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On a problem of minimal additive complements for not eventually periodic $S$ -difference sets

This paper provides an affirmative answer to a specific problem regarding minimal additive complements for not eventually periodic $S$ -difference sets, as posed by Ma and Chen.

Original authors: Min Tang, Wenjing He

Published 2026-07-30

📖 7 min read🧠 Deep dive

CC BY 4.0

Original authors: Min Tang, Wenjing He

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

Imagine you are standing in an infinite hallway made of integer tiles, stretching forever in both directions. You have a special set of "jumping stones" called $W$ . If you stand on any stone in $W$ and take a step from a specific collection of "helper" stones called $C$ , you want to be able to land on every single tile in the hallway. In the language of mathematics, if the sum of your helper stones and your jumping stones covers the entire number line, we say $C$ is an "additive complement" to $W$ .

But here is the twist: what if your collection of helpers is too big? What if you could throw away a few stones and still land on every tile? A "minimal additive complement" is the smallest possible group of helpers you can use—so small that if you remove even one single stone, you'll leave a gap in the hallway that no one can reach. Mathematicians have been fascinated by this puzzle for over a decade, trying to figure out which patterns of jumping stones allow for such a perfect, tiny team of helpers. The big question was: if your jumping stones follow a pattern that never repeats itself (a "not eventually periodic" set) but the gaps between them are always small and chosen from a specific list of numbers, can you always find this minimal team?

This paper, written by Min Tang and Wenjing He, answers that question with a resounding "yes." The authors tackle a specific, tricky version of the problem where the gaps between the stones in $W$ are chosen from a finite list of positive integers, $S$ , and every number in that list appears as a gap infinitely often. They prove that no matter what list $S$ you pick (as long as it has at least two different numbers), you can always construct a never-repeating sequence of stones that has a minimal additive complement. They don't just guess; they build a detailed, step-by-step recipe to create these sequences and prove mathematically that the resulting helper team is indeed the smallest possible one.

The Story of the Gap-Fillers

To understand what Tang and He did, let's picture the problem as a game of filling a giant, infinite mosaic.

The Players

The Pattern ( $W$ ): Imagine a line of stepping stones. The distance between one stone and the next is never random; it's always a number from a specific "menu" of sizes, let's call it $S$ . For example, your menu might be $\{3, 5\}$ . So, you might jump 3 steps, then 5, then 3, then 3, then 5 again. The rule is that you must use every size on the menu infinitely many times, and the pattern of jumps must never settle into a boring, repeating loop (like 3-5-3-5-3-5 forever). This is what mathematicians call an "INEP S-difference set" (Infinite, Not Eventually Periodic).
The Helpers ( $C$ ): These are the stones you place in the gaps. If you stand on a helper stone and jump to any stone in your pattern $W$ , you should be able to reach every integer on the number line.
The Goal: Find the minimal set of helpers. This means finding the smallest team of helpers where every single member is absolutely essential. If you fire one, the coverage breaks.

The Previous Mystery
Before this paper, mathematicians knew the answer for some specific menus. If your menu was just $\{1, 2\}$ , or if the numbers had special relationships (like one being a multiple of the other), they could build the solution. But for a general menu like $\{3, 7, 11\}$ , or any random mix of numbers, the question hung in the air: Does a minimal team always exist? Some earlier work suggested that if the gaps were too regular, you might not find a minimal team, but if they were chaotic enough, you might. The authors of this paper wanted to settle the score for any finite menu of gaps.

The Master Plan: Building the Bridge
Tang and He didn't just say "it exists." They built it. Their proof is like an architectural blueprint for constructing a bridge that spans an infinite canyon. They split their construction into two main scenarios, depending on the smallest number in their menu $S$ .

Scenario 1: The Menu Includes the Number 1
If your smallest gap is 1, the construction is a bit like laying down a long, winding path. The authors start with a small, manageable chunk of stones. Then, they use a clever inductive method (building step-by-step) to extend the path forever.

They create "blocks" of stones.
Inside these blocks, they use a mathematical tool (related to the "Frobenius Coin Problem," which asks how to make change with specific coin denominations) to ensure that the gaps between stones match the numbers in their menu $S$ .
They carefully place "helper stones" (the set $C$ ) at specific intervals.
The magic happens in the "transitions" between blocks. They arrange the gaps so that the helper stones can reach every single integer, but if you remove even one helper, a specific "hole" appears that no other helper can fill. They prove that the gaps between the stones in their construction get larger and larger in a specific way, ensuring the pattern never repeats itself, yet the minimal team of helpers still works perfectly.

Scenario 2: The Menu Starts with a Number Greater than 1
This is the trickier part. If your smallest gap is, say, 3 or 5, you can't just fill in the gaps with single steps. The authors had to get creative.

They realized that if the numbers in the menu don't share a common divisor (they are "relatively prime" in a group sense), you can still build the path.
They constructed a more complex structure where the "helper stones" come in small groups or clusters.
They used a sophisticated counting argument to show that even though the gaps are larger, the arrangement of the helper clusters creates a "net" that catches every integer.
Crucially, they proved that the "holes" left by removing a helper are unique to that specific helper. It's like a lock and key system: Helper A opens a specific lock, and no other helper has the key. If you take Helper A away, that lock stays shut, and the coverage fails.

The Verdict
The authors' construction is rigorous. They didn't simulate this on a computer or suggest it might be true; they provided a mathematical proof. They showed that for any finite set of positive integers $S$ with at least two elements, you can create a never-repeating sequence of gaps using only numbers from $S$ , and for that sequence, a minimal additive complement always exists.

They effectively closed the book on this specific version of the problem. The answer to the question "Is it true that for any finite set $S$ ..." is a definitive yes. The paper confirms that the chaotic, non-repeating nature of the gaps doesn't prevent the existence of a perfect, minimal team of helpers. In fact, the very chaos of the pattern is what allows the authors to engineer the solution, ensuring that every helper is indispensable and the entire number line is covered.

Technical Summary: On a Problem of Minimal Additive Complements for Not Eventually Periodic S-Difference Sets

Problem Statement
The paper addresses a specific problem in additive number theory concerning the existence of minimal additive complements for infinite sets of integers. Let $C$ and $W$ be subsets of integers. $C$ is an additive complement to $W$ if $C + W = \mathbb{Z}$ . $C$ is a minimal additive complement if no proper subset of $C$ satisfies this condition.

While the existence of minimal complements for finite sets and certain infinite sets (such as those with unbounded gaps or specific periodic structures) has been established by Nathanson, Chen, Yang, Kiss, Sándor, and others, the case of not eventually periodic sets with bounded gaps remains less understood. Specifically, Ma and Chen [14] posed the following problem (Problem 1.3 in their work):

Let $S$ be a finite set of positive integers with $|S| \ge 2$ . Does there exist an infinite, not eventually periodic set $W = \{w_1 < w_2 < \dots\} \subseteq \mathbb{N}$ such that:

$W$ has a minimal additive complement;

The difference between consecutive elements $w_{i+1} - w_i$ belongs to $S$ for all $i$ ;

Each element $s \in S$ occurs as a difference $w_{i+1} - w_i$ for infinitely many indices $i$ ?

Such a set $W$ is termed an INEP $S$ -difference set (Infinite, Not Eventually Periodic $S$ -difference set). Previous results had solved specific subcases (e.g., when elements of $S$ divide each other, or when $\gcd(s_1, s_2)=1$ with specific modular constraints), but a general solution for arbitrary $S$ was lacking.

Methodology
The authors, Tang and He, provide an affirmative answer to Problem 2 by constructing such sets and their corresponding minimal complements. The methodology relies on an inductive construction of the set $W$ and a carefully designed set $C$ .

Reduction via Scaling: The authors first utilize Lemma 2.2 to reduce the general case to the scenario where $\gcd(s_1, \dots, s_l) = 1$ . They prove that if a minimal complement exists for a scaled-down set, a minimal complement can be constructed for the scaled-up set $dW$.
Inductive Construction: The proof is split into two cases based on the smallest element of $S$ $S$ , denoted $s_1$ $s_{1}$ :
- Case 1 ( $s_1 = 1$ ): The authors construct sequences $\{d_i\}$ ${d_{i}}$ , $\{W_i\}$ ${W_{i}}$ , and $\{c_i\}$ ${c_{i}}$ inductively. The set $W$ $W$ is built in stages, where each stage $W_i$ $W_{i}$ adds three disjoint subsets ( $V^{(1)}_{i-1}, V^{(2)}_{i-1}, V^{(3)}_{i-1}$ $V_{i - 1}^{(1)}, V_{i - 1}^{(2)}, V_{i - 1}^{(3)}$ ) to the previous set $W_{i-1}$ $W_{i - 1}$ .
  - Step 1: Fills a specific interval using steps of size 1 (since $s_1=1$ ) and the largest element $s_l$ .
  - Step 2 & 3: Utilizes Lemma 2.1 (a result on the Frobenius coin problem) to express interval lengths as positive linear combinations of elements in $S$ . This allows the authors to fill gaps with steps strictly belonging to $S$ , ensuring the difference set property.
  - The set $C$ is constructed as $\{c_i\}_{i=1}^\infty$ , where $c_i$ are chosen to satisfy specific inequalities ensuring that $C$ covers $\mathbb{Z}$ when added to $W$ , but no element can be removed without breaking the coverage.
- Case 2 ( $s_1 \ge 2$ ): This case is more complex as the step size 1 is unavailable. The authors construct sequences $\{d_i\}$ ${d_{i}}$ , $\{W_i\}$ ${W_{i}}$ , and a set of parameters $\{c^{(1)}_i, \dots, c^{(s_1)}_i\}$ ${c_{i}^{(1)}, \dots, c_{i}^{(s_{1})}}$ .
  - They define a function $h(s_1; r)$ based on the remainder $r$ when a specific $s_j$ is divided by $s_1$ .
  - The construction involves creating blocks of elements with specific modular properties. The set $C$ is constructed as a union of $s_1$ arithmetic progressions (shifted) at each stage.
  - Similar to Case 1, Lemma 2.1 is employed to fill intervals with steps from $S$ . The construction ensures that the "gaps" between the constructed blocks are filled by the specific structure of $C$ , while the internal gaps of $W$ remain within $S$ .
Verification of Minimality: For both cases, the authors rigorously prove that the constructed $C$ is a minimal complement. This involves showing that for any $n \in \mathbb{Z}$ , $n \in W+C$ , and for any specific element $c \in C$ , there exists an integer $n$ such that $n \in W+c$ but $n \notin W + (C \setminus \{c\})$ . This is achieved by analyzing the specific "holes" created in the sumset when an element is removed, leveraging the inductive definitions of $d_i$ and the disjointness of the constructed sets.

Key Results
The paper establishes the following main theorem:

Theorem 1.1: Let $l \ge 2$ be a positive integer and $S = \{s_1, s_2, \dots, s_l\}$ be a set of positive integers with $1 \le s_1 < s_2 < \dots < s_l$ . There exists an INEP $S$ -difference set $W$ and there exists a minimal complement to $W$ .

This result generalizes previous theorems (Theorems E, F, and G in the introduction) which were limited to specific relationships between elements of $S$ (e.g., divisibility or coprimality conditions). The current work removes these restrictions, proving the existence for any finite set $S$ with $|S| \ge 2$ .

Significance and Claims
The authors claim that their work provides a unified proof framework that extends the scope of known results to arbitrary $S$ -difference sets. By solving Problem 2 posed by Ma and Chen, the paper settles the existence question for minimal additive complements in the context of infinite, not eventually periodic sets with bounded gaps defined by an arbitrary finite difference set $S$ .

The paper does not claim to solve the problem for sets where $|S|=1$ (which is trivial or impossible depending on periodicity) nor does it claim to characterize all such sets, but rather to prove the existence of at least one such set for any given valid $S$ . The work builds upon the constructive ideas of Theorems F and G but synthesizes them to handle the general case.

Conclusion
The paper successfully answers the open problem regarding the existence of minimal additive complements for not eventually periodic $S$ -difference sets. Through a sophisticated inductive construction utilizing the Frobenius coin problem (Lemma 2.1) and careful control of modular arithmetic, the authors demonstrate that for any finite set of positive integers $S$ with cardinality at least 2, one can construct an infinite, non-periodic set $W$ with differences in $S$ that admits a minimal additive complement.

The Story of the Gap-Fillers

Technical Summary: On a Problem of Minimal Additive Complements for Not Eventually Periodic S-Difference Sets

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