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• Worst Case number of rotations for BST to AVL algorithm?

  - by spacker_lechuck

  I have a basic algorithm below and I know that the worst case input BST is one that has degenerated to a linked list from inserts to only one side. How would I compute the worst case complexity in terms of number of rotations for this BST to AVL conversion algorithm? IF tree is right heavy { IF tree's right subtree is left heavy { Perform Double Left rotation } ELSE { Perform Single Left rotation } } ELSE IF tree is left heavy { IF tree's left subtree is right heavy { Perform Double Right rotation } ELSE { Perform Single Right rotation } }

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• sshd running but no PID file

  - by dunxd

  I'm recently started using monit to monitor the status of sshd on my CentOS 5.4 server. This works fine, but every so often monit reports that sshd is no longer running. This isn't true - I am still able to login to the server via ssh, however I note the following: There is no longer any PID file at /var/run/sshd.pid - after a reboot this file exists. Once it is gone, restarting sshd via service sshd restart does not create the PID file. sudo service sshd status reports openssh-daemon is stopped - again, restarting sshd does not change this, but a reboot does. sudo service sshd stop reports failed, presumably because of the missing PID file. Any idea what is going on? Update sudo netstat -lptun gives the following output relating to port 22 tcp 0 0 :::22 :::* LISTEN 20735/sshd Killing the process with this PID as suggested by @Henry and then starting sshd via service results in service sshd status recognising the process by PID again. Would still like to understand this better. RPM verify suggested by a couple of answerers shows this: sudo rpm -vV openssh openssh-server openssh-clients | grep 'S\.5' S.5....T c /etc/pam.d/sshd S.5....T c /etc/ssh/sshd_config /etc/pam.d/sshd has the following contents: #%PAM-1.0 auth include system-auth account required pam_nologin.so account include system-auth password include system-auth session optional pam_keyinit.so force revoke session include system-auth #session required pam_loginuid.so Should that last line be commented out? Update Here's the output of @YannickGirouard 's script: $ sudo ./sshd_test Searching for the process listening on port 22... Found the following PID: 21330 Command line for PID 21330: /usr/sbin/sshd Listing process(es) relating to PID 21330: UID PID PPID C STIME TTY TIME CMD root 21330 1 0 14:04 ? 00:00:00 /usr/sbin/sshd Listing RPM information about openssh packages: Name : openssh Relocations: (not relocatable) Version : 4.3p2 Vendor: CentOS Release : 72.el5_7.5 Build Date: Tue 30 Aug 2011 12:34:14 AM BST Install Date: Sun 06 Nov 2011 12:50:57 AM GMT Build Host: builder10.centos.org Group : Applications/Internet Source RPM: openssh-4.3p2-72.el5_7.5.src.rpm Size : 745390 License: BSD Signature : DSA/SHA1, Fri 02 Sep 2011 01:13:01 AM BST, Key ID a8a447dce8562897 URL : http://www.openssh.com/portable.html Summary : The OpenSSH implementation of SSH protocol versions 1 and 2 ------------------------------------------------------ Name : openssh-clients Relocations: (not relocatable) Version : 4.3p2 Vendor: CentOS Release : 72.el5_7.5 Build Date: Tue 30 Aug 2011 12:34:14 AM BST Install Date: Sun 06 Nov 2011 12:51:04 AM GMT Build Host: builder10.centos.org Group : Applications/Internet Source RPM: openssh-4.3p2-72.el5_7.5.src.rpm Size : 871132 License: BSD Signature : DSA/SHA1, Fri 02 Sep 2011 01:13:01 AM BST, Key ID a8a447dce8562897 URL : http://www.openssh.com/portable.html Summary : The OpenSSH client applications ------------------------------------------------------ Name : openssh-server Relocations: (not relocatable) Version : 4.3p2 Vendor: CentOS Release : 72.el5_7.5 Build Date: Tue 30 Aug 2011 12:34:14 AM BST Install Date: Sun 06 Nov 2011 12:51:04 AM GMT Build Host: builder10.centos.org Group : System Environment/Daemons Source RPM: openssh-4.3p2-72.el5_7.5.src.rpm Size : 492478 License: BSD Signature : DSA/SHA1, Fri 02 Sep 2011 01:13:01 AM BST, Key ID a8a447dce8562897 URL : http://www.openssh.com/portable.html Summary : The OpenSSH server daemon ------------------------------------------------------ However, I've since got things working by killing the process and starting afresh, as suggested by @Henry below, so perhaps I am no longer seeing the same thing. Will try again if I am seeing the issue again after next reboot. Update - 14 March Monit alerted me that sshd had disappeared, and again I am able to ssh onto the server. So now I can run the script $ sudo ./sshd_test Searching for the process listening on port 22... Found the following PID: 2208 Command line for PID 2208: /usr/sbin/sshd Listing process(es) relating to PID 2208: UID PID PPID C STIME TTY TIME CMD root 2208 1 0 Mar13 ? 00:00:00 /usr/sbin/sshd root 1885 2208 0 21:50 ? 00:00:00 sshd: dunx [priv] Listing RPM information about openssh packages: Name : openssh Relocations: (not relocatable) Version : 4.3p2 Vendor: CentOS Release : 72.el5_7.5 Build Date: Tue 30 Aug 2011 12:34:14 AM BST Install Date: Sun 06 Nov 2011 12:50:57 AM GMT Build Host: builder10.centos.org Group : Applications/Internet Source RPM: openssh-4.3p2-72.el5_7.5.src.rpm Size : 745390 License: BSD Signature : DSA/SHA1, Fri 02 Sep 2011 01:13:01 AM BST, Key ID a8a447dce8562897 URL : http://www.openssh.com/portable.html Summary : The OpenSSH implementation of SSH protocol versions 1 and 2 ------------------------------------------------------ Name : openssh-clients Relocations: (not relocatable) Version : 4.3p2 Vendor: CentOS Release : 72.el5_7.5 Build Date: Tue 30 Aug 2011 12:34:14 AM BST Install Date: Sun 06 Nov 2011 12:51:04 AM GMT Build Host: builder10.centos.org Group : Applications/Internet Source RPM: openssh-4.3p2-72.el5_7.5.src.rpm Size : 871132 License: BSD Signature : DSA/SHA1, Fri 02 Sep 2011 01:13:01 AM BST, Key ID a8a447dce8562897 URL : http://www.openssh.com/portable.html Summary : The OpenSSH client applications ------------------------------------------------------ Name : openssh-server Relocations: (not relocatable) Version : 4.3p2 Vendor: CentOS Release : 72.el5_7.5 Build Date: Tue 30 Aug 2011 12:34:14 AM BST Install Date: Sun 06 Nov 2011 12:51:04 AM GMT Build Host: builder10.centos.org Group : System Environment/Daemons Source RPM: openssh-4.3p2-72.el5_7.5.src.rpm Size : 492478 License: BSD Signature : DSA/SHA1, Fri 02 Sep 2011 01:13:01 AM BST, Key ID a8a447dce8562897 URL : http://www.openssh.com/portable.html Summary : The OpenSSH server daemon ------------------------------------------------------ Again, when I look for /var/run/sshd.pid I don't find it. $ cat /var/run/sshd.pid cat: /var/run/sshd.pid: No such file or directory $ sudo netstat -anp | grep sshd tcp 0 0 0.0.0.0:22 0.0.0.0:* LISTEN 2208/sshd $ sudo kill 2208 $ sudo service sshd start Starting sshd: [ OK ] $ cat /var/run/sshd.pid 3794 $ sudo service sshd status openssh-daemon (pid 3794) is running... Is it possible that sshd is restarting and not creating a pidfile for some reason?

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• Binary Search Tree Contains Function

  - by Suede

  I am trying to write a "contains" function for a binary search tree. I receive the following error at compile "Unhandled exception at 0x77291CB3 (ntdll.dll) in BST.exe: 0xC00000FD: Stack overflow (parameters: 0x00000001, 0x001E2FFC)." The following is my code. struct Node { int data; Node* leftChild; Node* rightChild; Node() : leftChild(NULL), rightChild(NULL) {} }; struct BST { Node* root; BST() : root(NULL) {} void insert(int value); bool contains(int value); }; void BST::insert(int value) { Node* temp = new Node(); temp->data = value; if(root == NULL) { root = temp; return; } Node* current; current = root; Node* parent; parent = root; current = (temp->data < current->data ? (current->leftChild) : (current->rightChild) while(current != NULL) { parent = current; current = (temp->data < current->data) ? (current->leftChild) : (current->rightChild) } if(temp->data < parent->data) { parent->leftChild = temp; } if(temp->data > parent->data) { parent->rightChild = temp; } } bool BST::contains(int value) { Node* temp = new Node(); temp->data = value; Node* current; current = root; if(temp->data == current->data) { // base case for when node with value is found std::cout << "true" << std::endl; return true; } if(current == NULL) { // base case if BST is empty or if a leaf is reached before value is found std::cout << "false" << std::endl; return false; } else { // recursive step current = (temp->data < current->data) ? (current->leftChild) : (current->rightChild); return contains(temp->data); } } int main() { BST bst; bst.insert(5); bst.contains(4); system("pause"); } As it stands, I would insert a single node with value '5' and I would search the binary search tree for a node with value '4' - thus, I would expect the result to be false.

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• BST insert operation. don't insert a node if a duplicate exists already

  - by jeev

  the following code reads an input array, and constructs a BST from it. if the current arr[i] is a duplicate, of a node in the tree, then arr[i] is discarded. count in the struct node refers to the number of times a number appears in the array. fi refers to the first index of the element found in the array. after the insertion, i am doing a post-order traversal of the tree and printing the data, count and index (in this order). the output i am getting when i run this code is: 0 0 7 0 0 6 thank you for your help. Jeev struct node{ int data; struct node *left; struct node *right; int fi; int count; }; struct node* binSearchTree(int arr[], int size); int setdata(struct node**node, int data, int index); void insert(int data, struct node **root, int index); void sortOnCount(struct node* root); void main(){ int arr[] = {2,5,2,8,5,6,8,8}; int size = sizeof(arr)/sizeof(arr[0]); struct node* temp = binSearchTree(arr, size); sortOnCount(temp); } struct node* binSearchTree(int arr[], int size){ struct node* root = (struct node*)malloc(sizeof(struct node)); if(!setdata(&root, arr[0], 0)) fprintf(stderr, "root couldn't be initialized"); int i = 1; for(;i<size;i++){ insert(arr[i], &root, i); } return root; } int setdata(struct node** nod, int data, int index){ if(*nod!=NULL){ (*nod)->fi = index; (*nod)->left = NULL; (*nod)->right = NULL; return 1; } return 0; } void insert(int data, struct node **root, int index){ struct node* new = (struct node*)malloc(sizeof(struct node)); setdata(&new, data, index); struct node** temp = root; while(1){ if(data<=(*temp)->data){ if((*temp)->left!=NULL) *temp=(*temp)->left; else{ (*temp)->left = new; break; } } else if(data>(*temp)->data){ if((*temp)->right!=NULL) *temp=(*temp)->right; else{ (*temp)->right = new; break; } } else{ (*temp)->count++; free(new); break; } } } void sortOnCount(struct node* root){ if(root!=NULL){ sortOnCount(root->left); sortOnCount(root->right); printf("%d %d %d\n", (root)->data, (root)->count, (root)->fi); } }

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• Binary Search Tree Implementation

  - by Gabe

  I've searched the forum, and tried to implement the code in the threads I found. But I've been working on this real simple program since about 10am, and can't solve the seg. faults for the life of me. Any ideas on what I'm doing wrong would be greatly appreciated. BST.h (All the implementation problems should be in here.) #ifndef BST_H_ #define BST_H_ #include <stdexcept> #include <iostream> #include "btnode.h" using namespace std; /* A class to represent a templated binary search tree. */ template <typename T> class BST { private: //pointer to the root node in the tree BTNode<T>* root; public: //default constructor to make an empty tree BST(); /* You have to document these 4 functions */ void insert(T value); bool search(const T& value) const; bool search(BTNode<T>* node, const T& value) const; void printInOrder() const; void remove(const T& value); //function to print out a visual representation //of the tree (not just print the tree's values //on a single line) void print() const; private: //recursive helper function for "print()" void print(BTNode<T>* node,int depth) const; }; /* Default constructor to make an empty tree */ template <typename T> BST<T>::BST() { root = NULL; } template <typename T> void BST<T>::insert(T value) { BTNode<T>* newNode = new BTNode<T>(value); cout << newNode->data; if(root == NULL) { root = newNode; return; } BTNode<T>* current = new BTNode<T>(NULL); current = root; current->data = root->data; while(true) { if(current->left == NULL && current->right == NULL) break; if(current->right != NULL && current->left != NULL) { if(newNode->data > current->data) current = current->right; else if(newNode->data < current->data) current = current->left; } else if(current->right != NULL && current->left == NULL) { if(newNode->data < current->data) break; else if(newNode->data > current->data) current = current->right; } else if(current->right == NULL && current->left != NULL) { if(newNode->data > current->data) break; else if(newNode->data < current->data) current = current->left; } } if(current->data > newNode->data) current->left = newNode; else current->right = newNode; return; } //public helper function template <typename T> bool BST<T>::search(const T& value) const { return(search(root,value)); //start at the root } //recursive function template <typename T> bool BST<T>::search(BTNode<T>* node, const T& value) const { if(node == NULL || node->data == value) return(node != NULL); //found or couldn't find value else if(value < node->data) return search(node->left,value); //search left subtree else return search(node->right,value); //search right subtree } template <typename T> void BST<T>::printInOrder() const { //print out the value's in the tree in order // //You may need to use this function as a helper //and create a second recursive function //(see "print()" for an example) } template <typename T> void BST<T>::remove(const T& value) { if(root == NULL) { cout << "Tree is empty. No removal. "<<endl; return; } if(!search(value)) { cout << "Value is not in the tree. No removal." << endl; return; } BTNode<T>* current; BTNode<T>* parent; current = root; parent->left = NULL; parent->right = NULL; cout << root->left << "LEFT " << root->right << "RIGHT " << endl; cout << root->data << " ROOT" << endl; cout << current->data << "CURRENT BEFORE" << endl; while(current != NULL) { cout << "INTkhkjhbljkhblkjhlk " << endl; if(current->data == value) break; else if(value > current->data) { parent = current; current = current->right; } else { parent = current; current = current->left; } } cout << current->data << "CURRENT AFTER" << endl; // 3 cases : //We're looking at a leaf node if(current->left == NULL && current->right == NULL) // It's a leaf { if(parent->left == current) parent->left = NULL; else parent->right = NULL; delete current; cout << "The value " << value << " was removed." << endl; return; } // Node with single child if((current->left == NULL && current->right != NULL) || (current->left != NULL && current->right == NULL)) { if(current->left == NULL && current->right != NULL) { if(parent->left == current) { parent->left = current->right; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->right; cout << "The value " << value << " was removed." << endl; delete current; } } else // left child present, no right child { if(parent->left == current) { parent->left = current->left; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->left; cout << "The value " << value << " was removed." << endl; delete current; } } return; } //Node with 2 children - Replace node with smallest value in right subtree. if (current->left != NULL && current->right != NULL) { BTNode<T>* check; check = current->right; if((check->left == NULL) && (check->right == NULL)) { current = check; delete check; current->right = NULL; cout << "The value " << value << " was removed." << endl; } else // right child has children { //if the node's right child has a left child; Move all the way down left to locate smallest element if((current->right)->left != NULL) { BTNode<T>* leftCurrent; BTNode<T>* leftParent; leftParent = current->right; leftCurrent = (current->right)->left; while(leftCurrent->left != NULL) { leftParent = leftCurrent; leftCurrent = leftCurrent->left; } current->data = leftCurrent->data; delete leftCurrent; leftParent->left = NULL; cout << "The value " << value << " was removed." << endl; } else { BTNode<T>* temp; temp = current->right; current->data = temp->data; current->right = temp->right; delete temp; cout << "The value " << value << " was removed." << endl; } } return; } } /* Print out the values in the tree and their relationships visually. Sample output: 22 18 15 10 9 5 3 1 */ template <typename T> void BST<T>::print() const { print(root,0); } template <typename T> void BST<T>::print(BTNode<T>* node,int depth) const { if(node == NULL) { std::cout << std::endl; return; } print(node->right,depth+1); for(int i=0; i < depth; i++) { std::cout << "\t"; } std::cout << node->data << std::endl; print(node->left,depth+1); } #endif main.cpp #include "bst.h" #include <iostream> using namespace std; int main() { BST<int> tree; cout << endl << "LAB #13 - BINARY SEARCH TREE PROGRAM" << endl; cout << "----------------------------------------------------------" << endl; // Insert. cout << endl << "INSERT TESTS" << endl; // No duplicates allowed. tree.insert(0); tree.insert(5); tree.insert(15); tree.insert(25); tree.insert(20); // Search. cout << endl << "SEARCH TESTS" << endl; int x = 0; int y = 1; if(tree.search(x)) cout << "The value " << x << " is on the tree." << endl; else cout << "The value " << x << " is NOT on the tree." << endl; if(tree.search(y)) cout << "The value " << y << " is on the tree." << endl; else cout << "The value " << y << " is NOT on the tree." << endl; // Removal. cout << endl << "REMOVAL TESTS" << endl; tree.remove(0); tree.remove(1); tree.remove(20); // Print. cout << endl << "PRINTED DIAGRAM OF BINARY SEARCH TREE" << endl; cout << "----------------------------------------------------------" << endl; tree.print(); cout << endl << "Program terminated. Goodbye." << endl << endl; } BTNode.h #ifndef BTNODE_H_ #define BTNODE_H_ #include <iostream> /* A class to represent a node in a binary search tree. */ template <typename T> class BTNode { public: //constructor BTNode(T d); //the node's data value T data; //pointer to the node's left child BTNode<T>* left; //pointer to the node's right child BTNode<T>* right; }; /* Simple constructor. Sets the data value of the BTNode to "d" and defaults its left and right child pointers to NULL. */ template <typename T> BTNode<T>::BTNode(T d) : left(NULL), right(NULL) { data = d; } #endif Thanks.

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• Why is systemd not setting my system time?

  - by Alex Chamberlain

  I'm running Arch Linux. Recently, when I turn on my PC, the system time is set to 1:00 1 January 1970 - presumably the 1:00 o'clock is from the timezone shift. Does anyone have any ideas why systemd isn't setting my system time correctly? Some useful output (I think)... [root@alex-desktop network.d]# timedatectl status Local time: Sun 2013-06-09 16:33:04 BST Universal time: Sun 2013-06-09 15:33:04 UTC RTC time: Sun 2013-06-09 15:18:50 Timezone: Europe/London (BST, +0100) NTP enabled: yes NTP synchronized: no RTC in local TZ: no DST active: yes Last DST change: DST began at Sun 2013-03-31 00:59:59 GMT Sun 2013-03-31 02:00:00 BST Next DST change: DST ends (the clock jumps one hour backwards) at Sun 2013-10-27 01:59:59 BST Sun 2013-10-27 01:00:00 GMT

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• A balanced binary search tree which is also a heap

  - by saeedn

  I'm looking for a data structure where each element in it has two keys. With one of them the structure is a BST and looking at the other one, data structure is a heap. With a little search, I found a structure called Treap. It uses the heap property with a random distribution on heap keys to make the BST balanced! What I want is a Balanced BST, which can be also a heap. The BST in Treap could be unbalanced if I insert elements with heap Key in the order of my choice. Is there such a data structure?

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• Binary Search Tree, cannot do traversal

  - by ihm

  Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

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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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• Can anyone help me find why this C program work on VS2005 but not on DEV-C++

  - by user333771

  Hello to everybody..and greetings from Greece I have a C program for an exercise and it has a strange issue The program runs just fine on VS 2005 but it crashes on DEV-C++ and the problem that the problem is that the exercise is always evaluated against DEV-C++ The program is about inserting nodes to a BST and this is where the problem lies... Well i would really appreciate some help. enter code here #include <stdio.h> #include <stdlib.h> #include <malloc.h> typedef struct tree_node { int value; int weight; struct tree_node *left; struct tree_node *right; } TREE_NODE; /* The Following function creates a Binary Search Treed */ TREE_NODE *create_tree(int list[], int size); TREE_NODE *search_pos_to_insert(TREE_NODE *root, int value, int *left_or_right); /* this is the problematic function */ void inorder(TREE_NODE *root); /* Inorder Traversing */ TREE_NODE *temp; int main() { TREE_NODE *root; /* Pointer to the root of the BST */ int values[] = {10, 5, 3, 4, 1, 9, 6, 7, 8, 2}; /* Values for BST */ int size = 10, tree_weight; root = create_tree(values, 10); printf("\n"); inorder(root); /* Inorder BST*/ system("PAUSE"); } TREE_NODE *search_pos_to_insert(TREE_NODE *root, int value, int *left_or_right) { if(root !=NULL) { temp = root; if(value >root->value) { *left_or_right=1; *search_pos_to_insert(root->right, value, left_or_right); } else { *left_or_right=0; *search_pos_to_insert(root->left, value, left_or_right); } } else return temp;/* THIS IS THE PROBLEM (1) */ } TREE_NODE *create_tree(int list[], int size) { TREE_NODE *new_node_pntr, *insert_point, *root = NULL; int i, left_or_right; /* First Value of the Array is the root of the BST */ new_node_pntr = (TREE_NODE *) malloc(sizeof(TREE_NODE)); new_node_pntr->value = list[0]; /* ¸íèåóå ôçí ðñþôç ôéìÞ ôïõ ðßíáêá. */ new_node_pntr->weight = 0; new_node_pntr->left = NULL; new_node_pntr->right = NULL; root = new_node_pntr; /* Now the rest of the arrat. */ for (i = 1; i < size; i++) { insert_point = search_pos_to_insert(root, list[i], &left_or_right); /* THIS IS THE PROBLEM (2) */ /* insert_point just won't get the return from temp */ new_node_pntr = (TREE_NODE *) malloc(sizeof(TREE_NODE)); new_node_pntr->value = list[i]; new_node_pntr->weight = 0; new_node_pntr->left = NULL; new_node_pntr->right = NULL; if (left_or_right == 0) insert_point->left = new_node_pntr; else insert_point->right = new_node_pntr; } return(root); } void inorder(TREE_NODE *root) { if (root == NULL) return; inorder(root->left); printf("Value: %d, Weight: %d.\n", root->value, root->weight); inorder(root->right); }

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• Nth largest element in a binary search tree

  - by Codenotguru

  How to find the Nth largest node in a BST? Do I keep a count variable while doing In Order Traversal of a BST? Return the element when the count = N???

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• Why doesn't stackoverflow take into consideration different timezones? [closed]

  - by mark

  Let's say that you ask a question at 20:00 BST, you wait for comments, then let's say you wait until 04:00 BST in the morning, you then go to sleep and in the morning you discover that you can not longer add comments. I did not find anything in the FAQ. Can someone please explain to me how does it work?

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• Why does Google mark one e-mail as spam while does not the other?

  - by nKn

  I've a Postfix installation which works fine, I don't get any trouble with mails sent through a mail client (in my case, Thunderbird or RoundCube) when the To: address is a GMail account. However, I recently needed to use the PHPMailer tool to send some e-mails to some GMail accounts, so I configured an account to be used via SASL authentication + TLS. I don't mean mass mailing, just 2-3 mails. If I send the e-mail from the Thunderbird or RoundCube clients, the mail is not marked as spam. However, if I use PHPMailer, it always gets catalogued as spam. So I compared both headers and I just can't find the reason why the second is marked as spam while the first one is just ok. The first header sent from a mail client which is not marked as spam: Delivered-To: [email protected] Received: by 10.76.153.102 with SMTP id vf6csp230573oab; Tue, 19 Aug 2014 11:08:19 -0700 (PDT) X-Received: by 10.60.23.39 with SMTP id j7mr45544050oef.20.1408471699715; Tue, 19 Aug 2014 11:08:19 -0700 (PDT) Return-Path: <[email protected]> Received: from mail.mydomain.com (X.ip-92-222-X.eu. [92.222.X.X]) by mx.google.com with ESMTPS id t5si27115082oej.10.2014.08.19.11.08.18 for <[email protected]> (version=TLSv1.2 cipher=ECDHE-RSA-AES128-GCM-SHA256 bits=128/128); Tue, 19 Aug 2014 11:08:19 -0700 (PDT) Received-SPF: pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) client-ip=92.222.X.X; Authentication-Results: mx.google.com; spf=pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) [email protected]; dkim=pass (test mode) [email protected] Received: by mail.mydomain.com (Postfix, from userid 111) id D8F69120293D; Tue, 19 Aug 2014 19:08:17 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471697; bh=wKMX9gkQ7tCLv8ezrG5t4bICm/SSLQsNfTdZMToksWw=; h=Date:From:To:Subject:From; b=qRNcYVdmk+n3D1uuv0FInTx7/LzH2ojck9DgCmabFPvfke233lkojUOjezCUGx7iV DL8EayZ28mzzzHpB7ETeMzop/5OS3BmvFtGKVD9gzc78cDIFXTDoRFAnkRWDR2IOxI SOn5tiyODTFpkbDgJOndzQ6qL5K0S9ASNGCZrNL4= X-Spam-Checker-Version: SpamAssassin 3.4.0 (2014-02-07) on vpsX.ovh.net X-Spam-Level: X-Spam-Status: No, score=-1.0 required=3.0 tests=ALL_TRUSTED,T_DKIM_INVALID autolearn=ham autolearn_force=no version=3.4.0 Received: from [192.168.1.111] (unknown [77.231.X.X]) (using TLSv1 with cipher ECDHE-RSA-AES128-SHA (128/128 bits)) (No client certificate requested) (Authenticated sender: [email protected]) by mail.mydomain.com (Postfix) with ESMTPSA id 910341202624 for <[email protected]>; Tue, 19 Aug 2014 19:08:17 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471697; bh=wKMX9gkQ7tCLv8ezrG5t4bICm/SSLQsNfTdZMToksWw=; h=Date:From:To:Subject:From; b=qRNcYVdmk+n3D1uuv0FInTx7/LzH2ojck9DgCmabFPvfke233lkojUOjezCUGx7iV DL8EayZ28mzzzHpB7ETeMzop/5OS3BmvFtGKVD9gzc78cDIFXTDoRFAnkRWDR2IOxI SOn5tiyODTFpkbDgJOndzQ6qL5K0S9ASNGCZrNL4= Message-ID: <[email protected]> Date: Tue, 19 Aug 2014 19:08:24 +0100 From: My Name <[email protected]> User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:24.0) Gecko/20100101 Thunderbird/24.6.0 MIME-Version: 1.0 To: My other account <[email protected]> Subject: . Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit . The second header sent from PHPMailer which is always marked as spam: Delivered-To: [email protected] Received: by 10.76.153.102 with SMTP id vf6csp230832oab; Tue, 19 Aug 2014 11:12:10 -0700 (PDT) X-Received: by 10.60.121.67 with SMTP id li3mr44086252oeb.17.1408471930520; Tue, 19 Aug 2014 11:12:10 -0700 (PDT) Return-Path: <[email protected]> Received: from mail.mydomain.com (X.ip-92-222-X.eu. [92.222.X.X]) by mx.google.com with ESMTPS id w8si27103806obn.30.2014.08.19.11.12.10 for <[email protected]> (version=TLSv1.2 cipher=ECDHE-RSA-AES128-GCM-SHA256 bits=128/128); Tue, 19 Aug 2014 11:12:10 -0700 (PDT) Received-SPF: pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) client-ip=92.222.X.X; Authentication-Results: mx.google.com; spf=pass (google.com: domain of [email protected] designates 92.222.X.X as permitted sender) [email protected]; dkim=pass (test mode) [email protected] Received: by mail.mydomain.com (Postfix, from userid 111) id 1999D120293D; Tue, 19 Aug 2014 19:12:09 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471929; bh=N1JuHq1S+8GrjHcEK3xn8P1JS+ygEBv5LKe0BiXuVJo=; h=Date:To:From:Reply-to:Subject:From; b=K7tcPyArzSTY91VEw6mAAFtDurSGwgTLGkfUZdC5mqsg0g/1LzmZkgwdjj4NdJa6M E2kDz3dwYN8FcZmbampJYFXxj4NQVtSnzjiWV40rpfOFqD2rXDGNIyB2QOjBZZ4WK3 7s4lyoJ/BrdQH4en8ctLVsDHed/KpHD4iGFEl67E= X-Spam-Checker-Version: SpamAssassin 3.4.0 (2014-02-07) on vpsX.ovh.net X-Spam-Level: X-Spam-Status: No, score=-1.0 required=3.0 tests=ALL_TRUSTED,T_DKIM_INVALID autolearn=ham autolearn_force=no version=3.4.0 Received: from rpi.mydomain.com (unknown [77.231.X.X]) (using TLSv1 with cipher ECDHE-RSA-AES256-SHA (256/256 bits)) (No client certificate requested) (Authenticated sender: [email protected]) by mail.mydomain.com (Postfix) with ESMTPSA id B42AF1202624 for <[email protected]>; Tue, 19 Aug 2014 19:12:08 +0100 (BST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mydomain.com; s=mail; t=1408471928; bh=N1JuHq1S+8GrjHcEK3xn8P1JS+ygEBv5LKe0BiXuVJo=; h=Date:To:From:Reply-to:Subject:From; b=iXPM0tS36swudPTT4FOHHtPi5Ll6LbR60kNqCinZ8utcWoFE31SFTpoMEq5aCM5ux wQMdFiN8c6vkjRGabmvqFTTIbwJsrToHo/4+Lt5HEBoQQE2Y3T+xGmnmGAHCS6stKB yb7SVmtrIAsVtSMKA8VYIbmu2oYqV3afYt7g0OMQ= Date: Tue, 19 Aug 2014 20:12:07 +0200 To: [email protected] From: Trying another account <[email protected]> Reply-to: Trying another account <[email protected]> Subject: . Message-ID: <[email protected]> X-Priority: 3 X-Mailer: PHPMailer 5.1 (phpmailer.sourceforge.net) MIME-Version: 1.0 Content-Transfer-Encoding: 8bit Content-Type: text/plain; charset="UTF-8" . I also tried: Adding a User-Agent header to match the first one. Removing the X-Mailer header. No one of them made a difference. Is there some significant difference which is making the second e-mail to be marked as spam by Google?

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• Binary Search Tree in Java

  - by John R

  I want to make a generic BST, that can be made up of any data type, but i'm not sure how I could add things to the tree, if my BST is generic. All of my needed code is below. I want my BST made up of Locations, and sorted by the x variable. Any help is appreciated. Major thanks for looking. public void add(E element) { if (root == null) root = element; if (element < root) add(element, root.leftChild); if (element > root) add(element, root.rightChild); else System.out.println("Element Already Exists"); } private void add(E element, E currLoc) { if (currLoc == null) currLoc = element; if (element < root) add(element, currLoc.leftChild); if (element > root) add(element, currLoc.rightChild); else System.out.println("Element Already Exists); } Other Code public class BinaryNode<E> { E BinaryNode; BinaryNode nextBinaryNode; BinaryNode prevBinaryNode; public BinaryNode() { BinaryNode = null; nextBinaryNode = null; prevBinaryNode = null; } } public class Location<AnyType> extends BinaryNode { String name; int x,y; public Location() { name = null; x = 0; y = 0; } public Location(String newName, int xCord, int yCord) { name = newName; x = xCord; y = yCord; } public int equals(Location otherScene) { return name.compareToIgnoreCase(otherScene.name); } }

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• Java Graphics not displaying on successive function calls, why?

  - by primehunter326

  Hi, I'm making a visualization for a BST implementation (I posted another question about it the other day). I've created a GUI which displays the viewing area and buttons. I've added code to the BST implementation to recursively traverse the tree, the function takes in coordinates along with the Graphics object which are initially passed in by the main GUI class. My idea was that I'd just have this function re-draw the tree after every update (add, delete, etc...), drawing a rectangle over everything first to "refresh" the viewing area. This also means I could alter the BST implementation (i.e by adding a balance operation) and it wouldn't affect the visualization. The issue I'm having is that the draw function only works the first time it is called, after that it doesn't display anything. I guess I don't fully understand how the Graphics object works since it doesn't behave the way I'd expect it to when getting passed/called from different functions. I know the getGraphics function has something to do with it. Relevant code: private void draw(){ Graphics g = vPanel.getGraphics(); tree.drawTree(g,ORIGIN,ORIGIN); } vPanel is what I'm drawing on private void drawTree(Graphics g, BinaryNode<AnyType> n, int x, int y){ if( n != null ){ drawTree(g, n.left, x-10,y+10 ); if(n.selected){ g.setColor(Color.blue); } else{ g.setColor(Color.gray); } g.fillOval(x,y,20,20); g.setColor(Color.black); g.drawString(n.element.toString(),x,y); drawTree(g,n.right, x+10,y+10); } } It is passed the root node when it is called by the public function. Do I have to have: Graphics g = vPanel.getGraphics(); ...within the drawTree function? This doesn't make sense!! Thanks for your help.

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• How to correctly use DERIVE or COUNTER in munin plugins

  - by Johan

  I'm using munin to monitor my server. I've been able to write plugins for it, but only if the graph type is GAUGE. When I try COUNTER or DERIVE, no data is logged or graphed. The plugin i'm currently stuck on is for monitoring bandwidth usage, and is as follows: /etc/munin/plugins/bandwidth2 #!/bin/sh if [ "$1" = "config" ]; then echo 'graph_title Bandwidth Usage 2' echo 'graph_vlabel Bandwidth' echo 'graph_scale no' echo 'graph_category network' echo 'graph_info Bandwidth usage.' echo 'used.label Used' echo 'used.info Bandwidth used so far this month.' echo 'used.type DERIVE' echo 'used.min 0' echo 'remain.label Remaining' echo 'remain.info Bandwidth remaining this month.' echo 'remain.type DERIVE' echo 'remain.min 0' exit 0 fi cat /var/log/zen.log The contents of /var/log/zen.log are: used.value 61.3251953125 remain.value 20.0146484375 And the resulting database is: <!-- Round Robin Database Dump --><rrd> <version> 0003 </version> <step> 300 </step> <!-- Seconds --> <lastupdate> 1269936605 </lastupdate> <!-- 2010-03-30 09:10:05 BST --> <ds> <name> 42 </name> <type> DERIVE </type> <minimal_heartbeat> 600 </minimal_heartbeat> <min> 0.0000000000e+00 </min> <max> NaN </max> <!-- PDP Status --> <last_ds> 61.3251953125 </last_ds> <value> NaN </value> <unknown_sec> 5 </unknown_sec> </ds> <!-- Round Robin Archives --> <rra> <cf> AVERAGE </cf> <pdp_per_row> 1 </pdp_per_row> <!-- 300 seconds --> <params> <xff> 5.0000000000e-01 </xff> </params> <cdp_prep> <ds> <primary_value> NaN </primary_value> <secondary_value> NaN </secondary_value> <value> NaN </value> <unknown_datapoints> 0 </unknown_datapoints> </ds> </cdp_prep> <database> <!-- 2010-03-28 09:15:00 BST / 1269764100 --> <row><v> NaN </v></row> <!-- 2010-03-28 09:20:00 BST / 1269764400 --> <row><v> NaN </v></row> <!-- 2010-03-28 09:25:00 BST / 1269764700 --> <row><v> NaN </v></row> <snip> The value for last_ds is correct, it just doesn't seem to make it into the actual database. If I change DERIVE to GAUGE, it works as expected. munin-run bandwidth2 outputs the contents of /var/log/zen.log I've been all over the (sparse) docs for munin plugins, and can't find my mistake. Modifying an existing plugin didn't work for me either.

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• How to find largest common sub-tree in the given two binary search trees?

  - by Bhushan

  Two BSTs (Binary Search Trees) are given. How to find largest common sub-tree in the given two binary trees? EDIT 1: Here is what I have thought: Let, r1 = current node of 1st tree r2 = current node of 2nd tree There are some of the cases I think we need to consider: Case 1 : r1.data < r2.data 2 subproblems to solve: first, check r1 and r2.left second, check r1.right and r2 Case 2 : r1.data > r2.data 2 subproblems to solve: - first, check r1.left and r2 - second, check r1 and r2.right Case 3 : r1.data == r2.data Again, 2 cases to consider here: (a) current node is part of largest common BST compute common subtree size rooted at r1 and r2 (b)current node is NOT part of largest common BST 2 subproblems to solve: first, solve r1.left and r2.left second, solve r1.right and r2.right I can think of the cases we need to check, but I am not able to code it, as of now. And it is NOT a homework problem. Does it look like?

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• findNode in binary search tree

  - by Weadadada Awda

  Does this look right? I mean I am trying to implement the delete function. Node* BST::findNode(int tofind) { Node* node = new Node; node = root; while (node != NULL) { if (node->val == tofind) { return node; } else if (tofind < node->val) { node = node->left; } else { node = node->right; } } } Here is the delete, it's not even close to done but, void BST::Delete(int todelete) { // bool found = false; Node* toDelete = new Node(); toDelete=findNode(todelete); if(toDelete->val!=NULL) { cout << toDelete->val << endl; } } This causes a segmentation fault just running that, any ideas?

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• Trying to serialize and unserialze a Binary search tree to a file

  - by ClearMist

  I am trying to find a way to use these methods to save and restore a binary search tree that contains names and phone numbers in each node. I am just very lost and how to go about doing this. 'public void save(String fileName) { // TODO : implement this method. // save bst to its original shape. } public void restore(String fileName) { // TODO : implement this method. // restore bst from a file, if file exists. // do nothing, otherwise. File fichier = new File(fileName); if (fichier.exists()) { } }'

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• Understanding SARGability (to make your queries run faster)

  - by simonsabin

  Rob Farley is doing a live meeting this month on understanding what SARGable means. It is at 1pm BST and so if you are in the UK will be a very useful hour spent. for more details go to http://www.sqlpass.org/Events/ctl/ViewEvent/mid/521.aspx?ID=341 The description of the session  is Understanding SARGability (to make your queries run faster) SARGable means Search ARGument able. It relates to the ability to search through an index for a value, but unfortunately, many database professionals don...(read more)

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• Mark Hurd and Balaji Yelamanchili present Oracle’s Business Analytics Strategy

  - by swalker

  Join Mark Hurd and Balaji Yelamanchili as they unveil the latest advances in Oracle’s strategy for placing analytics into the hands of every decision-makers—so that they can see more, think smarter, and act faster. Wednesday, April 4, 2012 at 1.0 pm UK BST / 2.0 pm CET Register HERE today for this online event Agenda Keynote: Oracle’s Business Analytics StrategyMark Hurd, President, Oracle, and Balaji Yelamanchili, Senior Vice President, Analytics and Performance Management, Oracle Plus Breakout Sessions: Achieving Predictable Performance with Oracle Hyperion Enterprise Performance Managemen Explore All Relevant Data—Introducing Oracle Endeca Information Discovery Run Your Business Faster and Smarter with Oracle Business Intelligence Applications on Oracle Exalytics In-Memory Machine Analyzing and Deciding with Big Data

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• Where I missed boot.properties.?

  - by Dyade, Shailesh M

  Today one of my customer was trying to start the WebLogic Server ( Production Instance) , though he was trying to start the server in a standard way, but it was failing due to below error :   ####<Oct 22, 2012 12:14:43 PM BST> <Warning> <Security> <BanifB1> <> <main> <> <> <> <1350904483998> <BEA-090066> <Problem handling boot identity. The following exception was generated: weblogic.security.internal.encryption.EncryptionServiceException: weblogic.security.internal.encryption.EncryptionServiceException: [Security:090219]Error decrypting Secret Key java.security.ProviderException: setSeed() failed> And it started failing into below causes. ####<Oct 22, 2012 12:16:45 PM BST> <Critical> <WebLogicServer> <BanifB1> <AdminServer> <main> <<WLS Kernel>> <> <> <1350904605837> <BEA-000386> <Server subsystem failed. Reason: java.lang.AssertionError: java.lang.reflect.InvocationTargetException java.lang.AssertionError: java.lang.reflect.InvocationTargetException weblogic.security.internal.encryption.EncryptionServiceException: weblogic.security.internal.encryption.EncryptionServiceException: [Security:090219]Error decrypting Secret Key java.security.ProviderException: setSeed() failed weblogic.security.internal.encryption.EncryptionServiceException: [Security:090219]Error decrypting Secret Key java.security.ProviderException: setSeed() failed at weblogic.security.internal.encryption.JSafeSecretKeyEncryptor.decryptSecretKey(JSafeSecretKeyEncryptor.java:121) Customer was facing this issue without any changes in the system, it was stable suddenly started seeing this issue last night. When we checked, customer was manually entering the username and password, config.xml had the entries encrypted However when verified, customer had the boot.properties at the Servers/AdminServer/security folder and DomainName/security didn't have this file. Adding boot.properies fixed the issue. Regards Shailesh Dyade 

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• Free Webinar: Monitoring your business, not just your servers – Getting the most out of SQL Monitor

  Wednesday July 25 2012, 6:00pm BST: Learn how you can use SQL Monitor to gather information and alert on extra performance data for your servers and applications, making this tool vital for keeping an eye on your business. In this free webinar David Bick, Product Manager at Red Gate, will give you an overview of SQL Monitor including the new custom metric functionality in v3. Repeatable deployment without fear of data lossUse your version control system with the SSMS plug-in SQL Source Control and SQL Compare for accurate deployments without the worry. Find out more.

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• Oracle FY15 Global Partner Kickoff: Register Now!

  - by Julien Haye

  Join Oracle PartnerNetwork for our FY15 Global Partner Kickoff and get up close with Oracle executives including Rich Geraffo, SVP, Worldwide Oracle Alliances & Channels. Watch online and listen as our sales and product executives outline Oracle's strategy and direction for FY15, and learn about the different ways you can accelerate sales & revenue through Oracle's full-stack offering. The EMEA Regional Event will be held on June 25th at 3:00pm BST / 4:00pm CET. Learn more and register now!

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• Finding if a Binary Tree is a Binary Search Tree

  - by dharam

  Today I had an interview where I was asked to write a program which takes a Binary Tree and returns true if it is also a Binary Search Tree otherwise false. My Approach1: Perform an inroder traversal and store the elements in O(n) time. Now scan through the array/list of elements and check if element at ith index is greater than element at (i+1)th index. If such a condition is encountered, return false and break out of the loop. (This takes O(n) time). At the end return true. But this gentleman wanted me to provide an efficient solution. I tried but I was unsuccessfult, because to find if it is a BST I have to check each node. Moreover he was pointing me to think over recusrion. My Approach 2: A BT is a BST if for any node N N-left is < N and N-right N , and the INorder successor of left node of N is less than N and the inorder successor of right node of N is greater than N and the left and right subtrees are BSTs. But this is going to be complicated and running time doesn't seem to be good. Please help if you know any optimal solution. Thanks.

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