Search Results

Search found 19480 results on 780 pages for 'do your own 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 40/780 | < Previous Page | 36 37 38 39 40 41 42 43 44 45 46 47  | Next Page >

  • 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

  • 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

  • Write a program that allows the user to enter a string and then prints the letters of the String sep

    - by WM
    The output is always a String, for example H,E,L,L,O,. How could I limit the commas? I want the commas only between letters, for example H,E,L,L,O. import java.util.Scanner; import java.lang.String; public class forLoop { public static void main(String[] args) { Scanner Scan = new Scanner(System.in); System.out.print("Enter a string: "); String Str1 = Scan.next(); String newString=""; String Str2 =""; for (int i=0; i < Str1.length(); i++) { newString = Str1.charAt(i) + ","; Str2 = Str2 + newString; } System.out.print(Str2); } }

    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

  • Problem in finalizing the link list in C#

    - by Yasin
    My code was almost finished that a maddening bug came up! when i nullify the last node to finalize the link list , it actually dumps all the links and make the first node link Null ! when i trace it , its working totally fine creating the list in the loop but after the loop is done that happens and it destroys the rest of the link by doing so, and i don't understand why something so obvious is becoming problematic! (last line) struct poly { public int coef; public int pow; public poly* link;} ; poly* start ; poly* start2; poly* p; poly* second; poly* result; poly* ptr; poly* start3; poly* q; poly* q2; private void button1_Click(object sender, EventArgs e) { string holder = ""; IntPtr newP = Marshal.AllocHGlobal(sizeof(poly)); q = (poly*)newP.ToPointer(); start = q; int i = 0; while (this.textBox1.Text[i] != ',') { holder += this.textBox1.Text[i]; i++; } q->coef = int.Parse(holder); i++; holder = ""; while (this.textBox1.Text[i] != ';') { holder += this.textBox1.Text[i]; i++; } q->pow = int.Parse(holder); holder = ""; p = start; //creation of the first node finished! i++; for (; i < this.textBox1.Text.Length; i++) { newP = Marshal.AllocHGlobal(sizeof(poly)); poly* test = (poly*)newP.ToPointer(); while (this.textBox1.Text[i] != ',') { holder += this.textBox1.Text[i]; i++; } test->coef = int.Parse(holder); holder = ""; i++; while (this.textBox1.Text[i] != ';' && i < this.textBox1.Text.Length - 1) { holder += this.textBox1.Text[i]; if (i < this.textBox1.Text.Length - 1) i++; } test->pow = int.Parse(holder); holder = ""; p->link = test; //the addresses are correct and the list is complete } p->link = null; //only the first node exists now with a null link! }

    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

  • Very weird C file-handling anomaly

    - by KáGé
    Hello, I got a very weird issue that I cant figure out in my school project, which is the simulation of a simple filesystem in a human-readable textfile. Unfortunately I don't yet have enough time to translate the comments in my code or make it less gibberish, so if you are bothered by that, you don't have to help, I understand. See the code HERE. Now in drive.h, at line 574 is this part: i = getline(); #ifdef DEBUG printf("Free space in all found at %d.\n\n", i); if(drive.disk != NULL){ printf("Disk OK\n\n"); } #endif //write in data state = seekline(i); Before this it finds place for the allocation database entry in the ALL sector (see the "image files" in the mounts folder, this issue was tested on mount_30.efs-dbf), then gets the line with i = getline() fine (getline is in lglobal.h, line 39), but after that any file manipulation (in this case seekline's fseek, but if I comment that out, then the first fprintf after that) crashes the program straight away. I think the file gets somehow corrupted (though the Disk OK message appears) but can't figure out how. I've tried putting i = getline(); into comment, but it didn't make any difference. I've also tried asking at local programming forums but they didn't really help either. The last few lines of the output before it crashes: Dir written. (drive.h line 562) Seekline entered: 268 (called at drive.h line 564) Getline entered. (called at drive.h line 574) Line got: 268. Free space in all found at 268. (drive.h line 576) Seekline entered: 268 (called at drive.h line 582, note that this exact call was run successfully less than 20 lines back. This one should set the pointer to the beginning of the line it is currently in) After this it crashes. Does anyone has any idea of what causes this and how could I fix it? Thank you.

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • Count the number of lines in a Java String

    - by Simon Guo
    Need some compact code for counting the number of lines in a string in Java. The string is to be separated by \r or \n. Each instance of those newline characters will be considered as a separate line. For example, "Hello\nWorld\nThis\nIs\t" should return 4. The prototype is private static int countLines(String str) {...} Can someone provide a compact set of statements? I have solution at here but it is too long, I think. Thank you.

    Read the article

  • Candidate Elimination Question---Please help!

    - by leon
    Hi , I am doing a question on Candidate Elimination Algorithm. I am a little confused with the general boundary G. Here is an example, I got G and S to the fourth case, but I am not sure with the last case. Sunny,Warm,Normal,Strong,Warm,Same,EnjoySport=yes Sunny,Warm,High,Strong,Warm,Same,EnjoySport=yes Rainy,Cold,High,Strong,Warm,Change,EnjoySport=no Sunny,Warm,High,Strong,Cool,Change,EnjoySport=yes Sunny,Warm,Normal,Weak,Warm,Same,EnjoySport=no What I have here is : S 0 :{0,0,0,0,0,0} S 1 :{Sunny,Warm,Normal,Strong,Warm,Same} S 2 , S 3 : {Sunny,Warm,?,Strong,Warm,Same} S 4 :{Sunny,Warm,?,Strong,?,?} G 4 :{Sunny,?,?,?,?,?,?,Warm,?,?,?,?} G 3 :{Sunny,?,?,?,?,?,?,Warm,?,?,?,?,?,?,?,?,?,Same} G 0 , G 1 , G 2 : {?,?,?,?,?,?} What would be the result of G5? Is it G5 empty? {}? or {???Strong??) ? Thanks

    Read the article

  • Implement a Cellular Automaton ? "Rule 110"

    - by ZaZu
    I was wondering how to use the Rule 110, with 55 lines and 14 cells. I have to then display that in an LED matrix display. Anyway my question is, how can I implement such automaton ?? I dont really know where to start, can someone please shed some light on how can I approach this problem ? Is there a specific METHOD I must follow ? Thanks --PROGRAM USED IS - C EDIT char array[54][14]; for(v=0;v<55;v++){ for(b=0;b<15;b++){ if(org[v][b-1]==0 && org[v][b]==0 && org[v][b+1] == 0) { array[v][b]=0; } array[v][b]=org[v][b]; } } Does that make sense ?? org stands for original

    Read the article

  • How do I implement a fibonacci sequence in java using try/catch logic?

    - by Lars Flyger
    I know how to do it using simple recursion, but in order to complete this particular assignment I need to be able to accumulate on the stack and throw an exception that holds the answer in it. So far I have: public static int fibo(int index) { int sum = 0; try { fibo_aux(index, 1, 1); } catch (IntegerException me) { sum = me.getIntValue(); } return sum; } fibo_aux is supposed to throw an IntegerException (which holds the value of the answer that is retireved via getIntValue) and accumulates the answer on the stack, but so far I can't figure it out. Can anyone help?

    Read the article

  • 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;

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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?

    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 | 36 37 38 39 40 41 42 43 44 45 46 47  | Next Page >