Algebraic Inverse Tree (AITs) are a novel data structure designed to represent relationships within the Collatz sequence. Using AITs, researchers can identify patterns, predAIT the steps to reach 1, and explore the underlying dynamics of the sequence. An Inverse Collatz Tree is a complete tree with a weight function that assigns a positive integer to each node. The construction of an AIT stops when the node with the sought-after value n is found. Hence, an AIT is always finite in the sense that its construction ceases upon locating the target node. Its size or depth is determined by the value of n, but it does not grow indefinitely beyond the necessary point.
6.1. Axioms and Proofs relative to AIT
Definition 6.1 (Cycle). Acyclein the context of a function , a graph, or a directed graph is a finite sequence with the following properties:
For a function C: and
For a graph or a directed graph: and for all , where in a directed graph, is an edge for all
The length of a cycle is the number of transitions (function applications or edges) it contains, which is k for the cycle defined above.
Definition 6.2. Let be a tree with V the set of vertices and E the set of edges.
An infinite path in T is an infinite sequence of vertices such that:
are distinct vertices in V
For all , the edge is in E
The path does not contain cycles, that is, there are no such that
Definition 6.3. A non-trivial cycle in a tree or a function is a sequence of distinct nodes or values (other than the trivial node or value 1) that repeats indefinitely, forming a loop.
Definition 6.4 (Infinite Path in a Graph). An infinite path in a graph is a sequence of edges which connects a sequence of vertices through a repeated application of the graph’s adjacency relation. Formally, an infinite path is a sequence where each is a vertex and each is an edge in the graph, such that for all i, the edge connects the vertex to , and the sequence does not terminate.
Definition 6.5 (Infinite Path in a Directed Graph). Let be a directed graph, where V is the set of vertices and is the set of edges. Aninfinite pathin G is a sequence of nodes that continues indefinitely, where each and for all .
Definition 6.6 (Rooted Tree). Arooted treeis a connected acyclic graph in which one vertex is distinguished as the root and every edge is directed away from the root. Afinite treeis a rooted tree in which the set of nodes V is finite.
Definition 6.7 (Algebraic Inverse Tree). An algebraic inverse tree (AIT) is a rooted tree structure defined recursively using the inverse Collatz function . Specifically:
The tree is rooted at node 1.
Each node n has children given by the elements of .
An edge exists if and only if h is a child of n based on .
An AIT can be finite with some maximum depth, or infinite in depth.
Definition 6.8 (Finite AIT). A finite AIT is a directed tree , where is a finite set of vertices and is a set of directed edges. Each vertex represents a natural number, and each directed edge corresponds to an application of the inverse Collatz function from v to u. The tree thus represents a finite number of inverse Collatz iterations starting from 1.
Definition 6.9 (Level of an AIT). The level of a node in an AIT is defined recursively - the root is at level 0, and children of a level l node are at level .
Axiom 5 (Rooted Tree Structure). An AIT forms a rooted tree structure with node 1 as the root.
Axiom 6 (Node-Number Correspondence).
Each node in an AIT corresponds to a unique natural number.
Axiom 7 (Edge Formation). An edge exists in an AIT if and only if .
Axiom 8 (Path Convergence).
Every path in an AIT converges to the root node 1.
Axiom 9 (Well-Founded Order). The nodes in an AIT are well-ordered by the standard < relation on natural numbers.
Theorem 6.1 (Unique Path to Root). Every node in an AIT has a unique path to the root node 1.
Proof. Follows from the Path Convergence and No Cycle axioms. ☐
Theorem 6.2 (Well-Ordered Nodes). The set of nodes in an AIT is well-ordered.
Proof. Follows from the Well-Founded Order axiom. ☐
Theorem 6.3 (Subtree Preservation). Every subtree of an AIT is also an AIT.
Proof. Follows from the axioms which recursively define AITs. ☐
Theorem 6.4 (Node Countability). The number of nodes in an AIT is countably infinite.
Proof. Follows from the bijection between nodes and natural numbers. ☐
Theorem 6.5 (Cardinality Equivalence). The cardinality of nodes in an AIT equals the cardinality of .
Proof. Follows from the bijective correspondence between nodes and . ☐
Definition 6.10. An infinite AIT, denoted , is a directed tree where the set of nodes has cardinality , the cardinality of the natural numbers.
Theorem 6.6.
can be constructed as the limit of an increasing sequence of finite AITs :
Proof. By the Subtree Axiom, each is a finite AIT. By the Countability Axiom, each contains countably infinite nodes.
As , the number of nodes across the increases without bound, resulting in a countably infinite limit tree .
By the Cardinality Equivalence Theorem, , matching the cardinality of . ☐
Definition 6.11 (Trivial Cycle). Atrivial cyclein the context of a Collatz sequence or a similar iterative process is a sequence of numbers where the iteration returns to the starting number after a finite number of steps. In the simplest case, this could be a one-step cycle where a number maps directly to itself under the given function. For the Collatz sequence, a trivial cycle would be the sequence (1, 4, 2, 1), which is the only known cycle for positive integers under the traditional Collatz function.
Definition 6.12 (Non-Trivial Cycle). Anon-trivial cyclein the context of the Collatz conjecture is a sequence of natural numbers , with , such that:
For each , the Collatz function C applied to yields , i.e., .
Applying the Collatz function to yields , i.e., .
The cycle is not the trivial cycle or any repetition thereof.
Moreover, it is quantified that for the cycle to be non-trivial, it must satisfy such that .
Theorem 6.7 (Rooted Tree Structure of AIT). An Inverse Collatz Tree (AIT) forms a rooted tree with nodes representing natural numbers.
Proof. We will prove this theorem by mathematical induction.
Base Case: The AIT starting from the root node 1 trivially forms a rooted tree, since it contains only the single node 1 with no children.
Inductive Hypothesis: Assume that for some , the AIT constructed from k forms a rooted tree.
Inductive Step: Consider the AIT constructed from . By Axiom 3, each node m in this AIT has children given by . By Axiom 1, is well-defined. By the injectivity of (Lemma 4), each node has a unique predecessor. Therefore, there is a unique path from to any other node. By Lemma 7, no non-trivial cycles can occur. Hence, the AIT from is a rooted tree.
By mathematical induction, we have shown that the AIT constructed from any natural number is a rooted tree. ☐
Theorem 6.8 (Parental Structure in AIT). Each node in an Inverse Collatz Tree (AIT) has at most two parents.
Proof. Let us assume, for contradiction, that there exists a node n in an AIT that has more than two parents.
By Axiom 3, the potential parents of n are given by .
By Axiom 2, if n is even, then , so n would have only one parent, contradicting the assumption.
If n is odd, then by the definition of , either:
, in which case gives two parents, or , in which case gives one parent. In both odd cases, it contradicts having more than two parents.
Therefore, by contradiction, we have shown that each node n in an AIT can have at most two parent nodes. ☐
Theorem 6.9. The Inverse Collatz Tree derived from the inverse Collatz function is a binary tree.
Proof. Let be the Inverse Collatz Tree, where V is the set of nodes and E is the set of edges. By Theorem 5, V contains every natural number and is therefore countably infinite.
By Axiom 3, an edge exists iff . By the definition of , each has either one or two parent nodes in V:
If , then so n has one parent node. If , then so n has two parent nodes. Since is injective by Lemma 4, distinct nodes cannot share the same parent.
Therefore, every node in V has at most two parents, satisfying the definition of a binary tree. Hence, AIT is a binary tree. ☐
Theorem 6.10. The non-injectiveness of the Collatz function implies that some values in the Collatz sequence might correspond to multiple nodes in an AIT, but this multiplicity does not affect the fundamental connection between the structures of AITs and the dynamics of the Collatz sequence.
Proof. By Lemma 6, the Collatz function C is not injective. This means there exist distinct such that .
Consider an AIT derived from . By Axiom 3, each node x in the AIT has children given by .
Now suppose for some . Then by the definition of , both m and n would be children of node y in the AIT.
Therefore, the non-injectiveness of C implies that some nodes in the AIT can correspond to multiple values in a Collatz sequence.
However, by Axiom 2, each node in the AIT still represents a valid natural number. And by Axiom 3, the tree structure connecting nodes to their preimages under reflects the inverse Collatz dynamics.
So while multiplicity arises, the core connection between AIT structure and Collatz sequence dynamics remains intact. The AIT veridically represents the underlying dynamics. ☐
Theorem 6.11. Let be a rooted binary tree with root r. For any node , there exists a unique path from u to r.
Proof. We will use proof by mathematical induction on the distance between nodes u and r, defined as the length of the shortest path between u and r, where length is the number of edges. This distance function satisfies the following properties:
Non-negativity:
Identity: if and only if
Symmetry:
Triangle inequality:
Base case: If such that , then by identity , so the only path is u itself.
Inductive hypothesis: Assume that for all with , there exists a unique simple path from v to r.
Inductive step: If , u has a unique parent v where and . By the inductive hypothesis, there is a unique path . Extending this by the edge gives a unique path from u to r.
By mathematical induction, every has a unique path to the root r. ☐
Lemma 6.12. Let be the Inverse Collatz Tree derived from the Collatz function C and its multivalued inverse . For every natural number n, there exists a corresponding node in .
Proof. We prove this lemma using the principle of strong mathematical induction on n.
Let be the proposition that for every natural number k less than or equal to n, there exists a node k in .
Base Case (): The root of is defined as 1, thus is true.
Inductive Hypothesis: Assume that for all m less than or equal to n, the proposition holds true, meaning that each k within this range exists as a node in .
Inductive Step: To prove , consider the next natural number . According to the definition of , there must be a parent node p such that k is in the preimage of p under C. Given the inductive hypothesis, node p exists within since p is less than or equal to n. The structure of , which is determined by , guarantees that for node p to be present, node k must also be included as a child of p. Therefore, node k must exist in , confirming that is true.
By the principle of strong induction, we conclude that the proposition is valid for all natural numbers n. This implies that every natural number is represented as a node in . ☐
Theorem 6.13. For any natural number n, it is possible to reach node n from the root node in a finite number of steps in the Inverse Collatz Tree .
Proof. We employ the principle of strong mathematical induction to demonstrate this theorem. Let represent the following proposition:
For any natural number m that is less than or equal to k, the node m can be reached from the root of in a finite number of steps.
Base Case (): Given that node 1 is the root of by definition, reaching it requires zero steps, thus is established.
Inductive Hypothesis: Suppose that is valid for all natural numbers j up to k, which implies that each node m with can be finitely reached from the root in its respective AIT.
Inductive Step: To validate , consider a node . Axiom 3 dictates that node n has predecessors that are described by . We consider two scenarios:
Case 1: If n is not congruent to 4 modulo 6, then its predecessor is larger than n, and by the inductive hypothesis, all nodes up to k are reachable in a finite sequence of steps. Therefore, node n is accessible through its predecessor.
Case 2: If n is congruent to 4 modulo 6, it has at least one predecessor smaller than n. According to the inductive hypothesis, this smaller predecessor can be reached in a finite number of steps. Since n is the child of this reachable predecessor, n is also reachable within a finite sequence of steps.
Given that is confirmed in both instances, by the method of strong induction, the proposition is valid for every natural number n. ☐
Lemma 6.14. In the Collatz Inverse Tree (AIT_1) constructed from the function C and its inverse , non-trivial cycles are impossible regardless of the validity of the Collatz Conjecture.
Proof. Assume, for the sake of contradiction, that there exists a non-trivial cycle L different from in .
By the definition of a cycle, there must exist a node such that n is an element of for some , with denoting the k-times iterated application of .
By Axiom 3, each node in has a unique sequence of predecessors determined by .
Lemma states that has a complete image of into .
If L were to exist, repeatedly applying to the predecessors of n would result in an infinite descending sequence in , contradicting the countability of established in Theorem .
The assumption of the existence of L leads to a contradiction, implying that such a cycle cannot be present in .
By Axiom 2, is injective, so a node cannot have multiple predecessors, a necessary condition for a non-trivial cycle.
Furthermore, the existence of a cycle would violate the tree structure of AITs.
Thus, the existence of non-trivial cycles in , or any AIT derived from , is a logical impossibility. The initial assumption is denied, proving the non-existence of such cycles. ☐
Lemma 6.15. Let be a finite Inverse Collatz Tree of depth d rooted at 1. Consider any finite sequence of nodes in , where each . Then there exists such that for all , , implying P is eventually strictly decreasing.
Proof. Assume for contradiction that no such exists. This means such that .
Consider . By our assumption, S is infinite.
However, by the Well-Ordering Principle (Theorem ), every non-empty subset of has a least element. Let be the least element of S.
By definition of S, . And since is the least in S, , otherwise would be a smaller element in S.
Therefore, , contradicting the fact that is in S.
By contradiction, our assumption is invalidated. Hence such that , and P is eventually decreasing. ☐
Lemma 6.16. Let be a finite Collatz Inverse Tree (AIT) with root node . Then, every finite path in T converges to the root node r.
Proof. We will prove the lemma by structural induction on the height h of the nodes in T.
Base case: For the root node r with , every finite path starting from r converges to r in 0 steps.
Inductive Hypothesis: Suppose that for every node u with , every finite path starting from u converges to r.
Inductive Step: Let w be a node with and parent u. Let be a finite path starting from w. By the inductive hypothesis, the subpath converges to r. Concatenating the step to Q gives us P, and thus P also converges to r.
By structural induction on the height of the nodes, we have proven that every finite path in the finite AIT T converges to the root node r. ☐