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  • Explain JAVA code

    - by MIW
    I need some help to explain the meaning from line 5 to line 9. Thanks String words = "Rain Rain go away"; String mutation1, mutation2, mutation3, mutation4; mutation1 = words.toUpperCase(); System.out.println ("** " + mutation1 + " Nursery Rhyme **"); mutation1 = words.concat ("\nCome again another day"); mutation2 = "Johnny Johnny wants to play"; mutation3 = mutation2.replace (mutation2.charAt(5), 'i'); mutation4 = mutation3.substring (7, 27); System.out.print ("\'" + mutation1 + "\n" + mutation4 + "\'\n"); 10.System.out.println ("Title length: " + words.length());

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  • Simplest way to automatically alter "const" value in Java during compile time

    - by Michael Mao
    Hi all: This is a question corresponds to my uni assignment so I am very sorry to say I cannot adopt any one of the following best practices in a short time -- you know -- assignment is due tomorrow :( link to Best way to alter const in Java on StackOverflow Basically the only task (I hope so) left for me is the performance tuning. I've got a bunch of predefined "const" values in my single-class agent source code like this: //static final values private static final long FT_THRESHOLD = 400; private static final long FT_THRESHOLD_MARGIN = 50; private static final long FT_SMOOTH_PRICE_IDICATOR = 20; private static final long FT_BUY_PRICE_MODIFIER = 0; private static final long FT_LAST_ROUNDS_STARTTIME = 90; private static final long FT_AMOUNT_ADJUST_MODIFIER = 5; private static final long FT_HISTORY_PIRCES_LENGTH = 10; private static final long FT_TRACK_DURATION = 5; private static final int MAX_BED_BID_NUM_PER_AUC = 12; I can definitely alter the values manually and then compile the code to give it another go around. But the execution time for a thorough "statistic analysis" usually requires over 2000 times of execution, which will typically lasts more than half an hour on my own laptop... So I hope there is a way to alter values using other ways than dig into the source code to change the "const" values there, so I can automatically distributed compiled code to other people's PC and let them run the statistic analysis instead. One other reason for a automatically value adjustment is that I can try using my own agent to defeat itself by choosing different "const" values. Although my values are derived from previous history and statistical results, they are far from the most optimized values. I hope there is a easy way so I can quickly adopt that so to have a good sleep tonight while the computer does everything for me... :) Any hints on this sort of stuff? Any suggestion is welcomed and much appreciated.

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  • Printing a sideways triangle in java

    - by Will
    I'm trying to print a sideways triangle in java. If the user enters 5, the output should be: * *** ***** *** * If the user enters 6, the output should be: * *** ***** ***** *** * I've gotten it to work for the case when the user enters 5, 3, or 1 but my code seems to work for those three cases only. I was wondering if anyone could help me get my code working for more cases. Here it is: public void printArrow( int n ) { int asterisks = 1; for ( int i = 0; i <= n/2; i++ ) { for ( int j = i; j < asterisks; j++ ) { System.out.print( "*" ); } asterisks += 3; System.out.println(); } asterisks = asterisks / 2 - 2; for ( int i = 0; i < n/2; i++ ) { for ( int k = i; k < asterisks; k++ ) { System.out.print( "*" ); } if ( i == 1 ) { System.out.print( "*" ); } asterisks -= 2; System.out.println(); } }

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  • Limit Connections with semaphores

    - by Robert
    I'm trying to limit the number of connections my server will accept using semaphores, but when running, my code doesn't seem to make this restriction - am I using the semaphore correctly? eg. I have hardcoded the number of permit as 2, but I can connect an unlimited number of clients... public class EServer implements Runnable { private ServerSocket serverSocket; private int numberofConnections = 0; private Semaphore sem = new Semaphore(2); private volatile boolean keepProcessing = true; public EServer(int port) throws IOException { serverSocket = new ServerSocket(port); } @Override public void run() { while (keepProcessing) { try { sem.acquire(); Socket socket = serverSocket.accept(); process(socket, getNextConnectionNumber()); } catch (Exception e) { } finally { sem.release(); } } closeIgnoringException(serverSocket); } private synchronized int getNextConnectionNumber() { return ++numberofConnections; } // processing related methods }

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  • 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.

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  • Spatial domain to frequency domain

    - by John Elway
    I know about Fourier Transforms, but I don't know how to apply it here, and I think that is over the top. I gave my ideas of the responses, but I really don't know what I'm looking for... Supposed that you form a low-pass spatial filter h(x,y) that averages all the eight immediate neighbors of a pixel (x,y) but excludes itself. a. Find the equivalent frequency domain filter H(u,v): My answer is to (a): 1/8*H(u-1, v-1) + 1/8*H(u-1, v) + 1/8*H(u-1, v+1) + 1/8*H(u, v-1) + 0 + 1/8*H(u, v+1) + 1/8*H(u+1, v-1) + 1/8*H(u+1, v) + 1/8*H(u-1, v-1) is this the frequency domain? b. Show that your result is again a low-pass filter. does this have to do with the coefficients being positive?

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  • total number of magic square from 9 numbers

    - by Peeyush
    9 numbers need to be arranged in a magic number square. A magic number square is a square of numbers that is arranged such that every row and column has the same sum.(condition for diagonal has been relaxed) For example: 1 2 3 3 2 1 2 2 2 How do we calculate total number of distinct magic square from 9 numbers. Two magic number squares are distinct if they differ in value at one or more positions. For example, there is only one magic number square that can be made of 9 instances of the same number. e.g. for these 9 numbers { 4, 4, 4, 4, 4, 4, 4, 4, 4 }, answer should be 1. Also the complexity should be optimal. Do we need to iterate through all the permutations , discarding if a[0]+a[1]+a[2] %3!=0 such combinations ? moreover how do we remove duplicate magic square?

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  • Prolog - generate correct bracketing

    - by Henrik Bak
    I'd like to get some help in the following exam problem, i have no idea how to do this: Input: a list of numbers, eg.: [1,2,3,4] Output: every possible correct bracketing. Eg.: (in case of input [1,2,3,4]): ((1 2) (3 4)) ((1 (2 3)) 4) (1 ((2 3) 4)) (1 (2 (3 4))) (((1 2) 3) 4) Bracketing here is like a method with two arguments, for example multiplication - then the output is the possible multiplication orders. Please help, i'm stuck with this one. Any help is appreciated, thanks!

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  • Reliable UDP

    - by suresh
    How can I develop a Linux kernel module in order to make UDP reliable? This is my college assignment and I don't how to proceed. how to do change the default UDP behaviour in linux kernel by loading a new kernel module? and how to program such kernel module?

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  • 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?

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  • 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?

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  • anagram!! problem with this code

    - by danielDhobbs
    hello people!! i have a problem with this code can you fix it for me? int anagram(char* word, int cur, int len){ int i, b = cur+1; char temp=0; char arrA[len]; printf("//%d**%d//", b, cur); for (i = 0 ; i < len ; i++) { arrA[i] = word[i]; } for (i = cur ; i < len ; i++) { if (b < len) { printf("%s\n", arrA); temp = arrA[cur]; arrA[cur] = arrA[b]; arrA[b] = temp; b++; } else if (b == len) anagram(arrA, b, len); } return 0; }

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  • Convert VB.NET code to C#

    - by Sorush Rabiee
    Hi people, I have three projects written with VB.NET (2005) and have to convert them to C# code. (I know that i don't need to convert codes of .net languages at all). I have no time to rewrite them, need a tool or script to convert. Note: they are console applications.

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  • Multiple actions upon a case statement in Haskell

    - by Schroedinger
    One last question for the evening, I'm building the main input function of my Haskell program and I have to check for the args that are brought in so I use args <- getArgs case length args of 0 -> putStrLn "No Arguments, exiting" otherwise -> { other methods here} Is there an intelligent way of setting up other methods, or is it in my best interest to write a function that the other case is thrown to within the main? Or is there an even better solution to the issue of cases. I've just got to take in one name.

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  • A shortest path problem with superheroes and intergalactic journeys

    - by Dman
    You are a super-hero in the year 2222 and you are faced with this great challenge: starting from your home planet Ilop you must try to reach Acinhet or else your planet will be destroyed by evil green little monsters. To do this you are given a map of the universe: there are N planets and M inter-planetary connections ( bidirectional ) that bind these planets. Each connection requires a certain time and a certain amount of fuel in order for you to cover the connection from one planet to another. The total time spent going from one planet to another is obtained by multiplying the time past to cover each connection between all the planets you go through. There are some "key planets", that allow you to refuel if you arrive on those certain "key planets". A "key planet" is the planet with the property that if it disappears the road between at least two planets would be lost.(In the example posted below with the input/output files such a "key planet" is 2 because without it the road to 7 would be lost) When you start your mission you are given the possibility of choosing between K ships each with its own maximum fuel capacity. The goal is to find the SHORTEST TIME CONSUMING path but also choose the ship with the minimum fuel capacity that can cover that shortest path(this means that if more ships can cover the shortest path you choose the one with the minimum fuel capacity). Because the minimum time can be a rather large number (over long long int) you are asked to provide only the last 6 digits of the number. For a better understanding of the task, here is an example of input/output files: INPUT: mission.in 7 8 6 1 4 6 5 9 8 7 10 1 2 7 8 1 4 14 9 1 5 3 1 2 3 1 2 2 7 7 1 3 4 2 2 4 6 4 1 5 6 3 7 On the first line (in order): N M K On the second line :the number for the starting planet and the finishing planet On the third line :K numbers that represent the capacities of the ships you can choose from Then you have M lines, all of them have the same structure: Xi Yi Ti Fi-which means that there is a connection between Xi and Yi and you can cover the distance from Xi to Yi in Ti time and with a Fi fuel consumption. OUTPUT:mission.out 000014 8 1 2 3 4 On the first line:the minimum time and fuel consumption; On the second line :the path Restrictions: 2 = N = 1 000 1 = M = 30 000 1 = K = 10 000 Any suggestions or ideas of how this problem might be solved would be most welcomed.

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  • Which method should I use ?

    - by Ivan
    I want to do this exercise but I don't know exactly which method should I use for an exercise like this and what data will I use to test the algorithm. The driving distance between Perth and Adelaide is 1996 miles. On the average, the fuel consumption of a 2.0 litre 4 cylinder car is 8 litres per 100 kilometres. The fuel tank capacity of such a car is 60 litres. Design and implement a JAVA program that prompts for the fuel consumption and fuel tank capacity of the aforementioned car. The program then displays the minimum number of times the car’s fuel tank has to be filled up to drive from Perth to Adelaide. Note that 62 miles is equal to 100 kilometres. What data will you use to test that your algorithm works correctly?

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  • Returning new object, overwrite the existing one in Java

    - by lupin
    Note: This is an assignment. Hi, Ok I have this method that will create a supposedly union of 2 sets. i mport java.io.*; class Set { public int numberOfElements; public String[] setElements; public int maxNumberOfElements; // constructor for our Set class public Set(int numberOfE, int setE, int maxNumberOfE) { this.numberOfElements = numberOfE; this.setElements = new String[setE]; this.maxNumberOfElements = maxNumberOfE; } // Helper method to shorten/remove element of array since we're using basic array instead of ArrayList or HashSet from collection interface :( static String[] removeAt(int k, String[] arr) { final int L = arr.length; String[] ret = new String[L - 1]; System.arraycopy(arr, 0, ret, 0, k); System.arraycopy(arr, k + 1, ret, k, L - k - 1); return ret; } int findElement(String element) { int retval = 0; for ( int i = 0; i < setElements.length; i++) { if ( setElements[i] != null && setElements[i].equals(element) ) { return retval = i; } retval = -1; } return retval; } void add(String newValue) { int elem = findElement(newValue); if( numberOfElements < maxNumberOfElements && elem == -1 ) { setElements[numberOfElements] = newValue; numberOfElements++; } } int getLength() { if ( setElements != null ) { return setElements.length; } else { return 0; } } String[] emptySet() { setElements = new String[0]; return setElements; } Boolean isFull() { Boolean True = new Boolean(true); Boolean False = new Boolean(false); if ( setElements.length == maxNumberOfElements ){ return True; } else { return False; } } Boolean isEmpty() { Boolean True = new Boolean(true); Boolean False = new Boolean(false); if ( setElements.length == 0 ) { return True; } else { return False; } } void remove(String newValue) { for ( int i = 0; i < setElements.length; i++) { if ( setElements[i] != null && setElements[i].equals(newValue) ) { setElements = removeAt(i,setElements); } } } int isAMember(String element) { int retval = -1; for ( int i = 0; i < setElements.length; i++ ) { if (setElements[i] != null && setElements[i].equals(element)) { return retval = i; } } return retval; } void printSet() { for ( int i = 0; i < setElements.length; i++) { if (setElements[i] != null) { System.out.println("Member elements on index: "+ i +" " + setElements[i]); } } } String[] getMember() { String[] tempArray = new String[setElements.length]; for ( int i = 0; i < setElements.length; i++) { if(setElements[i] != null) { tempArray[i] = setElements[i]; } } return tempArray; } Set union(Set x, Set y) { String[] newXtemparray = new String[x.getLength()]; String[] newYtemparray = new String[y.getLength()]; int len = newYtemparray.length + newXtemparray.length; Set temp = new Set(0,len,len); newXtemparray = x.getMember(); newYtemparray = x.getMember(); for(int i = 0; i < newYtemparray.length; i++) { temp.add(newYtemparray[i]); } for(int j = 0; j < newXtemparray.length; j++) { temp.add(newXtemparray[j]); } return temp; } Set difference(Set x, Set y) { String[] newXtemparray = new String[x.getLength()]; String[] newYtemparray = new String[y.getLength()]; int len = newYtemparray.length + newXtemparray.length; Set temp = new Set(0,len,len); newXtemparray = x.getMember(); newYtemparray = x.getMember(); for(int i = 0; i < newXtemparray.length; i++) { temp.add(newYtemparray[i]); } for(int j = 0; j < newYtemparray.length; j++) { int retval = temp.findElement(newYtemparray[j]); if( retval != -1 ) { temp.remove(newYtemparray[j]); } } return temp; } } // This is the SetDemo class that will make use of our Set class class SetDemo { public static void main(String[] args) { //get input from keyboard BufferedReader keyboard; InputStreamReader reader; String temp = ""; reader = new InputStreamReader(System.in); keyboard = new BufferedReader(reader); try { System.out.println("Enter string element to be added" ); temp = keyboard.readLine( ); System.out.println("You entered " + temp ); } catch (IOException IOerr) { System.out.println("There was an error during input"); } /* ************************************************************************** * Test cases for our new created Set class. * ************************************************************************** */ Set setA = new Set(0,10,10); setA.add(temp); setA.add("b"); setA.add("b"); setA.add("hello"); setA.add("world"); setA.add("six"); setA.add("seven"); setA.add("b"); int size = setA.getLength(); System.out.println("Set size is: " + size ); Boolean isempty = setA.isEmpty(); System.out.println("Set is empty? " + isempty ); int ismember = setA.isAMember("sixb"); System.out.println("Element sixb is member of setA? " + ismember ); Boolean output = setA.isFull(); System.out.println("Set is full? " + output ); //setA.printSet(); int index = setA.findElement("world"); System.out.println("Element b located on index: " + index ); setA.remove("b"); //setA.emptySet(); int resize = setA.getLength(); System.out.println("Set size is: " + resize ); //setA.printSet(); Set setB = new Set(0,10,10); setB.add("b"); setB.add("z"); setB.add("x"); setB.add("y"); Set setC = setA.union(setB,setA); System.out.println("Elements of setA"); setA.printSet(); System.out.println("Union of setA and setB"); setC.printSet(); } } The union method works a sense that somehow I can call another method on it but it doesn't do the job, i supposedly would create and union of all elements of setA and setB but it only return element of setB. Sample output follows: java SetDemo Enter string element to be added hello You entered hello Set size is: 10 Set is empty? false Element sixb is member of setA? -1 Set is full? true Element b located on index: 2 Set size is: 9 Elements of setA Member elements on index: 0 hello Member elements on index: 1 world Member elements on index: 2 six Member elements on index: 3 seven Union of setA and setB Member elements on index: 0 b Member elements on index: 1 z Member elements on index: 2 x Member elements on index: 3 y thanks, lupin

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  • A specific string format with a number and character together represeting a certain item

    - by sil3nt
    Hello there, I have a string which looks like this "a 3e,6s,1d,3g,22r,7c 3g,5r,9c 19.3", how do I go through it and extract the integers and assign them to its corresponding letter variable?. (i have integer variables d,r,e,g,s and c). The first letter in the string represents a function, "3e,6s,1d,3g,22r,7c" and "3g,5r,9c" are two separate containers . And the last decimal value represents a number which needs to be broken down into those variable numbers. my problem is extracting those integers with the letters after it and assigning them into there corresponding letter. and any number with a negative sign or a space in between the number and the letter is invalid. How on earth do i do this?

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