Search Results

Search found 1552 results on 63 pages for 'homework'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 15/63 | < Previous Page | 11 12 13 14 15 16 17 18 19 20 21 22  | Next Page >

  • Basic Recursion, Check Balanced Parenthesis

    - by pws5068
    Greetings all, I've written software in the past that uses a stack to check for balanced equations, but now I'm asked to write a similar algorithm recursively to check for properly nested brackets and parenthesis. Good examples: () [] () ([]()[]) Bad examples: ( (] ([)] Suppose my function is called: isBalanced. Should each pass evaluate a smaller substring (until reaching a base case of 2 left)? Or, should I always evaluate the full string and move indices inward?

    Read the article

  • weighted matching algorithm in Perl

    - by srk
    Problem : We have equal number of men and women.each men has a preference score toward each woman. So do the woman for each man. each of the men and women have certain interests. Based on the interest we calculate the preference scores. So initially we have an input in a file having x columns. First column is the person(men/woman) id. id are nothing but 0.. n numbers.(first half are men and next half woman) the remaining x-1 columns will have the interests. these are integers too. now using this n by x-1 matrix... we have come up with a n by n/2 matrix. the new matrix has all men and woman as their rows and scores for opposite sex in columns. We have to sort the scores in descending order, also we need to know the id of person related to the scores after sorting. So here i wanted to use hash table. once we get the scores we need to make up pairs.. for which we need to follow some rules. My trouble is with the second matrix of n by n/2 that needs to give information of which man/woman has how much preference on a woman/man. I need these scores sorted so that i know who is the first preferred woman/man, 2nd preferred and so on for a man/woman. I hope to get good suggestions on the data structures i use.. I prefer php or perl. Thank you in advance

    Read the article

  • Exam Questions that use .Demand or .LinkDemand COULD NOT BE ANY MORE CONFUSING OR AMBIGIOUS ????

    - by IbrarMumtaz
    I am 110% sure this is WRONG !!!! Q.12) You develop a library, and want to ensure that the functions in the library cannot be either directly or indirectly invoked by applications that are not running on the local intranet. What attribute would you add to each method? A. [UrlIdentityPermission(SecurityAction.RequestRefuse, Url="http://myintranet")] B. [UrlIdentityPermission(SecurityAction.LinkDemand, Url="http://myintranet")] (correct answer) C. [UrlIdentityPermission(SecurityAction.Demand, Url="http://myintranet")] D. [UrlIdentityPermission(SecurityAction.Assert, Url="http://myintranet")] Explanation Link-Demand should be used as it ensures that all callers in the call stack have the necessary permission. In this case it ensures that all callers in the call stack are on the local intranet. There is an indentical question on Transencer so I already had a clue what was goin but Transcender was much more informative that this drivel as it mentioned class level and not assembly level. It also mentioned that some callers maybe coming externally from the company intranet via authroised and authenticated credentials. With information is easy to see why .Demand on would be wong option to go for? So Transcender was right .... so I thgt fine, that makes sense. With think information still fresh in my brain I had a good idea was was going on in the question. To my surprise .Demand was wrong agin !!!! WHAT? I am really starting to hate this setting now? I cannot be any more p*ssed right now!!! :@ Thanks For Reading, Ibrar

    Read the article

  • 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.

    Read the article

  • Maddening Linked List problem

    - by Mike
    This has been plaguing me for weeks. It's something really simple, I know it. Every time I print a singly linked list, it prints an address at the end of the list. #include <iostream> using namespace std; struct node { int info; node *link; }; node *before(node *head); node *after(node *head); void middle(node *head, node *ptr); void reversep(node *head, node *ptr); node *head, *ptr, *newnode; int main() { head = NULL; ptr = NULL; newnode = new node; head = newnode; for(int c1=1;c1<11;c1++) { newnode->info = c1; ptr = newnode; newnode = new node; ptr->link = newnode; ptr = ptr->link; } ptr->link=NULL; head = before(head); head = after(head); middle(head, ptr); //reversep(head, ptr); ptr = head; cout<<ptr->info<<endl; while(ptr->link!=NULL) { ptr=ptr->link; cout<<ptr->info<<endl; } system("Pause"); return 0; } node *before(node *head) { node *befnode; befnode = new node; cout<<"What should go before the list?"<<endl; cin>>befnode->info; befnode->link = head; head = befnode; return head; } node *after(node *head) { node *afnode, *ptr2; afnode = new node; ptr2 = head; cout<<"What should go after the list?"<<endl; cin>>afnode->info; ptr2 = afnode; afnode->link=NULL; ptr2 = head; return ptr2; } void middle(node *head, node *ptr) { int c1 = 0, c2 = 0; node *temp, *midnode; ptr = head; while(ptr->link->link!=NULL) { ptr=ptr->link; c1++; } c1/=2; c1-=1; ptr = head; while(c2<c1) { ptr=ptr->link; c2++; } midnode = new node; cout<<"What should go in the middle of the list?"<<endl; cin>>midnode->info; cout<<endl; temp=ptr->link; ptr->link=midnode; midnode->link=temp; } void reversep(node *head, node *ptr) { node *last, *ptr2; ptr=head; ptr2=head; while(ptr->link!=NULL) ptr = ptr->link; last = ptr; cout<<last->info; while(ptr!=head) { while(ptr2->link!=ptr) ptr2=ptr2->link; ptr = ptr2; cout<<ptr->info; } } I'll admit that this is class work, but even the professor can't figure it out, and says that its probably something insignificant that we're overlooking, but I can't put my mind to rest until I find out what it is.

    Read the article

  • Malloc inside another function (ANSI C)

    - by Casper
    Hi I'll go straight to it. I'm working on an assignment, where I suddenly ran into trouble. I have to allocate a struct from within another function, obviously using pointers. I've been staring at this problem for hours and tried in a million different ways to solve it. This is some sample code (very simplified): ... some_struct s; printf("Before: %d\n", &s"); allocate(&s); printf("After: %d\n", &s"); ... /* The allocation function */ int allocate(some_struct *arg) { arg = malloc(sizeof(some_struct)); printf("In function: %d\n", &arg"); return 0; } This does give me the same address before and after the allocate-call: Before: -1079752900 In function: -1079752928 After: -1079752900 I know it's probably because it makes a copy in the function, but I don't know how to actually work on the pointer I gave as argument. I tried defining some_struct *s instead of some_struct s, but no luck. I tried with: int allocate(some_struct **arg) which works just fine (the allocate-function needs to be changed as well), BUT according to the assignment I may NOT change the declaration, and it HAS to be *arg.. And it would be most correct if I just have to declare some_struct s.. Not some_struct *s. I hope I make sense and some of you out there can help me :P Thanks in advice

    Read the article

  • How to embed html table into the bpdy of email

    - by Michael Mao
    Hi all: I am sending info to target email via PHP native mail() method right now. Everything else works fine but the table part troubles me the most. See sample output : Dear Michael Mao : Thank you for purchasing flight tickets with us, here is your receipt : Your tickets will be delivered by mail to the following address : Street Address 1 : sdfsdafsadf sdf Street Address 2 : N/A City : Sydney State : nsw Postcode : 2 Country : Australia Credit Card Number : *************1234 Your purchase details are recorded as : <table><tr><th class="delete">del?</th><th class="from_city">from</th><th class="to_city">to</th><th class="quantity">qty</th><th class="price">unit price</th><th class="price">total price</th></tr><tr class="evenrow" id="Sydney-Lima"><td><input name="isDeleting" type="checkbox"></td><td>Sydney</td><td>Lima</td><td>1</td><td>1030.00</td><td>1030</td></tr><tr class="oddrow" id="Sydney-Perth"><td><input name="isDeleting" type="checkbox"></td><td>Sydney</td><td>Perth</td><td>3</td><td>340.00</td><td>1020</td></tr><tr class="totalprice"><td colspan="5">Grand Total Price</td><td id="grandtotal">2050</td></tr></table> The source of table is directly taken from a webpage, exactly as the same. However, Gmail, Hotmail and most of other emails will ignore to render this as a table. So I am wondering, without using Outlook or other email sending agent software, how could I craft a embedded table for the PHP mail() method to send? Current code snippet corresponds to table generation : $purchaseinfo = $_POST["purchaseinfo"]; //if html tags are not to be filtered in the body of email $stringBuilder .= "<table>" .stripslashes($purchaseinfo) ."</table>"; //must send json response back to caller ajax request if(mail($email, 'Your purchase information on www.hardlyworldtravel.com', $emailbody, $headers)) echo json_encode(array("feedback"=>"successful")); else echo json_encode(array("feedback"=>"error")); Any hints and suggestions are welcomed, thanks a lot in advance.

    Read the article

  • Why is this an invalid Turing machine?

    - by Danny King
    Whilst doing exam revision I am having trouble answering the following question from the book, "An Introduction to the Theory of Computation" by Sipser. Unfortunately there's no solution to this question in the book. Explain why the following is not a legitimate Turing machine. M = { The input is a polynomial p over variables x1, ..., xn Try all possible settings of x1, ..., xn to integer values Evaluate p on all of these settings If any of these settings evaluates to 0, accept; otherwise reject. } This is driving me crazy! I suspect it is because the set of integers is infinite? Does this somehow exceed the alphabet's allowable size? Thanks!

    Read the article

  • Algorithm to generate a list of unique combinations based on a list of numbers

    - by ross
    I would like to efficiently generate a unique list of combinations of numbers based on a starting list of numbers. example start list = [1,2,3,4,5] but the algorithm should work for [1,2,3...n] result = [1],[2],[3],[4],[5] [1,2],[1,3],[1,4],[1,5] [1,2,3],[1,2,4],[1,2,5] [1,3,4],[1,3,5],[1,4,5] [2,3],[2,4],[2,5] [2,3,4],[2,3,5] [3,4],[3,5] [3,4,5] [4,5] Note. I don't want duplicate combinations, although I could live with them, eg in the above example I don't really need the combination [1,3,2] because it already present as [1,2,3]

    Read the article

  • read file in C++

    - by Amm Sokun
    I am trying to read a list of words from a file in C++. However, the last word is read twice. I cannot understand why it is so. Can someone help me out? int main () { ifstream fin, finn; vector<string> vin; vector<string> typo; string word; fin.open("F:\\coursework\\pz\\gattaca\\breathanalyzer\\file.in"); if (!fin.is_open()) cout<<"Not open\n"; while (fin) { fin >> word; cout<<word<<endl; vin.push_back(word); } fin.close(); }

    Read the article

  • Best way to do powerOf(int x, int n)?

    - by Mike
    So given x, and power, n, solve for X^n. There's the easy way that's O(n)... I can get it down to O(n/2), by doing numSquares = n/2; numOnes = n%2; return (numSquares * x * x + numOnes * x); Now there's a log(n) solution, does anyone know how to do it? It can be done recursively.

    Read the article

  • Calculate shortest path through a grocery store

    - by Bart
    Hi, I'm trying to find a way to find the shortest path through a grocery store, visiting a list of locations (shopping list). The path should start at a specified startposition and can end at multiple endpositions (there are multiple checkout counters). Also, I have some predefined constraints on the path, such as "item x on the shopping list needs to be the last, second last, or third last item on the path". There is a function that will return true or false for a given path. Finally, this needs to be calculated with limited cpu power (on a smartphone) and within a second or so. If this isn't possible, then an approximation to the optimal path is also ok. Is this possible? So far I think I need to start by calculating the distance between every item on the list using something like A* or Dijkstra's. After that, should I treat it like the travelling salesman problem? Because in my problem there is a specified startnode, specified endnodes, and some constraints, which are not in the travelling salesman problem. Any help would be appreciated :)

    Read the article

  • MUD in python......

    - by matt
    how do you make a MUD in python can anyone help or start me of i dont have a clue on how to do if anyone knows any other source that i could use then plz let me know?

    Read the article

  • ANSI C blackjack assignment, linux GCC compiler, i'm stuck...

    - by Bill Adams
    Here's what i have so far... I have yet to figure out how i'm going to handle the 11 / 1 situation with an ace, and when the player chooses an option for hit/stand, i get segfault. HELP!!! #include <stdio.h> #include <string.h> #include <stdlib.h> #include <time.h> #define DECKSIZE 52 #define VALUE 9 #define FACE 4 #define HANDSIZE 26 typedef struct { int value; char* suit; char* name; }Card; typedef struct { int value; char* suit; char* name; }dealerHand; typedef struct { int value; char* suit; char* name; }playerHand; Card cards[DECKSIZE]; dealerHand deal[HANDSIZE]; playerHand dealt[HANDSIZE]; char *faceName[]={"two","three", "four","five","six", "seven","eight","nine", "ten", "jack","queen", "king","ace"}; char *suitName[]={"spades","diamonds","clubs","hearts"}; void printDeck(){ int i; for(i=0;i<DECKSIZE;i++){ printf("%s of %s value = %d\n ",cards[i].name,cards[i].suit,cards[i].value); if((i+1)%13==0 && i!=0) printf("-------------------\n\n"); } } void shuffleDeck(){ srand(time(NULL)); int this; int that; Card temp; int c; for(c=0;c<10000;c++){ //c is the index for number of individual card shuffles should be set to c<10000 or more this=rand()%DECKSIZE; that=rand()%DECKSIZE; temp=cards[this]; cards[this]=cards[that]; cards[that]=temp; } } /*void hitStand(i,y){ // I dumped this because of a segfault i couldn't figure out. int k; printf(" Press 1 to HIT or press 2 to STAND:"); scanf("%d",k); if(k=1){ dealt[y].suit=cards[i].suit; dealt[y].name=cards[i].name; dealt[y].value=cards[i].value; y++; i++; } } */ int main(){ int suitCount=0; int faceCount=0; int i; int x; int y; int d; int p; int k; for(i=0;i<DECKSIZE;i++){ //this for statement builds the deck if(faceCount<9){ cards[i].value=faceCount+2; }else{ //assigns face cards as value 10 cards[i].value=10; } cards[i].suit=suitName[suitCount]; cards[i].name=faceName[faceCount++]; if(faceCount==13){ //this if loop increments suit count once cards[i].value=11; //all faces have been assigned, and also suitCount++; //assigns the ace as 11 faceCount=0; } //end building deck } /*printDeck(); //prints the deck in order shuffleDeck(); //shuffles the deck printDeck(); //prints the deck as shuffled This was used in testing, commented out to keep the deck hidden!*/ shuffleDeck(); x=0; y=0; for(i=0;i<4;i++){ //this for loop deals the first 4 cards, dealt[y].suit=cards[i].suit; //first card to player, second to dealer, dealt[y].name=cards[i].name; //as per standard dealing practice. dealt[y].value=cards[i].value; i++; y++; deal[x].suit=cards[i].suit; deal[x].name=cards[i].name; deal[x].value=cards[i].value; x++; } printf(" Dealer's hand is: %s of %s and XXXX of XXXX. (Second card is hidden!)\n",deal[0].name,deal[0].suit,deal[1].name,deal[1].suit); printf(" Player's hand is: %s of %s and %s of %s.\n",dealt[0].name,dealt[0].suit,dealt[1].name,dealt[1].suit); printf(" the current value of the index i=%d\n",i); //this line gave me the value of i for testing d=deal[0].value+deal[1].value; p=dealt[0].value+dealt[1].value; if(d==21){ printf(" The Dealer has Blackjack! House win!\n"); }else{ if(d>21){ printf(" The dealer is Bust! You win!\n"); }else{ if(d>17){ printf(" Press 1 to HIT or 2 to STAND"); scanf("%d",k); if(k==1){ dealt[y].suit=cards[i].suit; dealt[y].name=cards[i].name; dealt[y].value=cards[i].value; y++; i++; } }else{ if(d<17){ printf(" Dealer Hits!"); deal[x].suit=cards[i].suit; deal[x].name=cards[i].name; deal[x].value=cards[i].value; x++; i++; } } } } return 0; }

    Read the article

  • Implementing union of Set using basic java array

    - by lupindeterd
    Note: This is an assignment. Hi, Continuing with my Set implementation using Java basic array, I'm now struggling with the 3 to last function namely the union. import java.io.*; class Set { private int numberOfElements = 0; private String[] setElements = new String[5]; private int maxNumberOfElements = 5; // constructor for our Set class public Set(int numberOfE, int setE, int maxNumberOfE) { int numberOfElements = numberOfE; String[] setElements = new String[setE]; int maxNumberOfElements = maxNumberOfE; } // Helper method to shorten/remove element of array since we're using basic array instead of ArrayList or HashSet from collection interface :( static String[] removeAt(int k, String[] arr) { final int L = arr.length; String[] ret = new String[L - 1]; System.arraycopy(arr, 0, ret, 0, k); System.arraycopy(arr, k + 1, ret, k, L - k - 1); return ret; } int findElement(String element) { int retval = 0; for ( int i = 0; i < setElements.length; i++) { if ( setElements[i] != null && setElements[i].equals(element) ) { return retval = i; } retval = -1; } return retval; } void add(String newValue) { int elem = findElement(newValue); if( numberOfElements < maxNumberOfElements && elem == -1 ) { setElements[numberOfElements] = newValue; numberOfElements++; } } int getLength() { if ( setElements != null ) { return setElements.length; } else { return 0; } } String[] emptySet() { setElements = new String[0]; return setElements; } Boolean isFull() { Boolean True = new Boolean(true); Boolean False = new Boolean(false); if ( setElements.length == maxNumberOfElements ){ return True; } else { return False; } } Boolean isEmpty() { Boolean True = new Boolean(true); Boolean False = new Boolean(false); if ( setElements.length == 0 ) { return True; } else { return False; } } void remove(String newValue) { for ( int i = 0; i < setElements.length; i++) { if ( setElements[i].equals(newValue) ) { setElements = removeAt(i,setElements); } } } int isAMember(String element) { int retval = -1; for ( int i = 0; i < setElements.length; i++ ) { if (setElements[i] != null && setElements[i].equals(element)) { return retval = i; } } return retval; } void printSet() { for ( int i = 0; i < setElements.length; i++) { System.out.println("Member elements on index: "+ i +" " + setElements[i]); } } String[] getMember() { String[] tempArray = new String[setElements.length]; for ( int i = 0; i < setElements.length; i++) { if(setElements[i] != null) { tempArray[i] = setElements[i]; } } return tempArray; } Set union(Set x, Set y) { String[] newtemparray = new String[x.getLength]; newtemparray = x.getMember; return x; } } // This is the SetDemo class that will make use of our Set class class SetDemo { public static void main(String[] args) { //get input from keyboard BufferedReader keyboard; InputStreamReader reader; String temp = ""; reader = new InputStreamReader(System.in); keyboard = new BufferedReader(reader); try { System.out.println("Enter string element to be added" ); temp = keyboard.readLine( ); System.out.println("You entered " + temp ); } catch (IOException IOerr) { System.out.println("There was an error during input"); } /* ************************************************************************** * Test cases for our new created Set class. * ************************************************************************** */ Set setA = new Set(1,10,10); setA.add(temp); setA.add("b"); setA.add("b"); setA.add("hello"); setA.add("world"); setA.add("six"); setA.add("seven"); setA.add("b"); int size = setA.getLength(); System.out.println("Set size is: " + size ); Boolean isempty = setA.isEmpty(); System.out.println("Set is empty? " + isempty ); int ismember = setA.isAMember("sixb"); System.out.println("Element six is member of setA? " + ismember ); Boolean output = setA.isFull(); System.out.println("Set is full? " + output ); setA.printSet(); int index = setA.findElement("world"); System.out.println("Element b located on index: " + index ); setA.remove("b"); setA.emptySet(); int resize = setA.getLength(); System.out.println("Set size is: " + resize ); setA.printSet(); Set setB = new Set(0,10,10); Set SetA = setA.union(setB,setA); } } Ok the method in question will be the implementation of union. As such this: Set union(Set x, Set y) { String[] newtemparray = new String[x.getLength]; newtemparray = x.getMember; return x; } I got this error: symbol : variable getLength location: class Set String[] newtemparray = new String[x.getLength]; ^ d:\javaprojects\Set.java:122: cannot find symbol symbol : variable getMember location: class Set newtemparray = x.getMember; ^ 2 errors My approach for union would be: *) create temporary array of string with size of the object x length. *) store object x members to temporary array by looping the object and calling the getMember *) loop object y members and check if element exist against temporary array. *) discard if it exist/add if it is not there *) return obj x with the union array. thanks, lupin

    Read the article

  • Cracking the Playfair cipher

    - by wwrob
    I have the ciphertext and an encrypting program (with the key hardcoded in). How would I go about finding the key? Surely the availability of the encryptor must open up possibilities beyond brute-forcing it.

    Read the article

  • Chatbot client and class modification

    - by blake
    ChatBot Class Modification: Modify the reply() method of the ChatBot class to recognize additional words and phrases. Part 1: Everyone must complete this section. When the userInput parameter value is: The reply method should return: how do I quit enter quit how do I exit enter quit how do I stop enter quit how do I ____ do you really want to do that how are you I'm fine how ______ I don't know Add two additional words or phrases to recognize and respond to. ChatBot Client Modification: Modify the ChatBot client application to loop until the end-user enters "quit". Here is my service class / ** * Java Chatbot Service class * @author Blake * 3/5/2012 */ /** * Default constructor. */ public class Chatbot { private String name; /** Users name */ private String introbot; /** Name of the Chatbot */ private String reply; /** Replies to the input of the string name and string introbot */ /** * Constructs mutebot object * @param mutebow - returns name of mutebot */ public Chatbot() { name = "MuteBot"; } /** * Changes Name * @param name - new name */ public void setName (String n) { name = n; } /** * Accesses name * @return a brand new name */ public String getName() { return name; } /** * Accesses introbot * @return name of mutebot */ public String introbot() { String intro = "Hello! My name is " + name; return intro; } /** * Accesses replay(String newuserinput) * @return introbot reply to user input */ public String getreply(String newuserinput) { String reply = "I'm just learning to talk"; if (newuserinput.equalsIgnoreCase("What")) reply = "Why do you ask?"; else if (newuserinput.equalsIgnoreCase("Why") ) reply = "Why Not"; else if (newuserinput.equalsIgnoreCase("How")) reply = "I don't know!"; else if (newuserinput.equalsIgnoreCase("Where") ) reply = "Anne Arundel Community College"; else if (newuserinput.equalsIgnoreCase("When")) reply = "Tomorrow"; else if (newuserinput.equalsIgnoreCase("how do I quit")) reply = "enter quit"; else if (newuserinput.equalsIgnoreCase("how do I exit")) reply = "enter quit"; else if (newuserinput.equalsIgnoreCase("how do I stop")) reply = "enter quit"; else if (newuserinput.equalsIgnoreCase("how are you")) reply = "I'm fine"; else if (newuserinput.equalsIgnoreCase("how do you do")) reply = "I am doing well"; else if (newuserinput.equalsIgnoreCase("how do I get out")) reply = "By going through the door"; else if (newuserinput.indexOf("how do I" ) ==0) { String substring = newuserinput.substring(8); reply = "do you really want to do that" + substring; } else if (newuserinput.indexOf("how" ) ==0) { String substring = newuserinput.substring(10); reply = "I don't know" + substring ; } return reply; } } Here is my client/application class /** * Java Chatbot Client class * @author Blake * 3/5/2012 */ import java.util.Scanner; public class ChatbotClient { public static void main(String[] args) { Scanner input = new Scanner(System.in); Chatbot t = new Chatbot(); System.out.print("What is your name? "); String name = input.nextLine(); System.out.println(t.introbot()); System.out.print(name + "> "); String reply = input.nextLine(); System.out.println(t.getName() + "> " + t.getreply(reply)); //while (reply < quit) /*{ quit++ i = i + 1 }*/ } } I don't know what I am doing wrong with this part right here Modify the ChatBot client application to loop until the end-user enters "quit". I am trying to create a while loop which will continue until user says quit.

    Read the article

  • Some more multitasking java issues

    - by owca
    I had a task to write simple game simulating two players picking up 1-3 matches one after another until the pile is gone. I managed to do it for computer choosing random value of matches but now I'd like to go further and allow humans to play the game. Here's what I already have : http://paste.pocoo.org/show/200660/ Class Player is a computer player, and PlayerMan should be human being. Problem is, that thread of PlayerMan should wait until proper value of matches is given but I cannot make it work this way. When I type the values it sometimes catches them and decrease amount of matches but that's not exactly what I was up to :) Logics is : I check the value of current player. If it corresponds to this of the thread currently active I use scanner to catch the amount of matches. Else I wait one second (I know it's kinda harsh solution, but I have no other idea how to do it). Class Shared keeps the value of current player, and also amount of matches. By the way, is there any way I can make Player and Shared attributes private instead of public and still make the code work ? CONSOLE and INPUT-DIALOG is just for choosing way of inserting values. class PlayerMan extends Player{ static final int CONSOLE=0; static final int INPUT_DIALOG=1; private int input; public PlayerMan(String name, Shared data, int c){ super(name, data); input = c; } @Override public void run(){ Scanner scanner = new Scanner(System.in); int n = 0; System.out.println("Matches on table: "+data.matchesAmount); System.out.println("which: "+data.which); System.out.println("number: "+number); while(data.matchesAmount != 0){ if(number == data.which){ System.out.println("Choose amount of matches (from 1 to 3): "); n = scanner.nextInt(); if(data.matchesAmount == 1){ System.out.println("There's only 1 match left !"); while(n != 1){ n = scanner.nextInt(); } } else{ do{ n = scanner.nextInt(); } while(n <= 1 && n >= 3); } data.matchesAmount = data.matchesAmount - n; System.out.println(" "+ name+" takes "+n+" matches."); if(number != 0){ data.which = 0; } else{ data.which = 1; } } else{ try { Thread.sleep(1000); } catch(InterruptedException exc) { System.out.println("End of thread."); return; } } System.out.println("Matches on table: "+data.matchesAmount); } if(data.matchesAmount == 0){ System.out.println("Winner is player: "+name); stop(); } } }

    Read the article

  • array of structs in C

    - by Hristo
    I'm trying to create an array of structs and also a pointer to that array. I don't know how large the array is going to be, so it should be dynamic. My struct would look something like this: typedef struct _stats_t { int hours[24]; int numPostsInHour; int days[7]; int numPostsInDay; int weeks[20]; int numPostsInWeek; int totNumLinesInPosts; int numPostsAnalyzed; } stats_t; ... and I need to have multiple of these structs for each file (unknown amount) that I will analyze. I'm not sure how to do this. I don't like the following approach because of the limit of the size of the array: # define MAX 10 typedef struct _stats_t { int hours[24]; int numPostsInHour; int days[7]; int numPostsInDay; int weeks[20]; int numPostsInWeek; int totNumLinesInPosts; int numPostsAnalyzed; } stats_t[MAX]; So how would I create this array? Also, would a pointer to this array would look something this? stats_t stats[]; stats_t *statsPtr = &stats[0]; Thanks, Hristo

    Read the article

  • Lock-Free, Wait-Free and Wait-freedom algorithms for non-blocking multi-thread synchronization.

    - by GJ
    In multi thread programming we can find different terms for data transfer synchronization between two or more threads/tasks. When exactly we can say that some algorithem is: 1)Lock-Free 2)Wait-Free 3)Wait-Freedom I understand what means Lock-free but when we can say that some synchronization algorithm is Wait-Free or Wait-Freedom? I have made some code (ring buffer) for multi-thread synchronization and it use Lock-Free methods but: 1) Algorithm predicts maximum execution time of this routine. 2) Therad which call this routine at beginning set unique reference, what mean that is inside of this routine. 3) Other threads which are calling the same routine check this reference and if is set than count the CPU tick count (measure time) of first involved thread. If that time is to long interrupt the current work of involved thread and overrides him job. 4) Thread which not finished job because was interrupted from task scheduler (is reposed) at the end check the reference if not belongs to him repeat the job again. So this algorithm is not really Lock-free but there is no memory lock in use, and other involved threads can wait (or not) certain time before overide the job of reposed thread. Added RingBuffer.InsertLeft function: function TgjRingBuffer.InsertLeft(const link: pointer): integer; var AtStartReference: cardinal; CPUTimeStamp : int64; CurrentLeft : pointer; CurrentReference: cardinal; NewLeft : PReferencedPtr; Reference : cardinal; label TryAgain; begin Reference := GetThreadId + 1; //Reference.bit0 := 1 with rbRingBuffer^ do begin TryAgain: //Set Left.Reference with respect to all other cores :) CPUTimeStamp := GetCPUTimeStamp + LoopTicks; AtStartReference := Left.Reference OR 1; //Reference.bit0 := 1 repeat CurrentReference := Left.Reference; until (CurrentReference AND 1 = 0)or (GetCPUTimeStamp - CPUTimeStamp > 0); //No threads present in ring buffer or current thread timeout if ((CurrentReference AND 1 <> 0) and (AtStartReference <> CurrentReference)) or not CAS32(CurrentReference, Reference, Left.Reference) then goto TryAgain; //Calculate RingBuffer NewLeft address CurrentLeft := Left.Link; NewLeft := pointer(cardinal(CurrentLeft) - SizeOf(TReferencedPtr)); if cardinal(NewLeft) < cardinal(@Buffer) then NewLeft := EndBuffer; //Calcolate distance result := integer(Right.Link) - Integer(NewLeft); //Check buffer full if result = 0 then //Clear Reference if task still own reference if CAS32(Reference, 0, Left.Reference) then Exit else goto TryAgain; //Set NewLeft.Reference NewLeft^.Reference := Reference; SFence; //Try to set link and try to exchange NewLeft and clear Reference if task own reference if (Reference <> Left.Reference) or not CAS64(NewLeft^.Link, Reference, link, Reference, NewLeft^) or not CAS64(CurrentLeft, Reference, NewLeft, 0, Left) then goto TryAgain; //Calcolate result if result < 0 then result := Length - integer(cardinal(not Result) div SizeOf(TReferencedPtr)) else result := cardinal(result) div SizeOf(TReferencedPtr); end; //with end; { TgjRingBuffer.InsertLeft } RingBuffer unit you can find here: RingBuffer, CAS functions: FockFreePrimitives, and test program: RingBufferFlowTest Thanks in advance, GJ

    Read the article

  • How to get Current Owner name ?

    - by user325739
    Hi My question is i have 4 Text Box 1) Prepared By 2) Checked By 3) Approved BY 4) Created BY First i will login as Smitha then in " Preapred by " - Smitha name should come automatically n all other text box should be blank, then i will submit the form it goes to our respective HOD now , Nagaraj sir will login as Nagaraj.S then in "Checked by" - Nagara.S name should appear automatically n Approved by and Created by should be blank then he submits the form, then it goes to level 1 person now , Jagadish is in Level 1 , he will login by his user name then in "Approved By" text box his name should appear automatically here Prepared by value and Checked by value which is taken previously should not be altered , and Created by should be blank now he submits the form to level 2 person ie Karthick , then his name should appear in "Created By" by not altering any values can u help me on this ?

    Read the article

  • bash : recursive listing of all files problem

    - by Michael Mao
    Run a recursive listing of all the files in /var/log and redirect standard output to a file called lsout.txt in your home directory. Complete this question WITHOUT leaving your home directory. An: ls -R /var/log/ /home/bqiu/lsout.txt I reckon the above bash command is not correct. This is what I've got so far: $ ls -1R .: cal.sh cokemachine.sh dir sort test.sh ./dir: afile.txt file subdir ./dir/subdir: $ ls -R | sed s/^.*://g cal.sh cokemachine.sh dir sort test.sh afile.txt file subdir But this still leaves all directory/sub-directory names (dir and subdir), plus a couple of empty newlines How could I get the correct result without using Perl or awk? Preferably using only basic bash commands(this is just because Perl and awk is out of assessment scope)

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 11 12 13 14 15 16 17 18 19 20 21 22  | Next Page >