N prime numbers in java - java

So I've written this code in java which should output numbers on the screen from 1 to n(given by the user) and it should write "-prime" near the ones that are prime.
import java.util.Scanner;
public class primeMass {
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int n;
int i,j;
System.out.print("Dati n: ");
n = sc.nextInt();
for(i=1;i<=n;i++)
for(j=2;j<=n/2;j++)
{
if(i%j==0)
System.out.println(i);
else System.out.println(i +"-prime");
}}
}
If I input 6 for example i get :
Dati n: 6
1-prime
1-prime
2
2-prime
3-prime
3
4
4-prime
5-prime
5-prime
6
6
I'm new to this, and i'm really struggling with my algorithmic, could you tell me how should i change my program so it outputs correct values, and explain to me what i did wrong ? Thank you
UPDATE:
I've done it, thank you everyone for helping me out : this is the outcome:
import java.util.Scanner;
public class primeMass {
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int n;
int i,j;
boolean gasit = false;
System.out.print("Dati n: ");
n = sc.nextInt();
for(i=1;i<=n;i++) {
gasit=false;
for(j=2;j*j<=i;j++)
{
if(i%j==0) gasit=true;}
if(!gasit) {System.out.println(i+"-prime");}
else {
System.out.println(i);}
}
}
}

You print something on every iteration of the inner loop.
Instead, you should print something after all iterations have completed, e.g.
boolean found = false;
for(j=2;j<=n/2;j++) {
if(i%j==0) found = true;
}
if (!found) {
System.out.println(i + "-prime");
} else {
System.out.println(i);
}
Additionally, you shouldn't be going up to n/2: you perhaps mean i/2 (a number doesn't have any factors greater than itself); but you can make it even tighter, since you don't have to check for factors greater than sqrt(i). Or, stated another way, that j * j <= i.
So you can make your loop declaration:
for(j=2; j*j<=i; j++) {

The problem is that the second loop goes until n/2, it should go until i/2 to check if i is prime. A more optimzed version of the primality check algorithm goes until sqrt(i), as suggested in comments.
The next issue is that you are concluding, in a false way, that if in the first case if(i%j==0) you say not prime, otherwise you say it is prime, which is not true necessairly. You should iterate the whole interval of values between [2:i/2] to conclude that i is prime.

Related

how can i check if the number is perfect square by not using math.sqrt

i want to make a program which related to this question:
An integer that can be expressed as the square of another integer is called a perfect square, such as 4,9,16,25, etc. Write a progran that checks if a number is a perfect square.
I did built something goes like:
import java.util.Scanner;
class Q3{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int num = 0;
int a = 0;
System.out.println("Type a number to check if it has square");
num = sc.nextInt();
for(a = 1;a<num;a++){ }
if (a*a == num){
System.out.println("Ok");
break;
}
else if (a*a != num){
System.out.println("Not ok");
}
}
}
So it doesn’t give what i want when i run it. What should i change or add ?
I think your for loop interpretation might be wrong, I made up something that might just work. Give this code a try.. You can make the method return a boolean too if you want.
static void perfectSquare(int number) {
for (int i = 1; i < i * number; ++i) {
// 'i' is the divisor, making sure
// it is equal to the quotient
if ((number % i == 0) && (number / i == i)) {
System.out.println(i);
}
}
If you want to brute force every number then you are on the right track but you should only print "Not ok" if all numbers in the loop have failed otherwise you may have a perfect square but "Ok" will be hidden within many "Not ok" lines. Also there is nothing in your for loop so the if statement always checks if 0*0 == num.
This looks like it may be a homework question so I won't give a full answer for you but think about how you can make this more efficient.
If you have found an integer number that matches do you need to keep going?
Do you need to check every number? (a starting point may be following the principles of a binary search)
I ended up like this:
import java.util.Scanner;
class Q3{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int num = 0;
int a = 0;
int b = 0;
System.out.println("Type a number to check if it has square");
num = sc.nextInt();
for(a = 1;a<num;a++){
if (a*a == num){
b = 1;
break;
}
}
if(b==1){
System.out.println("Perfect Square");
}
else {
System.out.println("Not ok");
}
}
}
Thanks for support !

How to create an addition loop of random numbers?

After writing 1 on scanner I want a random dice number generated and after pressing 1 again I want another random number generated but now I want it added with previous number. I want to make a loop, I want to keep pressing 1 and keep adding random numbers till I reach a certain number.
Thank you.
import java.util.Random;
import java.util.Scanner;
public class dice {
public static void main(String[] args) {
// TODO Auto-generated method stub
for (int k = 0; k < 100; k++) {
Scanner scan = new Scanner(System.in);
int s = scan.nextInt();
System.out.println(s);
int previous = s;
if (s == 1) {
Random ran = new Random();
int n = ran.nextInt(6) + 1;
System.out.print(n);
int next;
while (true) {
next = scan.nextInt();
if (next == 1) {
System.out.println(previous);
}
previous = n + 10;
}
}
}
}
}
Define previous outside the for loop, and replace
int previous = s;
previous = n + 10;
with
previous += s;
previous += n + 10;
Scanner sc=new Scanner(System.in);
int sum=0;
for(;;)
{
if(sc.nextInt()==1)
{
int num = (int)(Math.random()*6); // using the pre-defined random function in java.lang.Math class
System.out.println("Dice Value: "+num);
sum+=num; // shorthand adding the number for each iteration
}
//if(sum>100)
// break;
//if statement to check if value of sum is greater/lesser than a specific number
}
System.out.println("Final Answer: "+sum)
Something like this might work (not yet tested): an infinite loop that can be terminated as per choice.
If you are looking for a way that the program works as soon as you physically press the '1' key on your keyboard, without having to press the enter key, something like a keyevent might work:
https://docs.oracle.com/javase/7/docs/api/java/awt/event/KeyEvent.html
Please do let me know if there are any errors or doubts :)

Using arrays to store primes

Here is my program which is supposed to create an array and initialize prime numbers to it. The prime numbers should then be printed but the program just keeps running.
import java.util.*;
public class primes
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.println("Enter number of primes ");
int x = scan.nextInt();
int[] prime = new int[x];
int div=2,hold=2;
int c=0;
while (prime[x-1]==0)
{
for(int a=2; div>a;a++)
{
if(div>a && div%a==0)
a=div;
else if(div==(a-1))
hold=div;
}
if(div==2||hold!=prime[c-1])
{
prime[c]=hold;
c++;
}
div++;
}
for(int f =0; f<x;f++)
System.out.print(" "+prime[f]+" ");
}
}
I tried changing my loops but I just don't know whats wrong
Like the others mentioned your logic is not right, try something like:
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter number of primes ");
int x = scan.nextInt();
List<Integer> primes = getPrimes(x);
Integer[] primeArray = primes.toArray(new Integer[primes.size()]);
for(int i :primes.toArray(primeArray)){ // you could just use for(int i :primes){ if you don't need array
System.out.print(i + " ");
}
}
private static List<Integer> getPrimes(int upperLimit) {
ArrayList primes = new ArrayList();
for (int i = 2; i < upperLimit; i++) {
boolean isPrime = true;
// Is it prime?
for (int j = 2; j < i; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
if (isPrime)
primes.add(i);
}
return primes;
}
The above will print out up to the numbers entered so if you type 5 it will print out 2 3 but not 5.
The following is an other example with Java 8, this one will print as many prime numbers based on the input, if you input 5 you will get 2 3 5 7 11
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter number of primes ");
int x = scan.nextInt();
long[] prime = primes(x).toArray();
Arrays.stream(prime).forEach(value -> System.out.print(value + " " ));
}
private static LongStream primes(long max) {
return LongStream.iterate(2, i -> i + 1)
.filter(PrimeNumber::isPrime)
.limit(max);
}
private static boolean isPrime(long x) {
return LongStream.rangeClosed(2, (long)(Math.sqrt(x)))
.allMatch(n -> x % n != 0);
}
Your code is wrong. First correct it, And i think you want to store prime numbers coming in range of 1 to N where N is user provided number. Use arrayList (growable) to store it.
It will keep on running because you have this: while (prime[x-1]==0). Where x is an input from the user. Say 5 for instance, then prime[5-1] initially is going to contain a 0 always, and you are running your while loop on this condition which is always going to turn true, thus never ending. Also, your prime number generation logic is not right!
I ran your code in debugger mode and I found the problem.
I tested your program with x=5.
At the end of the first while loop iteration you have :
prime[0] = 2
div = 3
hold = 2
c = 1
And here's the problem :
if(div==2||hold!=prime[c-1])
{
prime[c]=hold;
c++;
}
This part won't ever be reached anymore because :
div is never decrement, so it will always be superior to 2.
hold is
equal to prime[c-1], and never change value.
So prime will always stick to be : 2 0 0 0 0, and your while loop will never end.
I found what was wrong and rewrote the code, it works now. The program asks the user for the number primes they want to see and it prints them after storing them in a basic integer array.
import java.util.*;
public class Prime
{
public static void main(String [] args)
{
Scanner scan= new Scanner(System.in);
int i=0, hold=2, d=2;
boolean flag = true;
System.out.println("Enter the number of primes.");
int[] prime= new int[scan.nextInt()];
for(;flag;){
for(int a=2;d>a;a++){
if(d==(a)||d%a==0){
break;
}
if((d-1)==a){
hold = d;
}
}
d++;
if(hold==2 || hold!=prime[i-1]){
prime[i] = hold;
i++;
}
if(i==prime.length)
flag= false;
}
for(int x=0;x<prime.length;x++)
System.out.print(prime[x]+" ");
System.out.println("");
}
}

How to recurse back to start?

I'm a noob programmer, but I've been stuck on this one bit of code. How do you recurse back to start? I've tried several different methods but they all either take a ridiculous amount of code or don't work properly. I've been trying to implement this "simple" piece of code in all of my programming assignments, but it hasn't been working out. Thanks!
p.s. I've already finished the assignment. I'm just trying to make it more "complete".
public class OddProduct {
public static void main(String[] args) {
//Inputs from user
System.out.println("Enter an odd number");
Scanner input_odd = new Scanner(System.in);
int odd = input_odd.nextInt();
int oddproduct = 1;
//Multiplies all odd integers
for (int counter = 1; counter <= odd; counter = counter + 2){
oddproduct = oddproduct * counter;
}//end of for- loop
System.out.printf("\nThe product of all the odd integers up to %d is %d\n",
odd, oddproduct);
/* MY NOTES FOR RECURSE
if (odd%2 == 1){ proceed normally}
else if (odd%2 != 1) { HOW TO LOOP BACK???}
else { println = "Application closed"}
*/
}//end of main method
}//end of OddProduct class
Based upon your Notes I think this is what you require
Scanner input_odd = new Scanner(System.in);
int odd = 0;
while (odd % 2 != 1) { // fails first time && if user enters even number
System.out.println("Enter an odd number");
odd = input_odd.nextInt();
}

Trying to solve a palindrome using integer arrays

I am writing a program that would help me find whether the number entered is a palindrome or not but i am trying it using arrays. And i would like to know if that is even possible?? And if it is possible then what am i doing wrong.
I have marked the code where i think the problem lies but feel free to suggest anything.!!!!
Thanks!!!
import java.util.Scanner;
public class palindrome
{
public static void main(String args[])
{
int size = 10,i,j,flag=0;
int num[] = new int[size];
Scanner sc = new Scanner(System.in);
System.out.println("Enter the size of the number ");
size = sc.nextInt();
System.out.println("Enter the number ");
for(i=0;i<size;i++)
{
num[i]=sc.nextInt();
}
i=size-1;
for(j=0;j<(size/2);j++,i--)
{
if(i>(size/2))
{
if(num[i]==num[j])
{
flag = 1;
}
}
}
if(flag==1)
{
System.out.println("The number is a palindrome");
}
else
System.out.println("The number is not a palindrome ");
}
}
Edit: Guys the problem is actually solved because i was doing a blunder mistake i.e. i was asking the user to enter the number in the form of an arry but i was not actually entering the digits in the number one by one instead i was entering the whole number in the first iteration.
But still a lot of thanks for the replies. I would still try your ideas and let you guys know. Thanks
:)
Try
public boolean isPalindrome(int[] num){
for(int i = 0 ; i < num.length/2 ; i++) {
if(num[i]!=num[num.length-(i+1)]) return false;
}
return true;
}
Yes it's possible, moreover, it's possible by using ArrayList, String - whatever you like. In order to write down a correct implementation, first decompose your current solution:
// Extract a method, do not cram all code into main()
// note: all you need is array, to get size of the array, put value.length
private static boolean isPalindrome(int[] value) {
...
}
public static void main(String args[]) {
int userInput[];
...
if (isPalindrome(userInput)) {
System.out.println("The number is a palindrome");
}
else {
System.out.println("The number is not a palindrome");
}
}
Now, let's implement isPalindrome():
private static boolean isPalindrome(int[] value) {
if (null == value)
return true; //TODO: or false, or throw exception
for (int i = 0; i < value.length / 2; ++i)
if (value[i] != value[value.length - 1 - i])
return false;
return true;
}
The easiest and most intuitive way (imo) to check for palindromes is through recursion. The idea is simple:
Is the first and last char the same?
YES Remove first and last char and check first and last char of the new String
NO There is no palindrome.
When the input is only 1 char then it's trivial.
Have a look at this code:
private void isPalindrome(String number){
if(number.length() == 1){
System.out.println("yes");
}else if(number.charAt(0) == number.charAt(number.length()-1)){
isPalindrome(number.substring(1, number.length()-1));
}else{
System.out.println("no");
}
}
Testing with:
isPalindrome(String.valueOf(232)) Returns "yes"
isPalindrome(String.valueOf(23)) Return "no"
Of course this also works with Arrays just as easily. Replace the parameter with an array and search through the indices the same way. When cutting down the array just create a new smaller array without first and last index of the previous array.
Your class has several issues:
First you're not checking if a number is a palindrome or not. Your algorithm is flawed
Second, you're asking to enter a size but in the end, the user inputs it but you don't use it yourself. Instead, you're using that introduced value in the number array.
Here's how you should do it.
public class Palindrome {
private static boolean isPalindrome(int[] array) {
for (int i = 0, j = array.length-1; i < j; i++, j--) {
if (array[i] != array[j]) {
return false;
}
}
return true;
}
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
System.out.print("How many numbers do you want to enter? ");
int size = scanner.nextInt();
int[] numbers = new int[size];
for (int i = 0; i < size; i++) {
System.out.printf("Enter number %s: ", i+1);
numbers[i] = scanner.nextInt();
}
if (isPalindrome(numbers)) {
System.out.println("The number is a palindrome");
} else {
System.out.println("The number is not a palindrome");
}
}
}

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