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Largest prime factor of a number in Java

I am trying to find the Largest prime factor of a number while solving this problem here. I think that I am doing everything right, however one of the test case (#2) is failing and I can't think of any corner case where it might fail. Here's my code, please have a look and try to spot something.
public class ProblemThree
{
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
int T = scanner.nextInt();
for (int i = 0; i < T; i++)
{
System.out.println(largestPrime(scanner.nextLong()));
}
}
private static long largestPrime(long n)
{
while (n % 2 == 0)
{
n = n / 2; // remove all the multiples of 2
}
while (n % 3 == 0)
{
n = n / 3; // remove all the multiples of 2
}
// remove multiples of prime numbers other than 2 and 3
while (n >= 5)
{
boolean isDivisionComplete = true;
for (long i = 5; i < Math.ceil(Math.sqrt(n)); i++)
{
if (n % i == 0)
{
n = n / i;
isDivisionComplete = false;
break;
}
}
if (isDivisionComplete)
{
break;
}
}
return n;
}
}
Basically, what I am doing is:
Largest_Prime(n):
1. Repeatedly divide the no by any small number, say x where 0 < x < sqrt(n).
2. Then set n = n/x and repeat steps 1 and 2 until there is no such x that divides n.
3 Return n.

It seems you have some bug in your code as as when you input 16 largestPrime function return 1. and this is true for when input is the power of 3.

Detailed Algorithm description:
You can do this by keeping three variables:
The number you are trying to factor (A)
A current divisor store (B)
A largest divisor store (C)
Initially, let (A) be the number you are interested in - in this case, it is 600851475143. Then let (B) be 2. Have a conditional that checks if (A) is divisible by (B). If it is divisible, divide (A) by (B), reset (B) to 2, and go back to checking if (A) is divisible by (B). Else, if (A) is not divisible by (B), increment (B) by +1 and then check if (A) is divisible by (B). Run the loop until (A) is 1. The (3) you return will be the largest prime divisor of 600851475143.
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int a0 = 0; a0 < t; a0++){
long n = in.nextLong();
long A=n;
long B=2;
long C=0;
while(Math.pow(B,2)<=A)
{
if(A%B==0)
{
C=B;
A=A/B;
B=2;
}
else
B++;
}
if(A>=C)
C=A;
if(A==1)
{ C=2;
break;
}
System.out.println(C);
}
}

Why are you removing multiples of 2 and multiples of 3? This way if you have a number that is any combination of powers of 2 and 3 you will get your answer as 1 which is clearly wrong.
For this problem you can do the naive way of looping from 2 to sqrt(n) and store the largest number which divides n, when you finish your loop just return the highest divisor you found.

1 drop your loop for 2 and 3. If not, you dont get 2, 2x2, 3, 2x3, ... all multiples of 2 and 3
2 change your loop to stop at 2 (and not 5):
while (n >= 2)
{
3 stop if 2
if (n==2) return 2;
4 loop from 2
and
5 loop until sqrt(n), with <= and not only < (if not, you dont get prime X Prime)
for (long i = 2; i <= Math.ceil(Math.sqrt(n)); i++)

One easy way of extracting prime factors is like this:
/**
* Prime factors of the number - not the most efficient but it works.
*
* #param n - The number to factorise.
* #param unique - Want only unique factors.
* #return - List of all prime factors of n.
*/
public static List<Long> primeFactors(long n, boolean unique) {
Collection<Long> factors;
if (unique) {
factors = new HashSet<>();
} else {
factors = new ArrayList<>();
}
for (long i = 2; i <= n / i; i++) {
while (n % i == 0) {
factors.add(i);
n /= i;
}
}
if (n > 1) {
factors.add(n);
}
return new ArrayList<>(factors);
}
Those first loops are a problem. They will reduce all even numbers to 1 - thus missing 2 as the factor. Changing your code to use:
while (n > 2 && n % 2 == 0) {
n = n / 2; // remove all the multiples of 2
}
while (n > 3 && n % 3 == 0) {
n = n / 3; // remove all the multiples of 2
}
You still have further issues - e.g. you report the largest prime factor of 25 to be 25 and the largest prime factor of 49 to be 49.
Just run this code using yours and mine to see where yours fails:
for (long i = 1; i < 1000; i++) {
long largestPrime = largestPrime(i);
List<Long> primeFactors = primeFactors(i, true);
if (primeFactors.size() > 0) {
Collections.sort(primeFactors, Collections.reverseOrder());
long highestFactor = primeFactors.get(0);
if (largestPrime != highestFactor) {
System.out.println("Wrong! " + i + " " + largestPrime + " != " + primeFactors);
}
} else {
System.out.println("No factors for " + i);
}
}

How do I do Euler 5 without bruteforcing?

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.

Project Euler #3 - Solution runs forever

First time I've ever encountered this problem. It feels like it will never end.
My approach:
import java.util.TreeSet;
public class Euler3 {
public static void main(String[] args) {
long result = 0;
long startTime = System.nanoTime();
long numberGiven = 600851475143L;
TreeSet<Long> nums = new TreeSet<>();
for (long i = 2L; i < numberGiven; i++) {
if (numberGiven % i == 0 && isPrime(i)) {
nums.add(i);
}
}
result = nums.last();
System.out.print("Result: " + result +
".\nTime used for calculation in nanoseconds: " +
(System.nanoTime() - startTime) + ".");
}
public static boolean isPrime(long n) {
if (n <= 3) {
return n == 1 ? false : true;
} else if (n % 2 == 0 || n % 3 == 0) {
return false;
} else {
for (int i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0) {
return false;
}
}
return true;
}
}
}
Of course this works on smaller numbers, but as is probably intended doesn't seem effective on over 600 billion.
I'm wondering, without giving the answer away:
Is there some obvious alteration I could employ to reduce the
running time/checks necessary?
Although it clearly doesn't work effectively here, is this approach
otherwise acceptable or would someone who posted this challenge,
even with a smaller number, be looking for something different?
I tried with integers the first time and got an overflow related error, is anything similar happening under the hood that would actually prevent this from terminating?

For every number you're checking that's a factor, you're doing an internal loop to figure out if it's a prime. That means your algorithm is effectively performing n * m operations.
You can instead use the following mathematical "trick", which I think is the same used by the UNIX factor program.
Since every number over one is either prime or a unique product of a set of primes (with potential duplicates in the set), we can just start dividing the number by the first prime two (actually reducing the number in the process) until that's no longer possible (i.e., it becomes odd). At that point, the reduced number will not have two or any multiples of two as a factor.
Then we do the same by continuously dividing by three until that's no longer possible.
Now you'd think that may be onerous but, because you've stripped out all the 'two' factors, the number cannot possibly be a multiple of four (or any other even number for that matter). So we detect that and move up to the next divisor of five and start dividing by that.
So the division operation are only done for prime divisors, speeding things up considerably. In addition, once the divisor gets above the square root of the (reduced) number, no more factors are possible, so we exit. In that case, the reduced number gives us the final (hence highest) prime factor.
For example, consider the number 924:
Number Divisor Result
------ ------- ------
924 2* 462
462 2* 231
231 2 not divisible, go to 3
231 3* 77
77 3 not divisible, go to 4
77 4 not divisible, go to 5
77 5 not divisible, go to 6
77 6 not divisible, go to 7
77 7* 11
11* 7 stop since 7 * 7 > 11
So the prime factors of 924 are {2, 2, 3, 7, 11}.
Now I urge you to try that algorithm on your own before looking below since the entire point of Euler is to test your own abilities. I simply provide the solution for completeness:
public class Test
{
public static void main(String[] args) {
long startTime = System.nanoTime();
long number = 600851475143L;
// Start with a divisor of two,
// continue until over sqrt(number).
long divisor = 2L;
while (divisor * divisor <= number) {
if ((number % divisor) == 0) {
// If factor, output then reduce number.
System.out.println(divisor);
number = number / divisor;
} else {
// Otherwise, move to next divisor.
divisor++;
}
}
// Final number is final divisor.
System.out.println(number);
System.out.print("Time used for calculation in nanoseconds: " +
(System.nanoTime() - startTime) + ".");
}
}
That gives you the four prime factors in about five thousandths of a second (on my box, anyway):
71
839
1471
6857
Time used for calculation in nanoseconds: 458826.

The program can be simple like this, which runs under a second:
long val = 600851475143L;
long ans = 0;
for(long i = 2; i*i <= val; i++){
if(val % i == 0){
ans = i;
while(val % i == 0)//This step will make sure that i is prime
val /= i;
}
}
if(val != 1){//If val is not 1, so val is a prime
ans = val > ans ? val : ans;
}
System.out.println(ans);
Answer is 6857, and it is correct answer :)
Notice that we only check for all i values which i*i smaller than val.

the method to calculate prime number

I am trying to figure out how this Java method calculates a prime number but something is confusing me.
public static boolean isPrime(int number){
for(int divisor =2; divisor <= number / 2; divisor++){
if (number % divisor ==0){
return false;
}
}
return true;
}
As you see in the second line in the for loop it shows divisor <= number /2 instead of divisor <= number. Can anyone tell me the reason for that?

First, if you put divisor <= number, you would get no prime numbers at all, because every number is divisible by itself. If the loop does not exit before divisor becomes number, you would get to
number % divisor == 0
condition, and return false.
Whoever wrote this code made an observation that you can stop as soon as you have reached half the number, because if you did not find divisors in the lower half of the interval (2..number/2), there would be no divisors above half the number either, so you can declare the number prime without trying, unsuccessfully, the rest of the candidate divisors.
However, this is not the best you can do: a stronger condition can be used - you could compare divisor to square root of number. This works, because if you don't have a divisor that is less than or equal to the square root of number, there would be no divisors above the square root as well (it is a good idea to think why this is so).
int stop = Math.sqrt(number);
for(int divisor = 2; divisor <= stop ; divisor++) {
...
}

The reason is is that any number can't be divided by any divisor larger than it's half and give more than 1 (if we are talking integers, of course).

Any number would not be divisible by a number more than its half.
For example, the last number 10 would be divisible is 5. 10 is not divisible with 6, 7, 8 or 9.
This is why it's good to eliminate the obvious mismatches to improve the performance of the algorithm.

As others have noted, there are no factors of n greater than n/2. A better solution is comparing your iterating variable to the square root of n, as if there are no factors less than or equal to the square root, there can't be any greater than the square root (Note that it is more efficient to compare i*i <= n that i <= Math.sqrt(n)).
An even better approach is the AKS primality test. If the number is 2 or 3, then it obviously must be prime. Otherwise, it can be rewritten in the form (6k+i) where i = -1, 0, 1, 2, 3, 4. Any (6k + 2) or (6k + 4) is divisible by 2, and any (6k + 3) is divisible by three, so the prime numbers must either take the form (6k - 1) or (6k + 1).
public static boolean isPrime(long n) {
/* This code uses the AKS primality test
* http://en.wikipedia.org/wiki/AKS_primality_test
*/
if (n <= 3) return n > 1;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i*i <=n; i+=6) {
if (n % i == 0 || n % (i+2) == 0) return false;
}
return true;
}
}
I used this as a part of my solution to the PrimeCounter problem in Sedgewick's Intro to Programming in Java (although this is in the first chapter before methods are introduced).
public class PrimeCounter {
public static void main(String[] args) {
long n = 10000000;
long count = 0;
for (long i = 0; i <= n; i++) {
if (isPrime(i)) count++;
}
System.out.println("The number of primes less than "
+ n + " is " + count);
}
public static boolean isPrime(long n) {
/* This code uses the AKS primality test
* http://en.wikipedia.org/wiki/AKS_primality_test
*/
if (n <= 3) return n > 1;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i*i <=n; i+=6) {
if (n % i == 0 || n % (i+2) == 0) return false;
}
return true;
}
}

It's using the fact that if one of the factors is greater than N/2 the other must be smaller than 2.
In fact a massive (asymptotic) gain can be obtained by using the square-root.
That's because if one factor is greater than the square root the other is less.
Sorry Sebri Zouhaier. I'm changing allegiance.
The +1 isn't necessary so the best answer is below.
I'm sorry to change sides for such a tiny improvement!

A number N cannot have any divisors D that are > N/2 and < N. To see this, note that if D is a divisor of N, then it must equal N/D2 for some D2. So the divisors of N are those values of this sequence that are integers: N, N/2, N/3, ... This is a descending sequence. It should be obvious that there can't be any divisors between N and N/2.
In fact, it's common for programs that check for primality to stop at sqrt(N) instead of N/2. The reason is this: Suppose there's a divisor D such that D > sqrt(N). Then N/D = D2 also must be a divisor of N. And it must be the case that D2 < sqrt(N), because if both D and D2 were > sqrt(N), then D * D2 would have to be > N, which is wrong because D * D2 = N. This means that there is no need to check possible divisors D > sqrt(N); if such a divisor existed, we already would have found D2 earlier in the loop and proven that N was not prime.

One small caveat to some of the above answers is that 0 and 1 are not prime numbers. You could account for this (for positive integers) with an implementation such as
public static boolean isPrime(int number){
if (number == 0 || number == 1)
return false;
else
{
int stop = (int) Math.sqrt(number);
for (int divisor = 2; divisor <= stop ; divisor++)
{
if (number % divisor ==0)
return false;
}
return true;
}
}

public static Boolean isPrime(int num){ //method signature. returns Boolean, true if number isPrime, false if not
if(num==2){ //for case num=2, function returns true. detailed explanation underneath
return(true);
}
for(int i=2;i<=(int)Math.sqrt(num)+1;i++){ //loops through 2 to sqrt(num). All you need to check- efficient
if(num%i==0){ //if a divisor is found, its not prime. returns false
return(false);
}
}
return(true); //if all cases don't divide num, it is prime.
}
// Returns true iff n is prime. First checks if n is even, handling the
// cases of n=2 (prime) or n is even > 2 (not prime). Then checks if any
// odd #'s between 3 and sqrt(n), inclusive, are divisors of n, returning
// false if any are.
public static boolean isPrime(int n) {
if (n < 2) return false;
if (n % 2 == 0)
// n is an even, so return true iff n is exactly 2
return (n == 2);
for (int i=3; i*i<=n; i+=2)
if (n % i == 0)
// i divides evenly into n, so n is not prime
return false;
return true;
}

Tips to solve problem #41 of project euler

I'm trying to solve Problem 41of project Euler in Java, by counting the number from 99888888 to 80000000(which took very long time :( ), I got 98765431 as an answer, but I'm getting that answer not correct. Could anyone please tell me the reason of not getting the correct answer and how can I speed my program?

A pandigital number doesn't needs to contain all numbers from 1 to 9, but all from 1 to length.
So, you'll need to try all permutations from 1 to 9 starting with 1 digit and going up, filtering all prime numbers and, then, taking largest one.

The only possible prime pandigital numbers are those with length 1, 4, & 7 because every other pandigital number has the sum of its digits divisible by 3.
So, you'd only need to test for 7! = 5040 permutations.

To obtain a solution in a "reasonable" time, you need the following observations based on the special property that a number is divisible by 3 if the sum of its digits is divisible by 3:
divisible by
1+2+3+4 = 10 -
1+2+3+4+5 = 15 3
1+2+3+4+5+6 = 21 3
1+2+3+4+5+6+7 = 28 -
1+2+3+4+5+6+7+8 = 36 3
1+2+3+4+5+6+7+8+9 = 45 3
So, a "big" prime pandigital number has 4 or 7 digits. (a prime number bigger than 3 is not divisible by 3)
Because you want to obtain the biggest number, it's better to start with the 7-digit numbers and continue to check the 4-digit numbers only if the number was not found. For sure, a 4-digit number exists, because it is specified: 2143.
Now, a possible solution looks like this:
public class P41 {
public static void main(String[] args) {
boolean wasFound = false;
for (int nr = 7654321; nr >= 1234567; nr -= 2) { // even != prime
if (isPandigital(nr) && isOddPrime(nr)) {
System.out.println(nr);
wasFound = true;
break;
}
}
if (!wasFound) {
/* not <=, because 1234 is even */
for (int nr = 4321; nr > 1234; nr -= 2) {
if (isPandigital(nr) && isOddPrime(nr)) {
System.out.println(nr);
break;
}
}
}
}
private static boolean isOddPrime(int x) {
int sqrt = (int) Math.sqrt(x);
for (int i = 3; i <= sqrt; i += 2) {
if (x % i == 0) {
return false;
}
}
return true;
}
private static int getNumberOfDigits(int x) {
int count = 0;
while (x > 0) {
count++;
x /= 10;
}
return count;
}
private static boolean isPandigital(int x) {
int numberOfDigits = getNumberOfDigits(x);
Set<Integer> digits = new HashSet<Integer>();
for (int i = 1; i <= numberOfDigits; i++) {
digits.add(i);
}
for (int i = 1; i <= numberOfDigits; i++) {
digits.remove(x % 10);
x /= 10;
}
if (digits.size() == 0) {
return true;
} else {
return false;
}
}
}
Time: 8 ms.

Here is the problem statement:
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime. What is the largest n-digit pandigital prime that exists?
I wrote a program that started with 987654321 and counted down. I checked that the number was pandigital, and if it was, checked if it was prime.
It took 66 seconds on my Windows 8.1 computer to find the largest prime pandigital.
When I tested the other way around, first prime, then pandigital, the program took way longer than 66 seconds. I cancelled it.
When I applied GregS' tip about discounting all 9 digit and 8 digit pandigital numbers, and started counting down from 7654321, my brute force algorithm took 13 milliseconds.