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  • Tree structured troubleshooting survey which gives solution based on answers

    - by japie1001
    Is there some online tool which let me design a tree structured questionaire/survey which suggests a solution to a problem. Example troubleshooting survey for the problem of black monitor screen: Q1: Is the powercable connected to the monitor? 0 Yes 0 NO If answer is no: "The solution to your problem is to connect the power cable" If answer is yes: Contiue with Q2 Q2: Is the computer turned on: 0 yes 0 no etc Is there something like Google Forms where I can achieve this with online?

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  • How Spanning Tree Protocol detects Loops

    - by AMIT
    For last few days I've been reading about Spanning Tree Protocol ,L2 protocol and understood how it prevents loop in network ,various steps in STP but one thing i wanted to know how STP actually detects the loops in network so that it can prevent it.Somewhere I read STP uses BPDU as probe and detects loops I mean how it happen is when switch send a BPDU with Destination Address as multicast and receive same BPDU again mean there is loop in network . But is it how STP detects loops in network?

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  • Building a directory tree from a list of file paths

    - by Abignale
    I am looking for a time efficient method to parse a list of files into a tree. There can be hundreds of millions of file paths. The brute force solution would be to split each path on occurrence of a directory separator, and traverse the tree adding in directory and file entries by doing string comparisons but this would be exceptionally slow. The input data is usually sorted alphabetically, so the list would be something like: C:\Users\Aaron\AppData\Amarok\Afile C:\Users\Aaron\AppData\Amarok\Afile2 C:\Users\Aaron\AppData\Amarok\Afile3 C:\Users\Aaron\AppData\Blender\alibrary.dll C:\Users\Aaron\AppData\Blender\and_so_on.txt From this ordering my natural reaction is to partition the directory listings into groups... somehow... before doing the slow string comparisons. I'm really not sure. I would appreciate any ideas. Edit: It would be better if this tree were lazy loaded from the top down if possible.

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  • Google Closure Library - Adding non-TreeNode children to a TreeNode

    - by Andreas Jansson
    Hi, I'm using the Google Closure Library and goog.ui.tree in particular to build a tree structure GUI component. It works pretty well out of the box, but I'd like to add a few extra controls to each of the leaves (goog.ui.Checkboxes in particular). The problem is that Component.addChild has been overridden in BaseNode so that each added child is treated as a child tree node as opposed to a child component. In effect plenty of errors are thrown if you try to add anything else than an actual tree node as a child, as these children are traversed and BaseNode-specific functions are called on them. I must admit I'm quite a Closure newb, but I reckon there must be some workaround for this, right? Essentially all I want to do is have a bunch of checkboxes appear next to each leaf in my tree. Thanks, Andreas

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  • JQuery treeview - add node(s) in middle of tree

    - by Chris
    Hi all, I'm just getting started with JQuery and the treeview plugin so this should be a relatively easy question: The example code for adding branches to the tree: var newnodes = $("<li><span class='folder'>New Sublist</span><ul>" + "<li><span class='file'>Item1</span></li>" + "<li><span class='file'>Item2</span></li></ul></li>").appendTo("#browser"); $("#browser").treeview({ add: branches }); Works fine for me, but adds the new branch at the end of the tree - instead what I want is to be able to select a specific node and add to that branch. I've managed to get the node being added by using the id of the particular node instead of the whole treeview in - appendTo("nodeID") However I can't get the tree to render correctly, either with: $("nodeID").treeview({ add: branches }); or $("browser").treeview({ add: branches }); or calling it on both without arguments. Cheers in advance

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  • Setting TreeView.DataContext doesn't refresh the tree

    - by dan gibson
    I have a List that I've bound to a TreeView. Setting TreeView.DataContext works - everything displays correctly. I then change the list (add an item to it) and set TreeView.DataContext again (to the same value) but the tree does not refresh with the new items. How do I get the treeview to refresh? This is basically my code: public class xItemCollection : ObservableCollection<xItem> { } public class xItem : INotifyPropertyChanged { xItemCollection _Items; public xItem() { _Items = new xItemCollection(); } public xItemCollection Items {get{return _Items;}} } ... this.RootItem = new xItem(); treeView.DataContext = this; Adding items to the list works until the tree is rendered for the first time. After it is rendered, adding/removing items does not refresh the tree.

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  • Expression Tree

    - by nettguy
    My understanding of expression tree is : Expression trees are in-memory representation of expression like arithmetic or boolean expression.The expressions are stored into the parsed tree.so we can easily transalate into any other language. Linq to SQL uses expression tree.Normally in LINQ to SQL query the compiler translates it into parsed expression trees.These are passed to Sql Server as T-SQL Statements.The Sql server executes the T-SQL query and sends down the result back.That is why when you execute LINQ to SQL you gets IQueryable<T> not IEnumetrable<T>.Because IQuerybale contains public IQueryable:IEnumerable { Type Element {get;} Expression Expression {get;} IQueryaleProvider Provider {get;} } Questions : Microsoft uses Expression trees to play with LINQ-to-Sql.What are the different ways can i use expression trees to boost my code. Apart from LINQ to SQL,Linq to amazon ,who used expression trees in their applications? Linq to Object return IEnumerable,Linq to SQL return IQueryable ,What does LINQ to XML return?

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  • Design a GUI browser to view a tree

    - by iamrohitbanga
    I have a large tree. I want to be able to visualize it using a GUI tool. I want the ability to pan and zoom the tree image so that i can focus on part of the tree. Is there an existing tool to achieve this? If not i would like to write a small tool for myself to be able to do this. what is the simplest way of doing this? what computer language should i use? the image should look something like http://upload.wikimedia.org/wikipedia/commons/d/df/Binary_tree.png I should be able to zoom and pan the image.

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  • How to get a reference to node in DOM tree in Google Chrome debugger console

    - by .yahoo.co.jpaqwsykcj3aulh3h1k0cy6nzs3isj
    In the Google Chrome debugger, I often want to get a reference to a node in the DOM tree. I can click the "magnifying glass" button and then click the desired element in the browser window to select the corresponding node in the DOM tree displayed in the debugger. But how can I get a reference to that node in the console? If the element has an id, document.getElementById works, but if there is no id, is there a better alternative to XPath or manual traversal of the DOM tree using children? In case XPath is the best way, is there a better way than doing something like this: var evaluator = new XPathEvaluator(); var result = evaluator.evaluate("//div", document.documentElement, null, XPathResult.FIRST_ORDERED_NODE_TYPE, null); which is a pain to type out each time.

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  • Tree View Control problem with render Control function

    - by vikas
    I am using TreeView Control in Asp.net. I have placed this control inside a panel. The tree control is completely binded (we don’t want populate on demand) with an Xmldatasource during a callback and then I call Panel.renderControl to return the response (HTML) to the client side callback handler. Problem: 1. The tree expand/collapse (on click of plus sign) is causing postback whereas when I normally bind a tree with xml during postback and without using renderControl of container control, the expand/collapse (on click of plus sign) is being handled at client side.

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  • Gaining information from nodes of tree

    - by jainp
    I am working with the tree data structure and trying to come up with a way to calculate information I can gain from the nodes of the tree. I am wondering if there are any existing techniques which can assign higher numerical importance to a node which appears less frequently at lower level (Distance from the root of the tree) than the same nodes appearance at higher level and high frequency. To give an example, I want to give more significance to node Book, at level 2 appearing once, then at level 3 appearing thrice. Will appreciate any suggestions/pointers to techniques which achieve something similar. Thanks, Prateek

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  • Finding the heaviest length-constrained path in a weighted Binary Tree

    - by Hristo
    UPDATE I worked out an algorithm that I think runs in O(n*k) running time. Below is the pseudo-code: routine heaviestKPath( T, k ) // create 2D matrix with n rows and k columns with each element = -8 // we make it size k+1 because the 0th column must be all 0s for a later // function to work properly and simplicity in our algorithm matrix = new array[ T.getVertexCount() ][ k + 1 ] (-8); // set all elements in the first column of this matrix = 0 matrix[ n ][ 0 ] = 0; // fill our matrix by traversing the tree traverseToFillMatrix( T.root, k ); // consider a path that would arc over a node globalMaxWeight = -8; findArcs( T.root, k ); return globalMaxWeight end routine // node = the current node; k = the path length; node.lc = node’s left child; // node.rc = node’s right child; node.idx = node’s index (row) in the matrix; // node.lc.wt/node.rc.wt = weight of the edge to left/right child; routine traverseToFillMatrix( node, k ) if (node == null) return; traverseToFillMatrix(node.lc, k ); // recurse left traverseToFillMatrix(node.rc, k ); // recurse right // in the case that a left/right child doesn’t exist, or both, // let’s assume the code is smart enough to handle these cases matrix[ node.idx ][ 1 ] = max( node.lc.wt, node.rc.wt ); for i = 2 to k { // max returns the heavier of the 2 paths matrix[node.idx][i] = max( matrix[node.lc.idx][i-1] + node.lc.wt, matrix[node.rc.idx][i-1] + node.rc.wt); } end routine // node = the current node, k = the path length routine findArcs( node, k ) if (node == null) return; nodeMax = matrix[node.idx][k]; longPath = path[node.idx][k]; i = 1; j = k-1; while ( i+j == k AND i < k ) { left = node.lc.wt + matrix[node.lc.idx][i-1]; right = node.rc.wt + matrix[node.rc.idx][j-1]; if ( left + right > nodeMax ) { nodeMax = left + right; } i++; j--; } // if this node’s max weight is larger than the global max weight, update if ( globalMaxWeight < nodeMax ) { globalMaxWeight = nodeMax; } findArcs( node.lc, k ); // recurse left findArcs( node.rc, k ); // recurse right end routine Let me know what you think. Feedback is welcome. I think have come up with two naive algorithms that find the heaviest length-constrained path in a weighted Binary Tree. Firstly, the description of the algorithm is as follows: given an n-vertex Binary Tree with weighted edges and some value k, find the heaviest path of length k. For both algorithms, I'll need a reference to all vertices so I'll just do a simple traversal of the Tree to have a reference to all vertices, with each vertex having a reference to its left, right, and parent nodes in the tree. Algorithm 1 For this algorithm, I'm basically planning on running DFS from each node in the Tree, with consideration to the fixed path length. In addition, since the path I'm looking for has the potential of going from left subtree to root to right subtree, I will have to consider 3 choices at each node. But this will result in a O(n*3^k) algorithm and I don't like that. Algorithm 2 I'm essentially thinking about using a modified version of Dijkstra's Algorithm in order to consider a fixed path length. Since I'm looking for heaviest and Dijkstra's Algorithm finds the lightest, I'm planning on negating all edge weights before starting the traversal. Actually... this doesn't make sense since I'd have to run Dijkstra's on each node and that doesn't seem very efficient much better than the above algorithm. So I guess my main questions are several. Firstly, do the algorithms I've described above solve the problem at hand? I'm not totally certain the Dijkstra's version will work as Dijkstra's is meant for positive edge values. Now, I am sure there exist more clever/efficient algorithms for this... what is a better algorithm? I've read about "Using spine decompositions to efficiently solve the length-constrained heaviest path problem for trees" but that is really complicated and I don't understand it at all. Are there other algorithms that tackle this problem, maybe not as efficiently as spine decomposition but easier to understand? Thanks.

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  • Expression Tree : C#

    - by nettguy
    My understanding of expression tree is : Expression trees are in-memory representation of expression like arithmetic or boolean expression.The expressions are stored into the parsed tree.so we can easily transalate into any other language. Linq to SQL uses expression tree.Normally when our LINQ to SQL query compiler translates it to parsed expression trees.These are passed to Sql Server as T-SQL Statements.The Sql server executes the T-SQL query and sends down the result back.That is why when you execute LINQ to SQL you gets IQueryable<T> not IEnumetrable<T>.Because IQuerybale contains public IQueryable:IEnumerable { Type Element {get;} Expression Expression {get;} IQueryaleProvider Provider {get;} } Questions : Microsoft uses Expression trees to play with LINQ-to-Sql.What are the different ways can i use expression trees to boost my code. Apart from LINQ to SQL,Linq to amazon ,who used expression trees in their applications? Linq to Object return IEnumerable,Linq to SQL return IQueryable ,What does LINQ to XML return?

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  • WPF: Calling a method from a different "branch" of the tree

    - by sofri
    Hey, I'm doing a WPF Application. The tree looks like this: SurfaceWindow --- Startscreen ..........................-------- Page---------- Subpage I'm trying to call a method from the "Subpage" from the "Code Behind" of the Startscreen(Startscreen.xaml.cs). The method from the Subpage looks like this: public void showTheme(ThemeViewModel theme) { ... } If know that I can call it when I'm on the "Page" or the "SurfaceWindow", because it's in the same "branch" of the tree, and I just do something like this: ThemeViewModel theme = (ThemeViewModel)mvm.CurrentItem.ThemeViewModel; katalog.katalogblatt.showTheme(theme); But how do I do it when I'm not on the same branch of the tree and want to call the method?

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  • Secondary keys in a B-tree

    - by Phenom
    Let's say that there is a file that contains an unsorted list of student information, which includes a student ID number as well as other information. I want to make a program that retrieves student information based on student ID number. In order to make it efficient, I store the student IDs in a B-tree. So when I enter a student ID number, it searches the B-tree to see if its there or not. It also does one more thing. If it finds the student ID number, then it also returns where in the file that student's information is. This is the secondary key. The program uses this information to locate the rest of the student's information and prints it to screen. Can this be done? Is this how a b-tree works?

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  • Tree data structure in php

    - by Piyush
    in my application user starts a new tree or get added under a child-user and keep on adding users in branches in such a way- >there are 10 level of tree type structure. >root node contain 1 user and each node(user) can have max 5 child-user in this way tree will be like level 0 = 1 user , level 1 = 5 user, level 2 = 25 user , level 3 = 125 user and so on. I created one MySQL table having columns like- User_id , level, super_id, child1_id, child2_id, child3_id, child4_id, child5_id my question is How can I get all child-user(child to child also) of a particular user at any level do I need to add some more columns in my table??

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  • java - Depth First Search - Perform DFS on a tree

    - by DJDonaL3000
    Im trying to perform DFS on a Minimum Spanning Tree which contains 26 nodes. Nodes are named 'A' to 'Z' and the tree is undirected. I have an empty function called DFS here that I am trying to write, which (i presume) takes in the tree (a 2D array) a startNode (randomly selected node 'M') and the endNode (randomly selected node 'Z'). The weights of connected nodes are identified in the 2D array parameter, but how do I actually get started visiting nodes? All that is required is to print each nodeName in the order of the DFS traversal. Do I need to create a Node_class for each node in the 2d array??

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  • WPF Logical Tree - bottom up vs. top down

    - by Dor Rotman
    Hello, I've read the MSDN article about the layouts pass, that states: When a node is added or removed from the logical tree, property invalidations are raised on the node's parent and all its children. As a result, a top-down construction pattern should always be followed to avoid the cost of unnecessary invalidations on nodes that have already been validated. Now lets assume I do this. Won't the users see the control tree populate itself and the layout change several times during the control creation process? I want the whole control tree to just appear completely full. Thanks!

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  • Constructing a tree using Python

    - by stealthspy
    I am trying to implement a unranked boolean retrieval. For this, I need to construct a tree and perform a DFS to retrieve documents. I have the leaf nodes but I am having difficulty to construct the tree. Eg: query = OR ( AND (maria sharapova) tennis) Result: OR | | AND tennis | | maria sharapova I traverse the tree using DFS and calculate the boolean equivalent of certain document ids to identify the required document from the corpus. Can someone help me with the design of this using python? I have parsed the query and retrieved the leaf nodes for now.

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  • How to create an AST with ANTLR from a hierarchical key-value syntax

    - by Brabster
    I've been looking at parsing a key-value data format with ANTLR. Pretty straightforward, but the keys represent a hierarchy. A simplified example of my input syntax: /a/b/c=2 /a/b/d/e=3 /a/b/d/f=4 In my mind, this represents a tree structured as follows: (a (b (= c 2) (d (= e 3) (= f 4)))) The nearest I can get is to use the following grammar: /* Parser Rules */ start: (component NEWLINE?)* EOF -> (component)*; component: FORWARD_SLASH ALPHA_STRING component -> ^(ALPHA_STRING component) | FORWARD_SLASH ALPHA_STRING EQUALS value -> ^(EQUALS ALPHA_STRING value); value: ALPHA_STRING; /* Lexer Rules */ NEWLINE : '\r'? '\n'; ALPHA_STRING : ('a'..'z'|'A'..'Z'|'0'..'9')+; EQUALS : '='; FORWARD_SLASH : '/'; Which produces: (a (b (= c 2))) (a (b (d (= e 3)))) (a (b (d (= f 4)))) I'm not sure whether I'm asking too much from a generic tool such as ANTLR here, and this is as close I can get with this approach. That is, from here I consume the parts of the tree and create the data structure I want by hand. So - can I produce the tree structure I want directly from a grammar? If so, how? If not, why not - is it a technical limitation in ANTLR or is it something more CS-y to do with the type of language involved?

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  • Shift-reduce: when to stop reducing?

    - by Joey Adams
    I'm trying to learn about shift-reduce parsing. Suppose we have the following grammar, using recursive rules that enforce order of operations, inspired by the ANSI C Yacc grammar: S: A; P : NUMBER | '(' S ')' ; M : P | M '*' P | M '/' P ; A : M | A '+' M | A '-' M ; And we want to parse 1+2 using shift-reduce parsing. First, the 1 is shifted as a NUMBER. My question is, is it then reduced to P, then M, then A, then finally S? How does it know where to stop? Suppose it does reduce all the way to S, then shifts '+'. We'd now have a stack containing: S '+' If we shift '2', the reductions might be: S '+' NUMBER S '+' P S '+' M S '+' A S '+' S Now, on either side of the last line, S could be P, M, A, or NUMBER, and it would still be valid in the sense that any combination would be a correct representation of the text. How does the parser "know" to make it A '+' M So that it can reduce the whole expression to A, then S? In other words, how does it know to stop reducing before shifting the next token? Is this a key difficulty in LR parser generation?

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  • Single-port 2600 router with 2900XL switch

    - by Slava Maslennikov
    I have a setup, where the single port 2600 router is in port 0/2 in the switch, outside network is on port 0/1, and the rest (0/3-0/24) should be clients for the second network that would be managed by the 2600 router. I configured everything with two VLANs: 100 for outside (0/2-0/24), 200 for inside (0/1-0/2). 0/2 is a trunk port for the two VLANs. The issue that came about is that I can't have two VLANs on at once: software doesn't allow it. Now, I can ping the outside network devices (172.16.7.1, 172.16.7.103), and even google (8.8.8.8) from the router, but not the switch. Devices on connected get a DHCP lease properly but can't ping outside the network, just the router - 172.17.7.1 and the switch itself, 172.17.7.7. The configuration for both the router and the switch are here, as well as below. Router: rt.throom#sho run Building configuration... Current configuration : 1015 bytes ! version 12.1 no service single-slot-reload-enable service timestamps debug uptime service timestamps log uptime no service password-encryption ! hostname rt.throom ! enable password To053cret ! ! ! ! ! no ip subnet-zero ip dhcp excluded-address 172.17.7.1 172.17.7.2 ip dhcp excluded-address 172.17.7.3 172.17.7.4 ip dhcp excluded-address 172.17.7.5 ! ip dhcp pool VLAN200 network 172.17.7.0 255.255.255.0 default-router 172.17.7.1 dns-server 8.8.8.8 ! ip audit notify log ip audit po max-events 100 ! ! ! ! ! ! ! interface Ethernet0/0 no ip address ! interface Ethernet0/0.100 encapsulation dot1Q 100 ip address 172.16.7.15 255.255.255.0 ip nat outside ! interface Ethernet0/0.200 encapsulation dot1Q 200 ip address 172.17.7.1 255.255.255.0 ip nat inside ! router eigrp 20 network 172.16.0.0 network 172.17.0.0 no auto-summary no eigrp log-neighbor-changes ! no ip classless no ip http server ! access-list 1 permit 172.17.7.0 0.0.0.255 ! ! line con 0 line aux 0 line vty 0 4 login ! end Switch: sw.throom#sho run Building configuration... Current configuration: ! version 11.2 no service pad no service udp-small-servers no service tcp-small-servers ! hostname sw.throom ! enable password Oh5053cret ! ! no spanning-tree vlan 100 no spanning-tree vlan 200 ip subnet-zero ! ! interface VLAN1 no ip address no ip route-cache ! interface FastEthernet0/1 switchport access vlan 100 spanning-tree portfast ! interface FastEthernet0/2 switchport trunk encapsulation dot1q switchport mode trunk ! interface FastEthernet0/3 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/4 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/5 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/6 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/7 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/8 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/9 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/10 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/11 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/12 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/13 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/14 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/15 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/16 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/17 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/18 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/19 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/20 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/21 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/22 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/23 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/24 switchport access vlan 200 spanning-tree portfast ! ! line con 0 stopbits 1 line vty 0 4 login line vty 5 9 login ! end sho ip route gives: Gateway of last resort is 172.16.7.1 to network 0.0.0.0 172.17.0.0/24 is subnetted, 1 subnets C 172.17.7.0 is directly connected, Ethernet0/0.200 172.16.0.0/24 is subnetted, 1 subnets C 172.16.7.0 is directly connected, Ethernet0/0.100 S* 0.0.0.0/0 [1/0] via 172.16.7.1

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  • How can I execute an ANTLR parser action for each item in a rule that can match more than one item?

    - by Chris Farmer
    I am trying to write an ANTLR parser rule that matches a list of things, and I want to write a parser action that can deal with each item in the list independently. Some example input for these rules is: $(A1 A2 A3) I'd like this to result in an evaluator that contains a list of three MyIdentEvaluator objects -- one for each of A1, A2, and A3. Here's a snippet of my grammar: my_list returns [IEvaluator e] : { $e = new MyListEvaluator(); } '$' LPAREN op=my_ident+ { /* want to do something here for each 'my_ident'. */ /* the following seems to see only the 'A3' my_ident */ $e.Add($op.e); } RPAREN ; my_ident returns [IEvaluator e] : IDENT { $e = new MyIdentEvaluator($IDENT.text); } ; I think my_ident is defined correctly, because I can see the three MyIdentEvaluators getting created as expected for my input string, but only the last my_ident ever gets added to the list (A3 in my example input). How can I best treat each of these elements independently, either through a grammar change or a parser action change? It also occurred to me that my vocabulary for these concepts is not what it should be, so if it looks like I'm misusing a term, I probably am. EDIT in response to Wayne's comment: I tried to use op+=my_ident+. In that case, the $op in my action becomes an IList (in C#) that contains Antlr.Runtime.Tree.CommonTree instances. It does give me one entry per matched token in $op, so I see my three matches, but I don't have the MyIdentEvaluator instances that I really want. I was hoping I could then find a rule attribute in the ANTLR docs that might help with this, but nothing seemed to help me get rid of this IList. Result... Based on chollida's answer, I ended up with this which works well: my_list returns [IEvaluator e] : { $e = new MyListEvaluator(); } '$' LPAREN (op=my_ident { $e.Add($op.e); } )+ RPAREN ; The Add method gets called for each match of my_ident.

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