A probabilistic and morphological change analysis on 3x+1(researchgate)withDOI

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A probabilistic and morphological change
analysis on 3x+1
Working Paper · December 2018
DOI: 10.13140/RG.2.2.11276.51841
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Yucong Duan
Hainan University
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A probabilistic and morphological change analysis on 3x+1
Yucong Duan
College of Information Science and Technology , Hainan University, China
Email: duanyucong@hotmail.com
Abstract: From probabilistic perspective, we present an analysis on potential incurred chances of
division by “2” without reminder during a single process of the conjuncture of “3x+1” in different
sections of the binary expression of an input number. We revealed the difference of the tail section
and the other parts for a big input number. Based on the ratio of the difference of the change of
different sections of the expression with randomly assigned values in a single process, we identified
the decreasing tendency of the input number.
1. Introduction
We formalize the Collatz conjuncture[1],[2] as the following operations on input number n.
,
(n) 3 1,( )
(n)
(n) ,(n even)
2
n N
TO n n odd
T n
TE
 
   

 
 

When we consider the situation of inputting an odd number of x. From the binary coding, the value
of the lowest position of the binary powers 1 2 3 1
{ , , ,..., , }i i
b b b b b
matching x, which is
denoted as 1b
x , must be “1”.
1 2 3 1
1 2 3 1
1
,
binaryPowers(x) : { , , ,..., , }
( ) {x ,x , ,..., , }, {0,1}
binary( (x)) 1;
i i
i i
b b b b b b
b
x odd
b b b b b
binary x x x x x
odd x


 

 
 
2. Predicting the probabilistic values of binary positions
For any input odd number, the values of the corresponding binary positions at various powers will
be either “0” or “1”. In general, the possibility of randomly chosen values which map to be “0” or
“1” is equal.
We apply this hypothesis to values of 2b
x .

2
2
probability :
( 1) 1 / 2;
( 0) 1 / 2;
b
b
p x
p x
 
 
Therefore we will have the following sections of x to evaluate the incurred direct chances of division
by “2”, which is denoted as steps of “/2”. For example, the situation that 2 1b b
x x equals to binary
“100” indicates that there are two chances of directly being divided by “2” at the tail section of x.
2 1
2 1
2 1
( ) :
01 100 " 2 "
11 1010 " 1 "
b b
b b
b b
T x x
x x steps
x x steps
  
  
We continue to apply this hypothesis to values of 3 4b b
x x . The values of four equal possibilities
are listed as follows.
4 3
4 3
4 3
4 3
4 3
p(random( )):
p( 00) 1 / 4
p( 01) 1 / 4
p( 10) 1 / 4
p( 11) 1 / 4
b b
b b
b b
b b
b b
x x
x x
x x
x x
x x
 
 
 
 
Therefore we will have the continued sections of x to evaluate the incurred direct chances of
division by “2”. We calculate the comprehensive situations of combining values of 3 4b b
x x and
values of 2 1b b
x x as follows. We also calculate the average incurred steps and the ratio of the
scaling up influence on x by “*3” vs. the scaling down influence on x from the incurred divisions by
“2” in a single phrase of “3*x+1”.

4 3 2 1
2 1
4 3
4 3
4 3
4 3
(random( ) ( )) :
01 100 " 2 "
00 0000 " 4 "
00 0100 " 2 "
00 1000 " 3 "
00 1100 " 2 "
4 2 3 2 11
averageSteps ("/ 2 ")
4 4
b b b b
b b
b b
b b
b b
b b
A
A
T x x T x x
x x steps
x x steps
x x steps
x x steps
x x steps
scalingUp
scaleInfluence
scalin
  
 
 
 
 
  
 

3 6
11 11
2 *
4
gDown
 
4 3 2 1
2 1
4 3
4 3
4 3
4 3
(random( ) ( )) :
11 1010 " 1 "
10 0010 " 1 "
10 0110 " 1 "
10 1010 " 1 "
10 1110 " 1 "
1 1 1 1
averageSteps ("/ 2 ") 1
4
b b b b
b b
b b
b b
b b
b b
B
B
T x x T x x
x x steps
x x steps
x x steps
x x steps
x x steps
scalingUp
scaleInfluence
scaling
  
 
 
 
 
  
 

3 3
2 * 1 2Down
 
Here comes the average comprehensive chances of division by “2” in a single process of “3*x+1”
based on random chosen values of 4 3 2 1
random( )b b b b
x x x x .
:
averageSteps ("/ 2 ") averageSteps ("/ 2 ")
2
11
1
4
2
15
8
A B
ComprehensiveSteps





The comprehensive scaling influence in a single process of “3*x+1” is calculated as follows.

:
*
6 3
*
11 2
9
11
1
A B
ComprehensiveScaling
scaleInfluence scaleInfluence



Since the comprehensive scaling is less than “1”, it indicates that x will be reduced on average with
this ratio in a single interval of “3*x+1”. Continued application of this scaling ratio on any given x
will reduce it to a small number. Since many computation attempts have examined that quiet big
numbers can be reduced to “1” following the conjuncture, we can reasonably assume that any
given x will be reduced to small enough to fall into the examined scope of numbers.
lim
lim
lim
3* 1
,
(x)
:: x*
9
:: x* ( )
11
9
x* ( ) min
11
itedTimes
itedTimes
itedTimes
x
x odd
TC
exa edNumber 
 


 
3. An additional metaphor of the change tendency
If x is a very big number, the tail section of the its binary expression will seem to be propelled by
the increasing of an average amount of “0” equal toComprehensiveSteps at each “3*x+1”
interval. The other part of the binary sections of x, which we call mainbody, will be propelled by
“*3” at equal intervals. We can evaluate the difference of the progressing velocities of the tail
section vs. the mainbody as follows.

Re Pr (tail/ MainBody) :
*
* MainBody
15
2 *
8
1 * 3
5
4
1
tail mainBody
lative ogress
step ComprehensiveSteps
step
velocity velocity




 
We can draw the conclusion that the tail section is progressing faster than the mainbody. Another
metaphor is that the section comprising purely “0” will approach the highest “1” of the binary
expression at a velocity ratio of
5
Re Pr (tail/ MainBody):
4
lative ogress .
4. Reaching more precision
Above investigation explores the conclusion based on the evaluation of the probability of the last
four bits of the binary expression of the number x.
The conclusion will be maintained as long as the decreasing tendency is justified as not being
weakened after taking the rest unexplored bits into consideration. This means that the amount of
steps of division by “2” in a single process of “3*x+1” should not be reduced. Similarly the scale
influence of considering the four bits should be no bigger than the situation of considering more
bits.
4 3 2 1
4 3 2 1
(random( ) ( ))
( )
(random( ) ( ))
( )
averageSteps("/ 2 ")
averageSteps("/ 2 ")
b b b b
b b b b
T x x T x x
T allBits
T x x T x x
T allBits
or
scaleInfluence
scaleInfluence


Let’s investigate the situation of considering more bits than 4. If 4 3 2b b b
x x x contains at least one
“1”, there will be no chance to have more direct chances of division by “2” even if more bits after
the 4 bits are taking into consideration. This is because the processing of division by “2” in a single
process of “3*x+1” will stop by the first “1” which indicates the even number gained from previous
“*3+1” operation is turned into an odd. The formation of odd within the 4 bits at the tail section
indicates that even if more bits are considered there will be no increase in the amount of chances
of division by “2”.

4 3 2
( 1) ( 1) ( 1)
(odd) Yes
(divisionBy(2))
b b b
x x x
identification
stop
    
 

From this conclusion, we can get the following detail of the change influence of considering more
bits. We mark “NoChange” to the situation that further consideration of more bits will not
influence the current concluded amount of steps due to the existence of “1” in current considered
bits. The only situation which need to be further explored is that all existing bits are “0”.
4 3 2 1
4 3
4 3
4 3
4 3
random( ) ( )) :
(moreBits) 00 ? 0000
(moreBits) 00 ? 0100 0100
(moreBits) 00 ? 1000 1000
(moreBits) 00 ? 1100 1100
(
b b b b
b b
b b
b b
b b
consideringMoreBitsThan x x T x x
x x ToBeExplored
x x NoChange
x x NoChange
x x NoChange
 
  
  
  
4 3
4 3
4 3
4 3
moreBits) 10 ? 0010 0010
(moreBits) 10 ? 0110 0110
(moreBits) 10 ? 1010 1010
(moreBits) 10 ? 1110 1110
b b
b b
b b
b b
x x NoChange
x x NoChange
x x NoChange
x x NoChange
  
  
  
  
Therefore only when 4 3 2b b b
x x x contains no “1” or purely “0” it will need to considered for more
bits since the intermediate result is not justified as an odd. The division by “2” should stop only
when an old is reached. The reaching of an odd is justified when the first “1” is met through the
consecutive division by “2” taking more bits into consideration.
4
4 3 2
( 0) ( 0) ( 0)
(odd)
(explore(bitPosition 4))
(divisionBy(2)) beforeReaching(first" 1 " )
b
b b b
x
x x x
identification No
continue
continue 
    
 
 
 
When considering more bits after 4 3 2b b b
x x x , there will be chances of increasing the amount of
steps of division by “2” while no chances of decreasing the amount of steps of division by “2”.
(explore(bitPosition 4)) (steps)
(4 ) reservedEvaluation
continue decrease
evaluation bits
  
 
Therefore the worst case is that 5b
x equals to “1” which shares the same amount of evaluation
of steps as the current evaluation of considering 4 bits. The other situations will be better since if
5b
x equals to “0” and subsequent bits equal to “0” will bring additional chances of division by “2”.

5. A further refinement towards more precision
To evaluate the tendency which is required by proving the conjecture, the above discussion is
sufficient.
Although we don’t take the evaluation of the stop time[3], [4] as a critical step for the proof, we’d
like to explore further by taking more bits into consideration to reveal the more precise evaluation
of the average steps and scaling influence in a single process of “3x+1”.
Since the only situation which needs to be further explored is
4 3
00 0000b b
x x ToBeExplored  , we focus on the exploration of the various situation
of that the coming bits are bundled with different values of either “0” or “1” before the last bit of
“1” which locates in the highest position and marks the end of the expression body of the binary
x.
Through the investigation, we identified the following rules:
(i) If the value of the next bit is “0”, it equals to a claiming of an additional chance of division
by “2”. Since the chance of that the value of a random positon equals to “0” is the same
as that the value of a random positon equals to “1”, we identify the average additional
contribution to the change of the existing accumulated chances of division by “2” with the
following formula.
::
" 0 "
(additional) 1 / 2
progress(additional) 1
(additional) * progress(additional)
(1 / 2) * 1
1 / 2
updated existingBits additional
additional
Steps Steps Steps
additional
probability
Steps
probability
 







(ii) If the value of the next bit is “1”, it means that an odd number has been identified for the
current process of division by “2” in the current process of “3x+1”. Therefore there is no
more additional chance of division by “2” to be added when more bits are considered from
the positon of current bit.
Assume there are m bits between the existing bits considered bits and the bit of “1” at the highest
position in the binary expression of the input odd number x, the accumulated change can be
calculated as follows:

1...
(i)
1
::
madditional
i m
existingBits additional
i
Steps
Steps Steps


  
1...
1... 1 2 1
1
1
1
1
1
" 0 0 ...0 0 "
(additional(0 ))
(1 / 2) * (additional(0 ))
(1 / 2) * (1 / 2)
(1 / 2) * (1 / 2)
(1 / 2)
progress(additional(0 )) 1
m m m
m
i
j i
j
j
j i
j
j
i
i
i
additional
additional
probability
probability
Steps

 

 











1
1
1
(additional(0 )) * progress(additional(0 ))
(1 / 2) * 1
(1 / 2)
(1 / 2) * (1 (1 / 2) ) / (1 (1 / 2))
1 (1 / 2)
m
i m
i i
i
i m
i
i
i m
i
i
m
m
probability









  
 



If m is very big, the accumulated steps brought by taking additional m bits into consideration will
approach “1” step.
lim(1 (1 / 2) ) 1m
m 
 
Therefore the total influence by considering all additional randomly chosen values of the bits in
the binary expression of x can be calculated as follows.

"1" (m 1) 2 1 4 3 2 1
4 3
"1" (m 1) 2 1 4 3 2 1
(random( ... ) ( )) :
00 0000 " 4 "
...
1 00...00 0000 " 4 1 5 "
5 2 3 2
4
am a b a b b b b
b b
am a b a b b b b
m
m
A m
A
T x x x x x x x T x x
x x steps
x x x x x x x x x
steps
scaling
scaleInfluence
 
 


 
   
  
 

3 1
2 * 3 2
Up
scalingDown
 
4 3 2 1
2 1
4 3
4 3
4 3
4 3
(random( ) ( )) :
11 1010 " 1 "
10 0010 " 1 "
10 0110 " 1 "
10 1010 " 1 "
10 1110 " 1 "
1 1 1 1
4
b b b b
b b
b b
b b
b b
b b
B
B
T x x T x x
x x steps
x x steps
x x steps
x x steps
x x steps
scalingUp
scaleInfluence
scaling
  
 
 
 
 
  
 

3 3
2 * 1 2Down
 
Here comes the average comprehensive chances of division by “2” in a single process of “3*x+1”
based on random chosen values of "1" (m 1) 2 1 4 3 2 1
random( ... ) ( ))am a b a b b b b
x x x x x x x T x x 
.
"1" (m 1) 2 1 4 3 2 1
(random( ... ) ( ))) :
averageSteps ("/ 2 ") averageSteps ("/ 2 ")
2
3 1
2
2
am a b a b b b b m
A B
ComprehensiveSteps x x x x x x x T x x
steps
  





The comprehensive scaling influence by considering
"1" (m 1) 2 1 4 3 2 1
random( ... ) ( ))am a b a b b b b
x x x x x x x T x x  in a single process of “3*x+1” is
calculated as follows.

"1" (m 1) 2 1 4 3 2 1
(random( ... ) ( ))) :
*
1 3
*
2 2
3
1
2
am a b a b b b b m
A B
ComprehensiveScaling x x x x x x x T x x
scaleInfluence scaleInfluence
  


 
Since the comprehensive scaling is less than “1”, it indicates that statistically any big x will be
reduced on average with this ratio in a single interval of “3*x+1”. Continued application of this
scaling ratio on any given big x will reduce it to a smaller number which will fall into the category
of verified numbers which will be reduced to “1” with operations of “3x+1”.
lim
lim
lim
3* 1
,
(x)
:: x*
3
:: x* ( )
2
3
x* ( ) min
2
itedTimes
itedTimes
itedTimes
x
x odd
TC
exa edNumber 
 


 
6. Reference
[1] Zoet A, Følsgaard J M, Pedersen R K, et al. On the strategies used to attack unsolved
mathematical problems-a case study of the Collatz Conjecture[D] , 2016.
[2] Gaifman H. A note on models and submodels of arithmetic[C]//Conference in Mathematical
Logic—London’70. Springer Berlin Heidelberg, 1972: 128-144.
[3] Crandall R E. On the “3𝑥+ 1” problem[J]. Mathematics of Computation, 1978, 32(144): 1281-
1292.
[4] Lagarias J C. The 3x + 1 Problem and Its Generalizations[J]. American Mathematical Monthly,
1985, 92(1):3-23.
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