Compute the square root of BigDecimal x to a given scale, x >= 0. - Java java.math

Java examples for java.math:BigDecimal Calculation

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Description

Compute the square root of BigDecimal x to a given scale, x >= 0.

Demo Code

/*//from w w w  .ja  v  a2 s. c  om
 * Copyright 2013 Valentyn Kolesnikov
 *
 * Licensed 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.
 */
//package com.java2s;
import java.math.BigDecimal;
import java.math.BigInteger;

public class Main {
    private static final int SCALE = 18;

    /**
     * Compute the square root of x to a given scale, x >= 0. Use Newton's
     * algorithm.
     * 
     * @param x
     *            the value of x
     * @return the result value
     */
    public static BigDecimal sqrt(BigDecimal x) {
        // Check that x >= 0.
        if (x.signum() < 0) {
            throw new ArithmeticException("x < 0");
        }

        // n = x*(10^(2*SCALE))
        BigInteger n = x.movePointRight(SCALE << 1).toBigInteger();

        // The first approximation is the upper half of n.
        int bits = (n.bitLength() + 1) >> 1;
        BigInteger ix = n.shiftRight(bits);
        BigInteger ixPrev;

        // Loop until the approximations converge
        // (two successive approximations are equal after rounding).
        do {
            ixPrev = ix;

            // x = (x + n/x)/2
            ix = ix.add(n.divide(ix)).shiftRight(1);

            Thread.yield();
        } while (ix.compareTo(ixPrev) != 0);

        return new BigDecimal(ix, SCALE);
    }
}

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