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  • Maximum depth of a B-tree

    - by Phenom
    How do you figure out the maximum depth of a B-tree? Say you had a B-tree of order 1625, meaning each node has 1625 pointers and 1624 elements. What is the maximum depth of the tree if it contains 85,000,000 keys?

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  • Sorting by custom field and fetching whole tree from DB

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

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  • How to Populate a 'Tree' structure 'Declaratively'

    - by mackenir
    I want to define a 'node' class/struct and then declare a tree of these nodes in code in such a way that the way the code is formatted reflects the tree structure, and there's not 'too much' boiler plate in the way. Note that this isn't a question about data structures, but rather about what features of C++ I could use to arrive at a similar style of declarative code to the example below. Possibly with C++0X this would be easier as it has more capabilities in the area of constructing objects and collections, but I'm using Visual Studio 2008. Example tree node type: struct node { string name; node* children; node(const char* name, node* children); node(const char* name); }; What I want to do: Declare a tree so its structure is reflected in the source code node root = node("foo", [ node("child1"), node("child2", [ node("grand_child1"), node("grand_child2"), node("grand_child3" ]), node("child3") ]); NB: what I don't want to do: Declare a whole bunch of temporary objects/colls and construct the tree 'backwards' node grandkids[] = node[3] { node("grand_child1"), node("grand_child2"), node("grand_child3" }; node kids[] = node[3] { node("child1"), node("child2", grandkids) node("child3") }; node root = node("foo", kids);

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  • unique substrings using suffix tree

    - by user1708762
    For a given string S of length n- Optimal algorithm for finding all unique substrings of S can't be less than O(n^2). So, the best algorithm will give us the complexity of O(n^2). As per what I have read, this can be implemented by creating suffix tree for S. The suffix tree for S can be created in O(n) time. Now, my question is- How can we use the suffix tree for S to get all the unique substrings of S in O(n^2)?

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  • Upgrading from 13.04 to 13.10 stops

    - by BuZZ-dEE
    The upgrade to 13.10 stops after a lot of error messages about texlive packages, that I could close. The upgrade goes then further, but now it is stopped. What can I do to initiate the process again? The following are the last messages from the command window.: g multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 47: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 47: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 47: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 47: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 59: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 59: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 59: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 59: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 59: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 72: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 86: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 86: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 86: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 98: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 98: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 109: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 116: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 130: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 130: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 130: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 130: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 130: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 130: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 138: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 146: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 157: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 157: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 157: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 157: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 165: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 173: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 182: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/30-cjk-aliases.conf", line 182: Having multiple <family> in <alias> isn't supported and may not work as expected Fontconfig warning: "/etc/fonts/conf.d/99-language-selector-zh.conf", line 11: Having multiple values in <test> isn't supported and may not work as expected ERROR: error('unpack requires a bytes object of length 4',) (apport-gtk:16828): Gtk-CRITICAL **: gtk_main_quit: assertion 'main_loops != NULL' failed

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

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

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  • adding nodes to a binary search tree randomly deletes nodes

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

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  • Tree Surgeon 2.0 - The future on the T4 Express

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

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

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

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  • GWT: Change padding of tree rows?

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

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  • red black tree balancing?

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

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  • C++ find largest BST in a binary tree

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

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  • Why can't RB-Tree be a list?

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

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  • navigate all items in a wpf tree view

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

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  • Evaluate an expression tree

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

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  • Flex Tree Properties, Null Reference?

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

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  • R Tree 50,000 foot overview?

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

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  • Sorting by some field and fetching whole tree from DB

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

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  • python recursive iteration exceeding limit for tree implementation

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

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  • Simulation tree command in C

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

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  • Nepotism In The SQL Family

    - by Rob Farley
    There’s a bunch of sayings about nepotism. It’s unpopular, unless you’re the family member who is getting the opportunity. But of course, so much in life (and career) is about who you know. From the perspective of the person who doesn’t get promoted (when the family member is), nepotism is simply unfair; even more so when the promoted one seems less than qualified, or incompetent in some way. We definitely get a bit miffed about that. But let’s also look at it from the other side of the fence – the person who did the promoting. To them, their son/daughter/nephew/whoever is just another candidate, but one in whom they have more faith. They’ve spent longer getting to know that person. They know their weaknesses and their strengths, and have seen them in all kinds of situations. They expect them to stay around in the company longer. And yes, they may have plans for that person to inherit one day. Sure, they have a vested interest, because they’d like their family members to have strong careers, but it’s not just about that – it’s often best for the company as well. I’m not announcing that the next LobsterPot employee is one of my sons (although I wouldn’t be opposed to the idea of getting them involved), but actually, admitting that almost all the LobsterPot employees are SQLFamily members… …which makes this post good for T-SQL Tuesday, this month hosted by Jeffrey Verheul (@DevJef). You see, SQLFamily is the concept that the people in the SQL Server community are close. We have something in common that goes beyond ordinary friendship. We might only see each other a few times a year, at events like the PASS Summit and SQLSaturdays, but the bonds that are formed are strong, going far beyond typical professional relationships. And these are the people that I am prepared to hire. People that I have got to know. I get to know their skill level, how well they explain things, how confident people are in their expertise, and what their values are. Of course there people that I wouldn’t hire, but I’m a lot more comfortable hiring someone that I’ve already developed a feel for. I need to trust the LobsterPot brand to people, and that means they need to have a similar value system to me. They need to have a passion for helping people and doing what they can to make a difference. Above all, they need to have integrity. Therefore, I believe in nepotism. All the people I’ve hired so far are people from the SQL community. I don’t know whether I’ll always be able to hire that way, but I have no qualms admitting that the things I look for in an employee are things that I can recognise best in those that are referred to as SQLFamily. …like Ted Krueger (@onpnt), LobsterPot’s newest employee and the guy who is representing our brand in America. I’m completely proud of this guy. He’s everything I want in an employee. He’s an experienced consultant (even wrote a book on it!), loving husband and father, genuine expert, and incredibly respected by his peers. It’s not favouritism, it’s just choosing someone I’ve been interviewing for years. @rob_farley

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  • Windows Phone–A beautiful phone which I admire but I don’t recommend to friends and family

    - by Gopinath
    Microsoft’s Windows Phones are the most beautiful phones I’ve seen. Look at the photo which Microsoft shared on their Facebook page today. It’s gorgeous. Windows Phones come in vibrant colors and the user interface is very lively. When you keep an iPhone, Android Phone & a Windows Phone on a table, Windows Phone definitely stands out. Android and iOS interfaces are routine – a bunch of apps icons arranged in rows and multiple screens. Windows Phone is very different, the live tiles concept mesmerizes us. I love Windows Phone, but neither I buy one nor I recommend to family/friends! Why? Because it does not have all the Apps I need. Microsoft advertises that Windows Phone has 100K apps on its Windows Market Place. It’s true, there are 100K+ apps available for Windows Phone but not many of them are really useful and most of the popular Apps I use on Android are not available. When I say this to my friends at Microsoft, they don’t agree and one of them asked me list the apps that are not available. For him today I spent an hour quickly scanning through the apps installed on my Google Nexus and searched for same apps on Windows Market Place. As expected many of them are not available. Here is the list of my favorite Android apps that are not available for Windows Phone Mint – I use this app more than any of the Banking Apps I’ve installed on my mobile. It’s one app to keep a tab on all the expenses and income, the best money management and tracking app. Google Chrome – Web without Google Chrome is too boring, either on Desktop or on mobile. IE is too heavy and Firefox is loosing its grip. Chrome is the new darling of web. Pulse, Flipboard – Flipboard and Pulse are one of the best apps for reading news and following content of favorite blogs. Dropbox – Sync content across devices and provides access to your content on any device.It really does not matter what is your gadget – mobile, tablet or computer; Dropbox lets you access your content. GMail, Google Maps – Should I say how important are these two apps in our day to day life!! Vonage Extension – For around 30 bucks a month, Vonage provide landline service in USA + unlimited calls to India and many other countries + Vonage Extension App that lets Android/iOS mobile to make unlimited international calls for free. Without Vonage Extension app, I’m almost cutoff from my family and friends back home in India. Instagram – The most popular camera app used from a common man to celebrities. Raaga, Dhingana  – Music is part and parcel of life and these two apps are the most like popular apps to listen to Indian music. Quora – Quora is the place where most of the sensible discussions happen on web. Google Analytics, Google Adsense – I’m a blogger and these two apps mean a lot to me The list goes on and on! There are many useful apps that are not available on Windows Phone – TuneIn, MyTWC, Chrome To Phone, Google Voice, etc. Without all these apps, Windows Phone is just another old Nokia phone. Even though Windows Phone is the most beautiful phone, it needs Apps to attract customers. Without apps a smartphone is more or less a dumb feature phone which we loved to use before release of iPhone. Wish in an year or two the beautiful Windows Phone may have all the missing Apps. When it happens I’ll buy a phone for myself and recommend it to my family & friends. But till then I prefer to stay away.

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