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发表于 2025-9-13 23:41
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Collatz conjecture & Mobius ring
Author：Yang Yan Hong From Jade dragon county Yun Nan province China
Email：13312690681m@Sina.cn translate by : Tim Lei
【Summary】 Collatz conjecture also named Sizuo Kakutani assumption, or Hail assumption, the content as below : For a natural number, if it is even number, then divide by 2, if it is odd number, then multiply by 3 and add 1, with regard to get result, same operating again & again , Finally, it will falling to 1 result.
In the proof, Change number to binary, then to fold the number like folio，used 0,1 form, combine周易八卦 to analyse the number frame, to get when x=3, then 3x+1=10, then it will close Collatz conjecture.
【Key words】Sizuo Kakutani assumption, Sizuo Kakutani regulation, Life formula, Equal volume transformation, 周易、八卦、乾 . Mobius ring .
The problem of halting Turing machine
If when input 000000, the Turing machine can shutdown.
As Sizuo Kakutani assumption rule , if have number (S+1)，(S-1), follow the rule（S= integer）
Then : A=3（S+1)+1 T=3(S-1)+1
A+T=6S+2
From Sizuo Kakutani assumption rule ，if A+T equal even number, should be divided by 2
Mark by : L(S)=(A+T)/2=3S+1
When -5，-7，-17, Run 3X+1（Sizuo Kakutani assumption rule），when calculation run, it will fall into circulating ring, and from negative number operation rules, change Sizuo Kakutani assumption rule, Negative odd repeat implement 3X-1, Negative even number repeat divided by 2。
G=3(S”+1)-1
C=3(S”-1)-1
G+C=6S”-2
Mark by: F(S”)=(G+C)/2=3S”-1。
Then: A+G+T+C=2*[L(S)+F(S”)]
Use Integer Y to show, Y=log（N*1/N*X）
Then -Y=-log（N*1/N*X)
A+T=6Y+2
G+C=6(-Y)-2
L(S)+F(S”)=3logN+3log（X/N)+1+3logN+3log[1/(NX)]-1
=6logN+3log(1/N*1/N)
=6logN-6logN=000000
The same: A+T+G+C=L(S)+F(S”)
Theorem 1: after reduced by one half, property have no change.
Integer S Convert to binary system, then reduce by one half like folio，can get start bit : 0,1,10,11; get 4 types。 And 0=00; 1=01; it is knowable 00 , 01 , 10, 11 is parallel with integer 0,1,2,3 .
Arrange “八卦”，from up down (vertical ) to left & right (horizontal), can get 64 divinatory symbols. It will get AGCT genetic code is parallelism with 64 divinatory symbols .
○○ A
●○ C
●● G
○● T
AAA is “乾 ” in 64 divinatory symbols .
AAA
○○
○○
○○
GGG is “坤” in 64 divinatory symbols .
GGG
●●
●●
●●
Rank it, get “八卦”
○○○
A○
○○●
A1
●○○
C○
●○●
C1
○●○
T○
○●●
T1
●●○
G○
●●●
G1
“乾”symbol “Sky” in China . Mean Positive.
AAA is array in vertical :
○○
○○
○○
Equal to:
AOAO is array in horezontal
oooooo
And
AAA=AOAO
Because : AAA=AOAO
Therefore : x3=2X2（1）
3x=2x+2（2）
Because Life Formula（Equal volume transformation）
A=G T=C
So: A+T=G+C
The same: x3+3x=2x2+2x+2
Therefore: 3x+1=2x2+2x+3-x3
When: X=2, it is equality .
If x=3, can calculate
10=0
Checkout 10 , then ,10→5→16→8→4→2→1 .
Then: 10 will fall to 1 also .
- 10=0，
-10÷2=- 5，
- 5x3+1=- 14
-14÷2=-7
-7x3+1=-20
-20÷2=-10
-10÷2=-5
About infinity number, ∞ carry on count backwards
S=1/∞ record: S=000000
Because up down arrange (array in vertical) “乾”equal to left & right (array in horizontal“乾”;
Get 000000= 00
00
00
○○
○○
○○
The same : AOAO=AAA
The same : 2A2=A3 ①
2A+2=3A ②
+ ②
Then A3+3A=2A2+2A+2
3A+1=2A2+2A+3-A3 ③
If A=2 It is equality .
If A=3 then
3A+1=2A2+2A+3-A3
10=0
Mark by : [十= O]
It means run Sizuo Kakutani assumption rules calculation by computer，when infinity number, will overflow from internal memory，can not calculate & verify .
For example, 20 Convert to binary system is 10100 , after reduced by one half, the start bit is 1 or 0 . if have theorem 1 that after reduced by one half, property have no change. 20 Convert to binary system is 10100 , reduced by one half,
Remark by : S=E(s)
When start bit is 1 , S=E(s) , S =2 mod(3)
When start bit is 0 , S/2 , S =0 mod(2)
And S =0, S -1, S +1 , input to Sizuo Kakutani assumption rules, get :
A=3(0+1)+1 ; T=3(0-1)+1 ;
So A+T=2.
Therefore have rule, L(0)=1 .
When 20 is written in binary, it can be expressed mathematically as follows: For S = E(s), when the starting digit is 1, it is 1; when S = E(s) and s = 2 mod(3), the starting digit 0 is 0; when s/2 and s = 0 mod(2), substituting s = 0, s - 1, s + 1 into the Collatz rule, we can know that A = 3(0 + 1) + 1, T = 3(0 - 1) + 1. Since A + T = 2, there exists the Collatz rule L(0) = 1, that is, write 0 on one side of a paper tape and 1 on the other side, and then twist and connect them, and we can know that the Collatz operation rule is like a Möbius strip.
The Collatz conjecture converts a number into binary and folds it in half. There are four possibilities for the starting number: 1, 0, 10, 11 (in binary). Assuming that after the iterative operation of the Collatz conjecture, it finally falls back to the starting number, then from L(0) = 1, that is, 0 and 1 are equivalent. Since 10 in binary is equal to 2, we can know that A = 2, and equation ③ holds. Since 11 in binary is equal to 3 which is G, the binary representation of -5 should be expressed in two's complement, which should be 1111 1011 (the original code is 1000 0101, the one's complement is 1111 1010, and the two's complement is 1111 1011). When the binary representation of -5 is folded in half and represented by symbols, it is GCGG. Then we can know that X = -5 has entered a cycle of -10 → -5 → -7 → -20 → -10.
References:
[1] "The Book of Changes"
[2] "Asimov's New Guide to Science"
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