrastructure="openais" \ expected-quorum-votes="2" \ no-quorum-policy="ignore" \ stonith-enabled="false" \ last-lrm-refresh="1408954702" \ maintenance-mode="false" rsc_defaults $id="rsc-options" \ resource-stickiness="100" \ migration-threshold="3" I also have a less important question concerning this line: colocation NeverBothLans -inf: LanCluster101 LanCluster100 How do I tell it that this collocation only applies to '11node'.

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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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