# Goldbach’s Conjecture and the 2-Way Sieve

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Goldbach’s Conjecture and the 2-Way Sieve

Intro:
Mr. Hui Sai Chuen is an amateur mathematician born in May 1937 in the Canton province of China. After graduating with an engineering and construction degree, he proceeded to do research on architecture and material science before moving to Hong Kong in 1981. From there he enjoyed a successful career with various construction and interior design companies until his retirement in 1993. The ever-curious mathematician, Mr. Hui then dedicated his time to studying number theory, particularly Goldbach’s conjecture and its uses. He published his findings in several papers, two of which were the ‘2-Way Sieve’ and the ‘Double-Key Lock Cipher Theory’, both highly commended for its novelty in several national symposiums (the latter discovery was patented in China in 2004).
The 2-Way Sieve is a new sieve method created by Mr. Hui that can sieve odd prime number pairs. Although the formula has been verified by Professor Jong Ji of the South China Normal University and by the “Mathematics Journal” published by the Mathematics Research Institute of the Chinese Academy of Social Sciences in 2001, the theory is nonetheless somewhat novel, a new mathematical concept of a multifunctional sieve. Put simply, it can be explained as follows. For example:

Take any even number, w = 26 say.
We look at the set of its non-negative integers:
Forward Subset: S+ = 0, 1, 2, … , 24, 25, 26
Backward Subset: S- = 26, 25, 24, … , 2, 1, 0

S+= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
S-= 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

By combining the two subsets, we can see that the sum of any arbitrary pair of numbers is equal w = 26.
The prime factors of the subsets S+ and S- are:
√w = √26 = 5.099, which after rounding, the prime factors are 2, 3 and 5.
By sieving the numbers that can be divided exactly by 2, 3 and 5 in the subsets S+ and S-, we obtain the forward and reverse collection as follows:

S+= x 1 2 3 x 5 x 7 x x x 11 x 13 x x x 17 x 19 x x x 23 x x x
S-= x x x 23 x x x 19 x 17 x x x 13 x 11 x x x 7 x 5 x 3 2 1 x

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CRGreathouse
Homework Helper
I couldn't find anything on Mr. Hui and his cipher (or its patent):

The "forward subset" and "backward subset" {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26} and {26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0} are equal (as sets). If you want them to be distinct, you'd need to call them 27-tuples or sequences rather than sets.

Sets don't have natural prime factors. The gcd of the set is 1, and its lcm is undefined/infinite/zero depending on your definition of lcm(0, 26771144400).

This sieving method takes time $2\pi(\sqrt n)\sim \sqrt n/\log n$ and memory $2n$ which is too slow for any practical use.

I don't understand why you sieve twice, what the correspondance between the two sieves is, or why 2, 3, and 5 remain.

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Hurkyl
Staff Emeritus
Gold Member
Sieving is good for some applications. For example, if you seek all prime numbers in a large range, you should first sieve it by small primes. It's much faster than doing trial division by small numbers on each one, which is the first step you'd normally do in a primailty test.

CRGreathouse
Homework Helper
I've written large segmented sieves, one running past 100 trillion to find a particular Sloane sequence. Certainly they're much faster than testing sequential numbers. But what possible advantage could this sieve have? First, I'm not entirely sure what it's claiming to find; second, it seems to use several times more memory and processing power than is needed.

Hurkyl
Staff Emeritus
Gold Member
Oh, sorry, I didn't notice you said "THIS sieving method". I thought you were talking about sieving in general.

If I understood correctly, by yuxtaposing the forward and backward sequences, he was trying to obtain pairs of primes that add to 26 (3+23, 7+19, 13+13), in the example. Though it seems more of a visual aid than a practical method.

CRGreathouse
Time: 60n sqrt(n) / C + 2C sqrt(n)/ln(n) + 1.5n (roughly; with large C this becomes $O(C\sqrt n/\log n)$)