Java gcd gcd(int u, int v)

Description

gcd

License

Open Source License

Declaration

public static int gcd(int u, int v) 

Method Source Code

//package com.java2s;
/**//w w w.ja v a 2  s .c o m
 * $Id$
 *
 * Copyright (c) 2009 Thomas Beckmann 
 *
 * Permission is hereby granted, free of charge, to any person obtaining a copy
 * of this software and associated documentation files (the "Software"), to deal
 * in the Software without restriction, including without limitation the rights
 * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
 * copies of the Software, and to permit persons to whom the Software is
 * furnished to do so, subject to the following conditions:
 *
 * The above copyright notice and this permission notice shall be included in
 * all copies or substantial portions of the Software.
 *
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
 * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
 * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
 * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
 * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
 * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
 * SOFTWARE.
 */

public class Main {
    public static int gcd(int u, int v) {

        /* GCD(0,v) := v, GCD(u,0) := u */
        if (u == 0 || v == 0)
            return u | v;

        int shift;

        /* Let shift := lg K, where K is the greatest power of 2
        dividing both u and v. */
        for (shift = 0; ((u | v) & 1) == 0; ++shift) {
            u >>= 1;
            v >>= 1;
        }

        while ((u & 1) == 0)
            u >>= 1;

        /* From here on, u is always odd. */
        do {
            while ((v & 1) == 0) /* Loop X */
                v >>= 1;

            /* Now u and v are both odd, so diff(u, v) is even.
               Let u = min(u, v), v = diff(u, v)/2. */
            if (u < v) {
                v -= u;
            } else {
                int diff = u - v;
                u = v;
                v = diff;
            }
            v >>= 1;
        } while (v != 0);

        return u << shift;
    }

    public static long gcd(long u, long v) {

        /* GCD(0,v) := v, GCD(u,0) := u */
        if (u == 0 || v == 0)
            return u | v;

        long shift;

        /* Let shift := lg K, where K is the greatest power of 2
        dividing both u and v. */
        for (shift = 0; ((u | v) & 1) == 0; ++shift) {
            u >>= 1;
            v >>= 1;
        }

        while ((u & 1) == 0)
            u >>= 1;

        /* From here on, u is always odd. */
        do {
            while ((v & 1) == 0) /* Loop X */
                v >>= 1;

            /* Now u and v are both odd, so diff(u, v) is even.
               Let u = min(u, v), v = diff(u, v)/2. */
            if (u < v) {
                v -= u;
            } else {
                long diff = u - v;
                u = v;
                v = diff;
            }
            v >>= 1;
        } while (v != 0);

        return u << shift;
    }
}

Related

1. gcd(int n1, int n2)
2. gcd(int num1, int num2)
3. gcd(int p, int q)
4. gcd(int p, int q)
5. gcd(int u, int v)
6. gcd(int u, int v)
7. gcd(int x, int y)
8. gcd(int x1, int x2)
9. gcd(int[] array)