Halton sequence

Illustration of the first 8 points of the 2,3 Halton sequence

The Halton sequence is constructed according to a deterministic method that uses coprime numbers as its bases. As a simple example, let's take one dimension of the Halton sequence to be based on 2 and the other on 3. To generate the sequence for 2, we start by dividing the interval (0,1) in half, then in fourths, eighths, etc., which generates

​¹⁄₂, ​¹⁄₄, ​³⁄₄, ​¹⁄₈, ​⁵⁄₈, ​³⁄₈, ​⁷⁄₈, ​¹⁄₁₆, ​⁹⁄₁₆,...

Equivalently, the nth number of this sequence is the number n written in binary representation, inverted, and written after the decimal point. This is true for any base. As an example, to find the sixth element of the above sequence, we'd write 6 = 1*2² + 1*2¹ + 0*2⁰ = 110₂, which can be inverted and placed after the decimal point to give 0.011₂ = 0*2^-1 + 1*2^-2 + 1*2^-3 = ​³⁄₈. So the sequence above is the same as

0.1₂, 0.01₂, 0.11₂, 0.001₂, 0.101₂, 0.011₂, 0.111₂, 0.0001₂, 0.1001₂,...

To generate the sequence for 3, we divide the interval (0,1) in thirds, then ninths, twenty-sevenths, etc., which generates

​¹⁄₃, ​²⁄₃, ​¹⁄₉, ​⁴⁄₉, ​⁷⁄₉, ​²⁄₉, ​⁵⁄₉, ​⁸⁄₉, ​¹⁄₂₇,...

When we pair them up, we get a sequence of points in a unit square:

(​¹⁄₂, ​¹⁄₃), (​¹⁄₄, ​²⁄₃), (​³⁄₄, ​¹⁄₉), (​¹⁄₈, ​⁴⁄₉), (​⁵⁄₈, ​⁷⁄₉), (​³⁄₈, ​²⁄₉), (​⁷⁄₈, ​⁵⁄₉), (​¹⁄₁₆, ​⁸⁄₉), (​⁹⁄₁₆, ​¹⁄₂₇).

Even though standard Halton sequences perform very well in low dimensions, correlation problems have been noted between sequences generated from higher primes. For example, if we started with the primes 17 and 19, the first 16 pairs of points: (​¹⁄₁₇, ​¹⁄₁₉), (​²⁄₁₇, ​²⁄₁₉), (​³⁄₁₇, ​³⁄₁₉) ... (​¹⁶⁄₁₇, ​¹⁶⁄₁₉) would have perfect linear correlation. To avoid this, it is common to drop the first 20 entries, or some other predetermined quantity depending on the primes chosen. Several other methods have also been proposed. One of the most prominent solutions is the scrambled Halton sequence, which uses permutations of the coefficients used in the construction of the standard sequence. Another solution is the leaped Halton, which skips points in the standard sequence. Using, eg, only each 409th point (also other prime numbers not used in the Halton core sequence are possible), can achieve significant improvements.[1]

In pseudocode:

algorithm Halton-Sequence is
    inputs: index ${\displaystyle i}$
            base ${\displaystyle b}$
    output: result ${\displaystyle r}$

    ${\displaystyle f\leftarrow 1}$
    ${\displaystyle r\leftarrow 0}$

    while ${\displaystyle i>0}$ do
        ${\displaystyle f\leftarrow f/b}$
        ${\displaystyle r\leftarrow r+f*(i\operatorname {mod} b)}$
        ${\displaystyle i\leftarrow \lfloor i/b\rfloor }$

    return ${\displaystyle r}$

An alternative implementation that produces subsequent numbers of a Halton sequence for base b is given in the following generator function (in Python).[2] This algorithm uses only integer numbers internally, which makes it robust against round-off errors.

def halton(b):
    n, d = 0, 1
    while True:
        x = d - n 
        if x == 1:
            n = 1 
            d *= b 
        else:
            y = d // b
            while x <= y:
                y //= b
            n = (b + 1) * y - x
        yield n / d

• Constructions of low-discrepancy sequences

• Kuipers, L.; Niederreiter, H. (2005), Uniform distribution of sequences, Dover Publications, p. 129, ISBN 0-486-45019-8
• Niederreiter, Harald (1992), Random number generation and quasi-Monte Carlo methods, SIAM, p. 29, ISBN 0-89871-295-5.
• Halton, J. (1964), "Algorithm 247: Radical-inverse quasi-random point sequence", Communications of the ACM, 7: 701-701, doi:10.1145/355588.365104.
• Kocis, Ladislav; Whiten, William (1997), "Computational Investigations of Low-Discrepancy Sequences", ACM Transactions on Mathematical Software, 23: 266-296, doi:10.1145/264029.264064.