java中的斐波那契哈希函数

Question

I found this article https://probablydance.com/2018/06/16/fibonacci-hashing-the-optimization-that-the-world-forgot-or-a-better-alternative-to-integer-modulo/

I am trying to implement this function in Java to get numbers from [0,1].

My first approach was :

public static double index_for_hash2(double hash)
{
    double golden_ratio=  ((1+Math.sqrt(5))/2);
    double num = hash*golden_ratio;
    double final_number = num - (int)num;
    return final_number;
    

}

but for some numbers, I am taking numbers greater than 1.

Then I tried this :

public static double ind_for_hash(int hash) {   
    BigInteger A = new BigInteger("11400714819323198485"); 
    String hash2 = String.valueOf(hash);
    BigInteger B = new BigInteger(hash2);
    BigInteger mult = A.multiply(B);
    BigInteger mask = new BigInteger("13061404503765839921");
    BigInteger mult1 = mult.and(mask);
    System.out.println(mult1.bitLength());
    BigInteger result = mult1.shiftRight(61);
    System.out.println(result);
    return result.doubleValue();    
}

But all I am taking is a number (0,7) because I have 3 bits. If I shift 63 I will take only 1 bit so I will only have 0 and 1. I won't have a number in a ranger from (0,1). For input hash=2042988003 my output is: 1.1581403804622722E9 not in the wanted range(0,1).

Answer 1

public static double index_for_hash2(double hash)
{
    Double golden_ratio=  ((1+Math.sqrt(5))/2);
    Double num = hash*golden_ratio;
    String sixLetter = num.toString().substring(0,16);
    num = Double.parseDouble(sixLetter);
    int fin_num = num.intValue();
    double final_number = num - fin_num;
    return final_number;

}

I came up with this solution and it seems to work.