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  • Can a binary tree or tree be always represented in a Database as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • Can a binary tree or tree be always represented in a Database table as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person under him can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • Inorder tree traversal in binary tree in C

    - by srk
    In the below code, I'am creating a binary tree using insert function and trying to display the inserted elements using inorder function which follows the logic of In-order traversal.When I run it, numbers are getting inserted but when I try the inorder function( input 3), the program continues for next input without displaying anything. I guess there might be a logical error.Please help me clear it. Thanks in advance... #include<stdio.h> #include<stdlib.h> int i; typedef struct ll { int data; struct ll *left; struct ll *right; } node; node *root1=NULL; // the root node void insert(node *root,int n) { if(root==NULL) //for the first(root) node { root=(node *)malloc(sizeof(node)); root->data=n; root->right=NULL; root->left=NULL; } else { if(n<(root->data)) { root->left=(node *)malloc(sizeof(node)); insert(root->left,n); } else if(n>(root->data)) { root->right=(node *)malloc(sizeof(node)); insert(root->right,n); } else { root->data=n; } } } void inorder(node *root) { if(root!=NULL) { inorder(root->left); printf("%d ",root->data); inorder(root->right); } } main() { int n,choice=1; while(choice!=0) { printf("Enter choice--- 1 for insert, 3 for inorder and 0 for exit\n"); scanf("%d",&choice); switch(choice) { case 1: printf("Enter number to be inserted\n"); scanf("%d",&n); insert(root1,n); break; case 3: inorder(root1); break; default: break; } } }

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  • AVL tree in C language

    - by I_S_W
    Hey all; i am currently doing a project that requires the use of AVL trees , the insert function i wrote for the avl does not seem to be working , it works for 3 or 4 nodes at maximum ; i would really appreciate your help The attempt is below enter code here Tree insert(Tree t,char name[80],int num) { if(t==NULL) { t=(Tree)malloc(sizeof(struct node)); if(t!=NULL) { strcpy(t->name,name); t->num=num; t->left=NULL; t->right=NULL; t->height=0; } } else if(strcmp(name,t->name)<0) { t->left=insert(t->left,name,num); if((height(t->left)-height(t->right))==2) if(strcmp(name,t->left->name)<0) { t=s_rotate_left(t);} else{ t=d_rotate_left(t);} } else if(strcmp(name,t-name)0) { t-right=insert(t-right,name,num); if((height(t-right)-height(t-left))==2) if(strcmp(name,t-right-name)0){ t=s_rotate_right(t); } else{ t=d_rotate_right(t);} } t-height=max(height(t-left),height(t-right))+1; return t; }

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  • New to AVL tree implementation.

    - by nn
    I am writing a sliding window compression algorithm (LZ77) that searches for phrases in a "moving" dictionary. So far I have written a BST where each node is stored in an array and it's index in the array is also the value of the starting position in the window itself. I am now looking at transforming the BST to an AVL tree. I am a little confused at the sample implementations I have seen. Some only appear to store the balance factors whereas others store the height of each tree. Are there any performance advantage/disadvantages of storing the height and/or balance factor for each node? Apologies if this is a very simple question, but I'm still not visualizing how I want to restructure my BST to implement height balancing. Thanks.

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  • Convert a binary tree to linked list, breadth first, constant storage/destructive

    - by Merlyn Morgan-Graham
    This is not homework, and I don't need to answer it, but now I have become obsessed :) The problem is: Design an algorithm to destructively flatten a binary tree to a linked list, breadth-first. Okay, easy enough. Just build a queue, and do what you have to. That was the warm-up. Now, implement it with constant storage (recursion, if you can figure out an answer using it, is logarithmic storage, not constant). I found a solution to this problem on the Internet about a year back, but now I've forgotten it, and I want to know :) The trick, as far as I remember, involved using the tree to implement the queue, taking advantage of the destructive nature of the algorithm. When you are linking the list, you are also pushing an item into the queue. Each time I try to solve this, I lose nodes (such as each time I link the next node/add to the queue), I require extra storage, or I can't figure out the convoluted method I need to get back to a node that has the pointer I need. Even the link to that original article/post would be useful to me :) Google is giving me no joy. Edit: Jérémie pointed out that there is a fairly simple (and well known answer) if you have a parent pointer. While I now think he is correct about the original solution containing a parent pointer, I really wanted to solve the problem without it :) The refined requirements use this definition for the node: struct tree_node { int value; tree_node* left; tree_node* right; };

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  • How to functionally generate a tree breadth-first. (With Haskell)

    - by Dennetik
    Say I have the following Haskell tree type, where "State" is a simple wrapper: data Tree a = Branch (State a) [Tree a] | Leaf (State a) deriving (Eq, Show) I also have a function "expand :: Tree a - Tree a" which takes a leaf node, and expands it into a branch, or takes a branch and returns it unaltered. This tree type represents an N-ary search-tree. Searching depth-first is a waste, as the search-space is obviously infinite, as I can easily keep on expanding the search-space with the use of expand on all the tree's leaf nodes, and the chances of accidentally missing the goal-state is huge... thus the only solution is a breadth-first search, implemented pretty decent over here, which will find the solution if it's there. What I want to generate, though, is the tree traversed up to finding the solution. This is a problem because I only know how to do this depth-first, which could be done by simply called the "expand" function again and again upon the first child node... until a goal-state is found. (This would really not generate anything other then a really uncomfortable list.) Could anyone give me any hints on how to do this (or an entire algorithm), or a verdict on whether or not it's possible with a decent complexity? (Or any sources on this, because I found rather few.)

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  • Spanning-tree setup with incompatible switches

    - by wfaulk
    I have a set of eight HP ProCurve 2910al-48G Ethernet switches at my datacenter that are set up in a star topology with no physical loops. I want to partially mesh the switches for redundancy and manage the loops with a spanning-tree protocol. However, our connection to the datacenter is provided by two uplinks, each to a Cisco 3750. The datacenter's switches are handling the redundant connection using PVST spanning-tree, which is a Cisco-proprietary spanning-tree implementation that my HP switches do not support. It appears that my switches are not participating in the datacenter's spanning-tree domain, but are blindly passing the BPDUs between the two switchports on my side, which enables the datacenter's switches to recognize the loop and put one of the uplinks into the Blocking state. This is somewhat supposition, but I can confirm that, while my switches say that both of the uplink ports are forwarding, only one is passing any real quantity of data. (I am assuming that I cannot get the datacenter to move away from PVST. I don't know that I'd want them to make that significant of a change anyway.) The datacenter has also sent me this output from their switches (which I have expurgated of any identifiable info): 3750G-1#sh spanning-tree vlan nnn VLAN0nnn Spanning tree enabled protocol ieee Root ID Priority 10 Address 00d0.0114.xxxx Cost 4 Port 5 (GigabitEthernet1/0/5) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32mmm (priority 32768 sys-id-ext nnn) Address 0018.73d3.yyyy Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Gi1/0/5 Root FWD 4 128.5 P2p Gi1/0/6 Altn BLK 4 128.6 P2p Gi1/0/8 Altn BLK 4 128.8 P2p and: 3750G-2#sh spanning-tree vlan nnn VLAN0nnn Spanning tree enabled protocol ieee Root ID Priority 10 Address 00d0.0114.xxxx Cost 4 Port 6 (GigabitEthernet1/0/6) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32mmm (priority 32768 sys-id-ext nnn) Address 000f.f71e.zzzz Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Gi1/0/1 Desg FWD 4 128.1 P2p Gi1/0/5 Altn BLK 4 128.5 P2p Gi1/0/6 Root FWD 4 128.6 P2p Gi1/0/8 Desg FWD 4 128.8 P2p The uplinks to my switches are on Gi1/0/8 on both of their switches. The uplink ports are configured with a single tagged VLAN. I am also using a number of other tagged VLANs in my switch infrastructure. And, to be clear, I am passing the tagged VLAN I'm receiving from the datacenter to other ports on other switches in my infrastructure. My question is: how do I configure my switches so that I can use a spanning tree protocol inside my switch infrastructure without breaking the datacenter's spanning tree that I cannot participate in?

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  • Better, simpler example of 'semantic conflict'?

    - by rhubbarb
    I like to distinguish three different types of conflict from a version control system (VCS): textual syntactic semantic A textual conflict is one that is detected by the merge or update process. This is flagged by the system. A commit of the result is not permitted by the VCS until the conflict is resolved. A syntactic conflict is not flagged by the VCS, but the result will not compile. Therefore this should also be picked up by even a slightly careful programmer. (A simple example might be a variable rename by Left and some added lines using that variable by Right. The merge will probably have an unresolved symbol. Alternatively, this might introduce a semantic conflict by variable hiding.) Finally, a semantic conflict is not flagged by the VCS, the result compiles, but the code may have problems running. In mild cases, incorrect results are produced. In severe cases, a crash could be introduced. Even these should be detected before commit by a very careful programmer, through either code review or unit testing. My example of a semantic conflict uses SVN (Subversion) and C++, but those choices are not really relevant to the essence of the question. The base code is: int i = 0; int odds = 0; while (i < 10) { if ((i & 1) != 0) { odds *= 10; odds += i; } // next ++ i; } assert (odds == 13579) The Left (L) and Right (R) changes are as follows. Left's 'optimisation' (changing the values the loop variable takes): int i = 1; // L int odds = 0; while (i < 10) { if ((i & 1) != 0) { odds *= 10; odds += i; } // next i += 2; // L } assert (odds == 13579) Right's 'optimisation' (changing how the loop variable is used): int i = 0; int odds = 0; while (i < 5) // R { odds *= 10; odds += 2 * i + 1; // R // next ++ i; } assert (odds == 13579) This is the result of a merge or update, and is not detected by SVN (which is correct behaviour for the VCS). int i = 1; // L int odds = 0; while (i < 5) // R { odds *= 10; odds += 2 * i + 1; // R // next i += 2; // L } assert (odds == 13579) The assert fails because odds is 37. So my question is as follows. Is there a simpler example than this? Is there a simple example where the compiled executable has a new crash? As a secondary question, are there cases of this that you have encountered in real code? Again, simple examples are especially welcome.

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  • What is the difference between an Abstract Syntax Tree and a Concrete Syntax Tree?

    - by Jason Baker
    I've been reading a bit about how interpreters/compilers work, and one area where I'm getting confused is the difference between an AST and a CST. My understanding is that the parser makes a CST, hands it to the semantic analyzer which turns it into an AST. However, my understanding is that the semantic analyzer simply ensures that rules are followed. I don't really understand why it would actually make any changes to make it abstract rather than concrete. Is there something that I'm missing about the semantic analyzer, or is the difference between an AST and CST somewhat artificial?

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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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  • Conflict Minerals - Design to Compliance

    - by C. Chadwick
    Dr. Christina  Schröder - Principal PLM Consultant, Enterprise PLM Solutions EMEA What does the Conflict Minerals regulation mean? Conflict Minerals has recently become a new buzz word in the manufacturing industry, particularly in electronics and medical devices. Known as the "Dodd-Frank Section 1502", this regulation requires SEC listed companies to declare the origin of certain minerals by 2014. The intention is to reduce the use of tantalum, tungsten, tin, and gold which originate from mines in the Democratic Republic of Congo (DRC) and adjoining countries that are controlled by violent armed militia abusing human rights. Manufacturers now request information from their suppliers to see if their raw materials are sourced from this region and which smelters are used to extract the metals from the minerals. A standardized questionnaire has been developed for this purpose (download and further information). Soon, even companies which are not directly affected by the Conflict Minerals legislation will have to collect and maintain this information since their customers will request the data from their suppliers. Furthermore, it is expected that the public opinion and consumer interests will force manufacturers to avoid the use of metals with questionable origin. Impact for existing products Several departments are involved in the process of collecting data and providing conflict minerals compliance information. For already marketed products, purchasing typically requests Conflict Minerals declarations from the suppliers. In order to address requests from customers, technical operations or product management are usually responsible for keeping track of all parts, raw materials and their suppliers so that the required information can be provided. For complex BOMs, it is very tedious to maintain complete, accurate, up-to-date, and traceable data. Any product change or new supplier can, in addition to all other implications, have an effect on the Conflict Minerals compliance status. Influence on product development  It makes sense to consider compliance early in the planning and design of new products. Companies should evaluate which metals are needed or contained in supplier parts and if these could originate from problematic sources. The answer influences the cost and risk analysis during the development. If it is known early on that a part could be non-compliant with respect to Conflict Minerals, alternatives can be evaluated and thus costly changes at a later stage can be avoided. Integrated compliance management  Ideally, compliance data for Conflict Minerals, but also for other regulations like REACH and RoHS, should be managed in an integrated supply chain system. The compliance status is directly visible across the entire BOM at any part level and for the finished product. If data is missing, a request to the supplier can be triggered right away without having to switch to another system. The entire process, from identification of the relevant parts, requesting information, handling responses, data entry, to compliance calculation is fully covered end-to-end while being transparent for all stakeholders. Agile PLM Product Governance and Compliance (PG&C) The PG&C module extends Agile PLM with exactly this integrated functionality. As with the entire Agile product suite, PG&C can be configured according to customer requirements: data fields, attributes, workflows, routing, notifications, and permissions, etc… can be quickly and easily tailored to a customer’s needs. Optionally, external databases can be interfaced to query commercially available sources of Conflict Minerals declarations which obviates the need for a separate supplier request in many cases. Suppliers can access the system directly for data entry through a special portal. The responses to the standard EICC-GeSI questionnaire can be imported by the supplier or internally. Manual data entry is also supported. A set of compliance-specific dashboards and reports complement the functionality Conclusion  The increasing number of product compliance regulations, for which Conflict Minerals is just one example, requires companies to implement an efficient data and process management in this area. Consumer awareness in this matter increases as well so that an integrated system from development to production also provides a competitive advantage. Follow this link to learn more about Agile's PG&C solution

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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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  • Intelligent Conflict Detection and Resolution

    - by Doug Reid
    0 false 18 pt 18 pt 0 0 false false false /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-ascii-font-family:Cambria; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Cambria; mso-hansi-theme-font:minor-latin;} Conflict Detection and Resolution in Oracle GoldenGate11gR2 has gone through a significant overhaul. The improvements that have been made to this area are substantial and will make it easier for customers to implement complex, heterogeneous GoldenGate configurations. GoldenGate has provided methods for conflict detection and resolution for a number of past releases, but at Oracle we have the opportunity to take advantage of some of the great ideas in this area. Oracle has had feature rich conflict detection and resolution framework in other products, which has been implemented in Oracle GoldenGate 11gR2. These improvements are geared toward helping customers more easily implement advanced configurations that require conflict detection and resolution by providing a robust framework for conflict detection for all DML statements and resolution via pre-built methods, all with less code and simpler syntax than in prior releases. Conflict Detection and Resolution in Oracle GoldenGate 11gR2 is available for our supported heterogeneous platforms, which includes Oracle Database, MySQL, Sybase ASE, SQL Server, and DB2 Linux, Unix, Windows, z/OS, plus DB2 on i Series, which is newly supported in this release. Additional information on the Conflict Detection and Resolution capabilities can be found in our documentation. 

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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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  • 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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  • OpenDNS IP Conflict

    - by Paddy
    I have downloaded the OpenDNS client and it was working great till today. Shows me this error: dynamic IP update failed with message !yours or "Your IP address is taken by another user." I am pretty sure i'll get away with the problem by just restarting my connection as i'll be assigned a new dynamic ip. But I wonder how this is happening in the first place. Shouldn't Dynamic IP be unique for everyone?

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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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  • 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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