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  • Deletion procedure for a Binary Search Tree

    - by Metz
    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. 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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  • Return parent of node in Binary Tree

    - by user188995
    I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs. //Assume that nodes are represented by numbers from 1...n where 1=root and even //nos.=left child and odd nos=right child. public int parent(Node node){ if (node % 2 == 0){ if (root.left==node) return root; else return parent(root.left); } //same case for right } But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right. If it's the child or if the node that was queried else, recurses it with the child.

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  • Problems in Binary Search Tree

    - by user2782324
    This is my first ever trial at implementing the BST, and I am unable to get it done. Please help The problem is that When I delete the node if the node is in the right subtree from the root or if its a right child in the left subtree, then it works fine. But if the node is in the left subtree from root and its any left child, then it does not get deleted. Can someone show me what mistake am I doing?? the markedNode here gets allocated to the parent node of the node to be deleted. the minValueNode here gets allocated to a node whose left value child is the smallest value and it will be used to replace the value to be deleted. package DataStructures; class Node { int value; Node rightNode; Node leftNode; } class BST { Node rootOfTree = null; public void insertintoBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { Node newNode = new Node(); newNode.value = value; rootOfTree = newNode; newNode.rightNode = null; newNode.leftNode = null; } else { while (true) { if (value >= markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.rightNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.leftNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } } } } public void searchBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { System.out.println("Element Not Found"); } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { System.out.println("Element Not Found"); break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { System.out.println("Element Not Found"); break; } } if (value == markedNode.value) { System.out.println("Element Found"); break; } } } } public void deleteFromBST(int value) { Node markedNode = rootOfTree; Node minValueNode = null; if (rootOfTree == null) { System.out.println("Element Not Found"); return; } if (rootOfTree.value == value) { if (rootOfTree.leftNode == null && rootOfTree.rightNode == null) { rootOfTree = null; return; } else if (rootOfTree.leftNode == null ^ rootOfTree.rightNode == null) { if (rootOfTree.rightNode != null) { rootOfTree = rootOfTree.rightNode; return; } else { rootOfTree = rootOfTree.leftNode; return; } } else { minValueNode = rootOfTree.rightNode; if (minValueNode.leftNode == null) { rootOfTree.rightNode.leftNode = rootOfTree.leftNode; rootOfTree = rootOfTree.rightNode; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node rootOfTree.value = minValueNode.leftNode.value; // The value has been swapped if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { break; } else { markedNode = markedNode.rightNode; } } else { System.out.println("Element Not Found"); return; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { break; } else { markedNode = markedNode.leftNode; } } else { System.out.println("Element Not Found"); return; } } } // Parent of the required element found // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { if (markedNode.rightNode.rightNode == null && markedNode.rightNode.leftNode == null) { markedNode.rightNode = null; return; } else if (markedNode.rightNode.rightNode == null ^ markedNode.rightNode.leftNode == null) { if (markedNode.rightNode.rightNode != null) { markedNode.rightNode = markedNode.rightNode.rightNode; return; } else { markedNode.rightNode = markedNode.rightNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.rightNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { if (markedNode.leftNode.rightNode == null && markedNode.leftNode.leftNode == null) { markedNode.leftNode = null; return; } else if (markedNode.leftNode.rightNode == null ^ markedNode.leftNode.leftNode == null) { if (markedNode.leftNode.rightNode != null) { markedNode.leftNode = markedNode.leftNode.rightNode; return; } else { markedNode.leftNode = markedNode.leftNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.leftNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// } } } } } } public class BSTImplementation { public static void main(String[] args) { BST newBst = new BST(); newBst.insertintoBST(19); newBst.insertintoBST(13); newBst.insertintoBST(10); newBst.insertintoBST(20); newBst.insertintoBST(5); newBst.insertintoBST(23); newBst.insertintoBST(28); newBst.insertintoBST(16); newBst.insertintoBST(27); newBst.insertintoBST(9); newBst.insertintoBST(4); newBst.insertintoBST(22); newBst.insertintoBST(17); newBst.insertintoBST(30); newBst.insertintoBST(40); newBst.deleteFromBST(5); newBst.deleteFromBST(4); newBst.deleteFromBST(9); newBst.deleteFromBST(10); newBst.deleteFromBST(13); newBst.deleteFromBST(16); newBst.deleteFromBST(17); newBst.searchBST(5); newBst.searchBST(4); newBst.searchBST(9); newBst.searchBST(10); newBst.searchBST(13); newBst.searchBST(16); newBst.searchBST(17); System.out.println(); newBst.deleteFromBST(20); newBst.deleteFromBST(23); newBst.deleteFromBST(27); newBst.deleteFromBST(28); newBst.deleteFromBST(30); newBst.deleteFromBST(40); newBst.searchBST(20); newBst.searchBST(23); newBst.searchBST(27); newBst.searchBST(28); newBst.searchBST(30); newBst.searchBST(40); } }

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  • Problem with building tree bottom up

    - by Esmond
    Hi, I have problems building a binary tree from the bottom up. THe input of the tree would be internal nodes of the trees with the children of this node being the leaves of the eventual tree. So initially if the tree is empty the root would be the first internal node. Afterwards, The next internal node to be added would be the new root(NR), with the old root(OR) being one of the child of NR. And so on. The problem i have is that whenever i add a NR, the children of the OR seems to be lost when i do a inOrder traversal. This is proven to be the case when i do a getSize() call which returns the same number of nodes before and after addNode(Tree,Node) Any help with resolving this problem is appreciated edited with the inclusion of node class code. both tree and node classes have the addChild methods because i'm not very sure where to put them for it to be appropriated. any comments on this would be appreciated too. The code is as follows: import java.util.*; public class Tree { Node root; int size; public Tree() { root = null; } public Tree(Node root) { this.root = root; } public static void setChild(Node parent, Node child, double weight) throws ItemNotFoundException { if (parent.child1 != null && parent.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (parent.child1 != null) { parent.child2 = child; child.parent = parent; parent.c2Weight = weight; } else { parent.child1 = child; child.parent = parent; parent.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public static Tree addNode(Tree tree, Node node) throws ItemNotFoundException { Tree tree1; if (tree.root == null) { tree.root = node; } else if (tree.root.getSeq().equals(node.getChild1().getSeq()) || tree.root.getSeq().equals(node.getChild2().getSeq())) { Node oldRoot = tree.root; oldRoot.setParent(node); tree.root = node; } else { //form a disjoint tree and merge the 2 trees tree1 = new Tree(node); tree = mergeTree(tree, tree1); } System.out.print("addNode2 = "); if(tree.root != null ) { Tree.inOrder(tree.root); } System.out.println(); return tree; } public static Tree mergeTree(Tree tree, Tree tree1) { String root = "root"; Node node = new Node(root); tree.root.setParent(node); tree1.root.setParent(node); tree.root = node; return tree; } public static int getSize(Node root) { if (root != null) { return 1 + getSize(root.child1) + getSize(root.child2); } else { return 0; } } public static boolean isEmpty(Tree Tree) { return Tree.root == null; } public static void inOrder(Node root) { if (root != null) { inOrder(root.child1); System.out.print(root.sequence + " "); inOrder(root.child2); } } } public class Node { Node child1; Node child2; Node parent; double c1Weight; double c2Weight; String sequence; boolean isInternal; public Node(String seq) { sequence = seq; child1 = null; c1Weight = 0; child2 = null; c2Weight = 0; parent = null; isInternal = false; } public boolean hasChild() { if (this.child1 == null && this.child2 == null) { this.isInternal = false; return isInternal; } else { this.isInternal = true; return isInternal; } } public String getSeq() throws ItemNotFoundException { if (this.sequence == null) { throw new ItemNotFoundException("No such node"); } else { return this.sequence; } } public void setChild(Node child, double weight) throws ItemNotFoundException { if (this.child1 != null && this.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (this.child1 != null) { this.child2 = child; this.c2Weight = weight; } else { this.child1 = child; this.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public void setParent(Node parent){ this.parent = parent; } public Node getParent() throws ItemNotFoundException { if (this.parent == null) { throw new ItemNotFoundException("This Node has no parent"); } else { return this.parent; } } public Node getChild1() throws ItemNotFoundException { if (this.child1 == null) { throw new ItemNotFoundException("There is no child1"); } else { return this.child1; } } public Node getChild2() throws ItemNotFoundException { if (this.child2 == null) { throw new ItemNotFoundException("There is no child2"); } else { return this.child2; } } }

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  • What does "context-free" mean in the term "context-free grammar"?

    - by rick
    Given the amount of material that tries to explain what a context-free grammar (CFG) is, I found it surprising that very few (in my sample, less than 1 in 20) give an explanation on why such grammars are called "context-free". And, to my mind, none succeeds in doing so. My question is, why are context-free grammars called context-free? What is "the context"? I had an intuition that the context could be other language constructs surrounding the currently analyzed construct, but that seems not to be the case. Could anyone provide a precise explanation?

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  • Behaviour tree code example?

    - by jokoon
    http://altdevblogaday.org/2011/02/24/introduction-to-behavior-trees/ Obviously the most interesting article I found on this website. What do you think about it ? It lacks some code example, don't you know any ? I also read that state machines are not very flexible compared to behaviour trees... On top of that I'm not sure if there is a true link between state machines and the state pattern... is there ?

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  • Context-sensitive grammar for specific language

    - by superagio
    How can I construct a grammar that generates this language? Construct a grammar that generates L: L = {a^n b^m c^k|k>n, k>m} I believe my productions should go along this lines: S-> ABCC A-> a|aBC|BC B-> b|bBC C-> c|Cc CB->BC The idea is to start with 2 c and keep always one more c, and then with C-c|Cc ad as much c as i want. How can my production for C remember the numbers of m and n.

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  • On ocamlyacc, function application grammar and precedence

    - by Amadan
    I'm OCaml newbie and I'm trying to write a simple OCaml-like grammar, and I can't figure this out. My grammar allows something like this: let sub = fun x -> fun y -> x - y;; However, if I want to use the function so defined, I can write: (sub 7) 3 but I can't write sub 7 3, which really bugs me. For some reason, it gets interpreted as if I wrote sub (7 3) (which would treat 7 as a function with argument 3). The relevant sections are: /* other operators, then at the very end: */ %left APPLY /* ... */ expr: /* ... */ | expr expr %prec APPLY { Apply($1, $2) } Thanks!

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  • How to procedurally (create) grow an artistic (2D) tree in real-time (L-System?).

    - by lalan
    Recently I programmed an L-system module, It got me interested further. I am a Plants vs Zombies junkie as well, really liked the concept of Tree of Wisdom. Would love to create similar procedural art just for fun and learn more. Question: How should I approach the process of creating an artistic tree (2d perhaps with fixed camera/perspective) dynamically? Ideally I would like to start with a plant (only a stem with a leaf) and grow it dynamically using some influence (input/user action) over its structure. These influences may result in different type of branching, curves in branches, its spread, location of fruits, color of flowers, etc. Want it to be really full of life/spirit. :) Plants vs Zombies: Tree of wisdom It would be great to dynamically grow a similar tree, but with lot more variation and animations happening. My Background: Student / Programmer, have used few game engines (Ogre3d, cocos2d, unity). Haven't really programmed directly using openGL, trying to fix that :). I am ready to spend considerable time, Please let me know about the APIs? and how would an expert like you would take on this problem? Why 2D? I think it's easier to solve the problem only considering 2 dimensions. Artistic inspirations: Only the tree, with fruits and leaves, without the shrubs at the bottom The large tree (visible branches, green leaves, flowers, fruits, etc) on the left, behind monkey. PixelJunk's Eden (Art style inspiration). Procedurally Generated Apple Tree using Fractals Please let me know if it was easy for you to understand the question, I may elaborate further. I hope a discussion of various approach would be helpful for everyone. You guys are awesome.

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  • Do search engines directly penalize bad grammar?

    - by Nicolas Raoul
    Let's say I have a web page with user-contributed content, which is good content but with bad grammar, slang terms, inappropriate tone. I know that bad grammar is a also a problem because it drives away visitors and scares people from linking to it, but let's put that aside. Let's also put aside the fact that incorrectly spelt terms might be ignored by a crawler, potentially leading to less text-comparizon hits. QUESTION: Do search engines like Google directly recognize and penalize bad grammar? For instance because they might consider bad-grammar as a sign of low-quality content.

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  • Upgrade tree to 1.6?

    - by Pureferret
    I'm trying to upgrade my version of tree to 1.6 on ubuntu 12.04. I've d'loaded, ran make and make install in the terminal using the sudo command. ~/tree-1.6.0$ sudo make make: Nothing to be done for `all'. I've already run sudo make here ~/tree-1.6.0$ sudo make install install -d /usr/bin install -d /usr/man/man1 if [ -e tree ]; then \ install -s tree /usr/bin/tree; \ fi install doc/tree.1 /usr/man/man1/tree.1 What's this output though? It's not updated. I've checked the man page, and -du doesn't work. How am I supposed to update tree if not via the terminal?

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  • Looking for Antlr Grammar syntaxt highlight in VS2010

    - by David Mårtensson
    I am looking for some way to edit antlr grammar files directly within VS2010 with syntax highlight. I have used antlrworks a lot but it has the drawback that I have to start antlrworks separately and then browse to the file I want to edit, do the changed and save. For minor fixes I do not need all the tools in Antlrworks but I still would like the syntax highlight. But I have not been able to get VS2010 to open antlrworks with the right file and I have found no other way to get syntax highlight directly within VS2010 editor, it just opens as plain text. I can get visual studio to open antlrworks but it will open with only the last set of files it had open, not the one I clicked on. So my question(s) are: Is there a way to get antlrworks to open with the right file when I double click in it in visual studio project explorer? Is there any other way to get correct syntax highlight for antlr grammar files within visual studio (or with another editor, preferably not one that costs money, but if there are no free ones a commercial one might be an option).

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  • Natural language grammar and user-entered names

    - by Owen Blacker
    Some languages, particularly Slavic languages, change the endings of people's names according to the grammatical context. (For those of you who know grammar or studied languages that do this to words, such as German or Russian, and to help with search keywords, I'm talking about noun declension.) This is probably easiest with a set of examples (in Polish, to save the whole different-alphabet problem): Dorothy saw the cat — Dorota zobaczyla kota The cat saw Dorothy — Kot zobaczyl Dorote It is Dorothy’s cat — To jest kot Doroty I gave the cat to Dorothy — Dalam kota Dorotie I went for a walk with Dorothy — Poszlam na spacer z Dorota “Hello, Dorothy!” — “Witam, Doroto!” Now, if, in these examples, the name here were to be user-entered, that introduces a world of grammar nightmares. Importantly, if I went for Katie (Kasia), the examples are not directly comparable — 3 and 4 are both Kasi, rather than *Kasy and *Kasie — and male names will be wholly different again. I'm guessing someone has dealt with this situation before, but my Google-fu appears to be weak today. I can find a lot of links about natural-language processing, but I don'think that's quite what I want. To be clear: I'm only ever gonna have one user-entered name per user and I'm gonna need to decline them into known configurations — I'll have a localised text that will have placeholders something like {name nominative} and {name dative}, for the sake of argument. I really don't want to have to do lexical analysis of text to work stuff out, I'll only ever need to decline that one user-entered name. Anyone have any recommendations on how to do this, or do I need to start calling round localisation agencies ;o) Further reading (all on Wikipedia) for the interested: Declension Grammatical case Declension in Polish Declension in Russian Declension in Czech nouns and pronouns Disclaimer: I know this happens in many other languages; highlighting Slavic languages is merely because I have a project that is going to be localised into some Slavic languages.

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  • Configure spanning tree from HP to Cisco hardware

    - by Tim Brigham
    I have three switches I'd like to configure in a loop - a Cisco stack (3750s) and two HP 2900 series. Each is connected to the next with a 10 gig backplane of one form or another. How do I configure the spanning tree on these systems to make this function correctly? From the documents I've looked at it looks like I need to set both sets of hardware to use MST mode but I'm not sure past that point. The trunking, etc is all set up as needed. HP Switch 1 A4 connected to Cisco 1/0/1. HP Switch 2 B2 connected to Cisco 2/0/1. HP Switch 1 A2 connected to HP Switch 2 A1. HP Switch 1 show spanning-tree Multiple Spanning Tree (MST) Information STP Enabled : Yes Force Version : MSTP-operation IST Mapped VLANs : 1-4094 Switch MAC Address : 0021f7-126580 Switch Priority : 32768 Max Age : 20 Max Hops : 20 Forward Delay : 15 Topology Change Count : 352,485 Time Since Last Change : 2 secs CST Root MAC Address : 0018ba-c74268 CST Root Priority : 1 CST Root Path Cost : 200000 CST Root Port : 1 IST Regional Root MAC Address : 0021f7-126580 IST Regional Root Priority : 32768 IST Regional Root Path Cost : 0 IST Remaining Hops : 20 Root Guard Ports : TCN Guard Ports : BPDU Protected Ports : BPDU Filtered Ports : PVST Protected Ports : PVST Filtered Ports : | Prio | Designated Hello Port Type | Cost rity State | Bridge Time PtP Edge ----- --------- + --------- ---- ---------- + ------------- ---- --- ---- ... A1 | Auto 128 Disabled | A2 10GbE-CX4 | 2000 128 Forwarding | 0021f7-126580 2 Yes No A3 10GbE-CX4 | Auto 128 Disabled | A4 10GbE-SR | 2000 128 Forwarding | 0021f7-126580 2 Yes No HP Switch 2 show spanning-tree Multiple Spanning Tree (MST) Information STP Enabled : Yes Force Version : MSTP-operation IST Mapped VLANs : 1-4094 Switch MAC Address : 0024a8-cd6000 Switch Priority : 32768 Max Age : 20 Max Hops : 20 Forward Delay : 15 Topology Change Count : 19,623 Time Since Last Change : 32 secs CST Root MAC Address : 0018ba-c74268 CST Root Priority : 1 CST Root Path Cost : 202000 CST Root Port : A1 IST Regional Root MAC Address : 0024a8-cd6000 IST Regional Root Priority : 32768 IST Regional Root Path Cost : 0 IST Remaining Hops : 20 Root Guard Ports : TCN Guard Ports : BPDU Protected Ports : BPDU Filtered Ports : PVST Protected Ports : PVST Filtered Ports : | Prio | Designated Hello Port Type | Cost rity State | Bridge Time PtP Edge ----- --------- + --------- ---- ---------- + ------------- ---- --- ---- ... A1 10GbE-CX4 | 2000 128 Forwarding | 0021f7-126580 2 Yes No A2 10GbE-CX4 | Auto 128 Disabled | B1 SFP+SR | 2000 128 Blocking | a44c11-a67c80 2 Yes No B2 | Auto 128 Disabled | Cisco Stack 1 show spanning-tree ... (additional VLANs) VLAN0100 Spanning tree enabled protocol ieee Root ID Priority 1 Address 0018.bac7.426e Cost 2 Port 107 (TenGigabitEthernet2/1/1) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32868 (priority 32768 sys-id-ext 100) Address a44c.11a6.7c80 Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Te1/1/1 Desg FWD 2 128.53 P2p Te2/1/1 Root FWD 2 128.107 P2p

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  • Gson Deserialize to Java Tree

    - by MountainX
    I need to deserialize some JSON to a Java tree structure that contains TreeNodes and NodeData. TreeNodes are thin wrappers around NodeData. I'll provide the JSON and the classes below. I have looked at the usual Gson help sources, including here, but I can't seem to come up with the solution. Serialization works fine with Gson. The JSON below was produced by Gson. But deserialization is the problem I need help with. Can someone show me how to write the deserializer (or suggest an alternative approach using Gson best practices)? Here is my JSON. The "data" element corresponds to class NodeData, and the "subList" JSON element corresponds to Java class TreeNode. { "data": { "version": "032", "name": "root", "path": "/", "id": "1", "parentId": "0", "toolTipText": "rootNode" }, "subList": [ { "data": { "version": "032", "name": "level1", "labelText": "Some Label Text at Level1", "path": "/root", "id": "2", "parentId": "1", "toolTipText": "a tool tip for level1" }, "subList": [ { "data": { "version": "032", "name": "level1_1", "labelText": "Label level1_1", "path": "/root/level1", "id": "3", "parentId": "2", "toolTipText": "ToolTipText for level1_1" } }, { "data": { "version": "032", "name": "level1_2", "labelText": "Label level1_2", "path": "/root/level1", "id": "4", "parentId": "2", "toolTipText": "ToolTipText for level1_2" } } ] }, { "data": { "version": "032", "name": "level2", "path": "/root", "id": "5", "parentId": "1", "toolTipText": "ToolTipText for level2" }, "subList": [ { "data": { "version": "032", "name": "level2_1", "labelText": "Label level2_1", "path": "/root/level2", "id": "6", "parentId": "5", "toolTipText": "ToolTipText for level2_1" }, "subList": [ { "data": { "version": "032", "name": "level2_1_1", "labelText": "Label level2_1_1", "path": "/root/level2/level2_1", "id": "7", "parentId": "6", "toolTipText": "ToolTipText for level2_1_1" } } ] } ] } ] } Here are the Java classes: public class Tree { private TreeNode rootElement; private HashMap<String, TreeNode> indexById; private HashMap<String, TreeNode> indexByKey; private long nextAvailableID = 0; public Tree() { indexById = new HashMap<String, TreeNode>(); indexByKey = new HashMap<String, TreeNode>(); } public long getNextAvailableID() { return this.nextAvailableID; } ... [snip] ... } public class TreeNode { private Tree tree; private NodeData data; public List<TreeNode> subList; private HashMap<String, TreeNode> indexById; private HashMap<String, TreeNode> indexByKey; //this default ctor is used only for Gson deserialization public TreeNode() { this.tree = new Tree(); indexById = tree.getIdIndex(); indexByKey = tree.getKeyIndex(); this.makeRoot(); tree.setRootElement(this); } //makes this node the root node. Calling this obviously has side effects. public NodeData makeRoot() { NodeData rootProp = new NodeData(TreeFactory.version, "example", "rootNode"); String nextAvailableID = getNextAvailableID(); if (!nextAvailableID.equals("1")) { throw new IllegalStateException(); } rootProp.setId(nextAvailableID); rootProp.setParentId("0"); rootProp.setKeyPathOnly("/"); rootProp.setSchema(tree); this.data = rootProp; rootProp.setNode(this); indexById.put(rootProp.getId(), this); indexByKey.put(rootProp.getKeyFullName(), this); return rootProp; } ... [snip] ... } public class NodeData { protected static Tree tree; private LinkedHashMap<String, String> keyValMap; protected String version; protected String name; protected String labelText; protected String path; protected String id; protected String parentId; protected TreeNode node; protected String toolTipText;//tool tip or help string protected String imagePath;//for things like images; not persisted to properties protected static final String delimiter = "/"; //this default ctor is used only for Gson deserialization public NodeData() { this("NOT_SET", "NOT_SET", "NOT_SET"); } ... [snip] ... } Side note: The tree data structure is a bit strange, as it includes indexes. Obviously, this isn't a typical search tree. In fact, the tree is used mainly to create a hierarchical path element (String) in each NodeData element. (Example: "path": "/root/level2/level2_1".) The indexes are actually used for NodeData retrieval.

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  • How does a BSP tree work for Z sorting?

    - by Jenko
    I'm developing a 3D engine in software, and so I must compute Z sorting manually. I'm currently using the painters algorithm to sort triangles and then drawing them back-to-front. This causes artifacts that I'm trying to correct. Would using a dynamic BSP-tree ensure "correct Z sorting" of triangles? Why? Because the bounding volumes of triangles would be similar? Since I would have a single "world" BSP tree, would I have to remove and re-add any moved/scaled/rotated object into the tree? Is it possible to add triangles into a BSP tree without the expensive cutting process? Why do you need to cut triangles on the axis planes anyway? Is it faster to traverse a BSP tree from any angle, than to sort all tris each draw like the painters algorithm?

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  • Implement Tree/Details With Taskflow Regions Using EJB

    - by Deepak Siddappa
    This article describes on Display Tree/Details using taskflow regions.Use Case DescriptionLet us take scenario where we need to display Tree/Details, left region contains category hierarchy with items listed in a tree structure (ex:- Region-Countries-Locations-Departments in tree format) and right region contains the Employees list.In detail, Here User may drills down through categories using a tree until Employees are listed. Clicking the tree node name displays Employee list in the adjacent pane related to particular tree node. Implementation StepsThe script for creating the tables and inserting the data required for this application CreateSchema.sql Lets create a Java EE Web Application with Entities based on Regions, Countries, Locations, Departments and Employees table. Create a Stateless Session Bean and data control for the Stateless Session Bean. Add the below code to the session bean and expose the method in local/remote interface and generate a data control for that.Note:- Here in the below code "em" is a EntityManager. public List<Employees> empFilteredByTreeNode(String treeNodeType, String paramValue) { String queryString = null; try { if (treeNodeType == "null") { queryString = "select * from Employees emp ORDER BY emp.employee_id ASC"; } else if (Pattern.matches("[a-zA-Z]+[_]+[a-zA-Z]+[_]+[[0-9]+]+", treeNodeType)) { queryString = "select * from employees emp INNER JOIN departments dept\n" + "ON emp.department_id = dept.department_id JOIN locations loc\n" + "ON dept.location_id = loc.location_id JOIN countries cont\n" + "ON loc.country_id = cont.country_id JOIN regions reg\n" + "ON cont.region_id = reg.region_id and reg.region_name = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.contains("regionsFindAll_bc_countriesList_1")) { queryString = "select * from employees emp INNER JOIN departments dept \n" + "ON emp.department_id = dept.department_id JOIN locations loc \n" + "ON dept.location_id = loc.location_id JOIN countries cont \n" + "ON loc.country_id = cont.country_id and cont.country_name = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.contains("regionsFindAll_bc_locationsList_1")) { queryString = "select * from employees emp INNER JOIN departments dept ON emp.department_id = dept.department_id JOIN locations loc ON dept.location_id = loc.location_id and loc.city = '" + paramValue + "' ORDER BY emp.employee_id ASC"; } else if (treeNodeType.trim().contains("regionsFindAll_bc_departmentsList_1")) { queryString = "select * from Employees emp INNER JOIN Departments dept ON emp.DEPARTMENT_ID = dept.DEPARTMENT_ID and dept.DEPARTMENT_NAME = '" + paramValue + "'"; } } catch (NullPointerException e) { System.out.println(e.getMessage()); } return em.createNativeQuery(queryString, Employees.class).getResultList(); } In the ViewController project, create two ADF taskflow with page Fragments and name them as FirstTaskflow and SecondTaskflow respectively. Open FirstTaskflow,from component palette drop view(Page Fragment) name it as TreeList.jsff. Open SeconfTaskflow, from component palette drop view(Page Fragment) name it as EmpList.jsff and create two paramters in its overview parameters tab as shown in below image. Open TreeList.jsff , from data control palette drop regionsFindAll->Tree as ADF Tree. In Edit Tree Binding dialog, for Tree Level Rules select the display attributes as follows:-model.Regions - regionNamemodel.Countries - countryNamemodel.Locations - citymodel.Departments - departmentName In structure panel, click on af:Tree - t1 and select selectionListener with edit property. Create a "TreeBean" managed bean with scope as "session" as shown in below Image. Create new method as getTreeNodeSelectedValue and click ok. Open TreeBean managed bean and add the below code: private String treeNodeType; private String paramValue; public void getTreeNodeSelectedValue(SelectionEvent selectionEvent) { RichTree tree = (RichTree)selectionEvent.getSource(); RowKeySet addedSet = selectionEvent.getAddedSet(); Iterator i = addedSet.iterator(); TreeModel model = (TreeModel)tree.getValue(); model.setRowKey(i.next()); JUCtrlHierNodeBinding node = (JUCtrlHierNodeBinding)tree.getRowData(); //oracle.jbo.Row Row rw = node.getRow(); Object selectedTreeNode = node.getAttribute(0); Object treeListType = node.getBindings(); String treeNodeType = treeListType.toString(); this.setParamValue(selectedTreeNode.toString()); this.setTreeNodeType(treeNodeType); } public void setTreeNodeType(String treeNodeType) { this.treeNodeType = treeNodeType; } public String getTreeNodeType() { return treeNodeType; } public void setParamValue(String paramValue) { this.paramValue = paramValue; } public String getParamValue() { return paramValue; }<br /> Open EmpList.jsff , from data control palette drop empFilteredByTreeNode->Employees->Table as ADF Read-only Table. After selecting the  Employees result set, in Edit Action Binding dialog window pass the pageFlowScope parameters as shown in below Image. In empList.jsff page, click Binding tab and click on Create Executable binding and select Invoke action and follow as shown in below image. Edit executeEmpFiltered invoke action properties and set the Refresh to ifNeeded, So when ever the page needs the method will be executed. Create Main.jspx page with page template as Oracle Three Column Layout. Drop FirstTaskflow as Region in start facet and drop SecondTaskflow as Region in center facet, Edit task Flow Binding dialog window pass the Input Paramters as shown in below Image. Run the Main.jspx, tree will be displayed in left region and emp details will displyaed on the right region. Click on the Americas in tree node, all emp related to the Americas related will be displayed. Click on Americas->United States of America->South San Francisco->Accounting, only employee belongs to the Accounting department will be displayed.

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  • public family tree

    - by Remus Rigo
    Hi all Does anyone know a ancestry site that allows you to create a public profile or tree, so that other visitors can see your family tree. In all sites that I have found (dynastree.com, familylink.com, ancestry.com, genebase.com), if someone wants to see your family tree, they must be members or register. thanks

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  • Visualizing Parallel Branch and Bound Tree Exploration

    - by Akhil
    I have written a parallel program that does a depth first branch and bound exploration of a tree. I can dump the id's (id's are like this 0, 00, 01, 0000, 0001, etc.) of the nodes at frequent intervals to know the frontier of the tree that is being explored at that instant in the tree. The challenge is to visualize the tree exploration with time. Any ideas? e.g. I can draw trees(e.g. using graphViz) at different times and create a movie out of it. Looking for ideas to facilitate this visualization - some better ways to do so or easy tools that can help me make the visualization

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