Search Results

Search found 39440 results on 1578 pages for 'possible homework'.

Page 11/1578 | < Previous Page | 7 8 9 10 11 12 13 14 15 16 17 18  | Next Page >

• Possible SWITCH Optimization in DAX – #powerpivot #dax #tabular

  - by Marco Russo (SQLBI)

  In one of the Advanced DAX Workshop I taught this year, I had an interesting discussion about how to optimize a SWITCH statement (which could be frequently used checking a slicer, like in the Parameter Table pattern). Let’s start with the problem. What happen when you have such a statement? Sales :=     SWITCH (         VALUES ( Period[Period] ),         "Current", [Internet Total Sales],         "MTD", [MTD Sales],         "QTD", [QTD Sales],         "YTD", [YTD Sales],          BLANK ()     ) The SWITCH statement is in reality just syntax sugar for a nested IF statement. When you place such a measure in a pivot table, for every cell of the pivot table the IF options are evaluated. In order to optimize performance, the DAX engine usually does not compute cell-by-cell, but tries to compute the values in bulk-mode. However, if a measure contains an IF statement, every cell might have a different execution path, so the current implementation might evaluate all the possible IF branches in bulk-mode, so that for every cell the result from one of the branches will be already available in a pre-calculated dataset. The price for that could be high. If you consider the previous Sales measure, the YTD Sales measure could be evaluated for all the cells where it’s not required, and also when YTD is not selected at all in a Pivot Table. The actual optimization made by the DAX engine could be different in every build, and I expect newer builds of Tabular and Power Pivot to be better than older ones. However, we still don’t live in an ideal world, so it could be better trying to help the engine finding a better execution plan. One student (Niek de Wit) proposed this approach: Selection := IF (     HASONEVALUE ( Period[Period] ),     VALUES ( Period[Period] ) ) Sales := CALCULATE (     [Internet Total Sales],     FILTER (         VALUES ( 'Internet Sales'[Order Quantity] ),         'Internet Sales'[Order Quantity]             = IF (                 [Selection] = "Current",                 'Internet Sales'[Order Quantity],                 -1             )     ) )     + CALCULATE (         [MTD Sales],         FILTER (             VALUES ( 'Internet Sales'[Order Quantity] ),             'Internet Sales'[Order Quantity]                 = IF (                     [Selection] = "MTD",                     'Internet Sales'[Order Quantity],                     -1                 )         )     )     + CALCULATE (         [QTD Sales],         FILTER (             VALUES ( 'Internet Sales'[Order Quantity] ),             'Internet Sales'[Order Quantity]                 = IF (                     [Selection] = "QTD",                     'Internet Sales'[Order Quantity],                     -1                 )         )     )     + CALCULATE (         [YTD Sales],         FILTER (             VALUES ( 'Internet Sales'[Order Quantity] ),             'Internet Sales'[Order Quantity]                 = IF (                     [Selection] = "YTD",                     'Internet Sales'[Order Quantity],                     -1                 )         )     ) At first sight, you might think it’s impossible that this approach could be faster. However, if you examine with the profiler what happens, there is a different story. Every original IF’s execution branch is now a separate CALCULATE statement, which applies a filter that does not execute the required measure calculation if the result of the FILTER is empty. I used the ‘Internet Sales’[Order Quantity] column in this example just because in Adventure Works it has only one value (every row has 1): in the real world, you should use a column that has a very low number of distinct values, or use a column that has always the same value for every row (so it will be compressed very well!). Because the value –1 is never used in this column, the IF comparison in the filter discharge all the values iterated in the filter if the selection does not match with the desired value. I hope to have time in the future to write a longer article about this optimization technique, but in the meantime I’ve seen this optimization has been useful in many other implementations. Please write your feedback if you find scenarios (in both Power Pivot and Tabular) where you obtain performance improvements using this technique!

  Read the article

• is it possible to set bookmark in adobe reader [closed]

  - by bala

  Possible Duplicates: PDF reader that can save bookmarks & notes outside of the file Is it possible to add your own bookmarks/tabs to a PDF file? is it possible to set bookmark in adobe reader.

  Read the article

• Beginning Java (Working with Arrays; Class Assignment)

  - by Jason

  I am to the point where I feel as if I correctly wrote the code for this homework assignment. We were given a skeleton and 2 classes that we had to import (FileIOHelper and Student). /* * Created: *** put the date here *** * * Author: *** put your name here *** * * The program will read information about students and their * scores from a file, and output the name of each student with * all his/her scores and the total score, plus the average score * of the class, and the name and total score of the students with * the highest and lowest total score. */ // import java.util.Scanner; import java.io.*; // C:\Users\Adam\info.txt public class Lab6 { public static void main(String[] args) throws IOException { // Fill in the body according to the following comments Scanner key boardFile = new Scanner(System.in); // Input file name String filename = getFileName(keyboardFile); //Open the file // Input number of students int numStudents = FileIOHelper.getNumberOfStudents(filename); Student students[] = new Student[numStudents]; // Input all student records and create Student array and // integer array for total scores int totalScore[] = new int[students.length]; for (int i = 0; i < students.length; i++){ for(int j = 1; j < 4; j++){ totalScore[i] = totalScore[i] + students[i].getScore(j); } } // Compute total scores and find students with lowest and // highest total score int maxScore = 0; int minScore = 0; for(int i = 0; i < students.length; i++){ if(totalScore[i] >= totalScore[maxScore]){ maxScore = i; } else if(totalScore[i] <= totalScore[minScore]){ minScore = i; } } // Compute average total score int allScores = 0; int average = 0; for (int i = 0; i < totalScore.length; i++){ allScores = allScores + totalScore[i]; } average = allScores / totalScore.length; // Output results outputResults(students, totalScore, maxScore, minScore, average); } // Given a Scanner in, this method prompts the user to enter // a file name, inputs it, and returns it. private static String getFileName(Scanner in) { // Fill in the body System.out.print("Enter the name of a file: "); String filename = in.next(); return filename; // Do not declare the Scanner variable in this method. // You must use the value this method receives in the // argument (in). } // Given the number of students records n to input, this // method creates an array of Student of the appropriate size, // reads n student records using the FileIOHelper, and stores // them in the array, and finally returns the Student array. private static Student[] getStudents(int n) { Student[] myStudents = new Student[n]; for(int i = 0; i <= n; i++){ myStudents[i] = FileIOHelper.getNextStudent(); } return myStudents; } // Given an array of Student records, an array with the total scores, // the indices in the arrays of the students with the highest and // lowest total scores, and the average total score for the class, // this method outputs a table of all the students appropriately // formatted, plus the total number of students, the average score // of the class, and the name and total score of the students with // the highest and lowest total score. private static void outputResults( Student[] students, int[] totalScores, int maxIndex, int minIndex, int average ) { // Fill in the body System.out.println("\nName \t\tScore1 \tScore2 \tScore3 \tTotal"); System.out.println("--------------------------------------------------------"); for(int i = 0; i < students.length; i++){ outputStudent(students[i], totalScores[i], average); System.out.println(); } System.out.println("--------------------------------------------------------"); outputNumberOfStudents(students.length); outputAverage(average); outputMaxStudent(students[maxIndex], totalScores[maxIndex]); outputMinStudent(students[minIndex], totalScores[minIndex]); System.out.println("--------------------------------------------------------"); } // Given a Student record, the total score for the student, // and the average total score for all the students, this method // outputs one line in the result table appropriately formatted. private static void outputStudent(Student s, int total, int avg) { System.out.print(s.getName() + "\t"); for(int i = 1; i < 4; i++){ System.out.print(s.getScore(i) + "\t"); } System.out.print(total + "\t"); if(total < avg){ System.out.print("-"); }else if(total > avg){ System.out.print("+"); }else{ System.out.print("="); } } // Given the number of students, this method outputs a message // stating what the total number of students in the class is. private static void outputNumberOfStudents(int n) { System.out.println("The total number of students in this class is: \t" + n); } // Given the average total score of all students, this method // outputs a message stating what the average total score of // the class is. private static void outputAverage(int average) { System.out.println("The average total score of the class is: \t" + average); } // Given the Student with highest total score and the student's // total score, this method outputs a message stating the name // of the student and the highest score. private static void outputMaxStudent( Student student, int score ) { System.out.println(student.getName() + " got the maximum total score of: \t" + score); } // Given the Student with lowest total score and the student's // total score, this method outputs a message stating the name // of the student and the lowest score. private static void outputMinStudent( Student student, int score ) { System.out.println(student.getName() + " got the minimum total score of: \t" + score); } } But now I get an error at the line totalScore[i] = totalScore[i] + students[i].getScore(j); Exception in thread "main" java.lang.NullPointerException at Lab6.main(Lab6.java:42)

  Read the article

• C++ function not found during compilation

  - by forthewinwin

  For a homework assignment: I'm supposed to create randomized alphabetial keys, print them to a file, and then hash each of them into a hash table using the function "goodHash", found in my below code. When I try to run the below code, it says my "goodHash" "identifier isn't found". What's wrong with my code? #include <iostream> #include <vector> #include <cstdlib> #include "math.h" #include <fstream> #include <time.h> using namespace std; // "makeKey" function to create an alphabetical key // based on 8 randomized numbers 0 - 25. string makeKey() { int k; string key = ""; for (k = 0; k < 8; k++) { int keyNumber = (rand() % 25); if (keyNumber == 0) key.append("A"); if (keyNumber == 1) key.append("B"); if (keyNumber == 2) key.append("C"); if (keyNumber == 3) key.append("D"); if (keyNumber == 4) key.append("E"); if (keyNumber == 5) key.append("F"); if (keyNumber == 6) key.append("G"); if (keyNumber == 7) key.append("H"); if (keyNumber == 8) key.append("I"); if (keyNumber == 9) key.append("J"); if (keyNumber == 10) key.append("K"); if (keyNumber == 11) key.append("L"); if (keyNumber == 12) key.append("M"); if (keyNumber == 13) key.append("N"); if (keyNumber == 14) key.append("O"); if (keyNumber == 15) key.append("P"); if (keyNumber == 16) key.append("Q"); if (keyNumber == 17) key.append("R"); if (keyNumber == 18) key.append("S"); if (keyNumber == 19) key.append("T"); if (keyNumber == 20) key.append("U"); if (keyNumber == 21) key.append("V"); if (keyNumber == 22) key.append("W"); if (keyNumber == 23) key.append("X"); if (keyNumber == 24) key.append("Y"); if (keyNumber == 25) key.append("Z"); } return key; } // "makeFile" function to produce the desired text file. // Note this only works as intended if you include the ".txt" extension, // and that a file of the same name doesn't already exist. void makeFile(string fileName, int n) { ofstream ourFile; ourFile.open(fileName); int k; // For use in below loop to compare with n. int l; // For use in the loop inside the below loop. string keyToPassTogoodHash = ""; for (k = 1; k <= n; k++) { for (l = 0; l < 8; l++) { // For-loop to write to the file ONE key ourFile << makeKey()[l]; keyToPassTogoodHash += (makeKey()[l]); } ourFile << " " << k << "\n";// Writes two spaces and the data value goodHash(keyToPassTogoodHash); // I think this has to do with the problem makeKey(); // Call again to make a new key. } } // Primary function to create our desired file! void mainFunction(string fileName, int n) { makeKey(); makeFile(fileName, n); } // Hash Table for Part 2 struct Node { int key; string value; Node* next; }; const int hashTableSize = 10; Node* hashTable[hashTableSize]; // "goodHash" function for Part 2 void goodHash(string key) { int x = 0; int y; int keyConvertedToNumber = 0; // For-loop to produce a numeric value based on the alphabetic key, // which is then hashed into hashTable using the hash function // declared below the loop (hashFunction). for (y = 0; y < 8; y++) { if (key[y] == 'A' || 'B' || 'C') x = 0; if (key[y] == 'D' || 'E' || 'F') x = 1; if (key[y] == 'G' || 'H' || 'I') x = 2; if (key[y] == 'J' || 'K' || 'L') x = 3; if (key[y] == 'M' || 'N' || 'O') x = 4; if (key[y] == 'P' || 'Q' || 'R') x = 5; if (key[y] == 'S' || 'T') x = 6; if (key[y] == 'U' || 'V') x = 7; if (key[y] == 'W' || 'X') x = 8; if (key[y] == 'Y' || 'Z') x = 9; keyConvertedToNumber = x + keyConvertedToNumber; } int hashFunction = keyConvertedToNumber % hashTableSize; Node *temp; temp = new Node; temp->value = key; temp->next = hashTable[hashFunction]; hashTable[hashFunction] = temp; } // First two lines are for Part 1, to call the functions key to Part 1. int main() { srand ( time(NULL) ); // To make sure our randomization works. mainFunction("sandwich.txt", 5); // To test program cin.get(); return 0; } I realize my code is cumbersome in some sections, but I'm a noob at C++ and don't know much to do it better. I'm guessing another way I could do it is to AFTER writing the alphabetical keys to the file, read them from the file and hash each key as I do that, but I wouldn't know how to go about coding that.

  Read the article

• Decimal Base Conversion using PYTHON

  - by Butinyane More

  PROBLEM DESCRIPTION Given a decimal number, and a new base to represent it in. If the base is larger than 10, use capital letters for the digits(that is, A is 10, B is 11 and so forth). The decimal number given, and the new base, will both be integer values, separated by a space. The base to convert to will always be smaller than or equal to 30. Please create a program that will convert a decimal number to any base in this instance. When evaluating the program the sample input must something like: 18 2 and the program must output the following: 10010 Please i beg of you to send me a solution to this problem as soon as possible.

  Read the article

• C SIGSEGV Handler & Mprotect

  - by pws5068

  I'm constructing a program which uses mprotect() to restrict a block of memory from accessing. When the memory is requested, a SIGSEGV is thrown which I listen for using a signal() call. Once the SIGSEGV has been detected, I need to somehow access the pointer to the memory that was requested (that threw the fault) and the size of the segment requested. Is this possible? void fifoSigHandler(){ // Needs to only remove protection from requested block of virtual memory mprotect(fifoVm,(size_t)fifoVm_size,PROT_WRITE); printf("Caught Seg Fault"); } void fifo_init(void* vm, int vm_size, int n_frames, int page_size) { fifoVm = vm; fifoVm_size = vm_size; fifoFrames = n_frames; fifoPageSize = page_size; mprotect(fifoVm,(size_t)fifoVm_size,PROT_NONE); signal(SIGSEGV, fifoSigHandler); } Additionally, is there a way to determine the level of mprotect() a block of memory is currently assigned (PROT_NONE,PROT_READ, etc..)?

  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

• Modifying and Manipulating a interactive bezier curve

  - by rachel

  This is a homework question and I'm having a lot of trouble with it - I've managed to do some of it but still cant finish it - can i Please get some help. Q1. Bezier Curves The following example allows you to interactively control a bezier curve by dragging the control points Cubic.java Replace the call to draw the cubic shape (big.draw(cubic)), by your own function to draw a bezier by the recursive split method. Finally, add the ability to create a longer Bezier curve by adding more control points to create a second curve. Cubic.java import java.awt.*; import javax.swing.*; import java.awt.event.*; import java.applet.Applet; import java.awt.geom.*; import java.awt.image.BufferedImage; public class Cubic extends JApplet{ static protected JLabel label; CubicPanel cubicPanel; public void init(){ //Initialize the layout. getContentPane().setLayout(new BorderLayout()); cubicPanel = new CubicPanel(); cubicPanel.setBackground(Color.white); getContentPane().add(cubicPanel); label = new JLabel("Drag the points to adjust the curve."); getContentPane().add("South", label); } public static void main(String s[]) { JFrame f = new JFrame("Cubic"); f.addWindowListener(new WindowAdapter() { public void windowClosing(WindowEvent e) {System.exit(0);} }); JApplet applet = new Cubic(); f.getContentPane().add(applet, BorderLayout.CENTER); applet.init(); f.setSize(new Dimension(350,250)); f.setVisible(true); } } class CubicPanel extends JPanel implements MouseListener, MouseMotionListener{ BufferedImage bi; Graphics2D big; int x, y; Rectangle area, startpt, endpt, onept, twopt, rect; CubicCurve2D.Double cubic = new CubicCurve2D.Double(); Point2D.Double start, end, one, two, point; boolean firstTime = true; boolean pressOut = false; public CubicPanel(){ setBackground(Color.white); addMouseMotionListener(this); addMouseListener(this); start = new Point2D.Double(); one = new Point2D.Double(); two = new Point2D.Double(); end = new Point2D.Double(); cubic.setCurve(start, one, two, end); startpt = new Rectangle(0, 0, 8, 8); endpt = new Rectangle(0, 0, 8, 8); onept = new Rectangle(0, 0, 8, 8); twopt = new Rectangle(0, 0, 8, 8); } public void mousePressed(MouseEvent e){ x = e.getX(); y = e.getY(); if(startpt.contains(x, y)){ rect = startpt; point = start; x = startpt.x - e.getX(); y = startpt.y - e.getY(); updateLocation(e); } else if(endpt.contains(x, y)){ rect = endpt; point = end; x = endpt.x - e.getX(); y = endpt.y - e.getY(); updateLocation(e); } else if(onept.contains(x, y)){ rect = onept; point = one; x = onept.x - e.getX(); y = onept.y - e.getY(); updateLocation(e); } else if(twopt.contains(x, y)){ rect = twopt; point = two; x = twopt.x - e.getX(); y = twopt.y - e.getY(); updateLocation(e); } else { pressOut = true; } } public void mouseDragged(MouseEvent e){ if(!pressOut) { updateLocation(e); } } public void mouseReleased(MouseEvent e){ if(startpt.contains(e.getX(), e.getY())){ rect = startpt; point = start; updateLocation(e); } else if(endpt.contains(e.getX(), e.getY())){ rect = endpt; point = end; updateLocation(e); } else if(onept.contains(e.getX(), e.getY())){ rect = onept; point = one; updateLocation(e); } else if(twopt.contains(e.getX(), e.getY())){ rect = twopt; point = two; updateLocation(e); } else { pressOut = false; } } public void mouseMoved(MouseEvent e){} public void mouseClicked(MouseEvent e){} public void mouseExited(MouseEvent e){} public void mouseEntered(MouseEvent e){} public void updateLocation(MouseEvent e){ rect.setLocation((x + e.getX())-4, (y + e.getY())-4); point.setLocation(x + e.getX(), y + e.getY()); checkPoint(); cubic.setCurve(start, one, two, end); repaint(); } public void paintComponent(Graphics g){ super.paintComponent(g); update(g); } public void update(Graphics g){ Graphics2D g2 = (Graphics2D)g; Dimension dim = getSize(); int w = dim.width; int h = dim.height; if(firstTime){ // Create the offsecren graphics to render to bi = (BufferedImage)createImage(w, h); big = bi.createGraphics(); // Get some initial positions for the control points start.setLocation(w/2-50, h/2); end.setLocation(w/2+50, h/2); one.setLocation((int)(start.x)+25, (int)(start.y)-25); two.setLocation((int)(end.x)-25, (int)(end.y)+25); // Set the initial positions of the squares that are // drawn at the control points startpt.setLocation((int)((start.x)-4), (int)((start.y)-4)); endpt.setLocation((int)((end.x)-4), (int)((end.y)-4)); onept.setLocation((int)((one.x)-4), (int)((one.y)-4)); twopt.setLocation((int)((two.x)-4), (int)((two.y)-4)); // Initialise the CubicCurve2D cubic.setCurve(start, one, two, end); // Set some defaults for Java2D big.setColor(Color.black); big.setStroke(new BasicStroke(5.0f)); big.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON); area = new Rectangle(dim); firstTime = false; } // Clears the rectangle that was previously drawn. big.setColor(Color.white); big.clearRect(0, 0, area.width, area.height); // Set the colour for the bezier big.setPaint(Color.black); // Replace the following line by your own function to // draw the bezier specified by start, one, two, end big.draw(cubic); // Draw the control points big.setPaint(Color.red); big.fill(startpt); big.setPaint(Color.magenta); big.fill(endpt); big.setPaint(Color.blue); big.fill(onept); big.setPaint(new Color(0, 200, 0)); big.fill(twopt); // Draws the buffered image to the screen. g2.drawImage(bi, 0, 0, this); } /* Checks if the rectangle is contained within the applet * window. If the rectangle is not contained withing the * applet window, it is redrawn so that it is adjacent to the * edge of the window and just inside the window. */ void checkPoint(){ if (area == null) { return; } if((area.contains(rect)) && (area.contains(point))){ return; } int new_x = rect.x; int new_y = rect.y; double new_px = point.x; double new_py = point.y; if((rect.x+rect.width)>area.getWidth()){ new_x = (int)area.getWidth()-(rect.width-1); } if(point.x > area.getWidth()){ new_px = (int)area.getWidth()-1; } if(rect.x < 0){ new_x = -1; } if(point.x < 0){ new_px = -1; } if((rect.y+rect.width)>area.getHeight()){ new_y = (int)area.getHeight()-(rect.height-1); } if(point.y > area.getHeight()){ new_py = (int)area.getHeight()-1; } if(rect.y < 0){ new_y = -1; } if(point.y < 0){ new_py = -1; } rect.setLocation(new_x, new_y); point.setLocation(new_px, new_py); } }

  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

• is my function correct?

  - by sbsp

  This is part of an assignment so please dont post solutions, just point me in the right direction if possible? I am passing a pointer to a char array to my method, as well as a value for the actual height of the char array. I am looping through to see if all values are 0, if they are then return 0, esle return one The method is used as a test to see if i should free memory or not and set the pointer to null if it is full of 0's. The issue i am having is that the programme should have "some unfree" memory at the end, so i have no idea whether or not its doing it correctly - and gdb i struggle with immensley. Thanks for reading int shouldBeNull(char *charPointer, int sizeOfCharArray) { int isIn = 0; int i = 0; while(i < sizeOfCharArray){ if(*charPointer != '0'){ isIn = 1; break; } i++; charPointer++; } return isIn; }

  Read the article

• dynamic programming: speeding up this function

  - by aristotaly

  i have this program //h is our N static int g=0; int fun(int h){ if(h<=0){ g++; return g; } return g+fun(h-1)+fun(h-4); } is it possible to speed it up using dynamic programming i fugured out this function runs in O(2^n) it means that i suppose to reduce this time but the trouble is that i don get the idea of dinamic programming even a leading hint or a useful link to a resource will do it is a work assingment i do not ask for the solution :) just asking for the right direction

  Read the article

• Express any number as the sum of 4 prime numbers [Doubts]

  - by WarDoGG

  I was give a problem to express any number as sum of 4 prime numbers. Conditions: Not allowed to use any kind of database. Maximum execution time : 3 seconds Numbers only till 100,000 If the splitting is NOT possible, then return -1 What i did : using the sieve of eratosthenes, i calculated all prime numbers till the specified number. looked up a concept called goldbach conjecture which expresses an even number as the summation of 2 primes. However, i am stuck beyond that. Can anyone help me on this one as to what approach u might take ? The sieve of eratosthenes is taking 2 seconds to count primes till 100,000 :(((

  Read the article

• what is value of x for load and store

  - by Kevinniceguy

  This is some challenge On a single processor system, in which load and store are assumed to be atomic, what are all the possible values for x after both threads have completed in the following execution, assuming that x is initialised to O? Hint: you need to consider how this code might be compiled into machine language. for (int i = 0; i < 5; i++) : x = x + 1; for (int j = 0; j < 5; j++) : x = x + 1;

  Read the article

• I am designing a bus timetable using SQL. Each bus route has multiple stops, do I need a different t

  - by Henry

  I am trying to come up with the most efficient database as possible. My bus routes all have about 10 stops. The bus starts at number one until it reaches the 10th stop, then it comes back again. This cycle happens 3 times a day. I am really stuck as to how I can efficiently generate the times for the buses and where I should store the stops. If I put all the stops in one field and the times in another, the database won't be very dynamic. If I store all the stops one by one in a column and then the times in another column, there will be a lot of repeating happening further down as one stop has multiple times. Maybe I am missing something, I've only just started learning SQL and this is a task we have been set. Thanks in advance.

  Read the article

• Optimally reducing maximum flow

  - by ArIck

  Given a parameter k, I'm trying to delete k edges from a directed graph such that the maximum flow is reduced by as much as possible. The graph has a source s and a sink t, and the capacity of each edge is one. The graph may or may not contain cycles. My proposed solution would be to first perform a topological sorting on the graph, using an algorithm that "forgives" cycles -- perhaps by ignoring edges that lead us back to the source. Then (assuming k = 1): i = 0 for each vertex u order by topological(u) for each edge (u, v) order by topological(v) descending if topological(v) > topological(u) then delete (u, v) if ++i = k then return else // edge doesn't contribute to max flow, ignore Would this work, or am I totally off-track here?

  Read the article

• Codesample with bufferoverflow (gets method). Why does it not behave as expected?

  - by citronas

  This an extract from an c program that should demonstrate a bufferoverflow. void foo() { char arr[8]; printf(" enter bla bla bla"); gets(arr); printf(" you entered %s\n", arr); } The question was "How many input chars can a user maximal enter without a creating a buffer overflow" My initial answer was 8, because the char-array is 8 bytes long. Although I was pretty certain my answer was correct, I tried a higher amount of chars, and found that the limit of chars that I can enter, before I get a segmentation fault is 11. (Im running this on A VirtualBox Ubuntu) So my question is: Why is it possible to enter 11 chars into that 8 byte array?

  Read the article

• Finding the heaviest of N objects using M scales

  - by cpprulez

  We have N objects and M scales. It's up to us what the objects are, and we need to position the objects on the scales so that it is undoubtful which is the heaviest object. For example, if we have 3 objects: "a", "b", "c" and 2 scales, one possible solution is "a" "b", "b" = "c" (here "a" is the heaviest). I need an algorithm which generates such solutions given N and M. Also let's assume that "a" is always the heaviest object. I've lost a few hours figuring out how to do it, but no matter what I figure out, there's always cases which I miss. For example, another solution is: "a" + "c" = 2 * "b", "a" "c".

  Read the article

• Solve Physics exercise by brute force approach..

  - by Nils

  Being unable to reproduce a given result. (either because it's wrong or because I was doing something wrong) I was asking myself if it would be easy to just write a small program which takes all the constants and given number and permutes it with a possible operators (* / - + exp(..)) etc) until the result is found. Permutations of n distinct objects with repetition allowed is n^r. At least as long as r is small I think you should be able to do this. I wonder if anybody did something similar here..

  Read the article

• Why System.String's beahaves like a value type?

  - by Sorush Rabiee

  I want to write a 'Date' class that behaves like a Value Type. for example, Instead of writing a Clone method for setting properties safely, make the Date class to pass by value: public Date Birthday { get { return this.birthday; } set { this.birthday = value.Clone(); } //I want to write this.birthday = value; //without changing external value when this.Birthday changes } I know this is possible because System.String is a class and behaves like a value. for example: String s1 = "Hello"; String s2 = "Hi"; s1 = s2; s2="Hello"; Console.WriteLine(s1); //Prints 'Hi' First I thought writers of this class override '=' operator, but now I know that the '=' operator can not be overridden. so how they write String class?

  Read the article

• Java Programming Help

  - by user215049

  hi, im a bit confused on this and i dont know how to solve this question that i have been asked, id be grateful if you could assist me on this question, maybe try to tell me what needs to be done, and how. the question is: Write a method called countChars which takes an InputStream as a parameter, reads the stream and returns the number of characters it contains as an int. Any IOExceptions which might occur in the method should be passed back to the method's caller. Note that the method header should show that it is possible that an exception might occur. i attempted this question with the following code: public class countChars { public int countChars(int InputStream) { return InputStream; } } and i get an error message saying : Main.java:26: cannot find symbol symbol : method countChars(java.io.InputStream) location: class Main s1 = "" + countChars(f1); ^ 1 error

  Read the article

• user input question

  - by jdbeverly87

  My program checks to test if a word or phrase is a palindrome (reads the same both backward and forward, ex "racecar"). The issue I'm having is after someone enters in "racecar" getting it to actually test. In the below code, I marked where if I type in "racecar" and run, Java returns the correct answer so I know I'm right there. But what am I missing as far as entering it into the console. I think my code is ok, but maybe I have something missing or in the wrong spot? Not really looking for a new answer unless I'm missing something, but if possible perhaps a pro at this moving my code to the correct area bc I'm stuck! `import java.util.*; public class Palindrome { public static void main(String[] args) { String myInput; Scanner in = new Scanner(System.in); System.out.println("Enter a word or phrase: "); **//this asks user for input but doesn't check for whether or not it is a palindrome** myInput = in.nextLine(); in.close(); System.out.println("You entered: " + myInput); } { String s="racecar"; **//I can type a word here and it works but I need** int i; **//I need it to work where I ask for the input** int n=s.length(); String str=""; for(i=n-1;i>=0;i--) str=str+s.charAt(i); if(str.equals(s)) System.out.println(s+ " is a palindrome"); else System.out.println(s+ " is not a palindrome"); } }` I'm new at programming so I'm hoping what I got is ok. I know the palindrome test works I'm just needing helping having it test thru where I'm entering it into the console. Thanks

  Read the article

• Proving f (f bool) = bool

  - by Marcus Whybrow

  How can I in coq, prove that a function f that accepts a bool true|false and returns a bool true|false (shown below), when applied twice to a single bool true|false would always return that same value true|false: (f:bool -> bool) For example the function f can only do 4 things, lets call the input of the function b: Always return true Always return false Return b (i.e. returns true if b is true vice versa) Return not b (i.e. returns false if b is true and vice vera) So if the function always returns true: f (f bool) = f true = true and if the function always return false we would get: f (f bool) = f false = false For the other cases lets assum the function returns not b f (f true) = f false = true f (f false) = f true = false In both possible input cases, we we always end up with with the original input. The same holds if we assume the function returns b. So how would you prove this in coq? Goal forall (f:bool -> bool) (b:bool), f (f b) = f b.

  Read the article

• How to obtain dependency metrics from Java source code?

  - by Bram Schoenmakers

  For an assignment we have to extract some software metrics from the Hibernate project. We have to extract the afferent coupling and efferent coupling metrics (dependency fan-in, fan-out) from each revision of each package in Hibernate. Some tools were provided which are able to extract these metrics, such as ckjm and JDepend. Other tools I have checked were Sonar, javancss and AOP. There is also the Metrics Eclipse plugin which I didn't get to work either. What these tools have in common, as far as I can see, is that they all operate on bytecode (*.class files). This is a problem, because I have to build every revision from source in order to run, say, JDepend on it. Older revisions won't build because my development stack is too recent. What I would like to do is to do this kind of analysis on source files so that I don't have to build each revision. Is this possible? Or is there a good reason why all these tools only operate on bytecode?

  Read the article

• Which knowledge base/rule-based inference engine to choose for real time Runway incursion prevention

  - by Piligrim

  Hello, we are designing a project that would listen to dialog between airport controllers and pilots to prevent runway incursions (eg. one airplane is taking off while other is crossing the runway). Our professor wants us to use Jena for knowledge base (or anything else but it should be some sort of rule-based engine). Inference is not the main thing in Jena and there's not much documentation and examples of this. So we need an engine that would get messages from pilots as input and output possible risks of incursion or any other error in message protocol. It should be easy to write rules, and should be easy to provide engine with real time data. I image it something like this: A pilot sends a message that he lands on some runway, the system remembers that the runway is busy and no one should cross it If someone is given an instruction to cross this runway, the engine should fire a rule that something is wrong When the pilot sends a message that he left the runway and goes to the gate, the system clears the runway and lets other planes to use it. So is Jena, or prolog or any other rules engine suitable for this? I mean it is suitable, but do we really need to use it? I asked the prof. if we could just keep state of the runway and use some simple checks based on messages we receive and he said that it is not scalable and we need the knowledge base. Can someone give me any advise on which approach to use for this system? If you recommend k.b., then which one should we use? The project is written in java. Thank you.

  Read the article

• Array: mathematical sequence

  - by VaioIsBorn

  An array of integers A[i] (i 1) is defined in the following way: an element A[k] ( k 1) is the smallest number greater than A[k-1] such that the sum of its digits is equal to the sum of the digits of the number 4* A[k-1] . You need to write a program that calculates the N th number in this array based on the given first element A[1] . INPUT: In one line of standard input there are two numbers seperated with a single space: A[1] (1 <= A[1] <= 100) and N (1 <= N <= 10000). OUTPUT: The standard output should only contain a single integer A[N] , the Nth number of the defined sequence. Input: 7 4 Output: 79 Explanation: Elements of the array are as follows: 7, 19, 49, 79... and the 4th element is solution. I tried solving this by coding a separate function that for a given number A[k] calculates the sum of it's digits and finds the smallest number greater than A[k-1] as it says in the problem, but with no success. The first testing failed because of a memory limit, the second testing failed because of a time limit, and now i don't have any possible idea how to solve this. One friend suggested recursion, but i don't know how to set that. Anyone who can help me in any way please write, also suggest some ideas about using recursion/DP for solving this problem. Thanks.

  Read the article

< Previous Page | 7 8 9 10 11 12 13 14 15 16 17 18  | Next Page >