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  • Sentence Tree v/s Words List

    - by Rohit Jose
    I was recently tasked with building a Name Entity Recognizer as part of a project. The objective was to parse a given sentence and come up with all the possible combinations of the entities. One approach that was suggested was to keep a lookup table for all the know connector words like articles and conjunctions, remove them from the words list after splitting the sentence on the basis of the spaces. This would leave out the Name Entities in the sentence. A lookup is then done for these identified entities on another lookup table that associates them to the entity type, for example if the sentence was: Remember the Titans was a movie directed by Boaz Yakin, the possible outputs would be: {Remember the Titans,Movie} was {a movie,Movie} directed by {Boaz Yakin,director} {Remember the Titans,Movie} was a movie directed by Boaz Yakin {Remember the Titans,Movie} was {a movie,Movie} directed by Boaz Yakin {Remember the Titans,Movie} was a movie directed by {Boaz Yakin,director} Remember the Titans was {a movie,Movie} directed by Boaz Yakin Remember the Titans was {a movie,Movie} directed by {Boaz Yakin,director} Remember the Titans was a movie directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was a movie directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was a movie directed by {Boaz Yakin,director} The entity lookup table here would contain the following data: Remember the Titans=Movie a movie=Movie Boaz Yakin=director the Titans=Movie the Titans=Sports Team Another alternative logic that was put forward was to build a crude sentence tree that would contain the connector words in the lookup table as parent nodes and do a lookup in the entity table for the leaf node that might contain the entities. The tree that was built for the sentence above would be: The question I am faced with is the benefits of the two approaches, should I be going for the tree approach to represent the sentence parsing, since it provides a more semantic structure? Is there a better approach I should be going for solving it?

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  • vector rotations for branches of a 3d tree

    - by freefallr
    I'm attempting to create a 3d tree procedurally. I'm hoping that someone can check my vector rotation maths, as I'm a bit confused. I'm using an l-system (a recursive algorithm for generating branches). The trunk of the tree is the root node. It's orientation is aligned to the y axis. In the next iteration of the tree (e.g. the first branches), I might create a branch that is oriented say by +10 degrees in the X axis and a similar amount in the Z axis, relative to the trunk. I know that I should keep a rotation matrix at each branch, so that it can be applied to child branches, along with any modifications to the child branch. My questions then: for the trunk, the rotation matrix - is that just the identity matrix * initial orientation vector ? for the first branch (and subsequent branches) - I'll "inherit" the rotation matrix of the parent branch, and apply x and z rotations to that also. e.g. using glm::normalize; using glm::rotateX; using glm::vec4; using glm::mat4; using glm::rotate; vec4 vYAxis = vec4(0.0f, 1.0f, 0.0f, 0.0f); vec4 vInitial = normalize( rotateX( vYAxis, 10.0f ) ); mat4 mRotation = mat4(1.0); // trunk rotation matrix = identity * initial orientation vector mRotation *= vInitial; // first branch = parent rotation matrix * this branches rotations mRotation *= rotate( 10.0f, 1.0f, 0.0f, 0.0f ); // x rotation mRotation *= rotate( 10.0f, 0.0f, 0.0f, 1.0f ); // z rotation Are my maths and approach correct, or am I completely wrong? Finally, I'm using the glm library with OpenGL / C++ for this. Is the order of x rotation and z rotation important?

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  • Cloning from a given point in the snapshot tree

    - by Fat Bloke
    Although we have just released VirtualBox 4.3, this quick blog entry is about a longer standing ability of VirtualBox when it comes to Snapshots and Cloning, and was prompted by a question posed internally, here in Oracle: "Is there a way I can create a new VM from a point in my snapshot tree?". Here's the scenario: Let's say you have your favourite work VM which is Oracle Linux based and as you installed different packages, such as database, middleware, and the apps, you took snapshots at each point like this: But you then need to create a new VM for some other testing or to share with a colleague who will be using the same Linux and Database layers but may want to reconfigure the Middleware tier, and may want to install his own Apps. All you have to do is right click on the snapshot that you're happy with and clone: Give the VM that you are about to create a name, and if you plan to use it on the same host machine as the original VM, it's a good idea to "Reinitialize the MAC address" so there's no clash on the same network: Now choose the Clone type. If you plan to use this new VM on the same host as the original, you can use Linked Cloning else choose Full.  At this point you now have a choice about what to do about your snapshot tree. In our example, we're happy with the Linux and Database layers, but we may want to allow our colleague to change the upper tiers, with the option of reverting back to our known-good state, so we'll retain the snapshot data in the new VM from this point on: The cloning process then chugs along and may take a while if you chose a Full Clone: Finally, the newly cloned VM is ready with the subset of the Snapshot tree that we wanted to retain: Pretty powerful, and very useful.  Cheers, -FB 

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  • How could I represent 1.625 by 0 or a 1 (binary digit)?

    - by pepito
    This is an excerpt from wikipedia about 'full rate' speech coding standard. Full Rate or FR or GSM-FR or GSM 06.10 was the first digital speech coding standard used in the GSM digital mobile phone system. The bit rate of the codec is 13 kbit/s, or 1.625 bits/audio sample. And this one is an excerpt from wikipedia about bit. In computing parlance, bit is the abbreviation for a single binary digit, represented by a 0 or a 1. How could I represent 1.625 by 0 or a 1? Actually, that's my lecturer's question that I could not answer. Some links to papers are more than welcome. Thanks in advance.

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  • How to implement dynamic binary search for search and insert operations of n element (C or C++)

    - by iecut
    The idea is to use multiple arrays, each of length 2^k, to store n elements, according to binary representation of n.Each array is sorted and different arrays are not ordered in any way. In the above mentioned data structure, SEARCH is carried out by a sequence of binary search on each array. INSERT is carried out by a sequence of merge of arrays of the same length until an empty array is reached. More Detail: Lets suppose we have a vertical array of length 2^k and to each node of that array there attached horizontal array of length 2^k. That is, to the first node of vertical array, a horizontal array of length 2^0=1 is connected,to the second node of vertical array, a horizontal array of length 2^1= 2 is connected and so on. So the insert is first carried out in the first horizontal array, for the second insert the first array becomes empty and second horizontal array is full with 2 elements, for the third insert 1st and 2nd array horiz. array are filled and so on. I implemented the normal binary search for search and insert as follows: int main() { int a[20]= {0}; int n, i, j, temp; int *beg, *end, *mid, target; printf(" enter the total integers you want to enter (make it less then 20):\n"); scanf("%d", &n); if (n = 20) return 0; printf(" enter the integer array elements:\n" ); for(i = 0; i < n; i++) { scanf("%d", &a[i]); } // sort the loaded array, binary search! for(i = 0; i < n-1; i++) { for(j = 0; j < n-i-1; j++) { if (a[j+1] < a[j]) { temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; } } } printf(" the sorted numbers are:"); for(i = 0; i < n; i++) { printf("%d ", a[i]); } // point to beginning and end of the array beg = &a[0]; end = &a[n]; // use n = one element past the loaded array! // mid should point somewhere in the middle of these addresses mid = beg += n/2; printf("\n enter the number to be searched:"); scanf("%d",&target); // binary search, there is an AND in the middle of while()!!! while((beg <= end) && (*mid != target)) { // is the target in lower or upper half? if (target < *mid) { end = mid - 1; // new end n = n/2; mid = beg += n/2; // new middle } else { beg = mid + 1; // new beginning n = n/2; mid = beg += n/2; // new middle } } // find the target? if (*mid == target) { printf("\n %d found!", target); } else { printf("\n %d not found!", target); } getchar(); // trap enter getchar(); // wait return 0; } Could anyone please suggest how to modify this program or a new program to implement dynamic binary search that works as explained above!!

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  • How to generate and encode (for use in GA), random, strict, binary rooted trees with N leaves?

    - by Peter Simon
    First, I am an engineer, not a computer scientist, so I apologize in advance for any misuse of nomenclature and general ignorance of CS background. Here is the motivational background for my question: I am contemplating writing a genetic algorithm optimizer to aid in designing a power divider network (also called a beam forming network, or BFN for short). The BFN is intended to distribute power to each of N radiating elements in an array of antennas. The fraction of the total input power to be delivered to each radiating element has been specified. Topologically speaking, a BFN is a strictly binary, rooted tree. Each of the (N-1) interior nodes of the tree represents the input port of an unequal, binary power splitter. The N leaves of the tree are the power divider outputs. Given a particular power divider topology, one is still free to map the power divider outputs to the array inputs in an arbitrary order. There are N! such permutations of the outputs. There are several considerations in choosing the desired ordering: 1) The power ratio for each binary coupler should be within a specified range of values. 2) The ordering should be chosen to simplify the mechanical routing of the transmission lines connecting the power divider. The number of ouputs N of the BFN may range from, say, 6 to 22. I have already written a genetic algorithm optimizer that, given a particular BFN topology and desired array input power distribution, will search through the N! permutations of the BFN outputs to generate a design with compliant power ratios and good mechanical routing. I would now like to generalize my program to automatically generate and search through the space of possible BFN topologies. As I understand it, for N outputs (leaves of the binary tree), there are $C_{N-1}$ different topologies that can be constructed, where $C_N$ is the Catalan number. I would like to know how to encode an arbitrary tree having N leaves in a way that is consistent with a chromosomal description for use in a genetic algorithm. Also associated with this is the need to generate random instances for filling the initial population, and to implement crossover and mutations operators for this type of chromosome. Any suggestions will be welcome. Please minimize the amount of CS lingo in your reply, since I am not likely to be acquainted with it. Thanks in advance, Peter

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  • ubuntu 13.10 kvm binary is deprecated, please use qemu-system-x86_64

    - by ??1986
    I just upgrade from 13.04 to 13.10 and I have this issue when I run my KVM Unable to complete install: 'internal error: process exited while connecting to monitor: W: kvm binary is deprecated, please use qemu-system-x86_64 instead char device redirected to /dev/pts/10 (label charserial0) failed to initialize KVM: Device or resource busy Detail Error: Traceback (most recent call last): File "/usr/share/virt-manager/virtManager/asyncjob.py", line 96, in cb_wrapper callback(asyncjob, *args, **kwargs) File "/usr/share/virt-manager/virtManager/create.py", line 1983, in do_install guest.start_install(False, meter=meter) File "/usr/lib/python2.7/dist-packages/virtinst/Guest.py", line 1246, in start_install noboot) File "/usr/lib/python2.7/dist-packages/virtinst/Guest.py", line 1314, in _create_guest dom = self.conn.createLinux(start_xml or final_xml, 0) File "/usr/lib/python2.7/dist-packages/libvirt.py", line 2892, in createLinux if ret is None:raise libvirtError('virDomainCreateLinux() failed', conn=self) libvirtError: internal error: process exited while connecting to monitor: W: kvm binary is deprecated, please use qemu-system-x86_64 instead char device redirected to /dev/pts/8 (label charserial0) failed to initialize KVM: Device or resource busy

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  • Visual Tree Enumeration

    - by codingbloke
    I feel compelled to post this blog because I find I’m repeatedly posting this same code in silverlight and windows-phone-7 answers in Stackoverflow. One common task that we feel we need to do is burrow into the visual tree in a Silverlight or Windows Phone 7 application (actually more recently I found myself doing this in WPF as well).  This allows access to details that aren’t exposed directly by some controls.  A good example of this sort of requirement is found in the “Restoring exact scroll position of a listbox in Windows Phone 7”  question on stackoverflow.  This required that the scroll position of the scroll viewer internal to a listbox be accessed. A caveat One caveat here is that we should seriously challenge the need for this burrowing since it may indicate that there is a design problem.  Burrowing into the visual tree or indeed burrowing out to containing ancestors could represent significant coupling between module boundaries and that generally isn’t a good idea. Why isn’t this idea just not cast aside as a no-no?  Well the whole concept of a “Templated Control”, which are in extensive use in these applications, opens the coupling between the content of the visual tree and the internal code of a control.   For example, I can completely change the appearance and positioning of elements that make up a ComboBox.  The ComboBox control relies on specific template parts having set names of a specified type being present in my template.  Rightly or wrongly this does kind of give license to writing code that has similar coupling. Hasn’t this been done already? Yes it has.  There are number of blogs already out there with similar solutions.  In fact if you are using Silverlight toolkit the VisualTreeExtensions class already provides this feature.  However I prefer my specific code because of the simplicity principle I hold to.  Only write the minimum code necessary to give all the features needed.  In this case I add just two extension methods Ancestors and Descendents, note I don’t bother with “Get” or “Visual” prefixes.  Also I haven’t added Parent or Children methods nor additional “AndSelf” methods because all but Children is achievable with the addition of some other Linq methods.  I decided to give Descendents an additional overload for depth hence a depth of 1 is equivalent to Children but this overload is a little more flexible than simply Children. So here is the code:- VisualTreeEnumeration public static class VisualTreeEnumeration {     public static IEnumerable<DependencyObject> Descendents(this DependencyObject root, int depth)     {         int count = VisualTreeHelper.GetChildrenCount(root);         for (int i = 0; i < count; i++)         {             var child = VisualTreeHelper.GetChild(root, i);             yield return child;             if (depth > 0)             {                 foreach (var descendent in Descendents(child, --depth))                     yield return descendent;             }         }     }     public static IEnumerable<DependencyObject> Descendents(this DependencyObject root)     {         return Descendents(root, Int32.MaxValue);     }     public static IEnumerable<DependencyObject> Ancestors(this DependencyObject root)     {         DependencyObject current = VisualTreeHelper.GetParent(root);         while (current != null)         {             yield return current;             current = VisualTreeHelper.GetParent(current);         }     } }   Usage examples The following are some examples of how to combine the above extension methods with Linq to generate the other axis scenarios that tree traversal code might require. Missing Axis Scenarios var parent = control.Ancestors().Take(1).FirstOrDefault(); var children = control.Descendents(1); var previousSiblings = control.Ancestors().Take(1)     .SelectMany(p => p.Descendents(1).TakeWhile(c => c != control)); var followingSiblings = control.Ancestors().Take(1)     .SelectMany(p => p.Descendents(1).SkipWhile(c => c != control).Skip(1)); var ancestorsAndSelf = Enumerable.Repeat((DependencyObject)control, 1)     .Concat(control.Ancestors()); var descendentsAndSelf = Enumerable.Repeat((DependencyObject)control, 1)     .Concat(control.Descendents()); You might ask why I don’t just include these in the VisualTreeEnumerator.  I don’t on the principle of only including code that is actually needed.  If you find that one or more of the above  is needed in your code then go ahead and create additional methods.  One of the downsides to Extension methods is that they can make finding the method you actually want in intellisense harder. Here are some real world usage scenarios for these methods:- Real World Scenarios //Gets the internal scrollviewer of a ListBox ScrollViewer sv = someListBox.Descendents().OfType<ScrollViewer>().FirstOrDefault(); // Get all text boxes in current UserControl:- var textBoxes = this.Descendents().OfType<TextBox>(); // All UIElement direct children of the layout root grid:- var topLevelElements = LayoutRoot.Descendents(0).OfType<UIElement>(); // Find the containing `ListBoxItem` for a UIElement:- var container = elem.Ancestors().OfType<ListBoxItem>().FirstOrDefault(); // Seek a button with the name "PinkElephants" even if outside of the current Namescope:- var pinkElephantsButton = this.Descendents()     .OfType<Button>()     .FirstOrDefault(b => b.Name == "PinkElephants"); //Clear all checkboxes with the name "Selector" in a Treeview foreach (CheckBox checkBox in elem.Descendents()     .OfType<CheckBox>().Where(c => c.Name == "Selector")) {     checkBox.IsChecked = false; }   The last couple of examples above demonstrate a common requirement of finding controls that have a specific name.  FindName will often not find these controls because they exist in a different namescope. Hope you find this useful, if not I’m just glad to be able to link to this blog in future stackoverflow answers.

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  • How do I optimize searching for the nearest point?

    - by Rootosaurus
    For a little project of mine I'm trying to implement a space colonization algorithm in order to grow trees. The current implementation of this algorithm works fine. But I have to optimize the whole thing in order to make it generate faster. I work with 1 to 300K of random attraction points to generate one tree, and it takes a lot of time to compute and compare distances between attraction points and tree node in order to keep only the closest treenode for an attraction point. So I was wondering if some solutions exist (I know they must exist) in order to avoid the time loss looping on each tree node for each attraction point to find the closest... and so on until the tree is finished.

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  • How to get the size of a binary tree ?

    - by Andrei Ciobanu
    I have a very simple binary tree structure, something like: struct nmbintree_s { unsigned int size; int (*cmp)(const void *e1, const void *e2); void (*destructor)(void *data); nmbintree_node *root; }; struct nmbintree_node_s { void *data; struct nmbintree_node_s *right; struct nmbintree_node_s *left; }; Sometimes i need to extract a 'tree' from another and i need to get the size to the 'extracted tree' in order to update the size of the initial 'tree' . I was thinking on two approaches: 1) Using a recursive function, something like: unsigned int nmbintree_size(struct nmbintree_node* node) { if (node==NULL) { return(0); } return( nmbintree_size(node->left) + nmbintree_size(node->right) + 1 ); } 2) A preorder / inorder / postorder traversal done in an iterative way (using stack / queue) + counting the nodes. What approach do you think is more 'memory failure proof' / performant ? Any other suggestions / tips ? NOTE: I am probably going to use this implementation in the future for small projects of mine. So I don't want to unexpectedly fail :).

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  • Is it safe to delete rotated MySQL binary logs?

    - by Milan Babuškov
    I have a MySQL server with binary logging active. Once a days logs file is "rotated", i.e. MySQL seems to stop writing to it and creates and new log file. For example, I currently have these files in /var/lib/mysql -rw-rw---- 1 mysql mysql 10485760 Jun 7 09:26 ibdata1 -rw-rw---- 1 mysql mysql 5242880 Jun 7 09:26 ib_logfile0 -rw-rw---- 1 mysql mysql 5242880 Jun 2 15:20 ib_logfile1 -rw-rw---- 1 mysql mysql 1916844 Jun 6 09:20 mybinlog.000004 -rw-rw---- 1 mysql mysql 61112500 Jun 7 09:26 mybinlog.000005 -rw-rw---- 1 mysql mysql 15609789 Jun 7 13:57 mybinlog.000006 -rw-rw---- 1 mysql mysql 54 Jun 7 09:26 mybinlog.index and mybinlog.000006 is growing. Can I simply take mybinlog.000004 and mybinlog.000005, zip them up and transfer to another server, or I need to do something else before? What info is stored in mybinlog.index? Only the info about the latest binary log? UPDATE: I understand I can delete the logs with PURGE BINARY LOGS which updates mybinlog.index file. However, I need to transfer logs to another computer before deleting them (I test if backup is valid on another machine). To reduce the transfer size, I wish to bzip2 the files. What will PURGE BINARY LOGS do if log files are not "there" anymore?

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  • reconstructing a tree from its preorder and postorder lists.

    - by NomeN
    Consider the situation where you have two lists of nodes of which all you know is that one is a representation of a preorder traversal of some tree and the other a representation of a postorder traversal of the same tree. I believe it is possible to reconstruct the tree exactly from these two lists, and I think I have an algorithm to do it, but have not proven it. As this will be a part of a masters project I need to be absolutely certain that it is possible and correct (Mathematically proven). However it will not be the focus of the project, so I was wondering if there is a source out there (i.e. paper or book) I could quote for the proof. (Maybe in TAOCP? anybody know the section possibly?) In short, I need a proven algorithm in a quotable resource that reconstructs a tree from its pre and post order traversals. Note: The tree in question will probably not be binary, or balanced, or anything that would make it too easy. Note2: Using only the preorder or the postorder list would be even better, but I do not think it is possible. Note3: A node can have any amount of children. Note4: I only care about the order of siblings. Left or right does not matter when there is only one child.

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  • What is the most efficient/elegant way to parse a flat table into a tree?

    - by Tomalak
    Assume you have a flat table that stores an ordered tree hierarchy: Id Name ParentId Order 1 'Node 1' 0 10 2 'Node 1.1' 1 10 3 'Node 2' 0 20 4 'Node 1.1.1' 2 10 5 'Node 2.1' 3 10 6 'Node 1.2' 1 20 What minimalistic approach would you use to output that to HTML (or text, for that matter) as a correctly ordered, correctly intended tree? Assume further you only have basic data structures (arrays and hashmaps), no fancy objects with parent/children references, no ORM, no framework, just your two hands. The table is represented as a result set, which can be accessed randomly. Pseudo code or plain English is okay, this is purely a conceptional question. Bonus question: Is there a fundamentally better way to store a tree structure like this in a RDBMS? EDITS AND ADDITIONS To answer one commenter's (Mark Bessey's) question: A root node is not necessary, because it is never going to be displayed anyway. ParentId = 0 is the convention to express "these are top level". The Order column defines how nodes with the same parent are going to be sorted. The "result set" I spoke of can be pictured as an array of hashmaps (to stay in that terminology). For my example was meant to be already there. Some answers go the extra mile and construct it first, but thats okay. The tree can be arbitrarily deep. Each node can have N children. I did not exactly have a "millions of entries" tree in mind, though. Don't mistake my choice of node naming ('Node 1.1.1') for something to rely on. The nodes could equally well be called 'Frank' or 'Bob', no naming structure is implied, this was merely to make it readable. I have posted my own solution so you guys can pull it to pieces.

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  • What data is actually stored in a B-tree database in CouchDB?

    - by Andrey Vlasovskikh
    I'm wondering what is actually stored in a CouchDB database B-tree? The CouchDB: The Definitive Guide tells that a database B-tree is used for append-only operations and that a database is stored in a single B-tree (besides per-view B-trees). So I guess the data items that are appended to the database file are revisions of documents, not the whole documents: +---------|### ... | | +------|###|------+ ... ---+ | | | | +------+ +------+ +------+ +------+ | doc1 | | doc2 | | doc1 | ... | doc1 | | rev1 | | rev1 | | rev2 | | rev7 | +------+ +------+ +------+ +------+ Is it true? If it is true, then how the current revision of a document is determined based on such a B-tree? Doesn't it mean, that CouchDB needs a separate "view" database for indexing current revisions of documents to preserve O(log n) access? Wouldn't it lead to race conditions while building such an index? (as far as I know, CouchDB uses no write locks).

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  • Java: confirm method Binary division and find remainder is correct?

    - by cadwag
    I am parsing binary files and have to implement a CRC algorithm to ensure the file is not corrupted. Problem is that I can't seem to get the binary math working when using larger numbers. The example I'm trying to get working: BigInteger G = new BigInteger("11001", 2); BigInteger M = new BigInteger("1110010000", 2); BigInteger R = M.remainder(G); I am expecting: R = "0101" But I am getting: R = "1100" I am assuming the remainder of 0101 is correct since it is given to me in this book I am using as a reference for the CRC algorithm (it's not based in Java), but I can't seem to get it working. I can get small binary calculations to work that I have solved by hand, but not the larger ones. I'll admit that I haven't worked the larger ones by hand yet, that is my next step, but I wanted to see if someone could point out a glaring flaw I have in my code. Can anyone confirm or deny that my methodology is correct? Thanks

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  • what is the format of a binary image & how is it different from jpg, png images?

    - by Rahulsingh190
    I searched the internet for the basic formats of image files (e.g. .jpg, .png, .gif) as there is a specific format for .doc, .pdf etc. But didn't got anything relevant. And today I also came with an .bin image format. BIN signifies that the image is in the Binary format. So, what is the Internal format of .jpg image file. And How is it different from .bin (Binary) format. Because everything is Basically saved in Binary Form. And How is BITMAP Image different from .jpg format.

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  • C++ how to store integer into a binary file??

    - by blaxc
    i gt a struct with 2 integer, i want to store them in a binary file and read it again... here is my code... struct pw { int a; int b; }; void main(){ pw* p = new pw(); pw* q = new pw(); std::ofstream fout(ADMIN_FILE, ios_base::out | ios_base::binary | ios_base::trunc); std::ifstream fin(ADMIN_FILE, ios_base::in | ios_base::binary); p->a=123; p->b=321; fout.write((const char*)p, sizeof(pw)); fin.write((char*)q, sizeof(pw)); fin.close(); cout<< q->a << endl;} my output is 0. anyone can tell me what is the problem?

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  • Help me understand Inorder Traversal without using recursion

    - by vito
    I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion. This is what I've tried so far: def traverseInorder(node): lifo = Lifo() lifo.push(node) while True: if node is None: break if node.left is not None: lifo.push(node.left) node = node.left continue prev = node while True: if node is None: break print node.value prev = node node = lifo.pop() node = prev if node.right is not None: lifo.push(node.right) node = node.right else: break The inner while-loop just doesn't feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn't feel right. Where am I going wrong? I haven't tried postorder traversal, but I guess it's similar and I will face the same conceptual blockage there, too. Thanks for your time! P.S.: Definitions of Lifo and Node: class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right class Lifo: def __init__(self): self.lifo = () def push(self, data): self.lifo = (data, self.lifo) def pop(self): if len(self.lifo) == 0: return None ret, self.lifo = self.lifo return ret

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  • Algorithm for evaluating nested logical expression

    - by TravelingSalesman
    I have a logical expression that I would like to evaluate. The expression can be nested and consists of T (True) or F (False) and parenthesis. The parenthesis "(" means "logical OR". Two terms TF beside each others (or any other two combinations beside each others), should be ANDED (Logical AND). For example, the expression: ((TFT)T) = true I need an algorithm for solving this problem. I thought of converting the expression first to disjunctive or conjunctive normal form and then I can easily evaluate the expression. However, I couldn't find an algorithm that normalizes the expression. Any suggestions? Thank you. The problem statement can be found here: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=2&category=378&page=show_problem&problem=2967

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  • New Sample Demonstrating the Traversing of Tree Bindings

    - by Duncan Mills
    A technique that I seem to use a fair amount, particularly in the construction of dynamic UIs is the use of a ADF Tree Binding to encode a multi-level master-detail relationship which is then expressed in the UI in some kind of looping form – usually a series of nested af:iterators, rather than the conventional tree or treetable. This technique exploits two features of the treebinding. First the fact that an treebinding can return both a collectionModel as well as a treeModel, this collectionModel can be used directly by an iterator. Secondly that the “rows” returned by the collectionModel themselves contain an attribute called .children. This attribute in turn gives access to a collection of all the children of that node which can also be iterated over. Putting this together you can represent the data encoded into a tree binding in all sorts of ways. As an example I’ve put together a very simple sample based on the HT schema and uploaded it to the ADF Sample project. It produces this UI: The important code is shown here for a Region -> Country -> Location Hierachy: <af:iterator id="i1" value="#{bindings.AllRegions.collectionModel}" var="rgn"> <af:showDetailHeader text="#{rgn.RegionName}" disclosed="true" id="sdh1"> <af:iterator id="i2" value="#{rgn.children}" var="cnty">     <af:showDetailHeader text="#{cnty.CountryName}" disclosed="true" id="sdh2">       <af:iterator id="i3" value="#{cnty.children}" var="loc">         <af:panelList id="pl1">         <af:outputText value="#{loc.City}" id="ot3"/>           </af:panelList>         </af:iterator>       </af:showDetailHeader>     </af:iterator>   </af:showDetailHeader> </af:iterator>  You can download the entire sample from here:

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