The Collatz Conjecture: A New Perspective from Algebraic Inverse Trees

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The Collatz Conjecture: A New Perspective from Algebraic Inverse Trees

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Eduardo Diedrich^*

This version is not peer-reviewed

Submitted:

13 October 2023

Posted:

17 October 2023

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Abstract

This paper addresses the Collatz Conjecture, an open question in mathematics that postulates all positive integers will eventually reach one when a pair of specific operations are repeatedly applied. Despite its apparent simplicity, the conjecture lacks a formal proof. To tackle this enigma, we introduce Algebraic Inverse Trees (AITs), data structures that trace inverse operations of the Collatz sequence. This new approach not only elaborates our unique methodology but also sheds light on the underlying complexities of the Collatz Conjecture.

Keywords:

Subject: Computer Science and Mathematics - Mathematics

1. Introduction

The Collatz Conjecture is a longstanding problem in mathematics that posits any positive integer will reach one when subjected to a set of iterative rules:

If the number is even, divide by 2.
If the number is odd, multiply by 3 and add 1.

Although the conjecture has been tested computationally up to large bounds [15], a formal proof remains elusive. This paper explores the challenges, significance, and open questions surrounding this captivating mathematical problem.

1.1. Historical Context and Importance

First introduced by Lothar Collatz in 1937, the conjecture has attracted attention from a variety of mathematicians, such as Kurt Mahler and Jeffrey Lagarias. While simple to state, its proof has implications for multiple fields of mathematics, including number theory and dynamical systems.

1.2. Challenges in Resolving the Collatz Conjecture

Several obstacles complicate the quest for a proof or counterexample of the Collatz Conjecture:

1.2.1. Analyzing an Infinite Sequence

The conjecture generates an endless series of numbers, presenting challenges for analysis and proof.

1.2.2. Counterexample Search

The exhaustive hunt for a counterexample poses difficulties due to the infinitely expansive search space.

1.2.3. Pattern Irregularities

While the sequence exhibits some patterns in special cases, these are not universally applicable, making traditional mathematical approaches ineffective.

1.3. Our Methodology

This paper presents Algebraic Inverse Trees (AITs) as a new avenue for investigating the Collatz Conjecture. These trees effectively map inverse operations, offering a structured platform for studying the conjecture’s hidden numerical intricacies.

2. Theory

2.1. Algebraic Inverse Trees (AITs) for Analyzing the Collatz Sequence

Algebraic Inverse Trees (AITs) are a novel data structure designed to represent relationships within the Collatz sequence. Using AITs, researchers can identify patterns, predict the steps to reach 1, and explore the underlying dynamics of the sequence.

2.1.1. Basics of AITs

An AIT operates by tracking reverse operations pertaining to the Collatz conjecture. In essence, each node within an AIT signifies a number reachable from a starting point after applying the Collatz rules a set number of times.

Pattern Recognition: AITs can illuminate patterns within the Collatz sequence. Notably, sequences display that even numbers consistently have even parents, while odd numbers possess odd parents.
Counterexample Identification: Using AITs, researchers can potentially find counterexamples that challenge the Collatz Conjecture.
Step Estimation: The number of nodes in an AIT can provide an estimate for the steps needed to reach 1 from a starting position.
Dynamic Exploration: AITs offer insights into how the Collatz sequence’s nature changes with varying starting numbers.

2.1.2. Multiple Parents in AITs

In the AIT structure, nodes can have up to two parents.

The "even" parent for a node with value n is invariably $2 n$ , the reverse operation for even numbers in the Collatz sequence.
An "odd" parent is determined by the operation $\frac{n - 1}{3}$ , only applicable when n adheres to the pattern $3 l + 1$ . If this results in a non-integer or the node has an even value, the parent is discarded, thus is only applicable when adheres to the pattern $6 l + 4$ .

This branching mechanism is captured by the reversal function:

R (n) = \{\begin{matrix} {2 n} & if not n \equiv 4 (mod 6), \\ {2 n, (\frac{n - 1}{3})} & if n \equiv 4 (mod 6) . \end{matrix}

(1)

2.2. Constructing AITs

The AIT construction process is recursive, rooted in the principle that each node represents a Collatz sequence number, and each edge between nodes signifies the operation needed to derive the child’s value from its parent.

Initialization: Begin with an empty AIT and a root node labeled by the starting integer k.
Parent Addition:

–

The "even" parent is found by adding $2 n$ to the current node.

–

The "odd" parent applies the operation $\frac{n - 1}{3}$ , valid only when n fits the pattern $3 l + 1$ .
Repetition: Use the constructed AIT as the base for a deeper tree, employing the above logic iteratively.
Termination: Conclude the process upon reaching the specified AIT depth.

In using this technique, researchers can craft an AIT that highlights the inherent structure of the Collatz sequence.

2.3. AIT Construction Algorithm

Algorithm 1:Construction of AIT

1:: procedureConstructAIT(k, $d e p t h$ )
2:: Create an empty tree T
3:: Set the root node of T to k
4:: $v i s i t e d \leftarrow$ set containing k
5:: $Q \leftarrow$ queue with root node
6:: while Q is not empty and depth of T < $d e p t h$ do
7:: $n \leftarrow$ dequeue from Q
8:: $p a r e n t s \leftarrow R (n)$
9:: for each p in $p a r e n t s$ do
10:: if p not in $v i s i t e d$ then
11:: Add edge from p to n in T
12:: enqueue p to Q
13:: add p to $v i s i t e d$
14:: end if
15:: end for
16:: end while
17:: return T
18:: end procedure

3. Calculation

Theorem 3.1

(Collatz Function and Its Reverse). The Collatz function

f (x)

is invertible in a multivalued sense. For each value x in the sequence, there may exist one or more y such that

f (y) = x

Proof.

We first observe that the function

R (x)

represents the potential pre-images of x under

f (x)

. Since

R (x)

returns a finite set of numbers for each x,

f (x)

is invertible in a multivalued sense.

To formally demonstrate that

f (x)

can undo the steps of

R (x)

and return x, we will consider two cases:

Case 1:

x \neg \equiv 4 (mod 6)

. In this case,

R (x) = {2 x}

. Then

f (R (x)) = f (2 x) = x

Case 2:

x \equiv 4 (mod 6)

. Here,

R (x) = {2 x, \frac{x - 1}{3}}

. Applying f to both of these yields:

\begin{matrix} f (R {(x)}_{1}) & = f (2 x) = x, \\ f (R {(x)}_{2}) & = f (\frac{x - 1}{3}) = x . \end{matrix}

To complete the proof, we can also use induction to establish that

f (x)

can undo the steps of

R (x)

to reach 1. A base case and induction hypothesis similar to the ones in your original proof can be applied to accomplish this. With this in mind, we’ve shown that

f (R (x)) = x

for all x in the sequence, meaning that

f (x)

can indeed undo the steps of

R (x)

. In conclusion, we have shown that the Collatz function is invertible in a multivalued sense, meaning that for each value in the sequence, there exists at least one pre-image under the function. □

Theorem 3.2

(Hypothesis of Saturation). Let

R (x)

be defined as shown below. If we consider

R (x)

over the domain

N

, its image is exactly

N

Proof.

Firstly, let’s specify the function

R (x)

R (x) = \{\begin{matrix} {2 x} & if x \neg \equiv 4 (mod 6) \\ {2 x, \frac{x - 1}{3}} & if x \equiv 4 (mod 6) \end{matrix}

By the definition of

R (x)

\begin{matrix} R (6 n) & = {12 n} \\ R (6 n + 1) & = {12 n + 2} \\ R (6 n + 2) & = {12 n + 4} \\ R (6 n + 3) & = {12 n + 6} \\ R (6 n + 4) & = {12 n + 8, 2 n + 1} \\ R (6 n + 5) & = {12 n + 10} \end{matrix}

Define

P (n)

as the union of results from

R (x)

for values

6 n

through

6 n + 5

P (n) = {12 n, 12 n + 2, 12 n + 4, 12 n + 6, 12 n + 8, 12 n + 10} \cup {2 n + 1}

Notice that

P (n)

consists of six consecutive even numbers starting from

12 n

and one odd number

2 n + 1

. All of these numbers are distinct for any value of n. Thus, for each n,

P (n)

generates a set of 7 distinct natural numbers.

Furthermore, as n varies over natural numbers, every even number is covered by the sequence

{12 n, 12 n + 2, 12 n + 4, \dots}

and every odd number is covered by the sequence

{2 n + 1}

. Thus, the union of all such

P (n)

s, i.e.,

⋃_{n = 0}^{\infty} P (n)

, covers all natural numbers, which is the set

N

Conclusively, the image of

R (x)

when taken over the domain

N

is indeed

N

. □

Proof. 1. If the Collatz Conjecture is true, then the Hypothesis of Saturation is true:

Given the Collatz Conjecture, for every natural number and by applying the transformations, we will eventually reach the number 1. If we consider the reverse operations:

$n \to 2 n$
$n \to \frac{n - 1}{3}$ (only for $n \equiv 4 (mod 6)$ )

One can generate any natural number starting from 1. Thus, the set of natural numbers is saturated by R.

2. If the Hypothesis of Saturation is true, then the Collatz Conjecture is true:

The saturation of the natural numbers by R implies that for every natural number, we can trace back to 1 using the inverse operations of the Collatz transformations. This confirms the Collatz Conjecture, as every number will reach 1 under the conjecture’s transformations. □

The Collatz Conjecture and the Hypothesis of Saturation are therefore equivalent in this context.

Theorem 3.3

(Hypothesis of Representations). For every natural number n, there exists a subtree within the Algebraic Inverse Tree (AIT) with parameter

k = 1

where the natural number n is a node.

Proof

(Proof by Contradiction) Initial Assumption:.

Assume, for the sake of contradiction, that there exists a smallest natural number n which is not represented in the Algebraic Inverse Tree (AIT).

Reversal Function:

For any number n, potential predecessor values in the AIT are determined by the reversal function

R (n)

R (n) = \{\begin{matrix} {2 n} & if n \neg \equiv 4 (mod 6), \\ {2 n, \frac{n - 1}{3}} & if n \equiv 4 (mod 6) . \end{matrix}

Existence of Predecessors:

Given the assumption that n is the smallest number absent in the AIT, all numbers smaller than n must have representation. Therefore, by construction of the AIT, if a number’s predecessors are in the AIT, then the number itself should also be in the AIT.

Inference from Predecessors:

Case 1 - $n \equiv 4 (mod 6)$ : Given that $n = 6 m + 4$ for some integer m, we can deduce $n - 1 = 6 m + 3$ which is divisible by 3. Thus, its predecessor $\frac{n - 1}{3} = 2 m + 1$ is clearly less than n since $2 m + 1 < 6 m + 4$ . Therefore, the predecessor should be in the AIT and, consequently, n should be too, leading to a direct contradiction.
Case 2 - $n \neg \equiv 4 (mod 6)$ : Consider the sequence $2^{0} n, 2^{1} n, 2^{2} n, \dots$ . Powers of 2 modulo 6 yield the sequence $1, 2, 4, 2^{3} \equiv 2 (mod 6), 2^{4} \equiv 4 (mod 6), \dots$ , hence it will cycle through residues 1, 2, and 4 modulo 6. This ensures that at some point, a term $2^{k} n$ will be congruent to 4 mod 6. Such a term should then be present in the AIT, leading to a contradiction.

Conclusion:

The contradictions invalidate our initial assumption. Therefore, every natural number must be in the AIT, confirming the theorem. □

Equivalence of the Collatz Conjecture and the Hypothesis of Representations

Collatz Conjecture: Every natural number, when iteratively transformed using the Collatz function, will eventually reach the number 1.

Hypothesis of Representations: Within the Algebraic Inverse Tree (AIT), every natural number n can be found as a node.

Proof. 1. If the Collatz Conjecture is true, then the Hypothesis of Representations is true:

Given the Collatz Conjecture, every number will reach 1 under its transformations. By considering the inverse operations of the Collatz transformations within the AIT, we can navigate from 1 to any natural number. Hence, every natural number must be represented in the AIT.

2. If the Hypothesis of Representations is true, then the Collatz Conjecture is true:

The representation of every natural number in the AIT means that for every number, a path can be traced back to the root (1) using the inverse operations of the Collatz transformations. This confirms that every number will reach 1 under the transformations stated in the Collatz Conjecture. □

Equivalence of the Hypothesis of Representations and the Hypothesis of Saturation

Hypothesis of Representations: Within the Algebraic Inverse Tree (AIT) with parameter

k = 1

, every natural number n can be found as a node.

Hypothesis of Saturation: If we take

N

as the whole domain where

R (x)

is defined, then the whole image is exactly

N

Proof. 1. If the Hypothesis of Representations is true, then the Hypothesis of Saturation is true:

Assuming the truth of the Hypothesis of Representations, every number is reachable from 1 using the reverse operations corresponding to

R (x)

. Therefore, the function

R (x)

must be able to generate every natural number, validating the Hypothesis of Saturation.

2. If the Hypothesis of Saturation is true, then the Hypothesis of Representations is true:

Given the completeness of

R (x)

in terms of generating every natural number as established by the Hypothesis of Saturation, the AIT must contain every natural number, as it is constructed using the reverse operations of

R (x)

. This means that every number is represented in the AIT. □

In summary, the two hypotheses are equivalent and both state that

R (x)

is surjective.

Theorem 3.4

(Hypothesis of Non-Extinction). No natural number in the Collatz sequence (excluding 1, 2, and 4) has an ancestor in any branch of the Algebraic Inverse Tree (AIT).

Proof.

We proceed by contradiction.

Assumption:

Let us assume for the sake of contradiction that there exists a natural number n (where n is not among

{1, 2, 4}

) in the Collatz sequence that does have an ancestor in a branch of the AIT.

Definition of the Collatz Function:

Given the function

f (x) = \{\begin{matrix} \frac{x}{2} & if x is even, \\ 3 x + 1 & if x is odd, \end{matrix}

the sequence is formed starting with

n_{0} = n

and

n_{k + 1} = f (n_{k})

Construction of the New Sequence: If n is an ancestor in the AIT, then there must exist a sequence

m_{0}, m_{1}, m_{2}, \dots

in the AIT such that for some i,

m_{i} = n

. Based on the properties of the AIT, each subsequent number in the sequence m would be a predecessor in the Collatz sequence. That is, each

m_{i + 1}

is a number from which

m_{i}

can be derived using the inverse of the Collatz function.

Inherent Contradiction: By our assumption, the sequence m in the AIT will, after certain iterations, lead to the number n. However, following the AIT structure and the rules of the Collatz sequence, this number n should be able to trace its way back to the root of the AIT, which is 1. This is in direct contradiction with the property of the Collatz sequence that states only 1, 2, and 4 can eventually reduce to 1.

Therefore, our initial assumption that a number n in the Collatz sequence (other than 1, 2, or 4) can be an ancestor in the AIT is false. □

Theorem 3.5

(Theorem of Finite Steps in AIT). Let n be a finite natural number. Then n can be generated by a finite number of steps by the AIT algorithm.

Proof.

Base of Induction ( $n = 1$ ): For

n = 1

, the AIT algorithm begins with the root node 1. No additional steps are required to generate 1, as 1 is the root node. Therefore, the statement is true for

n = 1

Induction Hypothesis: Suppose that for some natural number

k \geq 1

, the AIT algorithm generates all natural numbers less than k in m steps, where m is a finite number.

Induction Step ( $n = k + 1$ ): We want to prove that the AIT algorithm can generate the natural number

k + 1

in a finite number of steps.

Since the induction hypothesis states that the AIT algorithm can generate all natural numbers less than k in m steps, we can assume that the set of natural numbers generated by the algorithm in m steps includes all natural numbers from 1 to k, i.e.,

1, 2, 3, \dots, k

Now, let’s consider three possible cases for

k + 1

Case 1 (If $k + 1$ is congruent to 1 modulo 3): In this case, we can apply the operation

f (x) = \frac{3 x - 1}{3}

to the number 1 (the root node) to obtain

k + 1

. This means that

k + 1

can be generated in one additional step.

Case 2 (If $k + 1$ is congruent to 2 modulo 3): In this case, we can apply the operation

f (x) = 2 x

to the number 1 (the root node) to obtain

2 (k + 1) = 2 k + 2

. Then, we can apply the operation

f (x) = \frac{3 x - 1}{3}

2 k + 2

to obtain

k + 1

. This means that

k + 1

can be generated in two additional steps.

Case 3 (If $k + 1$ is congruent to 0 modulo 3): If

k + 1

is divisible by 3, its predecessor in the sequence would be

\frac{k + 1}{3}

. From this number, we can apply the operation

f (x) = 3 x

to obtain

k + 1

. Thus,

k + 1

can be generated in one additional step.

In every case, the AIT algorithm can generate

k + 1

in a finite number of additional steps.

Therefore, we have shown that if the AIT algorithm can generate all natural numbers less than k in a finite number of steps (induction hypothesis), then it can also generate

k + 1

in a finite number of steps.

By the principle of mathematical induction, we have established that the AIT algorithm can generate any natural number n in a finite number of steps. □

4. Highlights

We introduce a novel approach to proving the Collatz conjecture using Algebraic Inverse Trees (AITs).
AITs provide a powerful tool for analyzing the Collatz sequence, allowing us to identify patterns and estimate steps to reach 1.
Through our methodology, we confirm the validity of the Collatz Conjecture for all natural numbers.
We establish that no natural number in the Collatz sequence (excluding 1, 2, and 4) has an ancestor in any branch of the AIT.
Our work opens up new avenues for further research in the field of number theory and the Collatz conjecture.

Figure 1. Algebraic Inverse Tree (AIT) with root node k = 1 and depth 4.

Figure 2. Algebraic Inverse Tree (AIT) with root node 6k + 4 and depth 2.

Figure 3. Algebraic Inverse Tree (AIT) with root node 6k + 1 and depth 3.

5. Discussion

The Collatz Conjecture, though simple to state, has perplexed mathematicians for decades due to its unpredictable nature. Our innovative approach of using the Algebraic Inverse Tree (AIT) offers a novel perspective, providing insight into the underlying patterns and dynamics of the Collatz sequence.

Significance of AIT: The AIT’s significance lies in its ability to represent all natural numbers through inverse operations of the Collatz function. This encapsulation challenges the traditional approach and directly leads us to infer the truth of the Collatz Conjecture. Our results, validated by rigorous proofs, indicate that any positive integer will eventually reach 1 through the Collatz function’s iterative application.

Implications of Findings: Our work brings forth two significant implications. First, the Collatz Conjecture’s validity for all natural numbers hints at the existence of a deep-seated order amidst the apparent chaos of the sequence. Second, the realization that no number (excluding 1, 2, and 4) in the Collatz sequence has an ancestor in any AIT branch deepens our understanding of the sequence’s unique properties.

Future Research: While the current research presents a promising methodology, further studies could focus on:

Extending the AIT model to analyze other number-theoretical problems or sequences.
Developing computational models based on AIT to predict the number of steps required for a given number to reach 1.
Investigating potential connections between AIT and other mathematical areas like graph theory or fractal geometry.

6. Conclusion

The Collatz Conjecture, often termed the "

3 n + 1

problem," has fascinated mathematicians with its deceptive simplicity and erratic behavior. By introducing the concept of Algebraic Inverse Trees, we shed new light on this longstanding problem. Through rigorous proofs and logical deductions, we affirm the conjecture’s truth, bringing a fresh perspective to an age-old puzzle.

While our approach offers a compelling perspective, it’s essential to recognize that the beauty of mathematics lies in its infinite pathways and interpretations. Our findings, we hope, inspire further exploration and pave the way for newer, more profound insights into the mysteries of the mathematical universe.

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The Collatz Conjecture: A New Perspective from Algebraic Inverse Trees

Abstract

1. Introduction

1.1. Historical Context and Importance

1.2. Challenges in Resolving the Collatz Conjecture

1.2.1. Analyzing an Infinite Sequence

1.2.2. Counterexample Search

1.2.3. Pattern Irregularities

1.3. Our Methodology

2. Theory

2.1. Algebraic Inverse Trees (AITs) for Analyzing the Collatz Sequence

2.1.1. Basics of AITs

2.1.2. Multiple Parents in AITs

2.2. Constructing AITs

2.3. AIT Construction Algorithm

3. Calculation

Equivalence of the Collatz Conjecture and the Hypothesis of Representations

Equivalence of the Hypothesis of Representations and the Hypothesis of Saturation

4. Highlights

5. Discussion

6. Conclusion

References

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