Hi I'm currently working on getting my Quick Sort program working but cannot figure out where I'm going wrong, i have spent hours trying to find out why it's not working but no luck,when I run my code nothing happens. I have four other sorting algorithms working in a similar fashion which is what's confusing me the most.
Below is my code for the Quick Sort program
import java.io.IOException;
public class QuickSort {
public static int[] compute(int[] array, int lower, int higher )throws IOException{
if(lower<higher){
int pivot=split(array,lower,higher);
if(pivot>1)
compute(array, lower, pivot-1);
if(pivot+1<higher)
compute(array, pivot+1, higher);
}
return array;
}
public static int split(int[] array, int lower, int higher){
while(true){
int pivot=array[lower];
while(array[lower]<pivot)
lower++;
while(array[higher]>pivot)
higher--;
if(lower<higher){
int temp=array[higher];
array[higher]=array[lower];
array[lower]=temp;
}
else{
return higher;
}
}
}
}
Here is my Test class that's running the code:
import java.io.IOException;
import java.util.Scanner;
public class Test extends ReadIn{
static BubbleSort bubble=new BubbleSort();
public static void main(String[] args) throws IOException{
System.out.println("Enter 1 for BubbleSort\nEnter 2 for Insertion Sort\nEnter 3 for Selection Sort\nEnter 4 for Merge Sort\nEnter 5 for QuickSort\nPlease input sorting algorithm to use for sorting:");
Scanner input=new Scanner(System.in);
int number=input.nextInt();
final long startTime = System.currentTimeMillis();
int[] Array=read();
if(number==1){
Array=BubbleSort.compute(Array);
for(int i=0;i<BubbleSort.compute(Array).length;i++){
System.out.println(Array[i]);
}
}
if(number==2){
Array=InsertionSort.compute(Array);
for(int i=0;i<InsertionSort.compute(Array).length;i++){
System.out.println(Array[i]);
}
}
if(number==3){
Array=SelectionSort.compute(Array);
for(int i=0;i<SelectionSort.compute(Array).length;i++){
System.out.println(Array[i]);
}
}
if(number==4){
Array=MergeSort.compute(Array);
for(int i=0;i<MergeSort.compute(Array).length;i++){
System.out.println(Array[i]);
}
}
if(number==5){
Array=QuickSort.compute(Array,0,Array.length-1);
for(int i=0;i<QuickSort.compute(Array,0,Array.length-1).length;i++){
System.out.print(Array[i]);
}
}
final long endTime = System.currentTimeMillis();
System.out.println("Total execution time: " + (endTime - startTime) );
}
}
When I press 5 nothing happens, the rest work perfectly.
I cant figure out whatsoever what the problem is so any help or input would be appreciated.
Your Quicksort code loops forever if there's duplicate numbers in the input, since the numbers can just keep swapping with each other. As mentioned in the comment, you should manually try out your code with a sample array, or check the code running with a debugger. See the example array below.
You might want to switch the while(true) loop to a recursive call to make the code a little clearer. Also, you can practice going through the different steps of Quicksort on my Quicksort online tutorial.
Say the array = {3,1,2,3} and the pivot is 3. Nothing will change and the code will loop forever.
while(true){
int pivot=array[lower];
while(array[lower]<pivot)
lower++;
while(array[higher]>pivot)
higher--;
if(lower<higher){ //this will just swap the 3's with each other.
int temp=array[higher];
array[higher]=array[lower];
array[lower]=temp;
}
else{
return higher;
}
}
While it looks like your program is doing nothing, it probably just loops infinitely inside this loop:
while(true){
int pivot=array[lower];
while(array[lower]<pivot)
lower++;
while(array[higher]>pivot)
higher--;
if(lower<higher){
int temp=array[higher];
array[higher]=array[lower];
array[lower]=temp;
// I recommend adding this line for you to see what's going on
System.out.println("lower="+lower+", higher="+higher+", pivot="+pivot);
}
else{
return higher;
}
}
for example, if you have equal values in your Array, say value 65 is there twice, that loop will run forever...
int[] Array={1,41,2,90,32,65,12,43,78,65,46,67};
Also, in your loop above I added and extra printout of higher, lower, and pivot values - it will help you see what is happening inside that devious loop.
In general writing a while(true) loop is not a good practice, because it's so easy to hang your system on it.
Related
I am learning Java programming and made a program which uses Binary Search to find a symbol in an array of char. However, I have a problem when I try to search for the symbol which is not in an array, my program becomes an endless cycle. I have no idea how to make an error sign if there is no such symbol in an array. Here is the code of my program
import java.lang.reflect.Array;
import java.util.Arrays;
public class Main {
public static void main(String[] args){
char[]arr = {'a','d','f','l','o','z'};
find(arr,'m');
}
public static void find(char[]arr,char ch){
int last = arr.length-1;
int mid=last;
while (arr[mid] != ch){
if (arr[mid]<ch){
mid = (last+mid)/2;
}
else{
last=mid;
mid=last/2;
}
}
System.out.print(mid);
}
}
Thank you in advance.
You have no case to break out of the while loop and the changed values assigned to the variables during the 2 if cases are incorrect. Plus, you should also check for when the element is found in the if-else ladder and use a variable (found) to indicate whether the element is found. Moreover, it'll be better if you use 3 variables for the lower bound (first), mid and upper bound (last). The code below is a revision of the find() method.
public static void find(char[]arr,char ch){
int first=0,mid=0,last=arr.length-1,found=0;
while (first<=last){
mid=(last+first)/2;
if(arr[mid]==ch){
System.out.print(ch+" found at index "+mid);
found=1;
break;
}
else if(arr[mid]<ch){
first=mid+1;
}
else if(arr[mid]>ch){
last=mid-1;
}
}
if(found==0)
System.out.print(ch+" was not found ");
}
This works fine and maybe you should use this method.
You have to make some changes in your find() method.First check that the value of mid should never exceed the value of last,so you have to change your while loop condition accordingly. Secondly you have to give a terminating condition if(arr[mid]==ch) to come out of the loop.
This can also be done by using two variables lo & hi ,easy to understand & implement.See the implementation here Binary Search.
Below is the solution using only only last & mid variables,as per your code requirement above:
public static void find(char[]arr,char ch){
int last = arr.length-1;
int mid=last/2;
while (mid<=last){
if(arr[mid] == ch) //if found, Print and exit the loop
{
System.out.println("found at:"+mid);
return;
}
if (arr[mid]<ch){
mid=((last+mid)/2)+1;
}
else{
last=mid-1;
mid=last/2;
}
}
System.out.println("Not found!!");
}
Im a beginner.
Here is a binary search code.Its showing array out of bounds error for main method.
please look into the program and kindly tell me my mistake.ill be grateful for ur service.
i have to write all this crap cause i cant post it as its asking for more details.
public class BinaryS
{
int n;
public BinaryS(int z)
{
n=z;
}
static int pos;
static boolean flag=false;
public void disp()
{
int arr[]={0,1,2,3,4};
int len=arr.length;
int first=0;
int last=len;
int mid=(int)(first+last/2);
//boolean flag=false;
while(mid>=0 && mid<=len)
{
if(n<arr[mid])
{
last=mid;
}
if(n>arr[mid])
{
first=mid;
}
if(n==arr[mid])
{
flag=true;
pos=mid+1;
}
}
if(flag==true){
System.out.println("the no."+n+"is found at"+pos);
}
else{
System.out.println("the no."+n+"is not found ");
}
}
public static void main(String args[])
{
BinaryS obj=new BinaryS(2);
obj.disp();
}
}
Currently your code does compile, and runs forever - because of this loop:
while(mid>=0 && mid<=len)
{
// Code which doesn't modify mid or len
}
Assuming it gets into that loop at all (which it does), the condition is never going to become false - so unless you return or break from within the loop (you don't) or an exception is thrown (it isn't) you're just going to keep going round the loop.
This is where you should:
Use a debugger to observe what's happening
Think about what the condition should actually be and how you want it to become false
Adjust your code to either change the condition, or change the loop body so that it modifies mid or len
I am trying to figure out a seemingly simple problem with this program. The goal is to rearrange the array in ascending order using a recursive method. I wrote the main method as well as the recursive method as shown below. The issue is that for some reason, I can't stop the method, despite having a condition to prevent it from continuing.
import java.util.Arrays;
public static void main(String[] args){
int[] a = {9,8,7,6,1,2,5,4};
recursiveSelectionSort(a, 0);
System.out.println(Arrays.toString(a));
}
public static void recursiveSelectionSort(int[] a, int start){
int minpos, temp;
//this condition is meant to stop the method
if (start==(a.length-1))
return;
minpos = start;
for (int i = start; i < a.length; i++) {
if (a[i] < a[minpos])
minpos = i;
}
temp = a[start];
a[start] = a[minpos];
a[minpos] = temp;
//this line causes the stack overflow error
recursiveSelectionSort(a, start++);
}
When I delete the recursiveSelectionSort(a,start++) statement, I get an output of {1,8,7,6,9,2,5,4}, which is correct. However, I want this process to continue so they will all eventually be sorted. When I keep the recursion statement, I get the error.
Don't use
recursiveSelectionSort(a, start++);
Doing it this way causes you to first pass start to recursiveSelectionSort and then increment start. start will remain the same in every recursive call.
Use
recursiveSelectionSort(a, start+1);
or if you really like to increment start (even if there is no reason to do so), use
recursiveSelectionSort(a, ++start);
to increment first.
I've been working ahead in my intro class and I've almost finished my last project, Keno. its a betting game that rewards money according to how many numbers you matched with the dealer. I'm having issues on where to put the betting aspect, they start with 100$ and are asked to wage a certain amount of money. I don't know which method that would go under for it to still work because my methods aren't voids, so i wont be able to return more than one data value.
My second issue, maybe the more important one, is that they need to be unique numbers. To do that i would need to search the array of numbers every time to see if they match, or use an array of booleans to keep track of the numbers. I don't know how i would do the second but i have a good idea of what i would do with the first. The issue is that im using a do while already, im not sure how i could add the for loop with a nested for loop in. Here is my code, sorry if its messy, i know my teacher hates my curly braces:
package Keno;
import cs1.Keyboard;
public class Keno {
public static void main(String[]args){
int userArr[]=user();
int compArr[]=computer();
int howMany=matchNums(compArr,userArr);
int moneyGained=betting(howMany);
System.out.println("You matched "+howMany+" numbers");
System.out.println("You have gained "+moneyGained+" dollars!");
}
public static int[] computer(){
int []compChoice=new int[20];
for(int x=0;x<compChoice.length;x++){
compChoice[x]=(int)(Math.random()*81);
}
return compChoice;
}
public static int[] user(){
int choice[]=new int[7];
System.out.println("Welcome to Keno!");
System.out.println("Choose 7 unique numbers ranging from 1-80");
System.out.println("*************************************************");
//assigns numbers to choice array
for(int x=0;x<choice.length;x++){
do{
int temp=x+1;
System.out.println("number "+temp+": ");
choice[x]=Keyboard.readInt();
}while(choice[x]<0||choice[x]>80);
}
System.out.println("Thanks!");
System.out.println("*************************************************");
return choice;
}
public static int matchNums(int arr1[], int arr2[]){
int count=0;
//checks each array slot individually to see if they match
for(int x=0;x<arr1.length;x++){
for(int y=0;y<arr2.length;y++){
if(arr1[x]==arr2[y]){
count++;
}
}
}
return count;
}
public static int betting(int matches){
int moneyGained=0;
if(matches==7){
moneyGained=12000;
}else if(matches==6){
moneyGained=200;
}else if(matches==5){
moneyGained=20;
}else if(moneyGained==4){
moneyGained=1;
}
return moneyGained;
}
}
The simplest way to add the betting/money concept would be to add an integer which represents how much money the player has (starting at 100). You will have to ask the player how much they want to wager, and then adjust their money accordingly.
public static void main(String[] args) {
int playerMoney = 100;
int wagered = getWager(); // dont forget it has to be 0 < wagered <= 100
// adjust players money according to the wager, and how much they won
For ensuring uniqueness, either one of your ideas would work. I like just checking for the numbers existence already in the array, but the boolean array of size 80 would work too. It just seems like a lot for only 7 numbers though.
I am a newbie Computer Science high school student and I have trouble with a small snippet of code. Basically, my code should perform a basic CLI search in an array of integers. However, what happens is I get what appears to be an infinite loop (BlueJ, the compiler I'm using, gets stuck and I have to reset the machine). I have set break points but I still don't quite get the problem...(I don't even understand most of the things that it tells me)
Here's the offending code (assume that "ArrayUtil" works, because it does):
import java.util.Scanner;
public class intSearch
{
public static void main(String[] args)
{
search();
}
public static void search()
{
int[] randomArray = ArrayUtil.randomIntArray(20, 100);
Scanner searchInput = new Scanner(System.in);
int searchInt = searchInput.nextInt();
if (findNumber(randomArray, searchInt) == -1)
{
System.out.println("Error");
}else System.out.println("Searched Number: " + findNumber(randomArray, searchInt));
}
private static int findNumber(int[] searchedArray, int searchTerm)
{
for (int i = 0; searchedArray[i] == searchTerm && i < searchedArray.length; i++)
{
return i;
}
return -1;
}
}
This has been bugging me for some time now...please help me identify the problem!
I don't know about the infinite loop but the following code is not going to work as you intended. The i++ can never be reached so i will always have the value 0.
for (int i = 0; searchedArray[i] == searchTerm && i < searchedArray.length; i++)
{
return i;
}
return -1;
You probably mean this:
for (int i = 0; i < searchedArray.length; i++)
{
if (searchedArray[i] == searchTerm)
{
return i;
}
}
return -1;
I don't know what is the class ArrayUtil (I can not import is using my Netbeans). When I try to change that line with the line int[] randomArray = {1 , 2, 3, 5, 7, 10, 1 , 5}; It works perfectly.
And you should change the loop condition. I will not tell you why but try with my array and you will see the bug soon. After you see it, you can fix it:)
There are 4 basic issues here.
1. Putting searchedArray[i] == searchTerm before i < searchedArray.length can result in an out-of-bounds exception. You must always prevent that kind of code.
2. Your intention seems to be the opposite of your code. Your method name implies finding a search term. But, your code implies that you want to continue your loop scan until the search term is not found, although your loop won't do that either. Think of "for (; this ;) { that } " as "while this do that".
3. Place a break point at the beginning of "search". Then, with a small array, step through the code line by line with the debugger and watch the variables. They don't lie. They will tell you exactly what's happening.
4. Please use a standard IDE and compiler, such as Eclipse and Sun's JDK 6 or 7. Eclipse with JDK 7 is a serious combination that doesn't exhibit a strange "infinite loop" as you describe above.