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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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  • Return the difference between the lowest and highest key

    - by stan
    This is a past exam paper i am attempting and have no way to check if the out put is correct as i am not capable of building one of these things the question is in the title class Tree{ Tree left; Tree right; int key; public static int span(Tree tree) { if ( tree == null ){ return null; } if( tree.left != null) int min = span(tree.left); } if( tree.right != null){ int max = span(tree.right); } return max - min; } } Could anyone suggest what i need to change to get 5/5 marks :D - the only thing we have to do is write the span method, the header was given for us Thanks

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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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  • 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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  • Installing and using acts-as-taggable-on

    - by seaneshbaugh
    This is going to be a really dumb question, I just know it, but I'm going to ask anyways because it's driving me crazy. How do I get acts-as-taggable-on to work? I installed it as a gem with gem install acts-as-taggable-on because I can't ever seem to get installing plugins to work, but that's a whole other batch of questions that are all probably really dumb. Anyways, no problems there, it installed correctly. I did ruby script/generate acts_as_taggable_on_migration and rake db:migrate, again no problems. I added acts_as_taggable to the model I want to use tags with, started up the server and then loaded the index for the model just to see if what I've got so far is working and got the following error: undefined local variable or method `acts_as_taggable' for #. I figure that just means I need to do something like require 'acts-as-taggable-on' to my model's file because that's typically what's necessary for gems. So I did that hit refresh and got uninitialized constant ActiveRecord::VERSION. I'm not even going to pretend to begin to know what that means went wrong. Did I go wrong somewhere or there something else I need to do. The installation instructions seem to me like they just assume you generally know what you're doing and don't even begin to explain what to do when things go wrong.

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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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  • [BST] Deletion procedure

    - by Metz
    Hi all. Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. Thank you. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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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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  • Failed to load viewstate.The control tree into which viewstate is being loaded...etc

    - by alaa9jo
    Two days ago,a colleague of mine tried to publish an asp.net website (which is built in VS2008 using framework 3.5) to our server,he configured everything in IIS (he made sure that the selected asp.net version is 2.0) and launched the website..at first it was working great but when he tried to click on a specific treeview...BOOM..: "Failed to load viewstate. The control tree into which viewstate is being loaded must match the control tree that was used to save viewstate during the previous request. For example, when adding controls dynamically, the controls added during a post-back must match the type and position of the controls added during the initial request." In that page there were these control: a TreeView and a Placeholder,when the user selects any node then it's controls will be created dynamically into that placeholder..for the first time it's working fine but when (s)he select another node then that issue appears. He called me to help him with this issue,for me this is the first time I see such an issue,scratch my head then I decided to eliminate the possibilities of this issue one by one,at the development machine it's working perfectly,he published the website at the local IIS and again..it's working perfectly,I took a copy of the website and published it into my laptop but no issues at all,so this is means that it's not an issue in the code. So there is something missing/wrong in our server [it has Windows Server 2003],we went to the server and checked on the web-config and the configurations on IIS...nothing wrong so far,so I decided to check if the framework 3.5 is installed or not and the answer: it wasn't installed Of course he assumed that it was installed and there was nothing to tell if it wasn't from the "ASP.Net version" in IIS because frameworks 3.0 and 3.5 will not be listed there [2.0 will be listed there instead],the only way to check if it was installed or not is to search for the framework in this path:[WINDOWS Folder]\Microsoft.NET\Framework or check if it was installed in Add or remove programs. The obvious solution for his case: We installed Framework 3.5 SP1 into our server,did a restart to the machine and it worked ! If anyone faced the same issue and solved it using the same solution or with a different one please post it here to share experience.

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  • How exactly is an Abstract Syntax Tree created?

    - by Howcan
    I think I understand the goal of an AST, and I've build a couple of tree structures before, but never an AST. I'm mostly confused because the nodes are text and not number, so I can't think of a nice way to input a token/string as I'm parsing some code. For example, when I looked at diagrams of AST's, the variable and its value were leaf nodes to an equal sign. This makes perfect sense to me, but how would I go about implementing this? I guess I can do it case by case, so that when I stumble upon an "=" I use that as a node, and add the value parsed before the "=" as the leaf. It just seems wrong, because I'd probably have to make cases for tons and tons of things, depending on the syntax. And then I came upon another problem, how is the tree traversed? Do I go all the way down the height, and go back up a node when I hit the bottom, and do the same for it's neighbor? I've seen tons of diagrams on ASTs, but I couldn't find a fairly simple example of one in code, which would probably help.

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  • Is there a program that will show a tree of the differences in two file trees?

    - by Huckle
    In windows I manually back up from time to time by formatting my external drive and copying the contents of my data partition over. Inevitably there is a difference in the number and size of the files copied because of system files, etc. Is there a program that would diff two directories recursively and compile the differences into a nice GUI tree that I could peruse (preferably filter) to ensure that everything I want made it over to the drive? It should only show files that are not in both directories. (Also, please ignore the inadequacy of my backup solution)

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  • d3 tree - parents having same children

    - by Larry Anderson
    I've been transitioning my code from JIT to D3, and working with the tree layout. I've replicated http://mbostock.github.com/d3/talk/20111018/tree.html with my tree data, but I wanted to do a little more. In my case I will need to create child nodes that merge back to form a parent at a lower level, which I realize is more of a directed graph structure, but would like the tree to accomodate (i.e. notice that common id's between child nodes should merge). So basically a tree that divides like normal on the way from parents to children, but then also has the ability to bring those children nodes together to be parents (sort of an incestual relationship or something :)). Asks something similar - How to layout a non-tree hierarchy with D3 It sounds like I might be able to use hierarchical edge bundling in conjunction with the tree hierarchy layout, but I haven't seen that done. I might be a little off with that though.

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  • DrScheme versus mzscheme: treatment of definitions

    - by speciousfool
    One long term project I have is working through all the exercises of SICP. I noticed something a bit odd with the most recent exercise. I am testing a Huffman encoding tree. When I execute the following code in DrScheme I get the expected result: (a d a b b c a) However, if I execute this same code in mzscheme by calling (load "2.67.scm") or by running mzscheme -f 2.67.scm, it reports: symbols: expected symbols as arguments, given: (leaf D 1) My question is: why? Is it because mzscheme and drscheme use different rules for loading program definitions? The program code is below. ;; Define an encoding tree and a sample message ;; Use the decode procedure to decode the message, and give the result. (define (make-leaf symbol weight) (list 'leaf symbol weight)) (define (leaf? object) (eq? (car object) 'leaf)) (define (symbol-leaf x) (cadr x)) (define (weight-leaf x) (caddr x)) (define (make-code-tree left right) (list left right (append (symbols left) (symbols right)) (+ (weight left) (weight right)))) (define (left-branch tree) (car tree)) (define (right-branch tree) (cadr tree)) (define (symbols tree) (if (leaf? tree) (list (symbol-leaf tree)) (caddr tree))) (define (weight tree) (if (leaf? tree) (weight-leaf tree) (cadddr tree))) (define (decode bits tree) (define (decode-1 bits current-branch) (if (null? bits) '() (let ((next-branch (choose-branch (car bits) current-branch))) (if (leaf? next-branch) (cons (symbol-leaf next-branch) (decode-1 (cdr bits) tree)) (decode-1 (cdr bits) next-branch))))) (decode-1 bits tree)) (define (choose-branch bit branch) (cond ((= bit 0) (left-branch branch)) ((= bit 1) (right-branch branch)) (else (error "bad bit -- CHOOSE-BRANCH" bit)))) (define (test s-exp) (display s-exp) (newline)) (define sample-tree (make-code-tree (make-leaf 'A 4) (make-code-tree (make-leaf 'B 2) (make-code-tree (make-leaf 'D 1) (make-leaf 'C 1))))) (define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0)) (test (decode sample-message sample-tree))

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  • Secondary IP (eth0:0) acts like main server IP

    - by George Tasioulis
    I have a CentOS server, configured with 4 consecutive IPs: eth0 5.x.x.251 eth0:0 5.x.x.252 eth0:1 5.x.x.253 eth0:2 5.x.x.254 The problem is that all traffic goes out to the internet with eth0:0 (5.x.x.252) as the source IP, instead of eth0. # curl ifconfig.me 5.x.x.252 How can I fix this, so that all traffic goes out via eth0, ie my main IP? PS: My server is VPS running on a Xen dom0, the latter being configured in routed mode networking. Thanks in advance! Server configuration # ifconfig eth0 Link encap:Ethernet HWaddr 00:x:x:x:x:AE inet addr:5.x.x.251 Bcast:5.x.x.255 Mask:255.255.255.255 inet6 addr: fe80::x:x:x:x/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 RX packets:14675569 errors:0 dropped:0 overruns:0 frame:0 TX packets:9463227 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:1000 RX bytes:4122016502 (3.8 GiB) TX bytes:25959110751 (24.1 GiB) Interrupt:23 eth0:0 Link encap:Ethernet HWaddr 00:x:x:x:x:AE inet addr:5.x.x.252 Bcast:5.x.x.255 Mask:255.255.255.224 UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 Interrupt:23 eth0:1 Link encap:Ethernet HWaddr 00:x:x:x:x:AE inet addr:5.x.x.253 Bcast:5.x.x.255 Mask:255.255.255.224 UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 Interrupt:23 eth0:2 Link encap:Ethernet HWaddr 00:x:x:x:x:AE inet addr:5.x.x.254 Bcast:5.x.x.255 Mask:255.255.255.224 UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 Interrupt:23 # cat /etc/hosts 127.0.0.1 localhost.localdomain localhost 5.x.x.251 [fqdn] [hostname] # cat ifcfg-eth0 DEVICE=eth0 BOOTPROTO=static ONBOOT=yes IPADDR=5.x.x.251 NETMASK=255.255.255.224 SCOPE="peer 5.x.y.82" # cat ifcfg-eth0:0 DEVICE=eth0:0 BOOTPROTO=static ONBOOT=yes IPADDR=5.x.x.252 NETMASK=255.255.255.224 # cat route-eth0 ADDRESS0=0.0.0.0 NETMASK0=0.0.0.0 GATEWAY0=5.x.y.82 # netstat -rn Kernel IP routing table Destination Gateway Genmask Flags MSS Window irtt Iface 5.x.y.82 0.0.0.0 255.255.255.255 UH 0 0 0 eth0 5.x.x.224 0.0.0.0 255.255.255.224 U 0 0 0 eth0 169.254.0.0 0.0.0.0 255.255.0.0 U 0 0 0 eth0 0.0.0.0 5.x.y.82 0.0.0.0 UG 0 0 0 eth0

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  • Looking for a software solution which acts like a hardware KVM

    - by Daisetsu
    I've seen programs like Synergy before which allow a keyboard and mouse to be used across multiple systems. Unfortunately Synergy treats both systems like they are the same thing (when you move the mouse too far to the right it jumps to the other computer, along with keyboard input). I only have 1 monitor so what I need to do here is have a quick way to view my other desktop that is easy to manage. Remote desktop is difficult because you have to take it out of fullscreen, then minimize it and then re-maximize it when you want to use it again. If there were a single button I could hit to switch to the other computer that would be optimal. Both computers are on the same lan.

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  • Linux disk usage analyser that acts like symlinks are real files

    - by Rory
    I am using git-annex, an extension to the DVCS git, which is designed for handling large files. It makes heavy use of symlinks. The actual large files are moved to the .git/annex directory and the original files are symlinked to there. I am running out of disk space, and need to clear up, and see what's using all my space. Usually I'd use a disk usage tool like ncdu, Baobab or Filelight. However they treat the symlink as essentially empty, and only count the file that it is pointing to as using any space. Which means when I use git-annex, it shows no space used in the main directories and lots of space used in the .git/annex directory. This is not helpful. Is there any (graphical or ncurses) based disk usage programme for linux (apt-get installable would be easie that is capable (through options or not) of counting a symlink as using up the space that the original file uses up? Many have options for different behaviour for hard links, so makes sense that some should h (I know counting symlinks as using space has flaws, like counting the space space twice, broken symlinks, etc. But that's OK for my purposes)

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