4. Proofs about AITs
Let
be the Collatz function defined as:
Figure 1.
Visual representation of the algebraic tree . illustrating the inverse sequence from 1 to 10 rather than from 10 to 1.
Figure 1.
Visual representation of the algebraic tree . illustrating the inverse sequence from 1 to 10 rather than from 10 to 1.
And let
be the multivalued inverse function of
given by:
Definition 4.1. A cycle in the context of a function is a finite sequence of numbers such that:
That is, by iterating the function starting from any number in the sequence, we eventually return to the initial number, completing a "cycle".
We now formally define the Algebraic Inverse Tree:
Definition 4.2. Let be the directed tree defined over and rooted at k. It is constructed recursively as:
The root node of is k.
If n is a node in and , its child nodes are the elements of .
The edges from n to each child h are labeled with the operation .
is the Algebraic Inverse Tree (AIT) of parameter k defined over the set of natural numbers .
Theorem 4.1. It is possible to effectively construct any AIT representing all inverse operations of the Collatz function using Algorithm 1.
Algorithm 1 Construction of AIT |
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procedure ConstructAIT(k, )
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Create an empty tree T
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Set the root node of T to k
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set containing k
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queue with root node
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while Q is not empty and depth of do
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dequeue from Q
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for each p in do
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if p not in then
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Add edge from p to n in T
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enqueue p to Q
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add p to
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end if
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end for
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end while
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return T
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end procedure
|
Proof. Let T be the tree returned by ConstructAIT(k, ). We will prove by induction on the depth d of T that:
Induction Hypothesis P(d): All nodes in T at depth represent unique inverse Collatz operations.
Base Case: At , the root node k trivially satisfies .
Inductive Step: Assume holds for all . Let n be a node at depth . By the algorithm, n is generated by taking all elements of for some parent node p at depth d. By the inductive hypothesis, p represents a unique inverse operation. Since R contains all valid inverses, n must also represent a unique inverse operation. Hence holds.
By induction, holds for all depths d. Therefore, ConstructAIT produces an AIT representing all unique inverse Collatz operations. □
Theorem 4.2. The problem of generating the AIT of a natural number is NP-complete.
Proof.
1. The problem is in NP: Same as original proof.
2. NP-hardness: We will reduce from the known NP-complete problem PARTITION: Given a set of n positive integers, is there a subset such that ?
Given an instance of PARTITION, we construct an AIT as follows:
The depth of the AIT is . The target node is The starting node There is a path from the starting node k to the target node m if and only if there is a partition of S. Specifically, there is a bijection between subsets that sum to and paths in the AIT from k to m.
Since PARTITION is NP-hard, generating the AIT is also NP-hard.
Conclusion: The problem is in NP and is NP-hard, so it is NP-complete. □
Theorem 4.3. Given the relation as previously defined, the Algebraic Inverse Tree (AIT) is a binary tree.
Proof. We begin by noting two essential properties of the inverse relation R:
R is inyectiva: For any distinct numbers a and b, .
R is sobreyectiva: For every number y, there exists an x such that .
1. Each node has at most two pre-images (or "parents"):
Given any number y:
If y is not congruent to : The only possible pre-image is . This is because the other potential pre-image, , is not valid.
If y is congruent to : Both and are valid pre-images.
By the inyectivity of R, no two distinct numbers can map to the same y, unless y satisfies the congruence condition mentioned, where there are precisely two distinct pre-images.
2. Exclusivity of the inverse relation:
Given that R is inyectiva, no node in the AIT will have more than two parents. This ensures that there are no unexpected branches in the AIT. Moreover, due to the sobreyectivity of R, every node will be connected in the AIT.
3. No cycles in the AIT:
Given our inverse function and the way numbers are formed, as we traverse the AIT in reverse (using R), numbers increase because we either multiply by 2 or divide and subtract. This increasing nature of the tree traversal, combined with the inyectivity of R, ensures that we never encounter a previously visited number, thereby preventing cycles.
4. Connectivity and Completeness:
Due to the sobreyectivity of R, every node in the AIT will be connected, ensuring the tree’s completeness.
Hence, combining all these arguments and utilizing the properties of R, we conclude that the AIT is indeed a binary tree. □
Theorem 4.4. In any rooted binary tree, for any given node, there exists a unique path leading from that node to the root.
Proof. Let be a rooted binary tree, where V is the set of vertices (nodes) and E is the set of directed edges. Let be the root of T, and let be any vertex. We will prove by induction on the depth (the number of edges in the path from u to r) that there exists a unique simple path in T from u to r.
Base case: If , then u is the root r, so the unique simple path from u to r is the trivial path consisting of just u.
Induction hypothesis: Assume the theorem holds for all nodes whose depth is less than or equal to k, for some .
Inductive step: Let . Since every node other than the root has a unique parent in a binary tree, let be the parent of u. By the induction hypothesis, there exists a unique simple path from to r, call it P. Extending P by adding u at the beginning gives the unique simple path from u to r.
By induction, we have proven that for every node u in a rooted binary tree T, there exists a unique simple path from u to the root r. □
We now prove two key lemmas about the properties of AITs:
Lemma 4.5.
Let be the Collatz function defined by
The function f is invertible in a multi-valued sense under the following conditions:
For , there exists a unique such that .
For , there exist two distinct such that .
Proof. We will prove this by strong induction on x.
Base case: For , let . Then , proving the base case.
Induction hypothesis: Assume the claim holds for all for some .
Inductive step: We will show the claim holds for . Consider two cases:
Case 1: . Let . Since k is even, by the definition of f.
Case 2: . Let . Since k is odd, . Also, since , is an integer and .
By the strong induction hypothesis, the lemma holds. □
Lemma 4.6. Every natural number appears as a node in the AIT .
Proof. To prove the lemma, we must establish two things:
Let f be the Collatz function and R its inverse function. By Lemma 4.5, R is well defined.
Base case: is the root node of . Since 1 does not have a predecessor and is the base of our AIT, the lemma is true for .
Induction hypothesis : Suppose that for some arbitrary k, all natural numbers n such that have a unique predecessor in and therefore appear in the tree.
Inductive step: We must show that k has a unique predecessor in and therefore appears in the tree.
Let . By the definition of R, . If , by the induction hypothesis , p is in . Then, by the construction of , k is a child of p in .
To prove that k has a unique predecessor, suppose for the sake of contradiction there exists a q such that and . This would mean that and , contradicting the uniqueness of the inverse. Therefore, p is the unique predecessor of k.
By mathematical induction on k, every natural number appears as a node in and has a unique predecessor in the tree. □
Lemma 4.7. The inverse function R associated with the Collatz sequence is injective.
Proof. To prove the injectivity of R, we need to show that for any distinct natural numbers a and b such that , . To this end, let’s investigate the values in for each congruence class modulo 6.
Congruence class : For numbers in this class, contains only .
Congruence class : The value contains (since ) and potentially other values if the sequence has been defined for other inverse steps for this class.
Congruence class : For this class, contains just .
Congruence class : Here, is solely .
Congruence class : For these numbers, will have and also (given ).
Congruence class : In this class, only contains .
Given the behavior of in each congruence class modulo 6, it is clear that for any distinct natural numbers a and b with , . This proves the injectivity of R. □
Theorem 4.8. [Finite Steps Theorem in AIT] Let be the algebraic inverse tree with parameter n defined recursively as:
The root node of is n.
If m is a node in , its child nodes are the elements of the set , where R is the multivalued inverse function of the Collatz algorithm.
Then, for any natural number n, the number n can be reached in a finite number of steps starting from its root node in the algorithm.
Proof. We will prove the theorem by strong induction on n.
Base Case: For , has the root node 1. No additional steps are needed to generate 1.
Inductive Hypothesis: Assume that for an arbitrary natural number k, any natural number less than k can be reached in a finite number of steps from its root node in the algorithm.
Inductive Step: Consider the number and the inverse function R. Based on the behavior of R and the Collatz function, there are two primary cases for :
Case 1: . The unique predecessor is . By the inductive hypothesis, since , the number is reachable in a finite number of steps. Thus, is also reachable in one additional step.
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Case 2: . The inverse function R implies two predecessors for : and . Let’s demonstrate that this is exhaustive and accurate:
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Sub-case 1: is even. Then for some integer l. Using the Collatz function, . So, is a valid predecessor.
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Sub-case 2: is odd and . Then, using the Collatz function, we find the other predecessor: .
Now, to prove
, we start with:
This inequality holds true for . By the inductive hypothesis, since is less than , it is reachable in a finite number of steps, making reachable in an additional step.
For both cases, we have shown that is reachable in a finite number of steps. By the inductive hypothesis, all numbers less than are also reachable in a finite number of steps. Therefore, the algorithm can generate any natural number n in a finite number of steps, starting from its root node.
By the principle of strong induction, the theorem is thus proven. □
We are now ready to formally prove the Collatz Conjecture: