Java gcd gcd(int u, int v)

Description

Gets the greatest common divisor of the absolute value of two numbers, using the "binary gcd" method which avoids division and modulo operations.

License

Apache License

Parameter

Parameter	Description
u	a non-zero number
v	a non-zero number

Return

the greatest common divisor, never zero

Declaration

public static int gcd(int u, int v) 

Method Source Code

//package com.java2s;
/*/*  w ww  .  jav a  2  s  .  c om*/
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *      http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

public class Main {
    /**
     * <p>
     * Gets the greatest common divisor of the absolute value of two numbers,
     * using the "binary gcd" method which avoids division and modulo
     * operations. See Knuth 4.5.2 algorithm B. This algorithm is due to Josef
     * Stein (1961).
     * </p>
     * 
     * @param u a non-zero number
     * @param v a non-zero number
     * @return the greatest common divisor, never zero
     * @since 1.1
     */
    public static int gcd(int u, int v) {
        if (u * v == 0) {
            return (Math.abs(u) + Math.abs(v));
        }
        // keep u and v negative, as negative integers range down to
        // -2^31, while positive numbers can only be as large as 2^31-1
        // (i.e. we can't necessarily negate a negative number without
        // overflow)
        /* assert u!=0 && v!=0; */
        if (u > 0) {
            u = -u;
        } // make u negative
        if (v > 0) {
            v = -v;
        } // make v negative
          // B1. [Find power of 2]
        int k = 0;
        while ((u & 1) == 0 && (v & 1) == 0 && k < 31) { // while u and v are
                                                         // both even...
            u /= 2;
            v /= 2;
            k++; // cast out twos.
        }
        if (k == 31) {
            throw new ArithmeticException("overflow: gcd is 2^31");
        }
        // B2. Initialize: u and v have been divided by 2^k and at least
        // one is odd.
        int t = ((u & 1) == 1) ? v : -(u / 2)/* B3 */;
        // t negative: u was odd, v may be even (t replaces v)
        // t positive: u was even, v is odd (t replaces u)
        do {
            /* assert u<0 && v<0; */
            // B4/B3: cast out twos from t.
            while ((t & 1) == 0) { // while t is even..
                t /= 2; // cast out twos
            }
            // B5 [reset max(u,v)]
            if (t > 0) {
                u = -t;
            } else {
                v = t;
            }
            // B6/B3. at this point both u and v should be odd.
            t = (v - u) / 2;
            // |u| larger: t positive (replace u)
            // |v| larger: t negative (replace v)
        } while (t != 0);
        return -u * (1 << k); // gcd is u*2^k
    }
}

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