Search Results

Search found 13891 results on 556 pages for 'maybe homework'.

Page 24/556 | < Previous Page | 20 21 22 23 24 25 26 27 28 29 30 31  | Next Page >

  • Intel IA-32 Assembly

    - by Kay
    I'm having a bit of difficulty converting the following java code into Intel IA-32 Assembly: class Person() { char name [8]; int age; void printName() {...} static void printAdults(Person [] list) { for(int k = 0; k < 100; k++){ if (list[k].age >= 18) { list[k].printName(); } } } } My attempt is: Person: push ebp; save callers ebp mov ebp, esp; setup new ebp push esi; esi will hold name push ebx; ebx will hold list push ecx; ecx will hold k init: mov esi, [ebp + 8]; mov ebx, [ebp + 12]; mov ecx, 0; k=0 forloop: cmp ecx, 100; jge end; if k>= 100 then break forloop cmp [ebx + 4 * ecx], 18 ; jl auxloop; if list[k].age < 18 then go to auxloop jmp printName; printName: auxloop: inc ecx; jmp forloop; end: pop ecx; pop ebx; pop esi; pop ebp; Is my code correct? NOTE: I'm not allowed to use global variables.

    Read the article

  • How do I check the validity of the Canadian Social Insurance Number in C#?

    - by user518307
    I've been given the assignment to write an algorithm in C# that checks the validity of a Canadian Social Insurance Number (SIN). Here are the steps to validate a SIN. Given an example Number: 123 456 782 Remove the check digit (the last digit): 123456782 Extract the even digits (2,4,6,8th digith): 12345678 Double them: 2 4 6 8 | | | | v v v v 4 8 12 16 Add the digits together: 4+8+1+2+1+6 = 22 Add the Odd placed digits: 1+3+5+7 = 16 Total : 38 Validity Algorithm If the total is a multiple of 10, the check digit should be zero. Otherwise, Subtract the Total from the next highest multiple of 10 (40 in this case) The check digit for this SIN must be equal to the difference of the number and the totals from earlier (in this case, 40-38 = 2; check digit is 2, so the number is valid) I'm lost on how to actually implement this in C#, how do I do this?

    Read the article

  • Linked List Sorting with Strings In C

    - by user308583
    I have a struct, with a Name and a single Node called nextName It's a Singly Linked list, and my task is to create the list, based on alphabetical order of the strings. So iff i enter Joe Zolt and Arthur i should get my list structured as Joe Than Joe Zolt Than Arthur Joe Zolt I'm having trouble implementing the correct Algorithm, which would put the pointers in the right order. This is What I have as of Now. Temp would be the name the user just entered and is trying to put into the list, namebox is just a copy of my root, being the whole list if(temp != NULL) { struct node* namebox = root; while (namebox!=NULL && (strcmp((namebox)->name,temp->name) <= 0)) { namebox = namebox->nextName; printf("here"); } temp->nextName = namebox; namebox = temp; root = namebox; This Works right now, if i enter names like CCC BBB than AAA I Get Back AAA BBB CCC when i print But if i put AAA BBB CCC , When i print i only get CCC, it cuts the previous off.

    Read the article

  • class header+ implementation

    - by igor
    what am I doing wrong here? I keep on getting a compilation error when I try to run this in codelab (turings craft) Instructions: Write the implementation (.cpp file) of the GasTank class of the previous exercise. The full specification of the class is: A data member named amount of type double. A constructor that no parameters. The constructor initializes the data member amount to 0. A function named addGas that accepts a parameter of type double . The value of the amount instance variable is increased by the value of the parameter. A function named useGas that accepts a parameter of type double . The value of the amount data member is decreased by the value of the parameter. A function named getGasLevel that accepts no parameters. getGasLevel returns the value of the amount data member. class GasTank{ double amount; GasTank(); void addGas(double); void useGas(double); double getGasLevel();}; GasTank::GasTank(){ amount=0;} double GasTank::addGas(double a){ amount+=a;} double GasTank::useGas(double a){ amount+=a;} double GasTank::getGasLevel(){ return amount;}

    Read the article

  • How should I generate the partitions / pairs for the Chinese Postman problem?

    - by Simucal
    I'm working on a program for class that involves solving the Chinese Postman problem. Our assignment only requires us to write a program to solve it for a hard-coded graph but I'm attempting to solve it for the general case on my own. The part that is giving me trouble is generating the partitions of pairings for the odd vertices. For example, if I had the following labeled odd verticies in a graph: 1 2 3 4 5 6 I need to find all the possible pairings / partitions I can make with these vertices. I've figured out I'll have i paritions given: n = num of odd verticies k = n / 2 i = ((2k)(2k-1)(2k-2)...(k+1))/2 So, given the 6 odd verticies above, we will know that we need to generate i = 15 partitions. The 15 partions would look like: 1 2 3 4 5 6 1 2 3 5 4 6 1 2 3 6 4 5 ... 1 6 ... Then, for each partition, I take each pair and find the shortest distance between them and sum them for that partition. The partition with the total smallest distance between its pairs is selected, and I then double all the edges between the shortest path between the odd vertices (found in the selected partition). These represent the edges the postman will have to walk twice. At first I thought I had worked out an appropriate algorithm for generating these partitions / pairs but it is flawed. I found it wasn't a simple permutation/combination problem. Does anyone who has studied this problem before have any tips that can help point me in the right direction for generating these partitions?

    Read the article

  • A[i] * A[j] = k in O(nlog(n))

    - by gleb-pendler
    A is an Array of n positive int numbers k given int Algorithm should find if there is a pair of numbers which product gives the result a. A[i] * A[j] = k b. A[i] = A[j] + k if there is such a couple the algorithm should return thier index. thanks in advance.

    Read the article

  • Which web Tier Framework for a public commercial website with heavy load ?

    - by Maxime ARNSTAMM
    Hello everyone, As a part of an enterprise architecture exercise, i need to find a java-based framework filling these constraints : heavy (i think) load : 5000 concurrent connections widely known : can't be too exotic, the contractors would be too high priced. relatively easy to use : developpement time must be reasonnable must be as compliant as possible with the css/html layout produced by a designer Must look like "web 2.0" from the marketing point of view. What i learned from my limited experience is : jsf : 1, don't know. 2, 3 ok. 4 not ok (at least not without huge effort) wicket : 1, not really. 2, 3 and 4 ok. gwt : 1, don't know. 2, 3 ok. 4 not ok (but more ok than jsf) others : not really "web 2.0" or not really known I'm really junior, so my ideas about those frameworks are probably wrong, that's why i come to you, stackoverflowees. Thanks for helping :)

    Read the article

  • javascript too much recursion?

    - by Ken
    Hi, I'm trying to make a script that automatically starts uploading after the data has been enter in the database(I need the autoId that the database makes to upload the file). When I run the javascript the scripts runs the php file but it fails calling the other php to upload the file. too much recursion setTimeout(testIfToegevoegd(),500); the script that gives the error send("/projects/backend/nieuwDeeltaak.php",'deeltaakNaam='+f.deeltaaknaam.value+'&beschrijving='+ f.beschrijving.value+'&startDatum='+f.startDatum.value+'&eindDatum='+f.eindDatum.value +'&deeltaakLeider='+f.leiderID.value+'&projectID='+f.projectID.value,id); function testIfToegevoegd(){ if(document.getElementById('resultaat').innerHTML == "<b>De deeltaak werd toegevoegd</b>"){ //stop met testen + upload file document.getElementById('nieuwDeeltaak').target = 'upload_target'; document.forms["nieuwDeeltaak"].submit() }else{ setTimeout(testIfToegevoegd(),500); } } testIfToegevoegd(); sorry for the dutch names we have to use them it is a school project. when I click the button that calls all this for a second time (after the error) it works fine.

    Read the article

  • VB.net Network Graph code/algorithm

    - by Jens
    For a school project we need to visualise a computer network graph. The number of computers with specific properties are read from an XML file, and then a graph should be created. Ad random computers are added and removed. Is there any open source project or algorithm that could help us visualising this in VB.net? Or would you suggest us to switch to java. Update: We eventually switched java and used the Jung libraries because this was easier for us to understand and implement.

    Read the article

  • Tricky Big-O complexity

    - by timeNomad
    public void foo (int n, int m) { int i = m; while (i > 100) i = i/3; for (int k=i ; k>=0; k--) { for (int j=1; j<n; j*=2) System.out.print(k + "\t" + j); System.out.println(); } } I figured the complexity would be O(logn). That is as a product of the inner loop, the outer loop -- will never be executed more than 100 times, so it can be omitted. What I'm not sure about is the while clause, should it be incorporated into the Big-O complexity? For very large i values it could make an impact, or arithmetic operations, doesn't matter on what scale, count as basic operations and can be omitted?

    Read the article

  • Java - Creating a Compiler Help

    - by Brian
    So for my programming class we have had a project to create a virtual machine including a memory unit, cpu, Input, Output, Instruction Register, Program Counter, MAR, MDR and so on. Now we need to create a compiler using Java Code that will take a .exe file written in some txt editor and convert it to java byte code and run the code. The code we will be writing in the .exe file is machine code along the lines of: IN X IN Y ADD X STO Y OUT Y STOP DC X 0 DC Y 0 I am just a beginner and only have 2 days to write this and am very lost and have no idea where to start....Any Help will be much appreciated. Thanks

    Read the article

  • using compareTo in Binary Search Tree program

    - by Scott Rogener
    I've been working on this program for a few days now and I've implemented a few of the primary methods in my BinarySearchTree class such as insert and delete. Insert seemed to be working fine, but once I try to delete I kept getting errors. So after playing around with the code I wanted to test my compareTo methods. I created two new nodes and tried to compare them and I get this error: Exception in thread "main" java.lang.ClassCastException: TreeNode cannot be cast to java.lang.Integer at java.lang.Integer.compareTo(Unknown Source) at TreeNode.compareTo(TreeNode.java:16) at BinarySearchTree.myComparision(BinarySearchTree.java:177) at main.main(main.java:14) Here is my class for creating the nodes: public class TreeNode<T> implements Comparable { protected TreeNode<T> left, right; protected Object element; public TreeNode(Object obj) { element=obj; left=null; right=null; } public int compareTo(Object node) { return ((Comparable) this.element).compareTo(node); } } Am I doing the compareTo method all wrong? I would like to create trees that can handle integers and strings (seperatly of course)

    Read the article

  • Why is my file being cleared if I don't save it?

    - by Kat
    My program is suppose to maintain a collection of Photos in a PhotoAlbum. It begins by reading a folder of photos and adds them to my PhotoAlbum. It then prints a menu that allows the user to list all the photos, add a photo, find a photo, save, and quit the program. Right now if I run my program it will add the 100 photos to the PhotoAlbum, but if I quit the program without saving, it clears the file I am reading from even if I haven't added a photo or done anything to the PhotoAlbum and I'm not sure why. Here is my method for printing to a file: private static void saveFile(PrintWriter writer) { String result; ArrayList<Photo> temp = album.getPhotoAlbum(); for (int i = 0; i < temp.size(); i++){ result = temp.get(i).toString() + "\n"; writer.println(result); } writer.close(); } And where the PrintWriter is instantiated: File file = new File(args[0] + File.separator + "album.dat"); try { PrintWriter fout = new PrintWriter(new FileWriter(file)); fileWriter = fout; } catch (IOException e){ System.out.println("ReadFromFile: Folder " + args[0] + " is not found."); System.exit(0); } And where it is called in my runMenu Method: private static void runMainMenu(Scanner scan) { String input; do { showMainMenu(); input = scan.nextLine().toLowerCase(); switch (input.charAt(0)) { case 'p': System.out.println(album.toString()); break; case 'a': album.addPhoto(readPhoto(scan, t)); break; case 'f': findMenu(scan); break; case 's': saveFile(fileWriter); System.exit(0); break; case 'q': break; default: System.out.println("Invalid entry: " + input.charAt(0)); break; } } while (!input.equalsIgnoreCase("q")); }

    Read the article

  • Finding height in Binary Search Tree

    - by mike
    Hey I was wondering if anybody could help me rework this method to find the height of a binary search tree. So far my code looks like this however the answer im getting is larger than the actual height by 1, but when I remove the +1 from my return statements its less than the actual height by 1? I'm still trying to wrap my head around recursion with these BST any help would be much appreciated. public int findHeight(){ if(this.isEmpty()){ return 0; } else{ TreeNode<T> node = root; return findHeight(node); } } private int findHeight(TreeNode<T> aNode){ int heightLeft = 0; int heightRight = 0; if(aNode.left!=null) heightLeft = findHeight(aNode.left); if(aNode.right!=null) heightRight = findHeight(aNode.right); if(heightLeft > heightRight){ return heightLeft+1; } else{ return heightRight+1; } }

    Read the article

  • Generalizing Fibonacci sequence with SICStus Prolog

    - by Christophe Herreman
    I'm trying to find a solution for a query on a generalized Fibonacci sequence (GFS). The query is: are there any GFS that have 885 as their 12th number? The initial 2 numbers may be restricted between 1 and 10. I already found the solution to find the Nth number in a sequence that starts at (1, 1) in which I explicitly define the initial numbers. Here is what I have for this: fib(1, 1). fib(2, 1). fib(N, X) :- N #> 1, Nmin1 #= N - 1, Nmin2 #= N - 2, fib(Nmin1, Xmin1), fib(Nmin2, Xmin2), X #= Xmin1 + Xmin2. For the query mentioned I thought the following would do the trick, in which I reuse the fib method without defining the initial numbers explicitly since this now needs to be done dynamically: fib2 :- X1 in 1..10, X2 in 1..10, fib(1, X1), fib(2, X2), fib(12, 885). ... but this does not seem to work. Is it not possible this way to define the initial numbers, or am I doing something terribly wrong? I'm not asking for the solution, but any advice that could help me solve this would be greatly appreciated.

    Read the article

  • Validating an integer or String without try-catch

    - by Phil
    Ok, I'm lost. I am required to figure out how to validate an integer and String, but for some stupid reason, I can't use the Try-Catch method. I know this is the easiest way and so all the solutions on the internet are using it. I'm writing in Java. The deal is this, I need someone to put in an numerical ID and String name. If either one of the two inputs are invalid I must tell them they made a mistake. Can someone help me?

    Read the article

  • Relational algebra help?!

    - by Tom
    im new to relational algebra and finding it difficult. Ive answered a few questions, however they where relatively simple. Could do with help with these though. Database Patient (PatientCode, PatientSurname, PatientFirstname, PatientSex, PatientAge, PatientOccupation, PatientHeight, PatientWeight, PatientAddress) Doctor (DoctorCode, DoctorSurName, DoctorFirstName, DoctorPrivateAddress, MobileNo, Doctor Specilisim) Operation (Operation Code, PatientCode, DoctorCode, Date, Time, Result, OperationType) Is_Seen_By (PatientCode, DoctorCode, Date, Time). Q1. Find the surname and gender of the patients that have been operated on by doctor "DR333" and results have not been successful. Q2. Find the code of the operations that have been done on the 18th of November 2010 and have been successful, please also list the name of the doctors which were involved with the operation.

    Read the article

  • How do I improve my performance with this singly linked list struct within my program?

    - by Jesus
    Hey guys, I have a program that does operations of sets of strings. We have to implement functions such as addition and subtraction of two sets of strings. We are suppose to get it down to the point where performance if of O(N+M), where N,M are sets of strings. Right now, I believe my performance is at O(N*M), since I for each element of N, I go through every element of M. I'm particularly focused on getting the subtraction to the proper performance, as if I can get that down to proper performance, I believe I can carry that knowledge over to the rest of things I have to implement. The '-' operator is suppose to work like this, for example. Declare set1 to be an empty set. Declare set2 to be a set with { a b c } elements Declare set3 to be a set with ( b c d } elements set1 = set2 - set3 And now set1 is suppose to equal { a }. So basically, just remove any element from set3, that is also in set2. For the addition implementation (overloaded '+' operator), I also do the sorting of the strings (since we have to). All the functions work right now btw. So I was wondering if anyone could a) Confirm that currently I'm doing O(N*M) performance b) Give me some ideas/implementations on how to improve the performance to O(N+M) Note: I cannot add any member variables or functions to the class strSet or to the node structure. The implementation of the main program isn't very important, but I will post the code for my class definition and the implementation of the member functions: strSet2.h (Implementation of my class and struct) // Class to implement sets of strings // Implements operators for union, intersection, subtraction, // etc. for sets of strings // V1.1 15 Feb 2011 Added guard (#ifndef), deleted using namespace RCH #ifndef _STRSET_ #define _STRSET_ #include <iostream> #include <vector> #include <string> // Deleted: using namespace std; 15 Feb 2011 RCH struct node { std::string s1; node * next; }; class strSet { private: node * first; public: strSet (); // Create empty set strSet (std::string s); // Create singleton set strSet (const strSet &copy); // Copy constructor ~strSet (); // Destructor int SIZE() const; bool isMember (std::string s) const; strSet operator + (const strSet& rtSide); // Union strSet operator - (const strSet& rtSide); // Set subtraction strSet& operator = (const strSet& rtSide); // Assignment }; // End of strSet class #endif // _STRSET_ strSet2.cpp (implementation of member functions) #include <iostream> #include <vector> #include <string> #include "strset2.h" using namespace std; strSet::strSet() { first = NULL; } strSet::strSet(string s) { node *temp; temp = new node; temp->s1 = s; temp->next = NULL; first = temp; } strSet::strSet(const strSet& copy) { if(copy.first == NULL) { first = NULL; } else { node *n = copy.first; node *prev = NULL; while (n) { node *newNode = new node; newNode->s1 = n->s1; newNode->next = NULL; if (prev) { prev->next = newNode; } else { first = newNode; } prev = newNode; n = n->next; } } } strSet::~strSet() { if(first != NULL) { while(first->next != NULL) { node *nextNode = first->next; first->next = nextNode->next; delete nextNode; } } } int strSet::SIZE() const { int size = 0; node *temp = first; while(temp!=NULL) { size++; temp=temp->next; } return size; } bool strSet::isMember(string s) const { node *temp = first; while(temp != NULL) { if(temp->s1 == s) { return true; } temp = temp->next; } return false; } strSet strSet::operator + (const strSet& rtSide) { strSet newSet; newSet = *this; node *temp = rtSide.first; while(temp != NULL) { string newEle = temp->s1; if(!isMember(newEle)) { if(newSet.first==NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; newSet.first = newNode; } else if(newSet.SIZE() == 1) { if(newEle < newSet.first->s1) { node *tempNext = newSet.first; node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = tempNext; newSet.first = newNode; } else { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; newSet.first->next = newNode; } } else { node *prev = NULL; node *curr = newSet.first; while(curr != NULL) { if(newEle < curr->s1) { if(prev == NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = curr; newSet.first = newNode; break; } else { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = curr; prev->next = newNode; break; } } if(curr->next == NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; curr->next = newNode; break; } prev = curr; curr = curr->next; } } } temp = temp->next; } return newSet; } strSet strSet::operator - (const strSet& rtSide) { strSet newSet; newSet = *this; node *temp = rtSide.first; while(temp != NULL) { string element = temp->s1; node *prev = NULL; node *curr = newSet.first; while(curr != NULL) { if( element < curr->s1 ) break; if( curr->s1 == element ) { if( prev == NULL) { node *duplicate = curr; newSet.first = newSet.first->next; delete duplicate; break; } else { node *duplicate = curr; prev->next = curr->next; delete duplicate; break; } } prev = curr; curr = curr->next; } temp = temp->next; } return newSet; } strSet& strSet::operator = (const strSet& rtSide) { if(this != &rtSide) { if(first != NULL) { while(first->next != NULL) { node *nextNode = first->next; first->next = nextNode->next; delete nextNode; } } if(rtSide.first == NULL) { first = NULL; } else { node *n = rtSide.first; node *prev = NULL; while (n) { node *newNode = new node; newNode->s1 = n->s1; newNode->next = NULL; if (prev) { prev->next = newNode; } else { first = newNode; } prev = newNode; n = n->next; } } } return *this; }

    Read the article

  • python programme.

    - by siva
    hi, i am siva this is frist time taken the python programming language i have a small problem please help me the question is **Write two functions, called countSubStringMatch and countSubStringMatchRecursive that take two arguments, a key string and a target string. These functions iteratively and recursively count the number of instances of the key in the target string. You should complete definitions for def countSubStringMatch(target,key): and def countSubStringMatchRecursive (target, key): **

    Read the article

  • Theory of computation - Using the pumping lemma for CFLs

    - by Tony
    I'm reviewing my notes for my course on theory of computation and I'm having trouble understanding how to complete a certain proof. Here is the question: A = {0^n 1^m 0^n | n>=1, m>=1} Prove that A is not regular. It's pretty obvious that the pumping lemma has to be used for this. So, we have |vy| = 1 |vxy| <= p (p being the pumping length, = 1) uv^ixy^iz exists in A for all i = 0 Trying to think of the correct string to choose seems a bit iffy for this. I was thinking 0^p 1^q 0^p, but I don't know if I can obscurely make a q, and since there is no bound on u, this could make things unruly.. So, how would one go about this?

    Read the article

  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

    Read the article

< Previous Page | 20 21 22 23 24 25 26 27 28 29 30 31  | Next Page >