I have the following statement and I do not understand very well what it asks for:
We say that a number n is rare when it verifies that for any number m <= n and such that both are cousins among themselves, it turns out that m is prime (two numbers are cousins each other when the greatest common divisor of both numbers is unity).
Write a program that lists all rare numbers between 3 and a value entered by the user.
After thinking how to shed it from the statement I got the solution, which I do not know if it is the correct one.
Can someone confirm that it is well and the result is what he really asks for?
import java.util.Scanner;
public class Cousins {
public static void main(String[] args) {
System.out.println("Enter a number to calculate the cousins:");
Scanner teclado = new Scanner(System.in);
int n = teclado.nextInt();
boolean cousins;
for (int m = 3; m < n; m++) {
cousins = mcd(n, m);
if (cousins == true) {
System.out.println(n + " " + m + " They are cousins among themselves.");
} else if (cousins == false) {
System.out.println(n + " " + m + " They are not cousins to each other.");
}
}
}
public static boolean mcd(int n, int m) {
boolean cousins = true;
for (int i = 2; i <= n; i++) {
if (n % i == 0 && m % i == 0) {
cousins = false;
}
}
return cousins;
}
}
Hmm, this seems not really correct to me. It would be correct, if your excercise would be to check if the numbers from 3 to the user-entered number (lets call it n) are cousins with n. This is what you are doing in your program:
You loop over the integeres from 3 to n and check if the current number and n are cousins. By the way in the loop inside the function mcd() I would add a second condition for finishing, like this:
for (int i = 2; i <= n && i <= m; i++)
But this is not really what you are supposed to do. You need tho check, whether the numbers from 3 to n are rare and not cousins with n.
To do so you basically need a second loop: For each number m between 3 and n you must iterate over the numbers i smaller than m to see if i and m are cousins. And for any number i, where this check returns true, see if this i is prime. If all i's that are cousins to m are prime, then m is rare.
Here is my try:
import java.util.Scanner;
public class Cousins {
public static void main(String[] args) {
System.out
.println("Enter a number to calculate the cousins:");
Scanner teclado = new Scanner(System.in);
int n = teclado.nextInt();
boolean cousins;
for (int m = 3; m < n; m++) {
boolean rare = true;
// Needless to check 1 as 1 is cousin to any number and is prime
for(int i = 2; i <= m; i++) {
if(mcd(i, m) && !numberIsPrime(i)) {
rare = false;
}
}
if(rare) {System.out.println(m + " is rare.");}
else {System.out.println(m + " is not rare.");}
}
}
public static boolean mcd(int n, int m) {
boolean cousins = true;
for (int i = 2; i <= n && i <= m; i++) {
if (n % i == 0 && m % i == 0) {
cousins = false;
}
}
return cousins;
}
public static boolean numberIsPrime(int n) {
boolean prime = true;
for(int i = 2; i < n; i++) {
if(n % i == 0) {prime = false;}
}
return prime;
}
}
EDIT: Sorry, my description above was a bit wrong, I have corrected it now.
EDIT: Regarding your comment: The output of the program that you have written (3,4,6,8, and so on) is the perfectly correct list of rare numbers between 1 and 100 according to the definition of rare that you have given. Ok, let's look at some examples. Let's take the number 5 and check if 5 is rare. To check this we must go through all integers i from 1 to 5. If any of them is a cousin of 5, but NOT prime, then 5 is not rare . Otherwise 5 is rare (according to your definition of rare). So let's do it:
i = 1: Is 1 a cousin of 5? Yes it is, because the only common divisor is 1. Is 1 prime? Yes. Ok. Continue with i=2.
i = 2: Is 2 a cousin of 5? Yes it is, because the only common divisor is 1. Is 2 prime? Yes Ok. Continue.
i = 3: Is 3 a cousin of 5? Yes it is, because the only common divisor is 1. Is 3 prime? Yes Ok. Continue.
i = 4: Is 4 a cousin of 5? Yes it is, because the only common divisor is 1. Is 4 prime? NO!!!
So we have found a number i < 5 which is a cousin of 5 but not prime. So 5 is not rare and that's why 5 is not part of the program output.
I hope it is clearer now.
I am writing code for counting the number of ways an integer can be represented as a sum of the consecutive integers. For Example
15=(7+8),(1+2+3+4+5),(4+5+6). So the number of ways equals 3 for 15.
Now the input size can be <=10^12. My program is working fine till 10^7(i think so, but not sure as i didnt check it on any online judge. Feel free to check the code for that)
but as soon as the i give it 10^8 or higher integer as input. it throws many runtime exceptions(it doesnt show what runtime error). Thanks in advance.
import java.io.*;
//sum needs to contain atleast 2 elements
public class IntegerRepresentedAsSumOfConsecutivePositiveIntegers
{
public static long count = 0;
public static void main(String[] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long num = Long.parseLong(br.readLine()); //Enter a number( <=10^12)
driver(num);
System.out.println("count = " + count);
}
public static void driver(long num)
{
long limit = num / 2;
for(long i = 1 ; i <= limit ; i++)
{
func(i,num);
}
}
public static void func(long i,long num)
{
if(i < num)
{
func(i + 1,num - i);
}
else if(i > num)
{
return;
}
else
{
count++;
}
}
}
Use some math: if arithmetic progression with difference 1 starts with a0 and contains n items, then its sum is
S = (2 * a0 + (n-1))/2 * n = a0 * n + n * (n-1) / 2
note that the second summand rises as quadratic function. So instead of checking all a0 in range S/2, we can check all n is smaller range
nmax = Ceil((-1 + Sqrt(1 + 8 * S)) / 2)
(I used some higher approximation).
Just test whether next expression gives integer positive result
a0 = (S - n * (n - 1) / 2) / n
Recursive function isn't suitable when you have big input size like your case.
The maximum depth of the java call stack is about 8900 calls and sometimes only after 7700 calls stack overflow occurs so it really depends on your program input size.
Try this algorithm I think it worked for your problem:
it will work fine until 10^9 after that it will take much more time to finish running the program.
long sum = 0;
int count = 0;
long size;
Scanner in = new Scanner(System.in);
System.out.print("Enter a number <=10^12: ");
long n = in.nextLong();
if(n % 2 != 0){
size = n / 2 + 1;
}
else{
size = n / 2;
}
for(int i = 1; i <= size; i++){
for(int j = i; j <= size; j++){
sum = sum + j;
if(sum == n){
sum = 0;
count++;
break;
}
else if(sum > n){
sum = 0;
break;
}
}
}
System.out.println(count);
Output:
Enter a number <=10^12: 15
3
Enter a number <=10^12: 1000000000
9
BUILD SUCCESSFUL (total time: 10 seconds)
There's a really excellent proof that the answer can be determined by solving for the unique odd factors (Reference). Essentially, for every odd factor of a target value, there exists either an odd series of numbers of that factor multiplied by its average to produce the target value, or an odd average equal to that factor that can be multiplied by double an even-sized series to reach the target value.
public static int countUniqueOddFactors(long n) {
if (n==1) return 1;
Map<Long, Integer> countFactors=new HashMap<>();
while ((n&1)==0) n>>>=1; // Eliminate even factors
long divisor=3;
long max=(long) Math.sqrt(n);
while (divisor <= max) {
if (n % divisor==0) {
if (countFactors.containsKey(divisor)) {
countFactors.put(divisor, countFactors.get(divisor)+1);
} else {
countFactors.put(divisor, 1);
}
n /= divisor;
} else {
divisor+=2;
}
}
int factors=1;
for (Integer factorCt : countFactors.values()) {
factors*=(factorCt+1);
}
return factors;
}
As #MBo noted, if a number S can be partitioned into n consecutive parts, then S - T(n) must be divisible by n, where T(n) is the n'th triangular number, and so you can count the number of partitions in O(sqrt(S)) time.
// number of integer partitions into (at least 2) consecutive parts
static int numberOfTrapezoidalPartitions(final long sum) {
assert sum > 0: sum;
int n = 2;
int numberOfPartitions = 0;
long triangularNumber = n * (n + 1) / 2;
while (sum - triangularNumber >= 0) {
long difference = sum - triangularNumber;
if (difference == 0 || difference % n == 0)
numberOfPartitions++;
n++;
triangularNumber += n;
}
return numberOfPartitions;
}
A bit more math yields an even simpler way. Wikipedia says:
The politeness of a positive number is defined as the number of ways it can be expressed as the sum of consecutive integers. For every x, the politeness of x equals the number of odd divisors of x that are greater than one.
Also see: OEIS A069283
So a simple solution with lots of room for optimization is:
// number of odd divisors greater than one
static int politeness(long x) {
assert x > 0: x;
int p = 0;
for (int d = 3; d <= x; d += 2)
if (x % d == 0)
p++;
return p;
}
Here is my program for outputting prime factorization of a given number. I am still just a beginner in java so I know it is not the most efficient code. The problem arises when I input relatively big numbers.
Input: 11 Output: 11
Input: 40 Output: 2 2 2 5
Input: 5427 Output: 3 3 3 3 67
Input: 435843 Output: 3 3 79 613
Input: 23456789 Output: none (there appears to be an infinite loop and the code should return 23456789 since it is a prime number on its own)
What might cause this issue?
import java.util.Scanner;
public class PrimeFactorization {
public static boolean isPrime(long n) {
boolean boo = false;
long counter = 0;
if (n == 1) {
boo = false;
} else if (n == 2) {
boo = true;
} else {
for (long i = 2; i < n; i++) {
if (n % i == 0) {
counter++;
}
}
if (counter == 0) {
boo = true;
}
}
return boo;
}
public static void primeFactorization(long num) {
for (long j = 1; j <= num; j++) {
if (isPrime(j)) {
if (num % j == 0) {
while (num % j == 0) {
System.out.printf(j + " ");
num = num / j;
}
}
}
if (num == 1) {
break;
}
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter any number:");
long num = scanner.nextLong();
System.out.print("Prime factorization of your number is: ");
primeFactorization(num);
scanner.close();
}
}
There's no actual error - you're just doing things a very inefficient way. Basically, you're checking every number between 1 and 23456789 for primeness, before dividing.
There's absolutely no point in doing this check. As you work your way up from 1 to 23456789, each time you uncover a factor, you know it has to be prime, because you've already divided out all smaller factors. So if you do all of the following, this will still work correctly, and much more quickly.
Remove the isPrime method completely.
Remove the line if (isPrime(j)) {, and the matching }
Change the loop so that j starts at 2, like for(long j = 2 ; j <= num ; j++) {
Remove if (num == 1) { break; } from the end of the loop. It serves no purpose at all.
No matter how efficient the code, factorizing large numbers takes a while - so long it may feel like the computer has hung. Given your code, even modestly large numbers will take a long time.
The main thing you can do to improve your code's efficiency to to note that for any pair of factors of a number, one of them will be no more than the square root of the number. You can use this fact to limit the loop to reduce the order of you algorithm for O(n) to O(log n).
long sqrt = Math.sqrt(number);
for (long i = 2; i < sqrt; i++) {
...
There are many other things you can do, but this change will have the greatest effect.
If number changes value during the loop (as for example in your second factorizing loop), you'll if course need to recalculate the end value:
for (...)
// if number changes
sqrt = Math.sqrt(number);
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.
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 );
}
}