Hey I am beginning to program using java and my teacher used this example in class for our homework which was to create a java program that prints out every prime number before reaching the upper limit that the user inputs. I am not understanding the second part and wondering if someone could help explaining it to me.
import java.util.Scanner;
public class Primes {
public static void main(String args[]) {
//get input for the upper limit
System.out.println("Enter the upper limit: ");
//read in the limit
int limit = new Scanner(System.in).nextInt();
//use for loop and isPrime method to loop through until the number reaches the limit
for(int number = 2; number<=limit; number++){
//print prime numbers only before the limit
if(isPrime(number)){
System.out.print(number + ", ");
}
}
}
//this part of the program determines whether or not the number is prime by using the modulus
public static boolean isPrime(int number){
for(int i=2; i<number; i++){
if(number%i == 0){
return false; //number is divisible so its not prime
}
}
return true; //number is prime now
}
}
I guess that what you mean by second part is the isPrime method.
What he is doing is using '%' operator which returns the integer remainder of the division between 'number' and 'i'. As a prime number is just divisor for itself and the number 1, if the remainder is 0 it means is not a prime number. Your teacher is looping with the 'i' variable until the limit number and checking if any of the numbers is prime by looking the result of the % operation.
Hope this is helpful for you!!
In the second part
if(number%i == 0)
% usually gives you a remainder if there is.
eg
5 % 2 gives 1
4 % 2 gives 0
for(int i=2; i<number; i++)
Here you are looping through from 2 to the number. Since all numbers are divisible by 1 you start from 2.
Also you stop before number (number -1) since you dont want to check if the number is divisible by itself (because it is).
If a number is divisible by any other number other than 1 and itself (number from 2 to number -1) then it is not a prime number.
Just a short (not efficient) reference for finding prime numbers if you need it:
int upperLimit = 30; //Set your upper limit here
System.out.println(2);
for(int i = 2; i < upperLimit; i++)
for(int j = 2; j < i; j++)
if(i % j == 0 && i != 2)
break;
else if(j == i - 1)
System.out.println(i);
Related
The problem is to check a random number n can be the sum of 2 random prime numbers. For example,
if n=34 the possibilities can be (3+31), (5+29), (17+17)...
So far I have managed to save prime numbers to the array, but have no clue how I could check, if n is the sum of 2 prime numbers.
This is part of my code:
public static void primeNumbers(int n) {
int i = 0, candidate = 2, countArray = 0, countPrime = 0;
boolean flag = true;
while (candidate <= n) {
flag = true;
for (i = 2; i < candidate; i++) {
if ((candidate % i) == 0) {
flag = false;
break;
}
}
if (flag) {
countPrime++;
}
candidate++;
}
int[] primeNumbers = new int[countPrime];
while (candidate <= n) {
flag = true;
for (i = 2; i < candidate; i++) {
if ((candidate % i) == 0) {
flag = false;
break;
}
}
if (flag) {
primeNumbers[countArray] = candidate;
}
candidate++;
countArray++;
}
for (i = 0; i <= primeNumbers.length; i++) {
}
}
First I counted how many prime numbers are between 1-n so I can declare and initialize my array for prime numbers. Then I save prime numbers to the array. But now I have no idea how I could check if n is the sum of 2 prime numbers.
Given that you already have list of "prime numbers less than the given number", It is a very easy task to check if two prime numbers can sum to given number.
for(int i=0; i<array.length; i++){
int firstNum = array[i];
int secondNum = givenNum - firstNum;
/* Now if it is possible to sum up two prime nums to result into given num, secondNum should also be prime and be inside array */
if(ArrayUtils.contains(array, secondNum)){
System.out.println("Yes, it is possible. Numbers are "+ firstNum + " and " + secondNum);
}
}
EDIT: ArrayUtils is part of Apache Commons Lang library
You can however use ArrayList instead to use contains method.
You do not have to check all prime numbers from 1 to the given number or you don't even need an array.
Algorithm one can be;
First of all write a function that checks whether a given number is prime.
Split the number into two parts, 0 and the remaining value(the number itself). Now start decreasing the number part by 1 and start adding 1 to 0 simultaneously. Stop when the number part which we are decreasing becomes 0 or both the parts are prime numbers.
Another algorithm can be like this;(It is similar to the accepted answer)
Start subtracting prime numbers from the given number starting from 2.Check whether the subtraction is also prime.
The time you get the subtraction as a prime number, stop , you have got the two prime numbers that sum up to the given number.
This is a basic prime number checker. How would I loop this code so it tells me every single number from 1 to 100 and if its prime or not. For example:
1 is not a prime number
2 is a prime number
3 is a prime number
and so on until 100. This is not a user generated program just straight up if I run it it gives me all 100 numbers
int check;
int number;
boolean prime = true;
for (int i=2; i<=check/2; i++)
{
number = check%i;
if (number == 0)
{
prime = false;
}
}
if (prime == true)
{
System.out.println(check + " is a prime number");
}
else
{
System.out.println(check + " is not a prime number");
}
}
You have almost all of the code correct, you just need to put it in the right place. For example, your println statements need to be inside the for loop and your for loop needs to start at 1 and increment by 1 to 100.
for(int x = 0; x < 101; x++)
{
if(x == 2)
System.out.println(x + " is a prime number");
if(x % 2 == 0)
System.out.println(x + " is not a prime number);
else
System.out.println(x + " is a prime number);
}
I just helped a friend with almost this exact problem, so this couldn't be better timing. I don't know if your scanner is going to be used, but this works if you are just looking for prime numbers from 1-100.
For starters, it seems your Scanner isn't being used anywhere. Maybe your code is incomplete?
You can replace the Scanner part where you get the input from the user with a for loop that will give the number to you. Like so:
for(int number = 1; number <= 100; number++) {
// code for checking if number is prime or not
}
Take a look at code below, let me know if something is unclear, ask, don't just read and copy:
package com.company;
public class Main {
public static void main(String[] args) {
out:
for (int i = 1; i <= 100 ; i++) {
for (int j = 2; j <= i / 2 ; j++) {
if (i % j == 0) {
System.out.println(i + " is not prime number.");
continue out;
}
}
System.out.println(i + " is prime number.");
}
}
}
So to just walk over the code real quick:
1.) Create labeled block called out:
out:
2.) Create two nested for loops to check for our numbers being prime or not
3.) If statement in the second for loop will check if our number is not prime (if it's divisible by any number from 1 to itself but not those two numbers)
4.) In case our evaluation in if statement is false we continue looping until our second loop condition fails, then we move outside of the loop and print that number is prime:
System.out.println(i + " is prime number.");
5.) In case our evaluation in if statement is true we enter the if block, we print that number is not prime and we move program control back to the labeled block we created earlier
6.) We continue doing this repeatedly until outter loop check condition fails and our program is done with it's work.
/* Prime numbers are numbers that are divisible by 1 and by themselves. So, if you take the modulus of a number 'N' with all the number from 1 till 'N' and increase a count each time the modulus is zero, you can use a simple if condition at the end to check whether they are prime or not. In this code, I start from checking with number 1 and go till 100*/
int count ;
for(int j=1;j<=100;j++)
{
count=0;
for(int i=1;i<=j;i++)
{
if(j%i==0)
count=count+1;
}
if(count>2)
System.out.println(j+"is not a prime number");
else
System.out.println(j+"is a prime number");
}
Okay my issue is less of how to figure out if a number is prime, because I think I figured that out, but more of how to get it to display properly.
Here's my code:
public static void main(String[] args) {
// Declare Variables
int randomNumbers = 0;
int sum = 0;
//Loop for number generation and print out numbers
System.out.print("The five random numbers are: ");
for (int i = 0; i <= 4; i++)
{
randomNumbers = (int)(Math.random()*20);
sum += randomNumbers;
if (i == 4) {
System.out.println("and " + randomNumbers + ".");
}
else {
System.out.print(randomNumbers + ", ");
}
}
//Display Sum
System.out.println("\nThe sum of these five numbers is " + sum + ".\n");
//Determine if the sum is prime and display results
for(int p = 2; p < sum; p++) {
if(sum % p == 0)
System.out.println("The sum is not a prime number.");
else
System.out.println("The sum is a prime number.");
break;
}
}
}
Now my problem is, if the number ends up being something like 9, it'll say it is a prime number, which it is not. I think the issue is that the break is stopping it after one loop so it's not incrementing variable p so it's only testing dividing by 2 (I think). But if I remove the break point it will print out "The sum is/is not a prime number" on every pass until it exits the loop. Not sure what to do here.
Your method for finding if your number is prime is the correct method.
To make it so that it does not consistently print out whether or not the number is prime, you could have an external variable, which represents whether or not the number is prime.
Such as
boolean prime = true;
for (int p = 2; p < sum; p++) {
if (sum % p == 0) {
prime = false;
break;
}
}
if (prime)
System.out.println("The sum is a prime number.");
else
System.out.println("The sum is not a prime number.");
By doing this method the program will assume the number is prime until it proves that wrong. So when it finds it is not prime it sets the variable to false and breaks out of the loop.
Then after the loop finishes you just have to print whether or not the number was prime.
A way that you could make this loop faster is to go from when p = 2 to when p = the square root of sum. So using this method your for loop will look like this:
double sq = Math.sqrt((double)sum);
for (int p = 2; p < sq; p++) {
//Rest of code goes here
}
Hope this helps
You need to store whether or not the number is prime in a boolean outside of the loop:
//Determine if the sum is prime and display results
boolean isPrime = true;
for(int p = 2; p < sum; p++) {
if(sum % p == 0){
isPrime = false;
break;
}
}
if(isPrime){
System.out.println("The sum is a prime number.");
} else {
System.out.println("The sum is not a prime number.");
}
You are right, currently your code tests dividing by two, and the break command is stopping after one loop.
After the first go of your loop (p==2), the break will always stop the loop.
The fastest fix to your code will change the loop part like this:
boolean isPrime=true;
for(int p = 2; p < sum; p++) {
if(sum % p == 0) {
isPrime=false;
System.out.println("The sum is not a prime number.");
break;
}
}
if (isPrime)
System.out.println("The sum is a prime number.");
This code can be improved for efficiency and for code elegance.
For efficiency, you don't need to check divisibility by all numbers smaller than sum, it's enough to check all numbers smaller by square-root of sum.
For better code, create a seperate function to test if a number is prime.
Here is an example that implements both.
// tests if n is prime
public static boolean isPrime(int n) {
if (n<2) return false;
for(int p = 2; p < Math.sqrt(n); p++) {
if(n % p == 0) return false; // enough to find one devisor to show n is not a prime
}
return true; // no factors smaller than sqrt(n) were found
}
public static void main(String []args){
...
System.out.println("sum is "+ sum);
if (isPrime(sum))
System.out.println("The sum is a prime number.");
else
System.out.println("The sum is not a prime number.");
}
Small prime numbers
Use Apache Commons Math primality test, method is related to prime numbers in the range of int. You can find source code on GitHub.
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-math3</artifactId>
<version>3.6.1</version>
</dependency>
// org.apache.commons.math3.primes.Primes
Primes.isPrime(2147483629);
It uses the Miller-Rabin probabilistic test in such a way that a
result is guaranteed: it uses the firsts prime numbers as successive
base (see Handbook of applied cryptography by Menezes, table 4.1 / page 140).
Big prime numbers
If you are looking for primes larger than Integer.MAX_VALUE:
Use BigInteger#isProbablePrime(int certainty) to pre-verify the prime candidate
Returns true if this BigInteger is probably prime, false if it's
definitely composite. If certainty is ≤ 0, true is returned.
Parameters: certainty - a measure of the uncertainty that the caller
is willing to tolerate: if the call returns true the probability that
this BigInteger is prime exceeds (1 - 1/2certainty). The execution
time of this method is proportional to the value of this parameter.
Next use "AKS Primality Test" to check whether the candidate is indeed prime.
So many answers have been posted so far which are correct but none of them is optimized. That's why I thought to share the optimized code to determine prime number here with you. Please have a look at below code snippet...
private static boolean isPrime(int iNum) {
boolean bResult = true;
if (iNum <= 1 || iNum != 2 && iNum % 2 == 0) {
bResult = false;
} else {
int iSqrt = (int) Math.sqrt(iNum);
for (int i = 3; i < iSqrt; i += 2) {
if (iNum % i == 0) {
bResult = false;
break;
}
}
}
return bResult;
}
Benefits of above code-:
It'll work for negative numbers and 0 & 1 as well.
It'll run the for loop only for odd numbers.
It'll increment the for loop variable by 2 rather than 1.
It'll iterate the for loop only upto square root of number rather than upto number.
Explanation-:
I have mentioned the four points above which I'll explain one by one. Code must be appropriately written for the invalid inputs rather the only for valid input. Whatever answers have been written so far are restricted to valid range of input in which number iNum >=2.
We should be aware that only odd numbers can be prime, Note-: 2 is the only one even prime. So we must not run for loop for even numbers.
We must not run for loop for even values of it's variable i as we know that only even numbers can be devided by even number. I have already mentioned in the above point that only odd numbers can be prime except 2 as even. So no need to run code inside for loop for even values of variable i in for.
We should iterate for loop only upto square root of number rather than upto number. Few of the answers has implemented this point but still I thought to mention it here.
With most efficient time Complexity ie O(sqrt(n)) :
public static boolean isPrime(int num) {
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
you can use the Java BigInteger class' isProbablePrime method to determine and print whether the sum is prime or not prime in an easy way.
BigInteger number = new BigInteger(sum);
if(number.isProbablePrime(1)){
System.out.println("prime");
}
else{
System.out.println("not prime");
}
You can read more about this method here https://docs.oracle.com/javase/7/docs/api/java/math/BigInteger.html#isProbablePrime%28int%29
I am a novice, please excuse my lack of organization.
Okay, so what I did was I made an array filled with all the prime numbers between 8 and 100. What I want to do now is make another array that finds all the prime numbers between 101-200. So allow me to explain how I did the first part:
//Prime1 is an dynamic integer array which stores all the prime numbers between 8 and 100
int arrayCounter = 0;
for(int primeTest = 8; primeTest<=100; primeTest++)
{
if(primeTest%2!=0 && primeTest%3!=0 && primeTest%5!=0 && primeTest%7!=0)
{
Prime1.add(primeTest); //adds the prime numbers to array
arrayCounter = arrayCounter +1;
}
else
{
arrayCounter = arrayCounter + 1;
}
}
Now back to the main issue, rather than writing "if(primeTest % "prime#" !=0)" I would like to be able to use modulus through the entire Prime1 array and see if all the values do not equal zero... Let me elaborate.
for(int primeTest2 = 101; primeTest2 <= 200; primeTest2++)
{
for(int arrayCounter2 = 0; arrayCounter2 < Prime1.size(); arrayCounter2++)
{
if(primeTest2 % Prime1.get(arrayCounter2) != 0 )
{
Prime2.add(primeTest2);
}
}
}
//please forgive any missing braces
^^So what happens here is that I take a value starting at 101 and modulus it with the first value of the Prime1 array. As you know, this may give me a false positive because 11 (the first prime number in the array) may still show true even with numbers which are not prime. This is why I need to be able to test a number with all the values in the array to ensure that it cannot be divided by any other prime number (meaning that it is prime).
Your method is extremely inefficient, nevertheless, here is how you can fix it:
for (int primeTest2 = 101; primeTest2 <= 200; primeTest2++)
{
boolean prime = true;
for (int arrayCounter2 = 0; arrayCounter2 < Prime1.size(); arrayCounter2++)
{
if (primeTest2 % Prime1.get(arrayCounter2) == 0)
{
prime = false;
break;
}
}
if (prime)
Prime2.add(primeTest2);
}
BTW, for the first set of prime numbers, it is sufficient to use 2, 3, 5, 7, 11, 13.
Take a boolean and set it to true. If the number can be divided by any of your primes from 8 to 100 without a remainder, than set it to false. If it is still true after testing every number, add the tested number to the Prime2 array, otherwise continue with the next number. Example:
for(int n = 101; n <= 200; n++)
{
boolean isPrime = true;
for(Integer p : Prime1)
if(n % p == 0 )
{
isPrime = false;
break;
}
if(isPrime)
Prime2.add(n);
}
But there are better alorithms out there to check if a number is prime or to calculate alle primes below n. For example the Sieve of Eratosthenes.
Can any body help to understand this java program?
It just prints prime numbers, as you enter how many you want and it works well.
class PrimeNumbers
{
public static void main(String args[])
{
int n, status = 1, num = 3;
Scanner in = new Scanner(System.in);
System.out.println("Enter the number of prime numbers you want");
n = in.nextInt();
if (n >= 1)
{
System.out.println("First "+n+" prime numbers are :-");
System.out.println(2);
}
for ( int count = 2 ; count <=n ; )
{
for ( int j = 2 ; j <= Math.sqrt(num) ; j++ )
{
if ( num%j == 0 )
{
status = 0;
break;
}
}
if ( status != 0 )
{
System.out.println(num);
count++;
}
status = 1;
num++;
}
}
}
I don't understand this for loop condition
for ( int j = 2 ; j <= Math.sqrt(num) ; j++ )
why we are taking sqrt of num...which is 3....why we assumed it as 3?
Imagine that n can be divided by a number k that is greater than sqrt(n). Then you have:
n = k * j
where j is a number which must be less than sqrt(n) (if both k and j are greater than sqrt(n) then their product would be greater than n).
So you only need to find the divisors that are less than or equals to sqrt(n) and you can find those that are greater than or equals to sqrt(n) by a simple division. In my example, once you have found j, you can find k = n / j.
The line in question is basically trying to find numbers that are factors of your given number (and eliminating them as not-primes). If you find no factors of a given number then you can say that the number is prime.
As far as finding factors goes, you only need to go up to sqrt(N) because if you go any higher you are looking at numbers you have already seen before. This is because every time you find a factor you actually find two factors. If a is a factor of N then N/a and a are both factors of N.
A number N is prime if the only integers that satisfy N = A*B are 1 and N (or N and 1).
Now to check a number N as prime we could search all A from 2 to N and B from 2 to N, to see if N=A*B. That would take O(N^2) time, and is really inefficient.
It turns out that we only need to divide N by a number and see if there is a remainder. This brings it down to O(N).
And further, we don't need to check all the way from A=2 to A=N. If A is greater than sqrt(N), then B must be less than sqrt(N). Therefore we only need to check A from 2 to sqrt(N) to see if N/A has a remainder.
If I'm right there is a theory in math, saying that nearly all prime numbers can be determined when checking numbers from 2 to square root of n instead of checking all numbers from 2 to n oder 2 to n/2.
The factor of a number can lie only from 1 to sqrt of num. So instead of checking from 1 till num for its factors, we check for all numbers from 1 to sqrt(num) so see if any of them divides num. If any divides num, it is not prime, else it is. This improves efficiency of code.
A quick but dirty solution..
import java.util.Scanner;
class testing
{
public static void main(String args[])
{
Scanner input = new Scanner(System.in);
int n, i, j, count = 1;
System.out.print("How many Numbers ? : ");
n = input.nextInt();
for(j = 1;count<=n;j++)
{
if(j==1 || j==2)
{
System.out.println(j);
count++;
continue;
}
for(i=2;i<=j/2;i++)
{
if(j%i==0)
break;
else if(i == j/2 && j%i != 0)
{
System.out.println(j);
count++;
}
}
}
}
}
public class PrimeNumber {
public static void main(String[] args) {
// TODO Auto-generated method stub
ArrayList a = new ArrayList();
for (int i = 1; i <= 100; ++i) {
if (isPrime(i))
a.add(i);
}
System.out.println("List : " + a);
}
public static boolean isPrime(int value) {
if (value <= 1)
return false;
if ((value % 2) == 0)
return (value == 2);
for (int i = 3; i <= value - 1; i++) {
if (value % i == 0) {
return false;
}
}
return true;
}
}