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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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  • Insert element into a tree from a list in Standard ML

    - by vichet
    I have just started to learn SML on my own and get stuck with a question from the tutorial. Let say I have: tree data type datatype node of (tree*int*tree) | null insert function fun insert (newItem, null) = node (null, newItem, null) | insert (newItem, node (left, oldItem, right)) = if (newItem <= oldItem) then node (insert(newItem,left),oldItem, right) else node (left, oldItem, insert(newItem, right) an integer list val intList = [19,23,21,100,2]; my question is how can I add write a function to loop through each element in the list and add to a tree? Your answer is really appreciated.

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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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  • Family Tree :- myheritage.com

    - by Nitesh Panchal
    Hello, The other day i just accidently visited the site myheritage.com. I was just wondering, how they must have created one? Can anybody tell me what can be their database design? and if possible, algorithm that we can use to generate such a tree? Generating simple binary tree is very easy using recursion. But if you have a look at the site(if you have time please make account on it and add few nodes to feel) when we add son to a father, it's mother is automatically added(if you don't add explicitly). Mother's family tree is also generated side by side and many such fancy things are happening. In a simple binary tree we have a root node and then many nodes below it. Thus we cannot show wife and husband in the tree and then show a line from wife and husband to child. In spare time, can anybody discuss what can be it's database design and the recursive algorithm that we can follow to generate it? I hope i am not asking too much from you :).

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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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  • Black Screen when Computer Boots

    - by BlueRaja
    I'm having a serious problem with my computer; I think I've narrowed it down to the motherboard, but I'd like a second opinion before I spend the money. Before I moved into my new apartment, my desktop was working fine; now, it just won't work. It will turn on, the fans will spin up, lights come on... but nothing appears on the screen. No POST, nothing. I've tried: A different monitor (both are VGA) A different video-card (both are DVI, PCIe) Three different, known-good VGA-DVI adapters The onboard video port (VGA) Reseating the memory, and trying only one stick Different, known-good wall-outlets Unplugging the HDD and CD-drive from both the motherboard and PSU Replacing the PSU Has anyone had this happen before? Perhaps it's a known problem with this motherboard? Any advice!? Here are my specs: A13G+ V3.0 motherboard 2 2-gig 800mhz DDR2 600-watt PSU two older Geforce video cards

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  • xp black screen

    - by ciss
    Hello, sorry for my english, i am from russia. so, my pc based on AMD athlon. In past i move harddrive with xp and insert this HardDrive to another PC (Intel core2duo) I do this because i need a lot of files from my pc to another. So after this i back my harddrive to original pc (amd) And i see blackscreen on win-loading! (additional info - i boot this hard on another pc ... i know this is bad =(

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  • Red Gate does Byte Night 2012

    - by red(at)work
    On the 5th of October 2012, a team of nine plucky Red Gaters braved the howling wind and the driving rain to sleep outside. No tents or mattresses were allowed – all we took for protection were sleeping bags, groundsheets, plastic sacks and Colin’s enormous fishing umbrella (a godsend in umbrella-y disguise). Why would we do such a thing? For Byte Night, an annual tech sector sleepout in support of Action for Children, who tackle the causes as well as the consequences of youth homelessness. Byte Night encourages technology professionals to do for one night a year what thousands of young people have to do every night – sleep rough.  We signed up for Byte Night in the warm, heady midst of the British summer, thinking it couldn’t possibly be all that bad. Even on the night itself – before the rain began to fall, sat in the comfort and warmth of a company canteen, drinking wine and eating chill and preparing to win the pub quiz – we were excited and optimistic about the night that lay ahead of us. All of that changed as soon as we stepped out into one of the worst rainstorms of the year. Brian, the team’s birthday boy, describes it best: Picture the scene: it’s 3 am on a Friday. I’m lying outside, fully clothed in a sleeping bag, wearing a raincoat, trussed up inside a large plastic pocket, on a ground sheet beneath a giant umbrella, wedged so tightly between two of my colleagues that I can’t move my arms. I’m wide awake, staring up at the grey sky beyond the edge of the umbrella; a limp, flickering white glow hints at a moon somewhere behind the drifting clouds. I haven’t slept since we first moved outside at 11 pm. Outside. Did I mention we were outside? I’m hung over. I need the loo. But there is no way on earth that I’m getting out of this sleeping bag. It’s cold. It’s raining. Not just raining, but chucking it down. It’s been doing this non-stop since 10pm. The rain sounds like a hyperactive drummer on the fishing umbrella, and the noise is loud and relentless. Puddles of water are forming all over the groundsheet, and, despite being ensconced inside the plastic pouch, I am wet. The fishing umbrella is protecting me from the worst of the driving rain, but not all of me is under it, and five hours of rain is no match for it. Everything is wet. My left side has become horribly damp. My trainers, which I placed next to my sleeping bag, are now completely soaked through. Mmm. That’ll be fun in the morning. My head is next to Colin’s head on one side, and a multi-pack of McCoy’s cheddar and onion crisps on the other. Don’t ask about the tub of hummus. That’s somewhere down by my ankles, abandoned to the night. Jess, who is lying next to me, rolls over onto her side. A mini waterfall cascades from her rain-pouch onto my face. Bah. I continue to stare into the heavens, willing the dawn to hurry up. Something lands on my face. It’s a mosquito. Great. Midnight, when this still seemed like fun – when we opened some champagne and my colleagues presented me with a caterpillar birthday cake, when everyone was drunk and jolly and full of stoic resolve – feels like a long time ago. Did I mention that today is my birthday? The remains of the caterpillar cake endure the same fate as the hummus, left out in the rain like a metaphor for sadness. It’s getting colder. I can see my breath. Silence has descended on the group, apart from the rustle of plastic. And the rain, obviously. Someone snores, and I envy whoever it is the sweet escape of sleep. I try to wriggle a bit further down inside my sleeping bag, but it doesn’t want to be wriggled into. Only 3 hours till dawn. 180 minutes. I begin to count them off, one at a time.  All nine of us got to go home in the morning, but thousands of children across the UK don’t have that luxury. If you’d like to sponsor the Red Gate Byte Night team, our JustGiving page can be found here.   Chris, before the outside bit actually happened. More photos from Byte Night Cambridge 2012 can be found here.

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  • TDD - Red-Light-Green_Light:: A critical view

    - by Renso
    Subject: The concept of red-light-green-light for TDD/BDD style testing has been around since the dawn of time (well almost). Having written thousands of tests using this approach I find myself questioning the validity of the principle The issue: False positive or a valid test strategy that can be trusted? A critical view: I agree that the red-green-light concept has some validity, but who has ever written 2000 tests for a system that goes through a ton of chnages due to the organic nature fo the application and does not have to change, delete or restructure their existing tests? If you asnwer to the latter question is" "Yes I had a situation(s) where I had to refactor my code and it caused me to have to rewrite/change/delete my existing tests", read on, else press CTRL+ALT+Del :-) Once a test has been written, failed the test (red light), and then you comlpete your code and now get the green light for the last test, the test for that functionality is now in green light mode. It can never return to red light again as long as the test exists, even if the test itself is not changed, and only the code it tests is changed to fail the test. Why you ask? because the reason for the initial red-light when you created the test is not guaranteed to have triggered the initial red-light result for the same reasons it is now failing after a code change has been made. Furthermore, when the same test is changed to compile correctly in case of a compile-breaking code change, the green-light once again has been invalidated. Why? Because there is no guarantee that the test code fix is in the same green-light state as it was when it first ran successfully. To make matters worse, if you fix a compile-breaking test without going through the red-light-green-light test process, your test fix is essentially useless and very dangerous as it now provides you with a false-positive at best. Thinking your code has passed all tests and that it works correctly is far worse than not having any tests at all, well at least for that part of the system that the test-code represents. What to do? My recommendation is to delete the tests affected, and re-create them from scratch. I have to agree. Hard to do and justify if it has a significant impact on project deadlines. What do you think?

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  • That’s a wrap! Almost, there’s still one last chance to attend a SQL in the City event in 2012

    - by Red and the Community
    The communities team are back from the SQL in the City multi-city US Tour and we are delighted to have met so many happy SQL Server professionals and Red Gate customers. We set out to run a series of back-to-back events in order to meet, talk to and delight as many SQL Server and Red Gate enthusiasts as possible in 5 different cities in 11 days. We did it! The attendees had a good time too and 99% of them would attend another SQL in the City event in 2013 – so it seems we left an impression. There were a range of topics on the event agenda, ranging from ‘The Whys & Hows of Continuous Integration’, ‘Database Maintenance Essentials’, ‘Red Gate tools – The Complete Lifecycle’, ‘Automated Deployment: Application And Database Releases Without The Headache’, ‘The Ten Commandments of SQL Server Monitoring’ and many more. Videos and slides from the events will be posted to the event website in November, after our last event of 2012. SQL in the City Seattle – November 5 Join us for free and hear from some of the very best names in the SQL Server world. SQL Server MVPs such as; Steve Jones, Grant Fritchey, Brent Ozar, Gail Shaw and more will be presenting at the Bell Harbor conference center for one day only. We’re even taking on board some of the recent attendee-suggestions of how we can improve the events (feedback from the 65% of attendees who came to our US tour events), first off we’re extending the drinks celebration in the evening! Rather than just a 30 minute drink and run, attendees will have up to 2 hours to enjoy free drinks, relax and network in a fantastic environment amongst some really smart like-minded professionals. If you’re interested in expanding your SQL Server knowledge, would like to learn more about Red Gate tools, get yourself registered for the last SQL in the City event of 2012. It’s free, fun and we’re very friendly! I look forward to seeing you in Seattle on Monday November 5. Cheers, Annabel.

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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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  • Black screen white cursor and can't boot from live disc that I installed from just yesterday (Ubuntu 12.04)

    - by Luke
    So, I've decided to change to Ubuntu from Windows 7 after reformatting. Install goes smoothly, I've set everything up, and it works for a day. It crashed when I had a full screen video, froze, so I rebooted, and now I can't get past the black screen with flashing white underscore cursor. If I wait a while, I get "Reboot and Select proper Boot device or Insert Boot Media in selected Boot device and press any key" I've tried that, removed any boot options but the DVD drive, even tried my Windows 7 boot CD as well. Nothing boots, so I can't do anything. This is on an Asus A52N laptop, and all I can access now is the BIOS (version K52N 217), as far as I can tell.

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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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  • C++ find largest BST in a binary tree

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

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  • navigate all items in a wpf tree view

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

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  • 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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