在Java中，是否有一种简单的方法来生成随机的N位，但是要确保所有N位中至少有一位等于1？

Question

As for N, it could be as large as 179.

For example, I was able to do it for n up to 32 bits:

new Random().nextInt(2^n-1)+1

nextLong() is not possible because I can't pass a value to it and it only generates a random value up to 48-bit.

Answer 1

Use BigInteger to create a random number. If it happens to be zero, try again.

public static BigInteger randomForBitsNonZero(int numBits, Random r) {
    BigInteger candidate = new BigInteger(numBits, r);
    while(candidate.equals(BigInteger.ZERO)) {
        candidate = new BigInteger(numBits, r); 
    }
    return candidate;
}

This will be randomly distrubted about the number of bits. It is very unlikely that the if statement will ever trigger for a significantly high numBits but the protection is good for compelteness and because you might have low numBits sometimes that would roll a 0 very often.

Answer 2

You can use Random.nextBytes() to generate as many random bytes as you want. The number of bytes you need is (number-of-bits + 1) / 8.

Then in order to ensure that at least one bit is set to 1, you can pick a bit, any bit, at random, and set it to 1, follows:

 int bitIndex = random.nextInt( N );
 int byteIndex = bitIndex / 8;
 bitIndex = bitIndex % 8;
 randomBytes[byteIndex] |= (1 << bitIndex);