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How to generate a random BigInteger value in Java?

Posted by: admin December 20, 2017 Leave a comment


I need to generate arbitrarily large random integers in the range 0 (inclusive) to n (exclusive). My initial thought was to call nextDouble and multiply by n, but once n gets to be larger than 253, the results would no longer be uniformly distributed.

BigInteger has the following constructor available:

public BigInteger(int numBits, Random rnd)

Constructs a randomly generated BigInteger, uniformly distributed over the range 0 to (2numBits – 1), inclusive.

How can this be used to get a random value in the range 0 – n, where n is not a power of 2?


Use a loop:

BigInteger r;
do {
    r = new BigInteger(n.bitLength(), rnd);
} while (r.compareTo(n) >= 0);

on average, this will require less than two iterations, and the selection will be uniform.

Edit: If your RNG is expensive, you can limit the number of iterations the following way:

int nlen = n.bitLength();
BigInteger nm1 = n.subtract(BigInteger.ONE);
BigInteger r, s;
do {
    s = new BigInteger(nlen + 100, rnd);
    r = s.mod(n);
} while (s.subtract(r).add(nm1).bitLength() >= nlen + 100);
// result is in 'r'

With this version, it is highly improbable that the loop is taken more than once (less than one chance in 2^100, i.e. much less than the probability that the host machine spontaneously catches fire in the next following second). On the other hand, the mod() operation is computationally expensive, so this version is probably slower than the previous, unless the rnd instance if exceptionally slow.


The following method uses the BigInteger(int numBits, Random rnd) constructor and rejects the result if it’s bigger than the specified n.

public BigInteger nextRandomBigInteger(BigInteger n) {
    Random rand = new Random();
    BigInteger result = new BigInteger(n.bitLength(), rand);
    while( result.compareTo(n) >= 0 ) {
        result = new BigInteger(n.bitLength(), rand);
    return result;

The drawback to this is that the constructor is called an unspecified number of times, but in the worst case (n is just slightly greater than a power of 2) the expected number of calls to the constructor should be only about 2 times.


The simplest approach (by quite a long way) would be to use the specified constructor to generate a random number with the right number of bits (floor(log2 n) + 1), and then throw it away if it’s greater than n. In the worst possible case (e.g. a number in the range [0, 2n + 1) you’ll throw away just under half the values you create, on average.


Why not constructing a random BigInteger, then building a BigDecimal from it ?
There is a constructor in BigDecimal : public BigDecimal(BigInteger unscaledVal, int scale) that seems relevant here, no ? Give it a random BigInteger and a random scale int, and you’ll have a random BigDecimal. No ?


Here is how I do it in a class called Generic_BigInteger available via:
Andy Turner’s Generic Source Code Web Page

 * There are methods to get large random numbers. Indeed, there is a
 * constructor for BigDecimal that allows for this, but only for uniform
 * distributions over a binary power range.
 * @param a_Random
 * @param upperLimit
 * @return a random integer as a BigInteger between 0 and upperLimit
 * inclusive
public static BigInteger getRandom(
        Generic_Number a_Generic_Number,
        BigInteger upperLimit) {
    // Special cases
    if (upperLimit.compareTo(BigInteger.ZERO) == 0) {
        return BigInteger.ZERO;
    String upperLimit_String = upperLimit.toString();
    int upperLimitStringLength = upperLimit_String.length();
    Random[] random = a_Generic_Number.get_RandomArrayMinLength(
    if (upperLimit.compareTo(BigInteger.ONE) == 0) {
        if (random[0].nextBoolean()) {
            return BigInteger.ONE;
        } else {
            return BigInteger.ZERO;
    int startIndex = 0;
    int endIndex = 1;
    String result_String = "";
    int digit;
    int upperLimitDigit;
    int i;
    // Take care not to assign any digit that will result in a number larger
    // upperLimit
    for (i = 0; i < upperLimitStringLength; i ++){
        upperLimitDigit = new Integer(
        startIndex ++;
        endIndex ++;
        digit = random[i].nextInt(upperLimitDigit + 1);
        if (digit != upperLimitDigit){
        result_String += digit;
    // Once something smaller than upperLimit guaranteed, assign any digit
    // between zero and nine inclusive
    for (i = i + 1; i < upperLimitStringLength; i ++) {
        digit = random[i].nextInt(10);
        result_String += digit;
    // Tidy values starting with zero(s)
    while (result_String.startsWith("0")) {
        if (result_String.length() > 1) {
            result_String = result_String.substring(1);
        } else {
    BigInteger result = new BigInteger(result_String);
    return result;


Just use modular reduction

new BigInteger(n.bitLength(), new SecureRandom()).mod(n)


Compile this F# code into a DLL and you can also reference it in your C# / VB.NET programs

type BigIntegerRandom() =
    static let internalRandom = new Random()

    /// Returns a BigInteger random number of the specified number of bytes.
    static member RandomBigInteger(numBytes:int, rand:Random) =
        let r = if rand=null then internalRandom else rand
        let bytes : byte[] = Array.zeroCreate (numBytes+1)
        bytes.[numBytes] <- 0uy
        bigint bytes

    /// Returns a BigInteger random number from 0 (inclusive) to max (exclusive).
    static member RandomBigInteger(max:bigint, rand:Random) =
        let rec getNumBytesInRange num bytes = if max < num then bytes else getNumBytesInRange (num * 256I) bytes+1
        let bytesNeeded = getNumBytesInRange 256I 1
        BigIntegerRandom.RandomBigInteger(bytesNeeded, rand) % max

    /// Returns a BigInteger random number from min (inclusive) to max (exclusive).
    static member RandomBigInteger(min:bigint, max:bigint, rand:Random) =
        BigIntegerRandom.RandomBigInteger(max - min, rand) + min