如何找到Java BigInteger的平方根？
是否有一个库可以找到BigInteger的平方根？我希望它离线计算-只有一次，而不是在任何循环内。所以，即使是计算上昂贵的解决方案也是可以的如何找到Java BigInteger的平方根？,java,biginteger,square-root,Java,Biginteger,Square Root,是否有一个库可以找到BigInteger的平方根？我希望它离线计算-只有一次，而不是在任何循环内。所以，即使是计算上昂贵的解决方案也是可以的 我不想找到一些算法和实现。一个现成的解决方案将是完美的。我无法验证它们的准确性，但在谷歌搜索时有几种国产解决方案。其中最好的似乎是： 还可以尝试Apache commons数学项目（一旦Apache从JCP博客文章后的轰炸中恢复过来）。对于最初的猜测，我将使用Math.sqrt（bi.doubleValue（）），您可以使用已经建议的链接，使答案更准确。我
我不想找到一些算法和实现。一个现成的解决方案将是完美的。我无法验证它们的准确性，但在谷歌搜索时有几种国产解决方案。其中最好的似乎是：
还可以尝试Apache commons数学项目（一旦Apache从JCP博客文章后的轰炸中恢复过来）。对于最初的猜测，我将使用
Math.sqrt（bi.doubleValue（））import java.math.BigInteger;
public class BigIntSqRoot {
public static BigInteger bigIntSqRootFloor(BigInteger x)
throws IllegalArgumentException {
if (x.compareTo(BigInteger.ZERO) < 0) {
throw new IllegalArgumentException("Negative argument.");
}
// square roots of 0 and 1 are trivial and
// y == 0 will cause a divide-by-zero exception
if (x .equals(BigInteger.ZERO) || x.equals(BigInteger.ONE)) {
return x;
} // end if
BigInteger two = BigInteger.valueOf(2L);
BigInteger y;
// starting with y = x / 2 avoids magnitude issues with x squared
for (y = x.divide(two);
y.compareTo(x.divide(y)) > 0;
y = ((x.divide(y)).add(y)).divide(two));
return y;
} // end bigIntSqRootFloor
public static BigInteger bigIntSqRootCeil(BigInteger x)
throws IllegalArgumentException {
if (x.compareTo(BigInteger.ZERO) < 0) {
throw new IllegalArgumentException("Negative argument.");
}
// square roots of 0 and 1 are trivial and
// y == 0 will cause a divide-by-zero exception
if (x == BigInteger.ZERO || x == BigInteger.ONE) {
return x;
} // end if
BigInteger two = BigInteger.valueOf(2L);
BigInteger y;
// starting with y = x / 2 avoids magnitude issues with x squared
for (y = x.divide(two);
y.compareTo(x.divide(y)) > 0;
y = ((x.divide(y)).add(y)).divide(two));
if (x.compareTo(y.multiply(y)) == 0) {
return y;
} else {
return y.add(BigInteger.ONE);
}
} // end bigIntSqRootCeil
} // end class bigIntSqRoot
import java.math.biginger；
公共类BigIntSqRoot{
公共静态biginger bigIntSqRootFloor（biginger x）
抛出IllegalArgumentException{
if（x.compareTo（BigInteger.ZERO）<0）{
抛出新的IllegalArgumentException（“否定参数”）；
}
//0和1的平方根是平凡的且
//y==0将导致被零除异常
如果（x.equals（biginger.ZERO）| x.equals（biginger.ONE））{
返回x；
}//如果结束，则结束
BigInteger二=BigInteger.valueOf（2L）；
大整数y；
//从y=x/2开始可以避免x平方的大小问题
对于（y=x），除以（2）；
y、 比较（x除以（y））>0；
y=（（x.divide（y））.add（y））.divide（two））；
返回y；
}//end bigIntSqRootFloor
公共静态biginger bigIntSqRootCeil（biginger x）
抛出IllegalArgumentException{
if（x.compareTo（BigInteger.ZERO）<0）{
抛出新的IllegalArgumentException（“否定参数”）；
}
//0和1的平方根是平凡的且
//y==0将导致被零除异常
if（x==biginger.ZERO | | x==biginger.ONE）{
返回x；
}//如果结束，则结束
BigInteger二=BigInteger.valueOf（2L）；
大整数y；
//从y=x/2开始可以避免x平方的大小问题
对于（y=x），除以（2）；
y、 比较（x除以（y））>0；
y=（（x.divide（y））.add（y））.divide（two））；
如果（x.compareTo（y.multiply（y））==0）{
返回y；
}否则{
返回y.add（biginger.ONE）；
}
}//end bigIntSqRootCeil
}//结束类bigIntSqRoot
public static void main(String args[]) {
BigInteger N = new BigInteger(
"17976931348623159077293051907890247336179769789423065727343008115"
+ "77326758055056206869853794492129829595855013875371640157101398586"
+ "47833778606925583497541085196591615128057575940752635007475935288"
+ "71082364994994077189561705436114947486504671101510156394068052754"
+ "0071584560878577663743040086340742855278549092581");
System.out.println(N.toString(10).length());
String sqrt = "";
BigInteger divisor = BigInteger.ZERO;
BigInteger toDivide = BigInteger.ZERO;
String Nstr = N.toString(10);
if (Nstr.length() % 2 == 1)
Nstr = "0" + Nstr;
for (int digitCount = 0; digitCount < Nstr.length(); digitCount += 2) {
toDivide = toDivide.multiply(BigInteger.TEN).multiply(
BigInteger.TEN);
toDivide = toDivide.add(new BigInteger(Nstr.substring(digitCount,
digitCount + 2)));
String div = divisor.toString(10);
divisor = divisor.add(new BigInteger(
div.substring(div.length() - 1)));
int into = tryMax(divisor, toDivide);
divisor = divisor.multiply(BigInteger.TEN).add(
BigInteger.valueOf(into));
toDivide = toDivide.subtract(divisor.multiply(BigInteger
.valueOf(into)));
sqrt = sqrt + into;
}
System.out.println(String.format("Sqrt(%s) = %s", N, sqrt));
}
private static int tryMax(final BigInteger divisor,
final BigInteger toDivide) {
for (int i = 9; i > 0; i--) {
BigInteger div = divisor.multiply(BigInteger.TEN).add(
BigInteger.valueOf(i));
if (div.multiply(BigInteger.valueOf(i)).compareTo(toDivide) <= 0)
return i;
}
return 0;
}
publicstaticvoidmain（字符串参数[]）{
BigInteger N=新的BigInteger(
"17976931348623159077293051907890247336179769789423065727343008115"
+ "77326758055056206869853794492129829595855013875371640157101398586"
+ "47833778606925583497541085196591615128057575940752635007475935288"
+ "71082364994994077189561705436114947486504671101510156394068052754"
+ "0071584560878577663743040086340742855278549092581");
System.out.println（N.toString（10.length（））；
字符串sqrt=“”；
BigInteger除数=BigInteger.0；
BigInteger toDivide=BigInteger.0；
字符串Nstr=N.toString（10）；
如果（Nstr.length（）%2==1）
Nstr=“0”+Nstr；
对于（int digitCount=0；digitCountMath.pow(bigInt.doubleValue(), (1/n));
public static BigInteger sqrt(BigInteger n) {
BigInteger a = BigInteger.ONE;
BigInteger b = new BigInteger(n.shiftRight(5).add(new BigInteger("8")).toString());
while(b.compareTo(a) >= 0) {
BigInteger mid = new BigInteger(a.add(b).shiftRight(1).toString());
if(mid.multiply(mid).compareTo(n) > 0) b = mid.subtract(BigInteger.ONE);
else a = mid.add(BigInteger.ONE);
}
return a.subtract(BigInteger.ONE);
}
public static BigInteger sqrt(BigInteger x) {
BigInteger div = BigInteger.ZERO.setBit(x.bitLength()/2);
BigInteger div2 = div;
// Loop until we hit the same value twice in a row, or wind
// up alternating.
for(;;) {
BigInteger y = div.add(x.divide(div)).shiftRight(1);
if (y.equals(div) || y.equals(div2))
return y;
div2 = div;
div = y;
}
}
static BigInteger fsqrt(BigInteger n)
{
string sn = n.ToString();
return guess(n, BigInteger.Parse(sn.Substring(0, sn.Length >> 1)), 0);
}
static BigInteger guess(BigInteger n, BigInteger g, BigInteger last)
{
if (last >= g - 1 && last <= g + 1) return g;
else return guess(n, (g + (n / g)) >> 1, g);
}
Console.WriteLine(fsqrt(BigInteger.Parse("783648276815623658365871365876257862874628734627835648726")));
static BigInteger fsqrt(BigInteger n)
{
if (n > 999)
{
string sn = n.ToString();
return guess(n, BigInteger.Parse(sn.Substring(0, sn.Length >> 1)), 0);
}
else return guess(n, n >> 1, 0);
}
public class BigIntSqRoot {
private static BigInteger two = BigInteger.valueOf(2L);
public static BigInteger bigIntSqRootFloor(BigInteger x)
throws IllegalArgumentException {
if (checkTrivial(x)) {
return x;
}
if (x.bitLength() < 64) { // Can be cast to long
double sqrt = Math.sqrt(x.longValue());
return BigInteger.valueOf(Math.round(sqrt));
}
// starting with y = x / 2 avoids magnitude issues with x squared
BigInteger y = x.divide(two);
BigInteger value = x.divide(y);
while (y.compareTo(value) > 0) {
y = value.add(y).divide(two);
value = x.divide(y);
}
return y;
}
public static BigInteger bigIntSqRootCeil(BigInteger x)
throws IllegalArgumentException {
BigInteger y = bigIntSqRootFloor(x);
if (x.compareTo(y.multiply(y)) == 0) {
return y;
}
return y.add(BigInteger.ONE);
}
private static boolean checkTrivial(BigInteger x) {
if (x == null) {
throw new NullPointerException("x can't be null");
}
if (x.compareTo(BigInteger.ZERO) < 0) {
throw new IllegalArgumentException("Negative argument.");
}
// square roots of 0 and 1 are trivial and
// y == 0 will cause a divide-by-zero exception
if (x.equals(BigInteger.ZERO) || x.equals(BigInteger.ONE)) {
return true;
} // end if
return false;
}
}
BigDecimal BDtwo = new BigDecimal("2");
BigDecimal BDtol = new BigDecimal(".000000001");
private BigDecimal bigIntSQRT(BigDecimal lNew, BigDecimal lOld, BigDecimal n) {
lNew = lOld.add(n.divide(lOld, 9, BigDecimal.ROUND_FLOOR)).divide(BDtwo, 9, BigDecimal.ROUND_FLOOR);
if (lOld.subtract(lNew).abs().compareTo(BDtol) == 1) {
lNew = bigIntSQRT(lNew, lNew, n);
}
return lNew;
}
private static BigInteger newtonIteration(BigInteger n, BigInteger x0)
{
final BigInteger x1 = n.divide(x0).add(x0).shiftRight(1);
return x0.equals(x1)||x0.equals(x1.subtract(BigInteger.ONE)) ? x0 : newtonIteration(n, x1);
}
public static BigInteger sqrt(final BigInteger number)
{
if(number.signum() == -1)
throw new ArithmeticException("We can only calculate the square root of positive numbers.");
return newtonIteration(number, BigInteger.ONE);
}
public static BigInteger sqrt(BigInteger n) {
BigInteger a = BigInteger.ONE;
BigInteger b = n.shiftRight(1).add(new BigInteger("2")); // (n >> 1) + 2 (ensure 0 doesn't show up)
while (b.compareTo(a) >= 0) {
BigInteger mid = a.add(b).shiftRight(1); // (a+b) >> 1
if (mid.multiply(mid).compareTo(n) > 0)
b = mid.subtract(BigInteger.ONE);
else
a = mid.add(BigInteger.ONE);
}
return a.subtract(BigInteger.ONE);
}
package com.example.so.math;
import java.math.BigInteger;
/**
*
* <p>https://stackoverflow.com/questions/4407839/how-can-i-find-the-square-root-of-a-java-biginteger</p>
* @author Ravindra
* @since 06August2017
*
*/
public class BigIntegerSquareRoot {
public static void main(String[] args) {
int[] values = {5,11,25,31,36,42,49,64,100,121};
for (int i : values) {
BigInteger result = handleSquareRoot(BigInteger.valueOf(i));
System.out.println(i+":"+result);
}
}
private static BigInteger handleSquareRoot(BigInteger modulus) {
int MAX_LOOP_COUNT = 100; // arbitrary for now.. but needs to be proportional to sqrt(modulus)
BigInteger result = null;
if( modulus.equals(BigInteger.ONE) ) {
result = BigInteger.ONE;
return result;
}
for(int i=2;i<MAX_LOOP_COUNT && i<modulus.intValue();i++) { // base-values can be list of primes...
//System.out.println("i"+i);
BigInteger bigIntegerBaseTemp = BigInteger.valueOf(i);
BigInteger bigIntegerRemainderTemp = bigIntegerBaseTemp.modPow(modulus, modulus);
BigInteger bigIntegerRemainderSubtractedByBase = bigIntegerRemainderTemp.subtract(bigIntegerBaseTemp);
BigInteger bigIntegerRemainderSubtractedByBaseFinal = bigIntegerRemainderSubtractedByBase;
BigInteger resultTemp = null;
if(bigIntegerRemainderSubtractedByBase.signum() == -1 || bigIntegerRemainderSubtractedByBase.signum() == 1) {
bigIntegerRemainderSubtractedByBaseFinal = bigIntegerRemainderSubtractedByBase.add(modulus);
resultTemp = bigIntegerRemainderSubtractedByBaseFinal.gcd(modulus);
} else if(bigIntegerRemainderSubtractedByBase.signum() == 0) {
resultTemp = bigIntegerBaseTemp.gcd(modulus);
}
if( resultTemp.multiply(resultTemp).equals(modulus) ) {
System.out.println("Found square root for modulus :"+modulus);
result = resultTemp;
break;
}
}
return result;
}
}
private static final BigInteger ZERO = BigInteger.ZERO;
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = BigInteger.valueOf(2);
private static final BigInteger THREE = BigInteger.valueOf(3);
/**
* This method computes sqrt(n) in O(n.bitLength()) time,
* and computes it exactly. By "exactly", I mean it returns
* not only the (floor of the) square root s, but also the
* remainder r, such that r >= 0, n = s^2 + r, and
* n < (s + 1)^2.
*
* @param n The argument n, as described above.
*
* @return An array of two values, where the first element
* of the array is s and the second is r, as
* described above.
*
* @throws IllegalArgumentException if n is not nonnegative.
*/
public static BigInteger[] sqrt(BigInteger n) {
if (n == null || n.signum() < 0) {
throw new IllegalArgumentException();
}
int bl = n.bitLength();
if ((bl & 1) != 0) {
++ bl;
}
BigInteger s = ZERO;
BigInteger r = ZERO;
while (bl >= 2) {
s = s.shiftLeft(1);
BigInteger crumb = n.testBit(-- bl)
? (n.testBit(-- bl) ? THREE : TWO)
: (n.testBit(-- bl) ? ONE : ZERO);
r = r.shiftLeft(2).add(crumb);
BigInteger d = s.shiftLeft(1);
if (d.compareTo(r) < 0) {
s = s.add(ONE);
r = r.subtract(d).subtract(ONE);
}
}
assert r.signum() >= 0;
assert n.equals(s.multiply(s).add(r));
assert n.compareTo(s.add(ONE).multiply(s.add(ONE))) < 0;
return new BigInteger[] {s, r};
}
package com.example.so.squareroot;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;
/**
* <p>https://stackoverflow.com/questions/4407839/how-can-i-find-the-square-root-of-a-java-biginteger</p>
* <p> Determine square-root of a number or its closest whole number (binary-search-approach) </p>
* @author Ravindra
* @since 07-July-2018
*
*/
public class BigIntegerSquareRootV2 {
public static void main(String[] args) {
List<BigInteger> listOfSquares = new ArrayList<BigInteger>();
listOfSquares.add(BigInteger.valueOf(5).multiply(BigInteger.valueOf(5)).pow(2));
listOfSquares.add(BigInteger.valueOf(11).multiply(BigInteger.valueOf(11)).pow(2));
listOfSquares.add(BigInteger.valueOf(15485863).multiply(BigInteger.valueOf(10000019)).pow(2));
listOfSquares.add(BigInteger.valueOf(533000401).multiply(BigInteger.valueOf(982451653)).pow(2));
listOfSquares.add(BigInteger.valueOf(11).multiply(BigInteger.valueOf(23)));
listOfSquares.add(BigInteger.valueOf(11).multiply(BigInteger.valueOf(23)).pow(2));
for (BigInteger bigIntegerNumber : listOfSquares) {
BigInteger squareRoot = calculateSquareRoot(bigIntegerNumber);
System.out.println("Result :"+bigIntegerNumber+":"+squareRoot);
}
System.out.println("*********************************************************************");
for (BigInteger bigIntegerNumber : listOfSquares) {
BigInteger squareRoot = determineClosestWholeNumberSquareRoot(bigIntegerNumber);
System.out.println("Result :"+bigIntegerNumber+":"+squareRoot);
}
}
/*
Result :625:25
Result :14641:121
Result :23981286414105556927200571609:154858924231397
Result :274206311533451346298141971207799609:523647125012112853
Result :253:null
Result :64009:253
*/
public static BigInteger calculateSquareRoot(BigInteger number) {
/*
* Can be optimized by passing a bean to store the comparison result and avoid having to re-calculate.
*/
BigInteger squareRootResult = determineClosestWholeNumberSquareRoot(number);
if( squareRootResult.pow(2).equals(number)) {
return squareRootResult;
}
return null;
}
/*
Result :625:25
Result :14641:121
Result :23981286414105556927200571609:154858924231397
Result :274206311533451346298141971207799609:523647125012112853
Result :253:15
Result :64009:253
*/
private static BigInteger determineClosestWholeNumberSquareRoot(BigInteger number) {
BigInteger result = null;
if(number.equals(BigInteger.ONE)) {
return BigInteger.ONE;
} else if( number.equals(BigInteger.valueOf(2)) ) {
return BigInteger.ONE;
} else if( number.equals(BigInteger.valueOf(3)) ) {
return BigInteger.ONE;
} else if( number.equals(BigInteger.valueOf(4)) ) {
return BigInteger.valueOf(2);
}
BigInteger tempBaseLow = BigInteger.valueOf(2);
BigInteger tempBaseHigh = number.shiftRight(1); // divide by 2
int loopCount = 11;
while(true) {
if( tempBaseHigh.subtract(tempBaseLow).compareTo(BigInteger.valueOf(loopCount)) == -1 ) { // for lower numbers use for-loop
//System.out.println("Breaking out of while-loop.."); // uncomment-for-debugging
break;
}
BigInteger tempBaseMid = tempBaseHigh.subtract(tempBaseLow).shiftRight(1).add(tempBaseLow); // effectively mid = [(high-low)/2]+low
BigInteger tempBaseMidSquared = tempBaseMid.pow(2);
int comparisonResultTemp = tempBaseMidSquared.compareTo(number);
if(comparisonResultTemp == -1) { // move mid towards higher number
tempBaseLow = tempBaseMid;
} else if( comparisonResultTemp == 0 ) { // number is a square ! return the same !
return tempBaseMid;
} else { // move mid towards lower number
tempBaseHigh = tempBaseMid;
}
}
BigInteger tempBasePrevious = tempBaseLow;
BigInteger tempBaseCurrent = tempBaseLow;
for(int i=0;i<(loopCount+1);i++) {
BigInteger tempBaseSquared = tempBaseCurrent.pow(2);
//System.out.println("Squared :"+tempBaseSquared); // uncomment-for-debugging
int comparisonResultTempTwo = tempBaseSquared.compareTo(number);
if( comparisonResultTempTwo == -1 ) { // move current to previous and increment current...
tempBasePrevious = tempBaseCurrent;
tempBaseCurrent = tempBaseCurrent.add(BigInteger.ONE);
} else if( comparisonResultTempTwo == 0 ) { // is an exact match!
tempBasePrevious = tempBaseCurrent;
break;
} else { // we've identified the point of deviation.. break..
//System.out.println("breaking out of for-loop for square root..."); // uncomment-for-debugging
break;
}
}
result = tempBasePrevious;
//System.out.println("Returning :"+result); // uncomment-for-debugging
return result;
}
}
BigInteger nine = BigInteger.valueOf(9);
BigInteger three = nine.sqrt();
sqrtAndRemainder() BigInteger[]
Returns an array of two BigIntegers containing the integer square root s
of this and its remainder this - s*s, respectively.
public static BigInteger sqrt(BigInteger n) {
BigInteger bottom = BigInteger.ONE;
BigInteger top = n;
BigInteger mid;
while(true) {
mid = top.add(bottom).divide(new BigInteger(""+2));
top = mid;
bottom = n.divide(top);
// System.out.println("top: "+top);
// System.out.println("mid: "+mid);
// System.out.println("bottom: "+bottom);
if(top.equals(bottom)) {
return top;
}
}
}