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  • 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; } }

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  • File processing-Haskell

    - by Martinas Maria
    How can I implement in haskell the following: I receive an input file from the command line. This input file contains words separated with tabs,new lines and spaces.I have two replace this elements(tabs,new lines and spaces) with comma(,) .Observation:more newlines,tabs,spaces will be replaced with a single comma.The result has to be write in a new file(output.txt). Please help me with this.My haskell skills are very scarse. This is what I have so far: processFile::String->String processFile [] =[] processFile input =input process :: String -> IO String process fileName = do text <- readFile fileName return (processFile text) main :: IO () main = do n <- process "input.txt" print n In processFile function I should process the text from the input file. I'm stuck..Please help.

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  • help with matlab and Discrete Fourier transform

    - by user504363
    Hi all I have previous experience with Matlab, but the problem that I face some problems in apply a problem in (DSP : Digital signal processing) which is not my study field, but I must finish that problems in days to complete my project. all i want is help me with method and steps of solving this problem in matlab and then I can write the code with myself. the problem is about the signal x(t) = exp(-a*t); 1) what's the Discrete Fourier transform of the sampled signal with sample rate fs 2) if a=1 and fs =1 , plot the amplitude spectrum of sampled signal 3) fix the sampling frequency at fs = 1(hz) [what's it mean ?] and plot the magnitude of the Fourier Transform of the sampled signal at various values of a thanks

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  • Unknown error in Producer/Consumer program, believe it to be an infinite loop.

    - by ray2k
    Hello, I am writing a program that is solving the producer/consumer problem, specifically the bounded-buffer version(i believe they mean the same thing). The producer will be generating x number of random numbers, where x is a command line parameter to my program. At the current moment, I believe my program is entering an infinite loop, but I'm not sure why it is occurring. I believe I am executing the semaphores correctly. You compile it like this: gcc -o prodcon prodcon.cpp -lpthread -lrt Then to run, ./prodcon 100(the number of randum nums to produce) This is my code. typedef int buffer_item; #include <stdlib.h> #include <stdio.h> #include <pthread.h> #include <semaphore.h> #include <unistd.h> #define BUFF_SIZE 10 #define RAND_DIVISOR 100000000 #define TRUE 1 //two threads void *Producer(void *param); void *Consumer(void *param); int insert_item(buffer_item item); int remove_item(buffer_item *item); int returnRandom(); //the global semaphores sem_t empty, full, mutex; //the buffer buffer_item buf[BUFF_SIZE]; //buffer counter int counter; //number of random numbers to produce int numRand; int main(int argc, char** argv) { /* thread ids and attributes */ pthread_t pid, cid; pthread_attr_t attr; pthread_attr_init(&attr); pthread_attr_setscope(&attr, PTHREAD_SCOPE_SYSTEM); numRand = atoi(argv[1]); sem_init(&empty,0,BUFF_SIZE); sem_init(&full,0,0); sem_init(&mutex,0,0); printf("main started\n"); pthread_create(&pid, &attr, Producer, NULL); pthread_create(&cid, &attr, Consumer, NULL); printf("main gets here"); pthread_join(pid, NULL); pthread_join(cid, NULL); printf("main done\n"); return 0; } //generates a randum number between 1 and 100 int returnRandom() { int num; srand(time(NULL)); num = rand() % 100 + 1; return num; } //begin producing items void *Producer(void *param) { buffer_item item; int i; for(i = 0; i < numRand; i++) { //sleep for a random period of time int rNum = rand() / RAND_DIVISOR; sleep(rNum); //generate a random number item = returnRandom(); //acquire the empty lock sem_wait(&empty); //acquire the mutex lock sem_wait(&mutex); if(insert_item(item)) { fprintf(stderr, " Producer report error condition\n"); } else { printf("producer produced %d\n", item); } /* release the mutex lock */ sem_post(&mutex); /* signal full */ sem_post(&full); } return NULL; } /* Consumer Thread */ void *Consumer(void *param) { buffer_item item; int i; for(i = 0; i < numRand; i++) { /* sleep for a random period of time */ int rNum = rand() / RAND_DIVISOR; sleep(rNum); /* aquire the full lock */ sem_wait(&full); /* aquire the mutex lock */ sem_wait(&mutex); if(remove_item(&item)) { fprintf(stderr, "Consumer report error condition\n"); } else { printf("consumer consumed %d\n", item); } /* release the mutex lock */ sem_post(&mutex); /* signal empty */ sem_post(&empty); } return NULL; } /* Add an item to the buffer */ int insert_item(buffer_item item) { /* When the buffer is not full add the item and increment the counter*/ if(counter < BUFF_SIZE) { buf[counter] = item; counter++; return 0; } else { /* Error the buffer is full */ return -1; } } /* Remove an item from the buffer */ int remove_item(buffer_item *item) { /* When the buffer is not empty remove the item and decrement the counter */ if(counter > 0) { *item = buf[(counter-1)]; counter--; return 0; } else { /* Error buffer empty */ return -1; } }

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  • java interfaces

    - by Codenotguru
    write an interface with one method,two classes that implement the interface, and a main method with an array holding an instance from both classes.Using this array variable call the method in a foreach loop.This is a interview question in java anybody?

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  • Finding if a path between 2 sides of a game board exists

    - by Meny
    Hi, i'm currently working on a game as an assignment for school in java. the game cuurently is designed for Console. the game is for 2 players, one attacking from north to south, and the other from west to east. the purpose of the game is to build a "bridge"/"path" between the 2 of your sides before your opponent does. for example: A B C D E F 1 _ _ X _ _ _ 1 2 O X X _ _ _ 2 3 O X O O O O 3 4 O X O _ _ _ 4 5 X X _ _ _ _ 5 6 X O _ _ _ _ 6 A B C D E F player that attacks from north to south won (path/bridge from C to A) my problem is, what algorithm would be good to check if the user have managed to create a path (will be checked at the end of each turn). you're help would be very appreciated.

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  • Dynamic programming: Largest diamond(rhombus) block

    - by Darksody
    I have a small program to do in Java. I have a 2D array filled with 0 and 1, and I must find the largest rhombus (as in square rotated by 90 degrees) and their numbers. Example: 0 1 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Result: 1 1 1 1 1 1 1 1 1 1 1 1 1 The problem is similar to this SO question. If you have any idea, post it here.

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  • Problem on a Floyd-Warshall implementation using c++

    - by Henrique
    I've got a assignment for my college, already implemented Dijkstra and Bellman-Ford sucessfully, but i'm on trouble on this one. Everything looks fine, but it's not giving me the correct answer. Here's the code: void FloydWarshall() { //Also assume that n is the number of vertices and edgeCost(i,i) = 0 int path[500][500]; /* A 2-dimensional matrix. At each step in the algorithm, path[i][j] is the shortest path from i to j using intermediate vertices (1..k-1). Each path[i][j] is initialized to edgeCost(i,j) or infinity if there is no edge between i and j. */ for(int i = 0 ; i <= nvertices ; i++) for(int j = 0 ; j <= nvertices ; j++) path[i][j] = INFINITY; for(int j = 0 ; j < narestas ; j++) //narestas = number of edges { path[arestas[j]->v1][arestas[j]->v2] = arestas[j]->peso; //peso = weight of the edge (aresta = edge) path[arestas[j]->v2][arestas[j]->v1] = arestas[j]->peso; } for(int i = 0 ; i <= nvertices ; i++) //path(i, i) = 0 path[i][i] = 0; //test print, it's working fine //printf("\n\n\nResultado FloydWarshall:\n"); //for(int i = 1 ; i <= nvertices ; i++) // printf("distancia ao vertice %d: %d\n", i, path[1][i]); //heres the problem, it messes up, and even a edge who costs 4, and the minimum is 4, it prints 2. //for k = 1 to n for(int k = 1 ; k <= nvertices ; k++) //for i = 1 to n for(int i = 1 ; i <= nvertices ; i++) //for j := 1 to n for(int j = 1 ; j <= nvertices ; j++) if(path[i][j] > path[i][k] + path[k][j]) path[i][j] = path[i][k] + path[k][j]; printf("\n\n\nResultado FloydWarshall:\n"); for(int i = 1 ; i <= nvertices ; i++) printf("distancia ao vertice %d: %d\n", i, path[1][i]); } im using this graph example i've made: 6 7 1 2 4 1 5 1 2 3 1 2 5 2 5 6 3 6 4 6 3 4 2 means we have 6 vertices (1 to 6), and 7 edges (1,2) with weight 4... etc.. If anyone need more info, i'm up to giving it, just tired of looking at this code and not finding an error.

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  • Implementing Dijkstra's Algorithm

    - by DeadMG
    I've been tasked (coursework @ university) to implement a form of path-finding. Now, in-spec, I could just implement a brute force, since there's a limit on the number of nodes to search (begin, two in the middle, end), but I want to re-use this code and came to implement Dijkstra's algorithm. I've seen the pseudo on Wikipedia and a friend wrote some for me as well, but it flat out doesn't make sense. The algorithm seems pretty simple and it's not a problem for me to understand it, but I just can't for the life of me visualize the code that would realize such a thing. Any suggestions/tips?

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  • Deferring signal handling in Linux

    - by EpsilonVector
    I'm trying to figure out how to block a signal in Linux kernel 2.4 (user space) from invoking its handler, but keep it available to be handled later, preferably as soon as I re activate the handling of said signal. The function sigprocmask seem to come up in all my search results, but I can't find a good, clear description that explains whether the blocked signal gets "saved" to be handled later, and if so where and how do I handle it when I'm ready for it. Can someone please clarify what's going on, preferably with a code example? Thanks in advance.

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  • Available message types in JMS?

    - by Caylem
    This is based on a past exam question. The question is asking to describe the four types of message available using JMS. The problem is it says the four, not just four. So it assumes their is only four, no more no less. However according to this site their seems to be five; streams maps text objects bytes *Another book states that XML is another potential type in future versions of JMS. Is XML already available? Am I missing something or is the question just wrong? Thanks.

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  • Help with a sort method

    - by Capsud
    Hi there, If i have an array of strings for example Static final String[] TEST = new String[] { "g","a","b","t","e" }; How would i go about sorting this in alphabetical order please?

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  • C++ Operator overloading - 'recreating the Vector'

    - by Wallter
    I am currently in a collage second level programing course... We are working on operator overloading... to do this we are to rebuild the vector class... I was building the class and found that most of it is based on the [] operator. When I was trying to implement the + operator I run into a weird error that my professor has not seen before (apparently since the class switched IDE's from MinGW to VS express...) (I am using Visual Studio Express 2008 C++ edition...) Vector.h #include <string> #include <iostream> using namespace std; #ifndef _VECTOR_H #define _VECTOR_H const int DEFAULT_VECTOR_SIZE = 5; class Vector { private: int * data; int size; int comp; public: inline Vector (int Comp = 5,int Size = 0) : comp(Comp), size(Size) { if (comp > 0) { data = new int [comp]; } else { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; } } int size_ () const { return size; } int comp_ () const { return comp; } bool push_back (int); bool push_front (int); void expand (); void expand (int); void clear (); const string at (int); int operator[ ](int); Vector& operator+ (Vector&); Vector& operator- (const Vector&); bool operator== (const Vector&); bool operator!= (const Vector&); ~Vector() { delete [] data; } }; ostream& operator<< (ostream&, const Vector&); #endif Vector.cpp #include <iostream> #include <string> #include "Vector.h" using namespace std; const string Vector::at(int i) { this[i]; } void Vector::expand() { expand(size); } void Vector::expand(int n ) { int * newdata = new int [comp * 2]; if (*data != NULL) { for (int i = 0; i <= (comp); i++) { newdata[i] = data[i]; } newdata -= comp; comp += n; delete [] data; *data = *newdata; } else if ( *data == NULL || comp == 0) { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; size = 0; } } bool Vector::push_back(int n) { if (comp = 0) { expand(); } for (int k = 0; k != 2; k++) { if ( size != comp ){ data[size] = n; size++; return true; } else { expand(); } } return false; } void Vector::clear() { delete [] data; comp = 0; size = 0; } int Vector::operator[] (int place) { return (data[place]); } Vector& Vector::operator+ (Vector& n) { int temp_int = 0; if (size > n.size_() || size == n.size_()) { temp_int = size; } else if (size < n.size_()) { temp_int = n.size_(); } Vector newone(temp_int); int temp_2_int = 0; for ( int j = 0; j <= temp_int && j <= n.size_() && j <= size; j++) { temp_2_int = n[j] + data[j]; newone[j] = temp_2_int; } //////////////////////////////////////////////////////////// return newone; //////////////////////////////////////////////////////////// } ostream& operator<< (ostream& out, const Vector& n) { for (int i = 0; i <= n.size_(); i++) { //////////////////////////////////////////////////////////// out << n[i] << " "; //////////////////////////////////////////////////////////// } return out; } Errors: out << n[i] << " "; error C2678: binary '[' : no operator found which takes a left-hand operand of type 'const Vector' (or there is no acceptable conversion) return newone; error C2106: '=' : left operand must be l-value As stated above, I am a student going into Computer Science as my selected major I would appreciate tips, pointers, and better ways to do stuff :D

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  • Dynamic programming: Find largest diamond (rhombus)

    - by Darksody
    I have a small program to do in Java. I have a 2D array filled with 0 and 1, and I must find the largest rhombus (as in square rotated by 90 degrees) and their numbers. Example: 0 1 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Result: 1 1 1 1 1 1 1 1 1 1 1 1 1 The problem is similar to this SO question. If you have any idea, post it here.

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  • How do I get the next token in a Cstring if I want to use it as an int? (c++)

    - by Van
    My objective is to take directions from a user and eventually a text file to move a robot. The catch is that I must use Cstrings(such as char word[];) rather than the std::string and tokenize them for use. the code looks like this: void Navigator::manualDrive() { char uinput[1]; char delim[] = " "; char *token; cout << "Enter your directions below: \n"; cin.ignore(); cin.getline (uinput, 256); token=strtok(uinput, delim); if(token == "forward") { int inches; inches=token+1; travel(inches); } } I've never used Cstrings I've never tokenized anything before, and I don't know how to write this. Our T.A.'s expect us to google and find all the answers because they are aware we've never been taught these methods. Everyone in my lab is having much more trouble than usual. I don't know the code to write but I know what I want my program to do. I want it to execute like this: 1) Ask for directions. 2) cin.getline the users input 3) tokenize the inputed string 4) if the first word token == "forward" move to the next token and find out how many inches to move forward then move forward 5) else if the first token == "turn" move to the next token. if the next token == "left" move to the next token and find out how many degrees to turn left I will have to do this for forward x, backward x, turn left x, turn right x, and stop(where x is in inches or degrees). I already wrote functions that tell the robot how to move forward an inch and turn in degrees. I just need to know how to convert the inputted strings to all lowercase letters and move from token to token and convert or extract the numbers from the string to use them as integers. If all is not clear you can read my lab write up at this link: http://www.cs.utk.edu/~cs102/robot_labs/Lab9.html If anything is unclear please let me know, and I will clarify as best I can.

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  • finding "distance" between two pixel's colors.

    - by igor
    Once more something relatively simple, but confused as to what they want. the method to find distance on cartesian coordinate system is distance=sqrt[(x2-x1)^2 + (y2-y1)^2] but how do i apply it here? //Requires: testColor to be a valid Color //Effects: returns the "distance" between the current Pixel's color and // the passed color // uses the standard method to calculate "distance" // uses the same formula as finding distance on a // Cartesian coordinate system double colorDistance(Color testColor) const;

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  • A textbox class only accept integers in Java

    - by alex
    I just want to do a textbox class onl accepts integers.. I have done something, but i think it's not enough. Can anyone help me, please? Thanks... import java.awt.TextField public class textbox extends TextField{ private int value; public textbox(){ super(); } public textbox(int value){ setDeger(value); } public int getValue() { return value; } public void setValue(int value) { this.value = value; } }

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  • Merging some sorted lists with unknown order sequence

    - by Gabriel
    I've some sorted lists with variable number of elements. I wold like to merge the lists into one big list which contains all other lists in same order, without duplicates. Example: 1. XS,M,L,XL 2. S,M,XXL 3. XXS,XS,S,L Result: XXS,XS,S,M,L,XL,XXL The function should notify, if there are elements which have ambiguous positions. Here, it would be XXL and I need to specify its position after XL.

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  • Implement dictionary using java

    - by ahmad
    Task Dictionary ADT The dictionary ADT models a searchable collection of key-element entries Multiple items with the same key are allowed Applications: word-definition pairs Dictionary ADT methods: find(k): if the dictionary has an entry with key k, returns it, else, returns null findAll(k): returns an iterator of all entries with key k insert(k, o): inserts and returns the entry (k, o) remove(e): remove the entry e from the dictionary size(), isEmpty() Operation Output Dictionary insert(5,A) (5,A) (5,A) insert(7,B) (7,B) (5,A),(7,B) insert(2,C) (2,C) (5,A),(7,B),(2,C) insert(8,D) (8,D) (5,A),(7,B),(2,C),(8,D) insert(2,E) (2,E) (5,A),(7,B),(2,C),(8,D),(2,E) find(7) (7,B) (5,A),(7,B),(2,C),(8,D),(2,E) find(4) null (5,A),(7,B),(2,C),(8,D),(2,E) find(2) (2,C) (5,A),(7,B),(2,C),(8,D),(2,E) findAll(2) (2,C),(2,E) (5,A),(7,B),(2,C),(8,D),(2,E) size() 5 (5,A),(7,B),(2,C),(8,D),(2,E) remove(find(5)) (5,A) (7,B),(2,C),(8,D),(2,E) find(5) null (7,B),(2,C),(8,D),(2,E) Detailed explanation: NO Specific requirements: please Get it done i need HELP !!!

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  • JAVA: storing input into array

    - by Jann
    I need to write a program where the program would generate random letter and i would need to store this random character into an array char[] arrayRandom = new char[10]; for (int i = 0; i < 8; i++) { randomNumLet = (generator.nextInt(20) + 1); System.out.print(arrayRandomLetter[randomNumLet] + " "); arrayRandomLetter[randomNumLet] = arrayRandom[i]; } is there anything wrong with my code? because when i run this and printed the array i get boxes for all the values in the array and there are some letter that this line of code cannot print System.out.print(arrayRandomLetter[randomNumLet] + " "); Thanks

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