|
|
楼主 |
发表于 2025-9-9 23:32
|
显示全部楼层
Proof of the Collatz Conjecture
Author:Yang Yanhong,Yulong County,Lijiang City,Yunnan Province
Email:13312690681m@Sina.cn Lei Zhigang
Abstract:The Collatz Conjecture involves an iterative operation on a number:if the number is even,divide it by 2;if it is odd,multiply it by 3 and add 1.Will this process eventually reach 1?This proof expresses a number in binary form,folds it,and analyzes the structure of the number in the form of 0 and 1,combined with the I Ching(Book of Changes)and the Bagua(Eight Trigrams),to conclude that the Collatz rule is equivalent to a Möbius strip.
Keywords:Collatz Conjecture,Collatz Rule,Life Formula,Equal Transformation;I Ching,Bagua,Qian(Heaven),Möbius Strip
Starting from the operation rule of the Collatz Conjecture,suppose there are numbersS+1
�
+
1
andS−1
�
−
1
that follow the same Collatz operation rule(S
�
is an integer),then:
A=3(S+1)+1
�
=
3
(
�
+
1
)
+
1
T=3(S−1)+1
�
=
3
(
�
−
1
)
+
1
A+T=6S+2
�
+
�
=
6
�
+
2
According to the Collatz operation rule,sinceA+T
�
+
�
is even,it should be divided by 2.This is denoted as the Collatz operation rule:
L(S)=A+T2=3S+1
�
(
�
)
=
�
+
�
2
=
3
�
+
1
​Starting from the 3X+1(Collatz operation rule)when solving-5,-7,-17,the calculation will enter a loop when repeatedly executed.Starting from the negative number operation rule,modify the Collatz operation rule:for negative odd numbers,repeatedly execute 3X-1;for even numbers,divide by 2.
G=3(S′′+1)−1
�
=
3
(
�
″
+
1
)
−
1
C=3(S′′−1)−1
�
=
3
(
�
″
−
1
)
−
1
G+C=6S′′−2
�
+
�
=
6
�
″
−
2
Denote the rule as:
F(S′′)=G+C2=3S′′−1
�
(
�
″
)
=
�
+
�
2
=
3
�
″
−
1
Then:
A+G+T+C=2[L(S)+F(S′′)]
�
+
�
+
�
+
�
=
2
[
�
(
�
)
+
�
(
�
″
)
]
Represent an integer Y as:
Y=log(N⋅1/N⋅X)
�
=
log
⁡
(
�
⋅
1
/
�
⋅
�
)
Then:
−Y=−log(N⋅1/N⋅X)
−
�
=
−
log
⁡
(
�
⋅
1
/
�
⋅
�
)
A+T=6Y+2
�
+
�
=
6
�
+
2
G+C=6(−Y)−2
�
+
�
=
6
(
−
�
)
−
2
L(S)+F(S′′)=3logN+3log(X/N)+1+3logN+3log[1/(NX)]−1
�
(
�
)
+
�
(
�
″
)
=
3
log
⁡
�
+
3
log
⁡
(
�
/
�
)
+
1
+
3
log
⁡
�
+
3
log
⁡
[
1
/
(
�
�
)
]
−
1
=6logN+3log(1/N⋅1/N)
=
6
log
⁡
�
+
3
log
⁡
(
1
/
�
⋅
1
/
�
)
=6logN−6logN=0
=
6
log
⁡
�
−
6
log
⁡
�
=
0
Thus:
A+T+G+C=L(S)+F(S′′)
�
+
�
+
�
+
�
=
�
(
�
)
+
�
(
�
″
)
That is,Theorem 1:The property remains unchanged after folding.
Write the number S in binary and then fold it.There are four forms corresponding to the starting number:0,1,10,11.By writing the Bagua in a left-right structure instead of the traditional top-bottom structure,the correspondence between the 64 hexagrams and the AGCT genetic code can be derived.
​○○A
●○C
●●G
○●T
AAA corresponds to Qian in the 64 hexagrams.
AAA
○○
○○
○○
GGG corresponds to Kun in the 64 hexagrams.
GGG
●●
●●
●●
In the horizontal row,there are the eight trigrams:
○○○
A○
○○●
A1
●○○
C○
●○●
C1
○●○
T○
○●●
T1
●●○
G○
●●●
G1
Qian represents Heaven
AAA
○○
○○
○○
Equivalent to
AOAO
oooooo
Qian represents Heaven
AAA=AOAO
Starting from the equationAAA=AOAO
AAA
=
AOAO
,
we derive the following:
x3=2x2(1)
�
3
=
2
�
2
(
1
)
3x=2x+2(2)
3
�
=
2
�
+
2
(
2
)
From the life formula(equivalent transformation),
A=TandG=C
�
=
�
and
�
=
�
Thus,
A+G=T+C
�
+
�
=
�
+
�
which implies
x3+3x=2x2+2x+2
�
3
+
3
�
=
2
�
2
+
2
�
+
2
Rearranging,we get
3x+1=2x2+2x+3−x3
3
�
+
1
=
2
�
2
+
2
�
+
3
−
�
3
Whenx=2
�
=
2
,the equation holds true.
Substitutingx=3
�
=
3
into the equation,
we get
10=0
10
=
0
Checking 10,we have the sequence:
10→5→16→8→4→2→1
10
→
5
→
16
→
8
→
4
→
2
→
1
Thus,10,when checked through the Collatz conjecture,also eventually reaches 1.
Ifx=3
�
=
3
,we can calculate:
10=0
10
=
0
Checking out 10,we get:
10→5→16→8→4→2→1
10
→
5
→
16
→
8
→
4
→
2
→
1
Therefore,10 will also fall to 1.
For negative numbers,consider the sequence starting from-10:
−10÷2=−5
−
10
÷
2
=
−
5
−5×3+1=−14
−
5
×
3
+
1
=
−
14
−14÷2=−7
−
14
÷
2
=
−
7
−7×3+1=−20
−
7
×
3
+
1
=
−
20
−20÷2=−10
−
20
÷
2
=
−
10
−10÷2=−5
−
10
÷
2
=
−
5
For an infinitely large number,∞
∞
,take the reciprocal:
S=1∞
�
=
1
∞
which is denoted as
S=000000
�
=
000000
The I Ching trigrams of Qian with an upper-lower structure and those with a left-right structure are equivalent.
It can be known that:
000000
○○
○○
○○
Thus,AOAO=AAA
That is,2A2=A3
2
�
2
=
�
3
①
2A+2=3A
2
�
+
2
=
3
�
②
Adding ① and ②:
A3+3A=2A2+2A+2
�
3
+
3
�
=
2
�
2
+
2
�
+
2
3A+1=2A2+2A+3−A3
3
�
+
1
=
2
�
2
+
2
�
+
3
−
�
3
③
WhenA=2
�
=
2
,equation ③ holds true.
WhenA=3
�
=
3
:
3A+1=2A2+2A+3−A3
3
�
+
1
=
2
�
2
+
2
�
+
3
−
�
3
10=0
10
=
0
This is denoted as[ten=O]
[
�
�
�
=
�
]
.
That is to say,when using a computer to calculate the Collatz conjecture,when the number reaches infinity,the memory overflows,and it cannot be verified by a computer.
For example,20 written in binary is 10100.When folded,there are two possible starting digits:1 or 0.
The number 20 written in binary can be mathematically expressed asS=E(s)
�
=
�
(
�
)
.
• When starting with 1,S=E(s)
�
=
�
(
�
)
wheres=2mod(3)
�
=
2
mod
(
3
)
.
• When starting with 0,s/2
�
/
2
wheres=0mod(2)
�
=
0
mod
(
2
)
.
By substitutings=0
�
=
0
,s−1
�
−
1
,ands+1
�
+
1
into the Collatz rule,we can determine thatA=3(0+1)+1
�
=
3
(
0
+
1
)
+
1
andT=3(0−1)+1
�
=
3
(
0
−
1
)
+
1
.Therefore,A+T=2
�
+
�
=
2
.
It follows that there exists a Collatz ruleL(0)=1
�
(
0
)
=
1
.This means that if you write 0 on one side of a strip of paper and 1 on the other side,and then twist and join the ends,you will create a Möbius strip,which illustrates the Collatz operation rule.
The Collatz conjecture involves converting a number to binary and folding it.There are four possible starting digits in binary:1,0,10,and 11.Assuming that the Collatz operation iteration eventually returns to the starting number,and givenL(0)=1
�
(
0
)
=
1
(i.e.,0 and 1 are equivalent),and since binary 10 equals 2,we can conclude thatA=2
�
=
2
,thus establishing Equation ③.
Given that binary 11 equals 3,which is denoted asG
�
,the binary representation of-5 should be in two's complement form,which is 1111 1011(the original code is 1000 0101,the one's complement is 1111 1010,and the two's complement is 1111 1011).
When-5 is folded in binary and represented symbolically,it isGCGG
�
�
�
�
.This indicates thatX=−5
�
=
−
5
has entered a loop of-10→-5→-7→-20→-10.
References:
[1]I Ching
[2]Asimov's New Guide to Science-The Formula of Life
​ |
|
IP卡
狗仔卡
发表于 2025-9-3 22:34
提升卡
置顶卡
沉默卡
喧嚣卡
变色卡
显身卡