/**
* The isPrime method iterates through every number in the range of 1 - 100 and adds every prime number to a
list in the file created by the user
* #param filename The file that numbers will be added to
*/
public static void isPrime(String filename) throws IOException
{
boolean status = true; // tells method if number is prime or not (if status = true, number is prime)
// opens up previously-created file for writing
FileWriter fwriter = new FileWriter(filename, true);
PrintWriter outputFile = new PrintWriter(fwriter);
for (int i = 1; i <= 100; i++)
{
// check if each number is prime
status = checkPrime(i);
if (status == true && i != 1)
{
outputFile.println(i);
}
}
outputFile.close();
}
/**
* The checkPrime method tests each number passed in as a parameter to see if it is prime
* #param num Integer to be tested
* #return Returns a boolean value of either true or false
*/
public static boolean checkPrime(int num)
{
int remainder; // used to test whether number is prime or not
for (int j = 2; j < num; j++)
{
remainder = num % j;
if (remainder == 0)
{
return false;
}
}
return true;
}
So this is a segment from my assignment. The code runs, but what I need help with is understanding the logic of why the 2nd method, checkPrime() works. I realize it's probably obvious, but I'm a new cs student, so if anyone is willing to explain why this works to me I'd appreciate it.
The % operator is the remainder operator. It is interating from 2 to num to see if the prime is divisible by any of those numbers. As soon as it is, it can't be a prime so it returns false. If finished and no divisor, it returns true.
But there is a more efficient way. Only divide the candidate number num by values staring with 2 up to the square root of num. Think about it, if there is no remainder up to that point, then no other number from the square root of num to num will divide it.
So first check for divisibility by 2. Then starting with 3 by just the odd numbers.
So the method would look like this.
public static boolean checkPrime(int num)
{
int remainder; // used to test whether number is prime or not
if(num % 2 == 0) {
return false;
}
for (int j = 3; j < (int)Math.sqrt(num)+1; j+=2)
{
remainder = num % j;
if (remainder == 0)
{
return false;
}
}
return true;
}
By definition, a prime number is a number that is only divisible by 1 or itself. For a number to be divisible the remainder of division should be 0.
Examples
5 is prime as it's only divisible by itself and 1:
5/5=1 remainder 0
5/4=1 remainder 1
5/3=1 remainder 2
5/2=2 remainder 1
5/1=5 remainder 0
While 4 is not prime because it's also divisible by 2:
4/4=1 remainder 0
4/3=2 remainder 1
4/2=2 remainder 0
4/1=4 remainder 0
The operator % is used to calculate the remainder 5%2=1.
Using this operator we can loop through all the numbers between 2 and num, for (int j = 2; j < num; j++) (we don't want to check 1 and num itself) and check the remainder, if (remainder == 0). Once remainder is equal to 0, the loop is broken and false is returned meaning the number is not prime. Otherwise, the loop finishes and the method returns true signalling that the number is prime.
I need to write a code which should calculate the first 200 prime numbers, but I can't hand it in as long as I can't explain everything. I used a piece of code from the internet as reference (http://crab.rutgers.edu/~dhong/cs325/chapter3/PrimeNumber.java).
Whole code:
public class Opdracht3 {
public static void main(String[] args) {
int limiet = 200;
int counter = 1;
int testpriem = 3;
boolean isPriem;
while (counter <= limiet) {
isPriem = true;
for (int i = 2; i <= testpriem / 2; i++) {
if (testpriem % i == 0) {
isPriem = false;
break;
}
}
if (isPriem) {
System.out.println(counter + ": " + testpriem);
counter++;
}
testpriem++;
}
}
}
Below part of code verifies if the number is a composite. If testpriem is composite, then comes out of the loop and starts over. Otherwise, it continues and prints the prime number testpriem.
The problem is here:
for (int i = 2; i <= testpriem / 2; i++) {
if (testpriem % i == 0) {
isPriem = false;
break;
}
}
I tested what happens to i, and one way or another it recognizes the divisor needed to calculate the composite. (With 4 divisor is 2, with 9 divisor is 3, with 221 divisor is 13) But I am flabbergasted as of why.
Any ideas?.
the % or ("remainder") operator in Java divides one operand by another and returns the remainder as its result. And of course if integer x is evenly divisible by another integer y (meaning x/y = some integer z with a remainder of zero), then x can not be prime.
First remember every number able to divide by its half or less. Consider number 7 it is possible divided by 1,2,3 because after 3, if try to divide number by 4 means 4x2 = 8 that is greater than 7. so it is optimum way to find divisor in this way. One more thing every number divided by 1 and number itself. so if number number is divided by 1 or itself then it is called prime so i starts from 2.
now consider testpriem =7 so you will get loop like
for (int i = 2; i <= 7 / 2(i.e 3); i++)
{
if(7 % i == 0)
{
isPriem = false;
break;
}
so first time it checks 7%2=1 so condition false. again check for 7%3 = 1 again condition false. and now this condition full fills here i <= 7 / 2(i.e 3) so loop stopped and it result 7 number is prime
Well, if testpriem % i == 0, it means that i divides testpriem, which means that testpriem is not a prime a number and i, is its first divider. % is the modulo operation, which is the rest of the division.
https://en.wikipedia.org/wiki/Modulo_operation
The break stops the for loop and moves to the next position in the while loop. So it does not restart the for loop for the current tested number.
The break is used for efficiency reasons. You could remove it and the algorithm would still work correctly but slower.
In my current Project Euler problem 5, I have a "working" solution. It works on smaller numbers (the example one in the question), but not on the actual problem, because I'm brute forcing it, and the program doesn't finish.
Here's the explanation of the problem:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible1 by all of the numbers from 1 to 20?
1: Divisible with no remainder
Here is my current code:
package Euler;
public class Euler5 {
public static void main(String[] args) {
int desiredNumber = 20;
boolean exitLoop = false;
long counter = 1;
while(exitLoop == false) {
long loopCounter = 0;
for(int i=1; i<=desiredNumber; i++) {
if(counter % i == 0) {
loopCounter++;
}
}
if(loopCounter == desiredNumber) {
exitLoop = true;
System.out.println(counter);
}
counter++;
}
}
}
You don't have a computer to answer this question. Look: if a number can be divided by each of the numbers from 1 to 20 it means that it should be a multiplication of primes in corresponding powers:
2**4 (from 16)
3**2 (from 9)
5
7
11
13
17
19
so the solution is
16 * 9 * 5 * 7 * 11 * 13 * 17 * 19 == 232792560
since the answer is quite large I doubt if brute force is a reasonable method here.
In general case (for some n >= 2) find out all the prime numbers that are not exeeding the n:
2, 3, ..., m (m <= n)
then, for each prime number a find out the power pa such that
a**pa <= n
but
a**(pa + 1) > n
the answer will be
2**p2 * 3**p3 * ... * m**pm
Possible Java implementation:
public static BigInteger evenlyDivisible(int n) {
if (n <= 0)
throw new IllegalArgumentException("n must be positive");
else if (n <= 2)
return BigInteger.valueOf(n);
ArrayList<Integer> primes = new ArrayList<Integer>();
primes.add(2);
for (int i = 3; i <= n; i += 2) {
boolean isPrime = true;
for (int p : primes) {
if (i % p == 0) {
isPrime = false;
break;
}
else if (p * p > i)
break;
}
if (isPrime)
primes.add(i);
}
BigInteger result = BigInteger.ONE;
for(int p : primes) {
// Simplest implemenation, check for round up errors however
int power = (int)(Math.log(n) / Math.log(p));
result = result.multiply(BigInteger.valueOf(p).pow(power));
}
return result;
}
...
System.out.println(evenlyDivisible(20)); // 232792560
The number you are seeking is the Least common multiple (LCM) of the numbers 1,2,3,...,20.
By splitting each numbers to the multiplication of its prime factors (easy for small numbers), finding LCM is fairly easy.
int isPrime(int number)
{
int i;
for(i=2; i*i <= number; i++)
{
if (number % i == 0)
{
return 0;
}
}
return 1;
}
Why is it the case that the boolean conditional of the control statement is: i*i <= number?
Deep mathematical explanation please!
If a number is not prime then it can be written in the form p * q. Without loss of generality we can say that p cannot be greater than q (since we can always write the multiplication this way round).
Therefore, in checking for primality, you only need to check the ps since the qs will be implicitly checked.
So p going no higher than q is the limit. Hence p * p.
(By the way, iterating over all the numbers (i++) is also inefficient - you only need to consider primes. Choosing the appropriate algorithm really depends on your use case and without further details I can't comment further.).
When not a prime, one of the factors is always <= sqrt(n), (if the potential factor is > sqrt(n), it will already have been found using the loop), so
i <= sqrt(n)
If you square both sides of the inequality:
i*i <= n
No solutions will be found outside this constraint.
Even better, you can start of at i = 3 and increase i by 2 every iteration!
if (number % 2 == 0) return 0;
for(i=3; i*i <= number; i+=2)
{
if (number % i == 0)
{
return 0;
}
}
I am working on a prime factorization program implemented in Java.
The goal is to find the largest prime factor of 600851475143 (Project Euler problem 3).
I think I have most of it done, but I am getting a few errors.
Also my logic seems to be off, in particular the method that I have set up for checking to see if a number is prime.
public class PrimeFactor {
public static void main(String[] args) {
int count = 0;
for (int i = 0; i < Math.sqrt(600851475143L); i++) {
if (Prime(i) && i % Math.sqrt(600851475143L) == 0) {
count = i;
System.out.println(count);
}
}
}
public static boolean Prime(int n) {
boolean isPrime = false;
// A number is prime iff it is divisible by 1 and itself only
if (n % n == 0 && n % 1 == 0) {
isPrime = true;
}
return isPrime;
}
}
Edit
public class PrimeFactor {
public static void main(String[] args) {
for (int i = 2; i <= 600851475143L; i++) {
if (isPrime(i) == true) {
System.out.println(i);
}
}
}
public static boolean isPrime(int number) {
if (number == 1) return false;
if (number == 2) return true;
if (number % 2 == 0) return false;
for (int i = 3; i <= number; i++) {
if (number % i == 0) return false;
}
return true;
}
}
Why make it so complicated? You don't need do anything like isPrime(). Divide it's least divisor(prime) and do the loop from this prime. Here is my simple code :
public class PrimeFactor {
public static int largestPrimeFactor(long number) {
int i;
for (i = 2; i <= number; i++) {
if (number % i == 0) {
number /= i;
i--;
}
}
return i;
}
/**
* #param args
*/
public static void main(String[] args) {
System.out.println(largestPrimeFactor(13195));
System.out.println(largestPrimeFactor(600851475143L));
}
}
edit: I hope this doesn't sound incredibly condescending as an answer. I just really wanted to illustrate that from the computer's point of view, you have to check all possible numbers that could be factors of X to make sure it's prime. Computers don't know that it's composite just by looking at it, so you have to iterate
Example: Is X a prime number?
For the case where X = 67:
How do you check this?
I divide it by 2... it has a remainder of 1 (this also tells us that 67 is an odd number)
I divide it by 3... it has a remainder of 1
I divide it by 4... it has a remainder of 3
I divide it by 5... it has a remainder of 2
I divide it by 6... it has a remainder of 1
In fact, you will only get a remainder of 0 if the number is not prime.
Do you have to check every single number less than X to make sure it's prime? Nope. Not anymore, thanks to math (!)
Let's look at a smaller number, like 16.
16 is not prime.
why? because
2*8 = 16
4*4 = 16
So 16 is divisible evenly by more than just 1 and itself. (Although "1" is technically not a prime number, but that's technicalities, and I digress)
So we divide 16 by 1... of course this works, this works for every number
Divide 16 by 2... we get a remainder of 0 (8*2)
Divide 16 by 3... we get a remainder of 1
Divide 16 by 4... we get a remainder of 0 (4*4)
Divide 16 by 5... we get a remainder of 1
Divide 16 by 6... we get a remainder of 4
Divide 16 by 7... we get a remainder of 2
Divide 16 by 8... we get a remainder of 0 (8*2)
We really only need one remainder of 0 to tell us it's composite (the opposite of "prime" is "composite").
Checking if 16 is divisible by 2 is the same thing as checking if it's divisible by 8, because 2 and 8 multiply to give you 16.
We only need to check a portion of the spectrum (from 2 up to the square-root of X) because the largest number that we can multiply is sqrt(X), otherwise we are using the smaller numbers to get redundant answers.
Is 17 prime?
17 % 2 = 1
17 % 3 = 2
17 % 4 = 1 <--| approximately the square root of 17 [4.123...]
17 % 5 = 2 <--|
17 % 6 = 5
17 % 7 = 3
The results after sqrt(X), like 17 % 7 and so on, are redundant because they must necessarily multiply with something smaller than the sqrt(X) to yield X.
That is,
A * B = X
if A and B are both greater than sqrt(X) then
A*B will yield a number that is greater than X.
Thus, one of either A or B must be smaller than sqrt(X), and it is redundant to check both of these values since you only need to know if one of them divides X evenly (the even division gives you the other value as an answer)
I hope that helps.
edit: There are more sophisticated methods of checking primality and Java has a built-in "this number is probably prime" or "this number is definitely composite" method in the BigInteger class as I recently learned via another SO answer :]
You need to do some research on algorithms for factorizing large numbers; this wikipedia page looks like a good place to start. In the first paragraph, it states:
When the numbers are very large, no efficient integer factorization algorithm is publicly known ...
but it does list a number of special and general purpose algorithms. You need to pick one that will work well enough to deal with 12 decimal digit numbers. These numbers are too large for the most naive approach to work, but small enough that (for example) an approach based on enumerating the prime numbers starting from 2 would work. (Hint - start with the Sieve of Erasthones)
Here is very elegant answer - which uses brute force (not some fancy algorithm) but in a smart way - by lowering the limit as we find primes and devide composite by those primes...
It also prints only the primes - and just the primes, and if one prime is more then once in the product - it will print it as many times as that prime is in the product.
public class Factorization {
public static void main(String[] args) {
long composite = 600851475143L;
int limit = (int)Math.sqrt(composite)+1;
for (int i=3; i<limit; i+=2)
{
if (composite%i==0)
{
System.out.println(i);
composite = composite/i;
limit = (int)Math.sqrt(composite)+1;
i-=2; //this is so it could check same prime again
}
}
System.out.println(composite);
}
}
You want to iterate from 2 -> n-1 and make sure that n % i != 0. That's the most naive way to check for primality. As explained above, this is very very slow if the number is large.
To find factors, you want something like:
long limit = sqrt(number);
for (long i=3; i<limit; i+=2)
if (number % i == 0)
print "factor = " , i;
In this case, the factors are all small enough (<7000) that finding them should take well under a second, even with naive code like this. Also note that this particular number has other, smaller, prime factors. For a brute force search like this, you can save a little work by dividing out the smaller factors as you find them, and then do a prime factorization of the smaller number that results. This has the advantage of only giving prime factors. Otherwise, you'll also get composite factors (e.g., this number has four prime factors, so the first method will print out not only the prime factors, but the products of various combinations of those prime factors).
If you want to optimize that a bit, you can use the sieve of Eratosthenes to find the prime numbers up to the square root, and then only attempt division by primes. In this case, the square root is ~775'000, and you only need one bit per number to signify whether it's prime. You also (normally) only want to store odd numbers (since you know immediately that all even numbers but two are composite), so you need ~775'000/2 bits = ~47 Kilobytes.
In this case, that has little real payoff though -- even a completely naive algorithm will appear to produce results instantly.
I think you're confused because there is no iff [if-and-only-if] operator.
Going to the square root of the integer in question is a good shortcut. All that remains is checking if the number within that loop divides evenly. That's simply [big number] % i == 0. There is no reason for your Prime function.
Since you are looking for the largest divisor, another trick would be to start from the highest integer less than the square root and go i--.
Like others have said, ultimately, this is brutally slow.
private static boolean isPrime(int k) throws IllegalArgumentException
{
int j;
if (k < 2) throw new IllegalArgumentException("All prime numbers are greater than 1.");
else {
for (j = 2; j < k; j++) {
if (k % j == 0) return false;
}
}
return true;
}
public static void primeFactorsOf(int n) {
boolean found = false;
if (isPrime(n) == true) System.out.print(n + " ");
else {
int i = 2;
while (found == false) {
if ((n % i == 0) && (isPrime(i))) {
System.out.print(i + ", ");
found = true;
} else i++;
}
primeFactorsOf(n / i);
}
}
For those answers which use a method isPrime(int) : boolean, there is a faster algorithm than the one previously implemented (which is something like)
private static boolean isPrime(long n) { //when n >= 2
for (int k = 2; k < n; k++)
if (n % k == 0) return false;
return true;
}
and it is this:
private static boolean isPrime(long n) { //when n >= 2
if (n == 2 || n == 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int k = 1; k <= (Math.floor(Math.sqrt(n)) + 1) / 6; k++)
if (n % (6 * k + 1) == 0 || n % (6 * k - 1) == 0) return false;
return true;
}
I made this algorithm using two facts:
We only need to check for n % k == 0 up to k <= Math.sqrt(n). This is true because for anything higher, factors merely "flip" ex. consider the case n = 15, where 3 * 5 = 5 * 3, and 5 > Math.sqrt(15). There is no need for this overlap of checking both 15 % 3 == 0 and 15 % 5 == 0, when we could just check one of these expressions.
All primes (excluding 2 and 3) can be expressed in the form (6 * k) + 1 or (6 * k) - 1, because any positive integer can be expressed in the form (6 * k) + n, where n = -1, 0, 1, 2, 3, or 4 and k is an integer <= 0, and the cases where n = 0, 2, 3, and 4 are all reducible.
Therefore, n is prime if it is not divisible by 2, 3, or some integer of the form 6k ± 1 <= Math.sqrt(n). Hence the above algorithm.
--
Wikipedia article on testing for primality
--
Edit: Thought I might as well post my full solution (*I did not use isPrime(), and my solution is nearly identical to the top answer, but I thought I should answer the actual question):
public class Euler3 {
public static void main(String[] args) {
long[] nums = {13195, 600851475143L};
for (num : nums)
System.out.println("Largest prime factor of " + num + ": " + lpf(num));
}
private static lpf(long n) {
long largestPrimeFactor = 1;
long maxPossibleFactor = n / 2;
for (long i = 2; i <= maxPossibleFactor; i++)
if (n % i == 0) {
n /= i;
largestPrimeFactor = i;
i--;
}
return largestPrimeFactor;
}
}
To find all prime factorization
import java.math.BigInteger;
import java.util.Scanner;
public class BigIntegerTest {
public static void main(String[] args) {
BigInteger myBigInteger = new BigInteger("65328734260653234260");//653234254
BigInteger originalBigInteger;
BigInteger oneAddedOriginalBigInteger;
originalBigInteger=myBigInteger;
oneAddedOriginalBigInteger=originalBigInteger.add(BigInteger.ONE);
BigInteger index;
BigInteger countBig;
for (index=new BigInteger("2"); index.compareTo(myBigInteger.add(BigInteger.ONE)) <0; index = index.add(BigInteger.ONE)){
countBig=BigInteger.ZERO;
while(myBigInteger.remainder(index) == BigInteger.ZERO ){
myBigInteger=myBigInteger.divide(index);
countBig=countBig.add(BigInteger.ONE);
}
if(countBig.equals(BigInteger.ZERO)) continue;
System.out.println(index+ "**" + countBig);
}
System.out.println("Program is ended!");
}
}
I got a very similar problem for my programming class. In my class it had to calculate for an inputted number. I used a solution very similar to Stijak. I edited my code to do the number from this problem instead of using an input.
Some differences from Stijak's code are these:
I considered even numbers in my code.
My code only prints the largest prime factor, not all factors.
I don't recalculate the factorLimit until I have divided all instances of the current factor off.
I had all the variables declared as long because I wanted the flexibility of using it for very large values of number. I found the worst case scenario was a very large prime number like 9223372036854775783, or a very large number with a prime number square root like 9223371994482243049. The more factors a number has the faster the algorithm runs. Therefore, the best case scenario would be numbers like 4611686018427387904 (2^62) or 6917529027641081856 (3*2^61) because both have 62 factors.
public class LargestPrimeFactor
{
public static void main (String[] args){
long number=600851475143L, factoredNumber=number, factor, factorLimit, maxPrimeFactor;
while(factoredNumber%2==0)
factoredNumber/=2;
factorLimit=(long)Math.sqrt(factoredNumber);
for(factor=3;factor<=factorLimit;factor+=2){
if(factoredNumber%factor==0){
do factoredNumber/=factor;
while(factoredNumber%factor==0);
factorLimit=(long)Math.sqrt(factoredNumber);
}
}
if(factoredNumber==1)
if(factor==3)
maxPrimeFactor=2;
else
maxPrimeFactor=factor-2;
else
maxPrimeFactor=factoredNumber;
if(maxPrimeFactor==number)
System.out.println("Number is prime.");
else
System.out.println("The largest prime factor is "+maxPrimeFactor);
}
}
public class Prime
{
int i;
public Prime( )
{
i = 2;
}
public boolean isPrime( int test )
{
int k;
if( test < 2 )
return false;
else if( test == 2 )
return true;
else if( ( test > 2 ) && ( test % 2 == 0 ) )
return false;
else
{
for( k = 3; k < ( test/2 ); k += 2 )
{
if( test % k == 0 )
return false;
}
}
return true;
}
public void primeFactors( int factorize )
{
if( isPrime( factorize ) )
{
System.out.println( factorize );
i = 2;
}
else
{
if( isPrime( i ) && ( factorize % i == 0 ) )
{
System.out.print( i+", " );
primeFactors( factorize / i );
}
else
{
i++;
primeFactors( factorize );
}
}
public static void main( String[ ] args )
{
Prime p = new Prime( );
p.primeFactors( 649 );
p.primeFactors( 144 );
p.primeFactors( 1001 );
}
}