2460-0245Journal of the Indonesian Mathematical SocietyJIMS2460-02452086-8952IndoMS10.22342/jims.v32i1.1656On K-AP And K-Avoid-AP Van Der Waerden GameSimKai An
Malaysiakaians@sunway.edu.my
RuzaliWan Muhammad Afif Wan
Malaysiawan.w126@imail.sunway.edu.my
WongKok Bin
Malaysiakbwong@um.edu.my
HoChee Kit
Malaysiackho@sunway.edu.my
NurwigantaraMu'amar Musa
Indonesiamuamar.musa.n@mail.ugm.ac.id
School of Mathematical SciencesSunway Universityhttps://ror.org/04mjt7f73MalaysiaInstitute of Mathematical Sciences, Faculty of ScienceUniversity of Malayahttps://ror.org/00rzspn62MalaysiaCorresponding author: Kai An Sim. Email: kaians@sunway.edu.my0103202605012026321MARCH1122002202404102025Copyright (c) 2026 Journal of the Indonesian Mathematical Society2026Journal of the Indonesian Mathematical Societyhttps://creativecommons.org/licenses/by-nc-nd/4.0/This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.

Let w(k; 2) be the van der Waerden number such that for every 2 colouring of [1, w(k; 2)] there is a monochromatic k-term arithmetic progression (AP). Consider the following two 2-players games: k-AP game and k-AVOID-AP game. These are two different games between two players, Player 1 and Player 2, on a sequence of integers [1, n] where n ∈ Z + . Each player's aim is to obtain or avoid forming a monochromatic k-term arithmetic progression. The player who first obtains a monochromatic k-term arithmetic progression wins or loses, thus ending the game. In this paper, we investigate these two games and propose a new parameter: the minimum number of turns ŵn(k) (and wn(k)) needed for any of the player to win in k-AP (and k-AVOID-AP game respectively). We propose the winning strategies for Player 1 and Player 2 and hence show that ŵn(3) = 5, ŵn(4) = 7 and wn(3) = 9. We also have shown that in a k-AVOID-AP game on [1, n], where n is sufficiently large, Player 2 always has a winning strategy if n is even.

van der Waerden numberk-term arithmetic progressionk-AP gamek-AVOID-AP gameFile created by JATS Editorhttps://jatseditor.comissue-created-year20261. Introduction

For an arithmetic progression , we say that A is a k-term arithmetic progression (AP) with diference d, or we write For a positive integer t, we denote the set by . An r-colouring of a set S is a surjective function . A monochromatic k-term arithmetic progression refers to a k-term arithmetic progression such that all of its elements are of the same colour. One of the most fundamental Ramsey-type theorem is the van der Waerden theorem. The van der Waerden’s theorem [1] states that there exists a least positive integer such that for any n , every r-colouring of [1, n] admits a monochromatic k-term arithmetic progression. For more results on van der Waerden numbers, see [2, 3, 4, 5, 6, 7].

Theorem 1.1.[1](van der Waerden’s Theorem) Let be integers. There exists a least positive integer such that for any every r-colouring of [1, n] admits a monochromatic k-term arithmetic progression.

Note that exact values of are only known for small value of k where and , see [5]. The exact value of is not known in general. We refer the readers to [4, 3, 8, 9, 6, 7].

There are many variants of Ramsey games that have been investigated, see [10, 11, 12, 13, 14, 15, 16, 17, 18]. Consider one of the games here, which is inspired by Ramsey Theory: Player 1 (blue) and Player 2 (red) alternate colouring edges of a complete graph with their colours. Let be a subgraph of . The first player to avoid a monochromatic H in his colour wins. This motivates us to investigate a similar van der Waerden game as follows.

A k-AP van der Waerden game (or k-AP game) is a game between two players, Player 1 and Player 2, on the set of integers . Initially, all the integers are uncoloured. Player 1 will first select an integer and colour it with red. Player 2 will then select another integer and colour it with blue. The two players continue this alternating colouring process on uncoloured integers. Each player’s aim is to form a monochromatic k-term arithmetic progression as soon as possible. The player who first obtains a monochromatic k-term arithmetic progression wins, thus ending the game. We assume all players play optimally where both players never purposefully avoid a monochromatic k-term arithmetic progression and at the same time, they would like to prevent the opponent from forming a monochromatic k-term arithmetic progression.

A k-AVOID-AP van der Waerden game (or k-AVOID-AP game) is a game opposite to that of the k-AP van der waerden game where both players’ aims are to avoid forming a monochromatic k-term arithmetic progression. The player who first obtains a monochromatic k-term arithmetic progression loses, thus ending the game. We assume all players play optimally where both players never purposefully form a monochromatic k-term arithmetic progression and at the same time, they would like to “force” a monochromatic k-term arithmetic progression for the opponent.

We begin by explaining some notation. A turn of the game is an action of colouring by either Player 1 or Player 2. If they are playing on the set of integers [1, n] and , then by Theorem 1.1, one of the players will win in game when all the integers are coloured. Let be the ith turn. Player 1 makes his move on the odd-numbered turns, whereas Player 2 makes his move on the even-numbered turns. Let denote the status of the integer i where Specifically, means that the number i has been coloured by Player 1 with red and means that the number i has been coloured by Player 2 with blue. A player is said to have a winning strategy if he will win the game by choosing certain integers at each of his turn regardless of what integer the other player chooses at the other player’s turn.

In a game on the set of integers , let to be the minimum number of turns for any of the players to win the game (i.e. the minimum number of turns to form a k-term arithmetic progression). This means that if is odd, then Player 1 wins and if is even, then Player 2 wins. In a game on the set of integers , let to be the minimum number of turns for any of the players to win the game (i.e. the minimum number of turns for the opponent to form a k-term arithmetic progression). So, if is odd, then Player 1 loses and if is even, then Player 2 loses. Clearly, if , then and

2. k-term arithmetic progression in a k-AP game

In a k-AP game, there are three possibilities:

Player 1 has a winning strategy.

Player 2 has a winning strategy.

Both players have no winning strategy.

First, we show that only (1) and (3) are possible.

Theorem 2.1.In a k-AP game on the set of integers , either Player 1 has a winning strategy or both players have no winning strategy.

Proof. Suppose Player 2 has a winning strategy on . At , Player 1 colours 1 with red. By using the strategy , Player 2 will need to colour at with blue. Now, Player 1 shall assume that he had not coloured 1 with red at , and Player 2 has coloured at turn with blue. Basically, Player 1 assume himself as the second player and assume Player 2 as the first player. Using the strategy , Player 1 will need to colour at with red. If ， then Player 1 simply chooses another uncoloured integer, say , and colours it with red. If , then Player 1 will colour with red.

At , Player 2 colours with blue according to the strategy , Player 1 will need to colour with red. We consider two cases:

Case 1: Suppose . If , then Player 1 simply choose another uncoloured integer, say and colour it with red. If , then Player 1 will colour with red.

Case 2: Suppose . Then both and 1 are already coloured by red. In this scenario, if , then Player 1 simply choose another uncoloured integer, say and colour it with red. If , then Player 1 will colour with red. The two players will continue this process until one of them wins.

Since Player 2 is using the strategy , at , Player 2 would have coloured m distinct integers and there is a k-term arithmetic progression among them. At , Player 1 have coloured m distinct integers Note that there should not be any k-term arithmetic progressions in

On the other hand, Player 1 is using the same strategy to win. at Player 1 have coloured distinct integers . Since there is no k-term arithmetic progressions in , at , Player 2 could not have won. But at , Player 2 have coloured m distinct integers and there is a k-term arithmetic progression in , a contradiction. Hence, Player 2 does not have a winning strategy and the theorem follows. □

We will show that in a k-AP game where , Player 1 will have a winning strategy. Furthermore, we show that in where and in where

Lemma 2.2.In a k-AP game on the set of integers

Proof. In order to form a monochromatic k-term arithmetic progression in the winner has to choose at least k integers. If Player 2 wins the game, then Player 1 must have chosen at least k integers; hence . If Player 1 wins the game, then Player 2 has to choose integers and hence . The result follows. □

Lemma 2.3.Suppose there is a sequence of increasing integers

where for all . If at turn (Player 2’s turn), and have been coloured with red and has not been coloured for all , then regardless of Player choices at or , Player 1 will obtain a 4-term arithmetic progression with red at .

Proof. Assume that at , Player 2 colours with blue. Suppose Then at , Player 1 colours with red. Now, and are two 4-term arithmetic progressions and all of these integers are coloured with red except and , which are not coloured. So, regardless of any choice Player 2 makes at , Player 1 will obtain a 4-term arithmetic progression with red at by colouring t or with red.

Suppose . Then at , Player 1 colours with red. Now, and are two 4-term arithmetic progressions and all of these integers are coloured with red except and , which are not coloured. So, regardless of any choice Player makes at , Player 1 will obtain a 4-term arithmetic progression with red at by colouring or with red. □

Theorem 2.4.Consider a k-AP game on

If and , then Player 1 has a winning strategy. Furthermore, .

If and , then Player 1 has a winning strategy. Furthermore, .

Proof. (a) By Lemma 2.2, it is clear that , Player 1 colours 3 with red. Suppose Player 2 colours 1 with blue . At T3, Player 1 colours 4 with red. A 3-term arithmetic progression with red colour is formed if 2 or 5 is coloured by Player 1 at regardless of which integer is coloured by Player 2 at , Player 1 will win.

Suppose Player 2 colours 2 with blue at . At , Player 1 colours 5 with red. A 3-term arithmetic progression with red colour is formed if 1 or 4 is coloured by Player 1 at . So, regardless of which integer is coloured by Player 2 at Player 1 will also win.

Suppose Player 2 colours an integer with blue at . Player 1 colours 1 with red if and colours 2 with red if . In either case, it is not hard to see that Player 1 will win at regardless of any choice Player 2 makes at .

(b) By Lemma 2.2, it is clear that . To show that the equality holds, we shall show that Player 1 has a winning strategy and will win at , regardless of the choices made by Player at and . At , Player 1 colours 13 with red. , Player 2 colours with blue. Now, we look at the following two diferent sequences:

Suppose . Then . At , Player 1 colours 17 with red. At , in the following sequence of integers

we see that integers 13 and 17 are coloured with red and the integers 5, 9, 11, 15, 19, 21 are not coloured. Lemma 2.3, Player 1 will obtain a 4-term arithmetic progression with red at .

Suppose . Then at , Player 1 colours 9 with red. At , in the following sequence of integers

we see that integers 9 and 13 are coloured with red and the integers 1, 5, 7, 11, 15, 17 are not coloured. By Lemma 2.3, Player 1 will obtain a 4-term arithmetic progression with red at .

Suppose . Then at , Player 1 colours 19 with red. At , in the following sequence of integers

we see that integers 13 and 19 are coloured with red and the integers 1, 7, 10, 16, 22, 25 are not coloured. By Lemma 2.3, Player 1 will obtain a 4-term arithmetic progression with red at .

To better describe Player 1’s strategy, we use the following graph for better ilustrations.

Figure 13. k-term arithmetic progression in k-AVOID-AP game

In this section, we shall consider k-AVOID-AP games.

Theorem 3.1.Let and set is even, otherwise set Then, in a k-AVOID-AP game on , Player 2 always has a winning strategy.

Proof. Note that N is always even. We can illustrate the interval as an array below:

Consider any turn where i is odd. If Player 1 chooses an integer at , then Player 2 is able to choose the integer on the following turn . Referring to our array (1) above, this means that Player 2 is choosing the integer that is mirrored by the separation between the first and second rows. For example, if Player 1 chooses 3, then Player 2 chooses , which is directly below 3 in array (1). The same is true if Player 1 chooses a number in the second row of the array; Player 2 will choose the number directly above it. Hence, Player 2 will always have an available ‘mirrored’ number to choose from after Player 1’s turns. As long as Player 1 does not obtain a k-term arithmetic progression, Player 2 will also not obtain a k-term arithmetic progression. By van der Waerden’s theorem, because , an arithmetic progression will eventually be formed. Since Player 1 always makes his colouring choices a turn earlier, Player 1 will obtain a monochromatic k-term arithmetic progression before Player 2 does, thus losing the game. Hence, Player 2 will always win. □

Corollary 3.2.In a k-AVOID-AP game on [1, n], if and n is even, then .

Proof. By Theorem 3.1, if n is even, then it is clear that Player 2 will win. Furthermore, even with an optimum strategy used by Player 1, Player 2 may keep avoiding a monochromatic k-term arithmetic progression in his colour until his second last turn. Hence Player 1 will form a k-term arithmetic progression in his colour in his last turn before Player 2 chooses the last integer. Hence, .□

Lemma 3.3.If Player 2 has a strategy to avoid k-term arithmetic progressions on [1, n], then Player 1 has a strategy to avoid -term arithmetic progressions on . Hence, .

Proof. Suppose Player 2 has a strategy to avoid k-term arithmetic progressions on . Then, in , Player 1 colours integer at with red, then in the rest of his turns, Player 1 can use strategy to avoid monochromatic k-term arithmetic progressions on . As a result, Player 1 may avoid -term arithmetic progressions in the game even with the integer coloured by him. □

In the next subsection, we will show that when , Player 2 still wins (Corollary 3.7).

3.1. 3-AVOID-AP game on [1, 9]

Lemma 3.4.In a 3-AVOID-AP game on [1, 9], Player 1 has a strategy to avoid defeat (or avoid forming a monochromatic 3-term arithmetic progression) until .

Proof. Let for and . At , Player 1 will colour 8 with red. If Player 2 colours 9 with blue at , then Player 1 colours 7 with red at . If Player 2 did not colour 9 at , then Player 1 will colour 9 with red at . Hence, after , we have the following two cases.

Case 1.. Player 2 colours an integer at . By the pigeonhole principle, at , we let Player 1 colour 1, or otherwise colour 2 if 1 has been coloured by Player 2. If Player 1 colours 1 at , then at , Player 1 can colour 2, 3 or 5. Any of these choices will not form a red monochromatic 3-term arithmetic progression. If Player 1 colours at , then after . Hence,

Player 1 can colour 3 or 4 at without forming a red monochromatic 3-term arithmetic progression.

Case 2.. Suppose at and , Player 2 colours integers in and respectively (i may be equal to j) . After , integers in are not coloured for some or . At , Player 1 colours an uncoloured integer in with red. Now, has two integers and one of them is coloured with red by Player 1 at . If the other integer has not been coloured by Player 2 at , then Player 1 colours the uncoloured integer in at . We are done. If Player 2 colours the integer in with blue at then at , Player 1 colours 5 or 6 (in which case does not result in a red monochromatic 3-term arithmetic progression) if both of these integers are uncoloured yet, or else colours any other uncoloured integers in or .

For all cases, Player 1 does not form a monochromatic 3-term arithmetic progression until . This completes the proof. □

Before we move to the next result, we introduce the parameter which refers to the number of possible moves available to Player 2 to avoid a monochromatic 3-term arithmetic progression after . Below we give an example:

The above represents the state of game on , after Player 2 colours a number during . Instead of writing the numbers for the interval, we instead represent the coloured numbers with the numbers 1 and 2 to represent each Players’ choices so far, and the uncoloured numbers are left as blank spaces. The above represents the state whereI and . From here, we see that Player 2 has two winning moves for the upcoming , by colouring either 4 or 7. In other words, either or at would guarantee a win for because neither of these choices lead to an arithmetic progression in blue. Hence, .

Note that we only calculate the value of λ after choice at the end of Meaning that in order for Player 2 to win, must be of value at least 2. If then Player 2 can only avoid blue 3-term arithmetic progression by colouring a particular integer i at . However, this may lead to the scenario where Player 1 colours i at and thus forcing Player 2 to lose (forming a blue 3-term arithmetic progression at .

Lemma 3.5. , Player 2 wins.

Proof. The parameter implies that there exist integers such that when coloured by Player 2 at , he can always avoid a monochromatic arithmetic progression and thus leads to the defeat of Player 1, i.e., Player 1 will form a red monochromatic 3-term arithmetic progression (because . Given that Player 1 can only colour a single integer at , there is still at least one uncoloured integer amongst for to choose at to remain undefeated. □

We are now well-equipped to prove our main result.

Theorem 3.6.In a 3-AVOID-AP game on [1, 9], Player 2 always has a winning strategy.

Proof. Suppose Player 1 colour integers with red and Player 2 colour integers with blue. In each of the cases below, if Player 2 did not form a blue 3-term arithmetic progression after his turn in , then by van der Waerden theorem where will definitely form a red 3-term arithmetic progression after which is the last turn. Hence, to show that has a winning strategy, we want to show that after , then Player 2 will have at least two possible choices in and remain undefeated (i.e. not form a blue 3-term arithmetic progression). Here, we investigate the possible choices of the two players in each turn.

Turn 1 : Player 1 picks an integer and colours it with red.

Turn 2 : Player 2 colours 1 with blue. If 1 was coloured by Player 1, then Player 2 colours 9 with blue. Now, relabel those integers for After relabelling, 1 is coloured with blue. Note that the relabelling can be reversed. If a monochromatic 3-term arithmetic progression after relabelling is formed after some turns, then by reversing the relabelling, we still have a monochromatic 3-term arithmetic progression after the reversing process. So, we may assume that 1 is coloured with blue.

Turn 3 : Player 1 colours his second integer.

Turn 4 : Before Player 2 colours his second integer, two integers are already coloured by Player 1 with red. We first consider the following cases that depend on the choices of Player 1:

Case 4.1: . Player 2 colours 8, and then at , Player 2 colours 6 if it is uncoloured or 9, otherwise. By checking the uncoloured integers after , we see that .

Case 4.2: . Player 2 colours 4, and then at colours 9. This gives us . Note that at , 9 must be uncoloured otherwise Player 1 loses.

Case 4.3: . Player 2 colours 7, and then at colours 6. This gives us . Note that at , 6 must be uncoloured otherwise Player 1 loses.

Case 4.4: . Player 2 colours 2, and then at , Player 2 colours 4 if it is uncoloured or otherwise. By checking the uncoloured integers after , we see that .

Case 4.5: . Player 2 colours 9, then in , Player 2 colours 6 if it is uncoloured or 8 otherwise. In either case, Player 2 colours 4 in .

Case 4.6: . Player 2 colours 3, and then at colours 8 if it is uncoloured or 9, otherwise. This gives us .

Case 4.7: . Player 2 colours 3, and then at colours 6 if it is uncoloured or 8, otherwise. This gives us .

Case 4.8: . Player 2 colours 2, and then at colours 6. This gives us . Note that at must be uncoloured otherwise Player 1 loses.

Case 4.9: . Player 2 colours 8 at . If Player 1 colours 2 at , then Player 2 colours 3 at . If Player 1 colours 4 at , then Player 2 colours at . If Player 1 colours 5 at then Player 2 colours 3 at If Player 1 colours 7 at , then Player 2 colours 4 at . In either case,

Case 4.10: . Player 2 colours and then at colours 2 if it is uncoloured or 3, otherwise. This gives us

Case 4.11: . Player 2 colours 4, and then at colours 8. This gives us . Note that at must be uncoloured otherwise Player 1 loses.

Case 4.12: . Player 2 colours 6 at . If Player 1 colours 2 at , then Player 2 colours 3 at . If Player 1 colours 3 at , then Player 2 colours 2 at . If Player 1 colours 4 at , then Player 2 colours 7 at If Player 1 colours 5 at , then Player 2 colours 7 at . In either case,

Case 4.13: . Player 2 colours 6 at . If Player 1 picks 2 at then Player 2 colours 3 at . If Player 1 colours 3 at then Player 2 colours 2 at . If Player 1 colours 4 at , then Player 2 colours 7 at If Player 1 colours 8 at , then Player 2 colours 3 at . In either case,

For the remaining cases, we may assume that

(i) any two or all integers in remain uncoloured by Player 1 before

(ii) any two or all integers in remain uncoloured by Player 1 before , and

(iii) 9 is uncoloured by Player 1 before .

So, at , Player 2 colours 9 with blue. At , Player 1 picks his third digit. In this scenario, would have to avoid integer 5 in all the remaining turns (i.e. and . Before , Player 1 has coloured 3 integers and these three integers cannot be {2, 3, 4} or {6, 7, 8} or {4, 6, 8}.

Furthermore, at least one of the integers in {4, 7, 8} and at least one of the integers in {2, 3, 6} are uncoloured before . Suppose that after are coloured with red. Now, at Player 2 colours 8 with blue. Since 2,3 are still uncoloured after , we have . So, we may assume that before , the three integers coloured by Player 1 cannot be {2, 3, 4} or {6, 7, 8} or {4, 6, 8} or {4, 6, 7}. Now, at , Player 2 colours 2 with blue if it has not been coloured by Player 1.

If two of the integers in {4, 7, 8} are uncoloured, then because after . If 7, 8 are coloured with red, then after , 4 and 6 are uncoloured, thus . If 4, 8 are coloured with red, then after , 6 and 7 are uncoloured, thus . If 4, 7 are coloured with red, then after , 6 and 8 are uncoloured, thus .

Suppose 2 has been coloured by Player 1 but 3 has not. If two of the integers in are uncoloured, then at , Player 2 colours 3 with blue. We have because after . Suppose that after are coloured with red. Now, at , Player 2 colours 4 with blue. Since 3,6 are still uncoloured after , we have . If are coloured with red after , then at , Player 2 colours 7 with blue. If 2,4,7 are coloured with red after , then at , Player 2 colours 8 with blue. In either case, 3 and 6 are still uncoloured after , so we have .

Suppose 2 and 3 have been coloured by Player 1, then this case as in Case 4.5 where Player will have a winning strategy.

This completes the proof.

Theorem 3.7.In a 3-AVOID-AP game on [1, 9], Player 2 always wins and 9.

Proof. This follows immediately from Lemma 3.4 and Theorem 3.6.

4. Conclusions

In this paper, we have found and . We also have shown that in a k-AVOID-AP game on , where n is suficiently large, Player 2 always has a winning strategy if n is even. New techniques might need to be developed in order to find the winning strategy for either Player 1 or 2 when n is odd. Besides that, for and are also remained unknown. For k-AP game, it will be interesting to determine if there is a strategy for Player 1 to win the game.

We only consider 2-players games in this paper. So, it will be interesting to consider r-players games where each player is given one colour. Hence, this will be an r-colouring of. Questions that can be asked are whether some of the results in this paper can be extended to r-players games.

Acknowledgement.

This project is supported by Fundamental Research Grant Scheme (FRGS)- FRGS/1/2020/STG06/SYUC/03/1 by Malaysia Ministry of Higher Education and Sunway University Publication Support Scheme.

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