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javagreatest-common-divisor
Albert
Albert

Reputation: 68360

Java: get greatest common divisor

I have seen that such a function exists for BigInteger, i.e. BigInteger#gcd. Are there other functions in Java which also work for other types (int, long or Integer)? It seems this would make sense as java.lang.Math.gcd (with all kinds of overloads) but it is not there. Is it somewhere else?

(Don't confuse this question with "how do I implement this myself", please!)

Upvotes: 116

Views: 261719

Answers (16)

Matt
Matt

Reputation: 7249

As far as I know, there isn't any built-in method for primitives. But something as simple as this should do the trick:

public int gcd(int a, int b) {
   if (b==0) return a;
   return gcd(b,a%b);
}

You can also one-line it if you're into that sort of thing:

public int gcd(int a, int b) { return b==0 ? a : gcd(b, a%b); }

It should be noted that there is absolutely no difference between the two as they compile to the same byte code.

Upvotes: 171

Tony Ennis
Tony Ennis

Reputation: 12309

For int and long, as primitives, not really. For Integer, it is possible someone wrote one.

Given that BigInteger is a (mathematical/functional) superset of int, Integer, long, and Long, if you need to use these types, convert them to a BigInteger, do the GCD, and convert the result back.

private static int gcdThing(int a, int b) {
    BigInteger b1 = BigInteger.valueOf(a);
    BigInteger b2 = BigInteger.valueOf(b);
    BigInteger gcd = b1.gcd(b2);
    return gcd.intValue();
}

Upvotes: 101

Jin Kwon
Jin Kwon

Reputation: 22025

Those GCD functions provided by Commons-Math and Guava have some differences.

  • Commons-Math throws an ArithematicException.class only for Integer.MIN_VALUE or Long.MIN_VALUE.
    • Otherwise, handles the value as an absolute value.
  • Guava throws an IllegalArgumentException.class for any negative values.

Upvotes: 0

Boleslovas Švitrigaila
Boleslovas Švitrigaila

Reputation: 263

Is it somewhere else?

Apache! - it has both gcd and lcm, so cool!

However, due to profoundness of their implementation, it's slower compared to simple hand-written version (if it matters).

Upvotes: 1

Mohsen Mousavi
Mohsen Mousavi

Reputation: 173

public int gcd(int num1, int num2) { 
    int max = Math.abs(num1);
    int min = Math.abs(num2);

    while (max > 0) {
        if (max < min) {
            int x = max;
            max = min;
            min = x;
        }
        max %= min;
    }

    return min;
}

This method uses the Euclid’s algorithm to get the "Greatest Common Divisor" of two integers. It receives two integers and returns the gcd of them. just that easy!

Upvotes: 2

John Doe
John Doe

Reputation: 1

I used this method that I created when I was 14 years old.

    public static int gcd (int a, int b) {
        int s = 1;
        int ia = Math.abs(a);//<-- turns to absolute value
        int ib = Math.abs(b);
        if (a == b) {
            s = a;
        }else {
            while (ib != ia) {
                if (ib > ia) {
                    s = ib - ia;
                    ib = s;
                }else { 
                    s = ia - ib;
                    ia = s;
                }
            }
        }
        return s;
    }

Upvotes: 0

Esann
Esann

Reputation: 77

we can use recursive function for find gcd

public class Test
{
 static int gcd(int a, int b)
    {
        // Everything divides 0 
        if (a == 0 || b == 0)
           return 0;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return gcd(a-b, b);
        return gcd(a, b-a);
    }

    // Driver method
    public static void main(String[] args) 
    {
        int a = 98, b = 56;
        System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));
    }
}

Upvotes: 3

Alexey
Alexey

Reputation: 9457

Unless I have Guava, I define like this:

int gcd(int a, int b) {
  return a == 0 ? b : gcd(b % a, a);
}

Upvotes: 14

Gitau Harrison
Gitau Harrison

Reputation: 3517

/*
import scanner and instantiate scanner class;
declare your method with two parameters
declare a third variable;
set condition;
swap the parameter values if condition is met;
set second conditon based on result of first condition;
divide and assign remainder to the third variable;
swap the result;
in the main method, allow for user input;
Call the method;

*/
public class gcf {
    public static void main (String[]args){//start of main method
        Scanner input = new Scanner (System.in);//allow for user input
        System.out.println("Please enter the first integer: ");//prompt
        int a = input.nextInt();//initial user input
        System.out.println("Please enter a second interger: ");//prompt
        int b = input.nextInt();//second user input


       Divide(a,b);//call method
    }
   public static void Divide(int a, int b) {//start of your method

    int temp;
    // making a greater than b
    if (b > a) {
         temp = a;
         a = b;
         b = temp;
    }

    while (b !=0) {
        // gcd of b and a%b
        temp = a%b;
        // always make a greater than b
        a =b;
        b =temp;

    }
    System.out.println(a);//print to console
  }
}

Upvotes: 0

MT0
MT0

Reputation: 168796

If you are using Java 1.5 or later then this is an iterative binary GCD algorithm which uses Integer.numberOfTrailingZeros() to reduce the number of checks and iterations required.

public class Utils {
    public static final int gcd( int a, int b ){
        // Deal with the degenerate case where values are Integer.MIN_VALUE
        // since -Integer.MIN_VALUE = Integer.MAX_VALUE+1
        if ( a == Integer.MIN_VALUE )
        {
            if ( b == Integer.MIN_VALUE )
                throw new IllegalArgumentException( "gcd() is greater than Integer.MAX_VALUE" );
            return 1 << Integer.numberOfTrailingZeros( Math.abs(b) );
        }
        if ( b == Integer.MIN_VALUE )
            return 1 << Integer.numberOfTrailingZeros( Math.abs(a) );

        a = Math.abs(a);
        b = Math.abs(b);
        if ( a == 0 ) return b;
        if ( b == 0 ) return a;
        int factorsOfTwoInA = Integer.numberOfTrailingZeros(a),
            factorsOfTwoInB = Integer.numberOfTrailingZeros(b),
            commonFactorsOfTwo = Math.min(factorsOfTwoInA,factorsOfTwoInB);
        a >>= factorsOfTwoInA;
        b >>= factorsOfTwoInB;
        while(a != b){
            if ( a > b ) {
                a = (a - b);
                a >>= Integer.numberOfTrailingZeros( a );
            } else {
                b = (b - a);
                b >>= Integer.numberOfTrailingZeros( b );
            }
        }
        return a << commonFactorsOfTwo;
    }
}

Unit test:

import java.math.BigInteger;
import org.junit.Test;
import static org.junit.Assert.*;

public class UtilsTest {
    @Test
    public void gcdUpToOneThousand(){
        for ( int x = -1000; x <= 1000; ++x )
            for ( int y = -1000; y <= 1000; ++y )
            {
                int gcd = Utils.gcd(x, y);
                int expected = BigInteger.valueOf(x).gcd(BigInteger.valueOf(y)).intValue();
                assertEquals( expected, gcd );
            }
    }

    @Test
    public void gcdMinValue(){
        for ( int x = 0; x < Integer.SIZE-1; x++ ){
            int gcd = Utils.gcd(Integer.MIN_VALUE,1<<x);
            int expected = BigInteger.valueOf(Integer.MIN_VALUE).gcd(BigInteger.valueOf(1<<x)).intValue();
            assertEquals( expected, gcd );
        }
    }
}

Upvotes: 1

Tom Tucker
Tom Tucker

Reputation: 11916

Jakarta Commons Math has exactly that.

ArithmeticUtils.gcd(int p, int q)

Upvotes: 11

Robot Monk
Robot Monk

Reputation: 117

Some implementations here are not working correctly if both numbers are negative. gcd(-12, -18) is 6, not -6.

So an absolute value should be returned, something like

public static int gcd(int a, int b) {
    if (b == 0) {
        return Math.abs(a);
    }
    return gcd(b, a % b);
}

Upvotes: 6

Morad
Morad

Reputation: 774

Use Guava LongMath.gcd() and IntMath.gcd()

Upvotes: 12

linuxjava
linuxjava

Reputation: 141

You can use this implementation of Binary GCD algorithm

public class BinaryGCD {

public static int gcd(int p, int q) {
    if (q == 0) return p;
    if (p == 0) return q;

    // p and q even
    if ((p & 1) == 0 && (q & 1) == 0) return gcd(p >> 1, q >> 1) << 1;

    // p is even, q is odd
    else if ((p & 1) == 0) return gcd(p >> 1, q);

    // p is odd, q is even
    else if ((q & 1) == 0) return gcd(p, q >> 1);

    // p and q odd, p >= q
    else if (p >= q) return gcd((p-q) >> 1, q);

    // p and q odd, p < q
    else return gcd(p, (q-p) >> 1);
}

public static void main(String[] args) {
    int p = Integer.parseInt(args[0]);
    int q = Integer.parseInt(args[1]);
    System.out.println("gcd(" + p + ", " + q + ") = " + gcd(p, q));
}

}

From http://introcs.cs.princeton.edu/java/23recursion/BinaryGCD.java.html

Upvotes: 7

Mr-Al7lawe
Mr-Al7lawe

Reputation: 1

The % going to give us the gcd Between two numbers, it means:- % or mod of big_number/small_number are =gcd, and we write it on java like this big_number % small_number.

EX1: for two integers

  public static int gcd(int x1,int x2)
    {
        if(x1>x2)
        {
           if(x2!=0)
           {
               if(x1%x2==0)     
                   return x2;
                   return x1%x2;
                   }
           return x1;
           }
          else if(x1!=0)
          {
              if(x2%x1==0)
                  return x1;
                  return x2%x1;
                  }
        return x2;
        }

EX2: for three integers

public static int gcd(int x1,int x2,int x3)
{

    int m,t;
    if(x1>x2)
        t=x1;
    t=x2;
    if(t>x3)
        m=t;
    m=x3;
    for(int i=m;i>=1;i--)
    {
        if(x1%i==0 && x2%i==0 && x3%i==0)
        {
            return i;
        }
    }
    return 1;
}

Upvotes: -3

Xorlev
Xorlev

Reputation: 8653

Or the Euclidean algorithm for calculating the GCD...

public int egcd(int a, int b) {
    if (a == 0)
        return b;

    while (b != 0) {
        if (a > b)
            a = a - b;
        else
            b = b - a;
    }

    return a;
}

Upvotes: 36

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