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A Study to the 3n+1 Problem with State

Transition Model

Shu-Ming Chang

Department of Applied Mathematics National Chiao Tung University

1001 University Road, Hsinchu, Taiwan 300, ROC smchang@math.nctu.edu.tw

Lung-Lung Liu

International College Ming Chuan University

5 De-Ming Road, Gui-Shan, Taoyuan County, Taiwan 333, ROC llliu@mail.mcu.edu.tw

Corresponding Author: Lung-Lung Liu Telephone: 886-3-3507001 ext. 3386 Fax: 886-3-3599883

Abstract: In this study to the 3n+1 problem, we observed the “behaviors” of a subset of 32 odd

numbers in binary, and we found that: (1) the numbers can be classified into 4 groups according to their bit sequences, (2) to the numbers, at any instance right after the 3n+1 and all possible n/2 operations were done, the results can still be classified into the 4 groups, and (3) if “a number in a specific group” can be defined as a “state” and the “operations for a result” can be applied as a “transition”, then it is feasible that a state transition model can be built. We did a quick analysis to support the observations and then try to build such a model. We have been clear that we are not to solve this open problem since the very beginning. Instead, we are to propose an alternate way to find and understand the non-determinism in this problem. The result of the study indicated that: (1) a general state transition model can be built, (2) at the instance of a state transition, the chance that the next state is non-deterministic is less than 1 among the 8 possible cases, and (3) a finite state machine is not able to finish the 3n+1 and n/2 operations when a given number n is large.

Keywords: 3n+1, Collatz conjecture, state transition model, non-determinism Manuscript

Click here to download Manuscript: A Study to the 3n1 with state transition model tocs version 2.docxClick here to view linked References

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1 Introduction

A description of the 3n+1 problem can be found in [1]. However, the problem can be simply stated as follows: (1) Let n be a given positive number. (2) If n is even, then assign it with n/2. Otherwise (i.e., n is odd), assign it with 3n+1. (3) Repeat the above mentioned even/odd checking and assigning operations, the result will finally be 1. (4) Is it true for any given positive number n? For example, let n be 7. The sequence of numbers being encountered is 7→ 22→ 11→ 34→ 17→ 52→ 26→ 13→ 40→ 20→ 10→ 5→ 16→ 8→ 4→ 2→ 1. It is true since the result is 1. But what if n is given as 7,777,777? Is it still true that the result will finally be 1? It is hard to tell.

In fact, this 3n+1 problem is an open question and it is also known as the Collatz conjecture (i.e., a mathematical statement seemed to be true but not yet been proved) [3]. When n is large, even a computer may not be able to tell the result in time. General studies and discussions are in [3] and [6]. An annotated

bibliography is collected as [4], which is publically available. The latest results about the study are those such as [5], which discussed the undecidable variants, and [2], which discussed the nature of 3n+1 map. We learned in the literature that binary (or ternary) numbers were applied before, and we think that there can be possible follow-up works. We are not to solve the problem, but instead, we would like to study in some way which may be clearer than others. Actually, the study is with the state transition model.

The study is with three steps: (1) to observe the results of 3n+1 and n/2 operations of given numbers in binary, or their bit sequences, (2) to induce the possible rules in the change of the bit sequences with analytical support, and (3) to try to build a state transition model and find the embedded non-determinism. In the following, observations are described in section 2, and a quick analysis to support the observations is provided in section 3. With the support, a state transition model is gradually constructed. It is specified and discussed in section 4. The analysis of working values (i.e., numbers encountered in the 3n+1 and n/2 operations) is provided in section 5. That the study is logically sound is demonstrated with proofs in section 6, and conclusions are given in section 7.

2 Observations with a Subset of Numbers in

Binary

To study the 3n+1 problem, a binary number system (e.g., only bits 0 and 1 are used) is convenient for observation. First, evens and odds can be easily recognized by checking their last bits. Second, the needed arithmetic operations 3n+1 and n/2 to any given number n can also be simplified by using basic “shift” and “add” operations. For example, by appending a 0 to the end of n (shifting left), add n, and then add 1, it is the result of 3n+1. On the other hand, by dropping the last bit of n (shifting right), it is the result of n/2 when n is even.

Let a number n be “qualified” if finally it will be 1 after the possible 3n+1 and n/2 operations according to the checking and assigning rules. It is clear that binary numbers 1, 10, 100, 1000 … are qualified. It is interesting that numbers 101, 10101, 1010101, 101010101 … are qualified, too. To each of these numbers, the 3n operation will obviously result a number with all 1s, and the next +1 operation

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makes it qualified. For example, giving 10101 as n, the result of 3n is 101010+10101=111111, and that of 3n+1 is 1000000. It is trivial that if n is qualified then 2n is qualified, too. Hence, we have to pay attention to odd numbers only.

Binary numbers 11, 111, 1111, 11111 …, which are with more than 1 “trailing 1s”, are with a unique feature in the 3n+1 and n/2 operations. If we check the result of the operations performed on any of these numbers, then we will see that the number of remaining “trailing 1s” is decreased by 1, compared with that of the original number. Besides, only one n/2 operation can be performed since after that the result is odd, and the result is greater than n. For example, giving 1111 as the number n, then (3n+1)/2 is 10111.

Binary numbers 1001, 10001, 100001, 1000001 …, which are with more than 1 “in-between 0s” of the two 1s at the end, are with another unique feature in the 3n+1 and n/2 operations. If we check the result of the operations performed on any of these numbers, then we will see that the number of remaining “in-between 0s” is decreased by 2, compared with that of the original number. Besides, only two n/2 operations can be performed since after that the result is odd, and the result is less than n. For example, giving 10001 as the number n, then ((3n+1)/2)/2 is 1101.

However, there are many other numbers in addition to those mentioned. To prevent that no any number is missed in the observations, we use a systematic discipline to count all of the elements in a set of odd numbers. The following is an example set of such numbers between 000001 and 111111. In this set, there are 32 odd numbers in binary, and they are classified into 4 groups according to their last bit sequences. The groups are named as Z1, S11, S101, and S1Z01 for being mnemonic: Z represents a bit sequence of 0, and S represents a bit sequence of either 0 or 1. In short, (1) the number 1 is in group Z1, (2) the numbers with more than 1 “trailing 1s” are in group S11, (3) those with only 1 “in-between 0” are in group S101, and (4) those with more than 1 “in-between 0s” are in group S1Z01. Numbers in each of the 4 groups are then arranged in ascending orders. This discipline guarantees that no number will be missed, since it is with a

combinatorial count on the number of “trailing 1s” and then the “in-between 0s”. In the previous example, the numbers of elements in the groups Z1, S11, S101, and S1Z01 are 1, 16, 8, and 7, respectively. Or, in general, the numbers are 1, c/2, c/4, and c/4-1, where c is the total number of elements in the set.

With the discipline, the details of observation on the numbers in the set are recorded, and they are depicted in Table 1.

Note that in Table 1, there are 32 entries and they are classified in 4 groups. For every entry, n is the number in binary, n’ is the result of 3n+1 and then all possible n/2 operations. The Group’ is the group that n’ belongs to (to be applied later in Section 4). Since n’ is greater than n for the entries in the S11 group, one extra bit is added to properly represent n’ in binary. (However, the classification rules remain). The underlined bits of the binary numbers are just for easy checking. In summary, what we have observed are:

1a: For n in S11, only 1 n/2 operation is done, and n’>n.

1b: For n in S11, the number of trailing 1s of n’ is 1 less than that of n.

2a: For n in S101, at least 3 (may be more) n/2 operations are done, and n’<n. 2b: Only for n in S101, n’=1 is possible.

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3a: For n in S1Z01, only 2 n/2 operations are done, and n’<n.

3b: For n in S1Z01, the number of “in-between 0s” of n’ is 2 less than that of n. Table 1: An Observation of odd numbers between 000001 and 111111

Group n (binary) value 3n+1 and then possible n/2 n’ (binary) Group’

Z1 000001 1 - - - S11 000011 001011 010011 011011 100011 101011 110011 111011 000111 010111 100111 110111 001111 101111 011111 111111 3 11 19 27 35 43 51 59 7 23 39 55 15 47 31 63 10→5 34→17 58→29 82→41 106→53 130→65 154→77 178→89 22→11 70→35 118→59 166→83 46→23 142→71 94→47 190→95 0000101 0010001 0011101 0101001 0110101 1000001 1001101 1011001 0001011 0100011 0111011 1010011 0010111 1000111 0101111 1011111 S101 S1Z01 S101 S1Z01 S101 S1Z01 S101 S1Z01 S11 S11 S11 S11 S11 S11 S11 S11 S101 000101 001101 010101 011101 100101 101101 110101 111101 5 13 21 29 37 45 53 61 16→8→4→2→1 40→20→10→5 64→32→16→8→4→2→1 88→44→22→11 112→56→28→14→7 136→68→34→17 160→80→40→20→10→5 184→92→46→23 000001 000101 000001 001011 000111 010001 000101 010111 Z1 S101 Z1 S11 S11 S1Z01 S101 S11 S1Z01 001001 011001 101001 111001 010001 110001 100001 9 25 41 57 17 49 33 28→14→7 76→38→19 124→62→31 172→86→43 52→26→13 148→74→37 100→50→25 000111 010011 011111 101011 001101 100101 011001 S11 S11 S11 S11 S101 S101 S1Z01 Note: n’ is the result of 3n+1 and then possible n/2, Group’ is the Group for n’

3 A Quick Analysis to What Have Observed

The purpose of doing a quick analysis is to arithmetically support what we observed. That is, although the behaviors such as the decreasing of trailing 1s or in-between 0s seemed to be features only, they have to be significant based on general arithmetic. We are to do that according to the numbers in groups S11, S101, and S1Z01, respectively.

3.1 Numbers in group S11

Let n be a number represented as S11 in binary, where S is a sequence of either 0 or 1 representing the value s. Then, n=s22+21+20, and

3n+1 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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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=3(s22+3)+1 =3(s22)+9+1 =12s+10 =2(6s+5).

This indicated that 3n+1 can be divided by 2 only once, since after the operation of division by 2 the result is 6s+5 which is odd. The result is greater than n, since (3n+1)/2>n. (Observation 1a has a support.)

Let s be some value t2k-1 where t >0 and k>0. In this case, S is the sequence staring with T, representing t in binary, and then k trailing 1s. Naturally, the number of trailing 1s of n in binary is k+2. We have

6s+5 =6(t2k-1)+5 =6(t2k)-6+5 =3t(2k+1)-1.

The number of trailing 1s of the value 3t(2k+1)-1 in binary is k+1, which is 1 less than that of n. (Observation 1b has a support.)

3.2 Numbers in group S101

Let n be a number represented as S101 in binary, where S is a sequence of either 0 or 1 representing the value s. Then, n=s23+22+20, and

3n+1

=3(s23+5)+1 =3(s23)+15+1 =23(3s) +23(2) = 23(3s+2).

This indicated that 3n+1 can be straightly divided by 2 at least 3 times, and after the 3 operations of division by 2 the result is 3s+2. The result is less than n, since (3n+1)/8<n for n>0. (Observation 2a has a support.)

If s is odd, then there will be no further operation of division by 2. However, if s is 0 or even, then there may be a number of consecutive operations of division by 2, even for a possible final result as 1. Cases happen when s=∑ 2i, i=1,3,5,7… , or s=2,10,42,170 …, since 3s+2=8,32,128,512…, which are 23,25,27,29…,

respectively. (Observation 2b has a support.)

3.3 Numbers in group S1Z01

Let n be a number represented as S1Z01 in binary, where S is a sequence of either 0 or 1 representing the value s, and Z is a sequence of k 0s, k>0. Then,

n=s2k+3+2k+2+1, and 3n+1 =3(s2k+3+2k+2+1)+1 =3(s2k+3)+3(2k+2)+3+1 =22(3s2k+1)+22(3)(2k)+22 =22(3s2k+1+(3)(2k)+1).

This indicated that 3n+1 can be divided by 2 only twice, since after the two operations of division by 2 the result is 3s2k+1+(3)(2k)+1 which is odd. The result is less than n, since (3n+1)/4<n for n>1. (Observation 3a has a support.)

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The number of “in-between 0s” of n in binary is k+1. For checking the number of that of 3s2k+1+(3)(2k)+1 in binary, we can have

3s2k+1+(3)(2k)+1 =3s2k+1+(2)(2k)+(2k)+1 =3s2k+1+(2k+1)+2k+1 =(3s+1)(2k+1)+2k+1.

This value is represented in binary as a bit sequence starting by the sequence representing the value 3s+1, followed by 1, k-1 0s, and then terminated by 1. The number of “in-between 0s” of the result is k-1, which is 2 less than that of n. (Observation 3b has a support.)

4 The Construction of a State Transition Model

Models are always helpful in doing analytical works, since complicated details may be symbolically simplified. A state transition model is a typical example. When the possible states of a system are clear, there is the mechanism to describe the behaviors in the system. According to Table 1, a given odd number n, which belongs to one of the 4 groups (i.e., Z1, S11, S101, and S1Z01), will be changed to another odd number n’ after the 3n+1 and all possible n/2 operations were done. We also see that n’ can still be classified into some one of the 4 groups. We are to apply the numbers, groups, and the changes of numbers in groups to gradually draft a state transition model.

4.1 The First State Transition Diagram and State Transition Table

Let z1, s101, s11, and s1z01 be states. Mnemonically, z1 stands for a given odd number being in group Z1, s101 stands for that in group S101, s11 stands for that in group S11, and s1z01 stands for that in group S1Z01. There is an extra starting state “s0” and the terminal state is z1. A given odd number n is initially at the “s0”

state. Then, after the 3n+1 and all possible n/2 operations, the next state is determined. This is the state transition. A state transition diagram can thus be depicted as Figure 1. Although simple, it has indicated that there are many

possible paths from s0 to z1. (The state transition is to terminate ultimately if there

is no infinite loop.)

Figure 1: A State Transition Diagram

The state transition diagram seemed to be good for specifying a general behavior s1z01 s0 s11 s101 z1 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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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of the odd numbers encountered in the computation process. There are only 4 states (excluding s0) introduced to the study to this hard problem.

We are interested in the determination of next state by directly checking n but not n’, or the result of the 3n+1 and all possible n/2 operations, based on the previous observations and analytical works. That is, we can try to predict the coming next states as earlier as possible. Initially, a state transition table which is equivalent to the state transition diagram in Figure 1 is drafted as Table 2. According to the columns, the entries in the table specified the current states, possible input

numbers from the 4 classified groups, and the corresponding next states. Note that there are entries with multiple next states hence the state transition is

non-deterministic. For example, at state s11, with input number n in group S11, the next state can be s11, s101, or s1z01. In the following, we are to try to refine the table such that the state transition can be deterministic (or more deterministic). Table 2: The Draft State Transition Table

s /n Z1 S11 S101 S1Z01 s0 z1 s11 s101 s1z01 z1 - - - - s11 - s11, s101, s1z01 - - s101 - - z1, s11, s101, s1z01 - s1z01 - - - s11, s101, s1z01

4.2 Refining Columns S1Z01 and S11 in the State Transition Table

At state s1z01, the number of “in-between 0s” of n’ is 2 less than that of n. Hence the next state can be determined by checking n directly as: (1) s11 if n is with bit sequence S1001, (2) s101 if n is with that of S10001, and (3) s1z01 if n is with that of S1Z0001. The S1Z01 column in Table 2 is then refined to 3 columns as indicated in Table 3. Note that the state transition is deterministic.

Table 3: Refined columns to the original S1Z01 column s /n S1001 S10001 S1Z0001 s0 s1z01 s1z01 s1z01 z1 - - - s11 - - - s101 - - - s1z01 s11 s101 s1z01

At state s11, the number of “trailing 1s” of n’ is 1 less than that of n. Hence the next state for n with more than 2 “trailing 1s” is still s11. The next state for n with exact 2 “trailing 1s” can be either s101 or s1z01 (i.e., ended with bit sequence 01). We recall that when n is in group S11, n’ or the result of (3n+1)/2, is 6s+5 where s is the value of bit sequence S. In addition, the value of 6s+5 is odd. It is clear that the next state is s101 if n’ is with the bit sequence of X101, where X is a bit sequence of either 0 or 1. The next state is 1z01 otherwise. Let x be the value of the bit sequence X. We are trying to match bit sequences S11 and X101 for proper S and X pairs where the value of S11 is n and the value of X101 is n’, which actually is 6s+5. We can set up the equation and find the solutions as follows: 6s+5=x23+22+20 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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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6s+5=8x+5 6s=8x s=4x/3.

Since the value of 6s+5 is odd, s must be even. The proper pairs of s and x are hence: (0, 0), (4, 3), (8, 6) … (4k, 3k) for all k≥0. For example, the next state of number 19 at state s11 can be predicted as s101. In this example, n is 19 or 10011 in binary, S is the underlined bit sequence 100 and s is 4. The obtained n’ is 29 or 11101 in binary, where X is the underlined bit sequence 11 and x is 3. This (s, x) pair fits well.

The S11 column in Table 2 is then refined to 3 columns as indicated in Table 4. The first column is for n with bit sequences S111, and the second column is for n with bit sequences S0011. The third column is for n with other bit sequences (or S1011, equivalently). Note that the state transition is deterministic.

Table 4: Refined columns to the original S11 column

s /n S111 S0011 S1011 s0 s11 s11 s11 z1 - - - s11 s11 s101 s1z01 s101 - - - s1z01 - - -

4.3 Refining Column S101 in the State Transition Table

At state s101, it is more complicated. The key concern is that there may be more than 3 straight n/2 operations after the 3n+1 operation. We recall that when n is in group S101, then (3n+1)/8=3s+2 where s is the value of bit sequence S. If 3s+2 is odd, then n’ is 3s+2, and if 3s+2 is even, then n’ is (3s+2)/2m for a certain m (the number of more n/2 operations), m>0. We are to try to set up equations and solve them again, as we did previously.

When 3s+2 is odd, we are to match (1o) bit sequences S101 and X11 for proper S and X pairs where the value of S101 is n and that of X11 is n’, and (2o) bit sequences S101 and X101 for proper S and X pairs where the value of S101 is n and that of X101 is n’. We know that there is no such match for S101 and Z1, since no s can fit for 3s+2=1. We expect that all the remainders for s are good to match bit sequences S101 and X1Z01.

For matching (1o), we can set up the equation and find the solutions as follows: 3s+2=x22+21+20

3s+2=4x+3 3s=4x+1 s=(4x+1)/3.

Since 3s+2 is odd, s must be odd. The proper pairs of s and x are hence: (3, 2), (7, 5), (11, 8) … (4k+3, 3k+2) for all k≥0.

For matching (2o), we can set up the equation and find the solutions as follows: 3s+2=x23+22+20 3s+2=8x+5 3s=8x+3 s=(8x+3)/3. 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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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Since 3s+2 is odd, s must be odd. The proper pairs of s and x are hence: (1, 0), (9, 3), (17, 6) … (8k+1, 3k) for all k≥0.

The remainders for s to fit 3s+2 (which is odd) are collected as 5, 13, 21 … 8k+5, for all k>0. A complete and clear table for odd s values of the match or remainder is organized as Table 5. Note that the bit sequence of n is S101 where S is the bit sequence of value s.

Table 5: The complete odd s values of match and remainder match/remainder s (odd)

match (1o) 3, 7, 11, …, 4k+3 for all k≥0 match (2o) 1, 9, 17, …, 8k+1 for all k≥0 remainder 5, 13, 21, …, 8k+5 for all k≥0

When 3s+2 is even, there must be a certain number m (m>0) of more n/2 operations for an odd result. Hence, n’ is (3s+2)/2m. We are to match (1e) bit sequences S101 for proper S where the value of S101 is n and n’ is 1, with the certain m, (2e) bit sequences S101 and X11 for proper S and X pairs where the value of S101 is n and that of X11 is n’, with the certain m, and (3e) bit sequences S101 and X101 for proper S and X pairs where the value of S101 is n and that of X101 is n’, with the certain m. We expect that all the remainders for s are good to match bit sequences S101 and X1Z01.

For matching (1e), we can set up the equation and find the solutions as follows: (3s+2)/2m=1

(3s+2) =2m 3s=2m-2 s=(2m-2)/3.

We have had the common idea about the “qualified” numbers, so the proper pairs of s and m are: (0, 1), (2, 3), (10, 5) … (q, 2k+1), where q= ∑ 2i, i=1,2(k-1)+1,2, for all k≥0.

For matching (2e), we can set up the equation and find the solutions as follows: (3s+2)/2m=x22+21+20

(3s+2)/2m=4x+3 3s=(4x+3)2m-2 s=((4x+3)2m-2)/3.

Since 3s+2 is even, s must be even or 0. The proper pairs of s and x with the

certain value of m are: (4, 1), (12, 4), (20, 7) … (8k+4, 3k+1) for all k≥0 with m=1; (14, 2), (30, 5), (46, 8) … (16k+14, 3k+2) with m=2; (18, 1), (50, 4), (82, 7) … (32k+18, 3k+1) with m=3; and there are more.

For matching (3e), we can set up the equation and find the solutions as follows: (3s+2)/2m=x23+22+20

(3s+2)/2m=8x+5 3s=(8x+5)2m-2 s=((8x+5)2m-2)/3.

Since 3s+2 is even, s must be even or 0. The proper pairs of s and x with the certain value of m are: (8, 1), (24, 4), (40, 7) … (16k+8, 3k+1) for all k≥0 with m=1; (6, 0), (38, 3), (70, 6) …(32k+6, 3k) for all k≥0 with m=2; (34, 1), (98, 4), (162, 7) … (64k+34, 3k+1) for all k≥0 with m=3; and there are more.

The remainders for s to fit 3s+2 are partially collected as: 16, 22 …, and they are

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determined after m matches were done, where 2m+2>s. An incomplete table for even s values of the match and remainders is suggested as Table 6.

Table 6: An incomplete even s values of match and remainder match/remainder m s (even)

match (1e) - 0, 2, 10, …, ∑i, i=0, 21, 23, 25, … match (2e) 1 2 3 4 : 4, 12, 20, …, 8k+4, for all k≥0 14, 30, 46, …, 16k+14, for all k≥0 18, 50, 82, …, 32k+18, for all k≥0 58, 122, 126, …, 64k+58, for all k≥0 : match (3e) 1 2 3 4 : 8, 24, 40, …, 16k+8, for all k≥0 6, 38, 70, …, 32k+6, for all k≥0 34, 98, 162, …, 64k+34, for all k≥0 26, 154, 282, …, 128k+26, for all k≥0 : remainder - 16, 22, …

The S101 column in Table 2 can be, if we insist, refined to 4 columns as indicated in Table 7. The first column is for n with bit sequences 10101 where 10 represents any repetition (including none) of the bit sequence 10. The second, third, and the last columns are for n with different bit sequences indicated. The notations (A*), (B*), and (C*) represent bit sequences of s values obtained by matching (2e), (3e), and remainders, in Table 6, respectively. Note that the state transition is non-deterministic, unless all of the m matches were done and the remainders were properly collected.

Table 7: Refined columns to the original S101 column 10101 S11101, (A*)0101 S001101, (B*)0101 S101101, (C*)0101 start s101 s101 s101 s101 z1 - - - - s11 - - - - s101 z1 s11 s101 s1z01 s1z01 - - - -

4.4 A Short Discussion on the Benefits of the Model

The study of using binary and constructing the state transition model did not solve the problem, since we were blocked at providing definite bit sequences of (A*), (B*), and (C*). However, we did simplify the understanding of its complexity. Here are the benefits: (1) Only 4 states (excluding the “start” state) are defined, the model is relatively simple. (2) Input numbers are classified into 4 groups according to their bit sequences, and the count of them is clear and complete. (3) It is indicated in the state transition diagram that there guarantees a path from the start state to the terminal state z1, if no infinite loop exists. (4) The

non-determinism of state transition is narrowed down to binary numbers ended with bit sequence 0101, and it is less than 1 among all of the 8 cases. (The case for 10101 is deterministic.) 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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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5 Working Value Analysis

In a known 3n+1 and n/2 computation process, temporary numbers encountered may be large before the process terminated at n=1 finally. Let these numbers be mentioned as working values in the following. When we are to use a finite state machine, such as a conventional digital computer, to verify a computation process, the working value is a concern. For example, if at an instance in the computation process the working value is beyond the capacity of the finite state machine, then the computation cannot continue. In fact, numbers larger than any specific number are always there.

As we checked in the analytical works, the working value grows only at the s11 state, since the next computation process is (3n+1)/2, which is greater than n. Hence, the working values will keep on going up at an instance that the bit sequence of an encountered number is with consecutive trailing 1s. There will be straight (3n+1)/2 computations in this case. If the number is n and there are k straight (3n+1)/2 computations to be performed, then the working value is

expected as (3kn+Σi=0,k-2(2/3)i3k-1+2k-1)/2k. The analysis is briefed as follows. Let ni

be the number encountered at the ith (3n+1)/2 computation, then n0=n n1=(3n0+1)/2=(3n+1)/2 n2=(3n1+1)/2=(3(3n+1)/2+1)/2=(9n+3+2)/4 n3=(3n2+1)/2=(3(9n+3+2)/4+1)/2=(27n+9+6+4)/8 n4=(3n3+1)/2=(3(27n+9+6+4)/8+1)/2=(81n+27+18+12+8)/16 : nk=(3nk-1+1)/2=(3kn+Σi=0,k-2(2/3)i3k-1+2k-1)/2k.

When n is a very large number and k is also significant in magnitude, we have to think of the working value to the capacity of a specific finite state machine.

6 Logical Soundness of the Study

We have exercised the observations, quick analysis works, trying to build the state transition tables, and working value analysis. In the following, we are to

demonstrate the logical soundness of the study. We list the facts (predicates which are assumed to be always true) first and then give the theorems (rules for

inferences) with proofs.

Fact 1. The 3n+1 problem is an open question. Fact 2. Nature numbers can be represented in binary.

Fact 3. The state transition model can be applied for analysis.

Fact 4. A practical finite state machine is limited by its memory size.

Theorem 1. The diagram provided in Figure 1 represents a state transition model.

[Proof] Since a state transition model can be described as M={S, I, T, s0}, where S is the set of states, I is the set of input, T is the set of transitions, and s0 in S is

the starting state, we can fit the diagram into such a model. Let S be the set {z1, s11, s101, s1z01, s0}, I be the set of odd numbers in binary, and T be the set of

transitions {s0→z1, s0→s101, s0→s11, s0→s1z01, s101→s101, s101→s11,

s101→z1 …} as indicated in the diagram. Then, there is the state transition model.

Theorem 2. Given a number n, if n=2mΣ4k where m, k≥0, then the 3n+1 and n/2 computation process will terminate successfully.

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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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[Proof] Since 3(Σ4k)+1=4k+1 for all k, 3n+1=(2m)(Σ4k)=2m+2k+2, which is qualified (i.e., we can see that after m+2k+2 times of n/2 operation, the result is 1).

Theorem 3. Given an odd number n, whenever n mod 2k=2k-1 where k≥2, then an instantly continuous increasing of working values (in the computation process) is expectable.

[Proof] As it was analyzed in the previous section, the expected largest working value is (3kn+Σi=0,k-2(2/3)i3k-1+2k-1)/2k.

Theorem 4. By using a finite state machine such as a computer, the computation

process trying to solve the 3n+1 problem does not always terminate successfully. [Proof] There are two cases that the computation process will not terminate successfully: (1) there is no such match that n=2mΣ4k where m, k≥0, in the whole life of the computation process, and (2) the memory size of the finite state machine is not capable to handle the working values. In the first case, we cannot be sure whether there is the match or not, although Theorem 2 stands. However, in the second case, the finite state machine may not be able to handle the working values due to the limitation of its memory size, as Theorem 3 stands.

7 Conclusions

This study to the 3n+1 problem covers the works in observation, analysis,

modeling, and logical soundness. As a result, we have indicated that (1) numbers in binary are helpful in the general observation, (2) there are features for numbers with “trailing 1s” or “in-between 0s” in the computation process, (3) a state transition model can be build to handle the major check points in the computation process with possible state transition rules, (4) the state transition is

non-deterministic in less than 1 of 8 possible cases, or it is non-deterministic if there is a finite set of numbers, (5) working values can be expected at an instance in the computation process, and (6) the study is logically sound.

The significance of the study is the trying to construct a state transition model. There are only 4 states (excluding the starting state), and the input numbers can be classified into 4 groups, depending on their bit sequences in binary. The

complexity of the original problem can be limited in the specification of state transitions. Actually, the state transition of numbers with bit sequences S11 and S1z01 is deterministic, and only that with bit sequence S101 is non-deterministic. The degree of non-determinism is narrowed down again. Theoretically, a general state transition table can be provided with 3 check lists for the possibly infinite number of input candidates in the S101 group. However, this is beyond the current scope.

The future works include the look-ahead analysis and the memory/memory-less analysis. This study seems to have exercised the 1-step look-ahead feature since next state or bit sequence of the result obtained after the 3n+1 and n/2 operations is mostly expectable. (The state transition in current model is memory-less.) Whether a model for 2-step or k-step analysis is possible or not is an interesting issue. In this case, a finite state machine with stack memory (with push/pop operations) for keeping previous information is required. Nevertheless, we can foresee that the stack size is the key concern.

References

[1] R.K. Guy, Unsolved problems in number theory, 3rd ed., New York, Springer-Verlag, 2004, pp. 330-332. 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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

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[2] B. Kraft, K. Monks, On conjugacies of the 3x+1 map induced by continuous endomorphisms of the shift dynamic system, Discrete Mathematics 310 (2010) 1875-1883.

[3] J.C. Lagarias, The 3x + 1 problem and its generalizations, Amer. Math. Monthly 92 (1985) 3-23. [4] J.C. Lagarias, “The 3x + 1 Problem: An Annotated Bibliography, II (2000-2009), available at http://arxiv.org/abs/math/0608208, accessed on 12/06/2010.

[5] E. Lehtonen, Two undecidable variants of Collatz’s problems, Theoretical Computer Science 407 (2008) 596-600.

[6] G.J. Wirsching, The dynamical system generated by the 3n + 1 function, Berlin, Springer, 1998, (Lecture Notes in Mathematics series: 1681), ISBN 978-3-540-63970-1.

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數據

Figure 1: A State Transition Diagram
Table 3: Refined columns to the original S1Z01 column  s /n  S1001  S10001  S1Z0001  s 0 s1z01  s1z01  s1z01  z1  -  -  -  s11  -  -  -  s101  -  -  -  s1z01  s11  s101  s1z01
Table 6: An incomplete even s values of match and remainder  match/remainder  m  s (even)

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