Variations of the Game 3-Euclid

Abstract

We present two variations of the game 3-Euclid. The games involve a triplet of positive integers. Two players move alternately. In the first game, each move is to subtract a positive integer multiple of the smallest integer from one of the other integers as long as the result remains positive. In the second game, each move is to subtract a positive integer multiple of the smallest integer from the largest integer as long as the result remains positive. The player who makes the last move wins. We show that the two games have the same 𝒫-positions and positions of Sprague-Grundy value 1. We present three theorems on the periodicity of 𝒫-positions and positions of Sprague-Grundy value 1. We also obtain a theorem on the partition of Sprague-Grundy values for each game. In addition, we examine the misère versions of the two games and show that the Sprague-Grundy functions of each game and its misère version differ slightly.

1. Introduction

The game Euclid, introduced by Cole and Davie [1], is a two-player game based on the Euclidean algorithm. A position in Euclid is a pair of positive integers. The two players move alternately. Each move is to subtract from one of the entries a positive integer multiple of the other without making the result negative. The game stops when one of the entries is reduced to zero. The player who makes the last move wins. In the literature, the term Euclid has been also used for a variation presented by Grossman [2] in which the game stops when the two entries are equal. More details and discussions on Euclid and Grossman’s game can be found in [3–9]. Some restrictions of Grossman’s game can be found in [10–12]. The misère version of Grossman’s game was studied in [13].

Collins and Lengyel [11] presented an extension of Grossman’s game to three dimensions that they called 3-Euclid. In 3-Euclid, a position is a triplet of positive integers. Each move is to subtract from one of the integers a positive integer multiple of one of the others as long as the result remains positive. Generally, from a position (𝑎,𝑏,𝑐), where 𝑎≤𝑏≤𝑐, there are three types of moves in 3-Euclid: (i) 1-2 moves: subtracting a multiple of 𝑎 from 𝑏; (ii) 1–3 moves: subtracting a multiple of 𝑎 from 𝑐; (iii) 2-3 moves: subtracting a multiple of 𝑏 from 𝑐.

In this paper, we present two restrictions of Collins and Lengyel’s game. In the first restriction that we call 𝐺₁, a move is to subtract a positive integer multiple of the smallest integer from one of the other integers as long as the result remains positive. In the second restriction that we call 𝐺₂, each move is to subtract a positive integer multiple of the smallest integer from the largest integer as long as the result remains positive. The two games therefore stop when the three integers are equal. Thus, the game 𝐺₁ is a restriction of the game 3-Euclid allowing the 1-2 and 1–3 moves while the game 𝐺₂ is a restriction of 3-Euclid allowing only the 1–3 move. We investigate 𝒫-positions and positions of Sprague-Grundy value 1 of the two games 𝐺₁ and 𝐺₂.

Recall that a position 𝑝 is said to be an 𝒩-position (we write 𝑝∈𝒩) if the next player to move from 𝑝 has a strategy to win. Otherwise, 𝑝 is said to be a 𝒫-position (we write 𝑝∈𝒫). Note that every move from a 𝒫-position terminates at some 𝒩-position while from an 𝒩-position, there exists a move terminating at some 𝒫-position.

Throughout this paper, the Sprague-Grundy value of the position 𝑝 is denoted by 𝒢(𝑝). Not surprisingly, the Sprague-Grundy functions for 3-Euclid, 𝐺₁, and 𝐺₂ are pairwise distinct. For example, calculations show that for the position (2,3,7) the Sprague-Grundy value is 1 for 3-Euclid, 2 for 𝐺₁, and 3 for 𝐺₂. The 𝒫-positions of 3-Euclid also differ to those of 𝐺₁ and 𝐺₂. For example, (3,6,7) is a 𝒫-position in 3-Euclid, but it has Sprague-Grundy value 1 for 𝐺₁ and 𝐺₂. Curiously, 𝐺₁ and 𝐺₂ have the same 𝒫-positions and the same positions of Sprague-Grundy value 1. This is the main result in this paper.

This paper is organized as follows. In Section 2, we show that the two games 𝐺₁ and 𝐺₂ have the same 𝒫-positions before proving a periodicity result for the 𝒫-positions. We also give some classes of 𝒫-positions of the two games. In Section 3, we show that the two games 𝐺₁ and 𝐺₂ also have the same positions of Sprague-Grundy value 1 and then present some connections between positions of Sprague-Grundy value 1 and 𝒫-positions. We give two theorems on the periodicity of the positions of Sprague-Grundy value 1. Some special cases of positions of Sprague-Grundy value 1 are also discussed. Section 4 discusses the existence of values 𝑐 satisfying the condition 𝒢(𝑎,𝑏,𝑐)=𝑠 for some given 𝑎,𝑏,𝑠. This is analogous to a result of Collins and Lengyel [11]. Section 5 examines the misère versions of the two games 𝐺₁ and 𝐺₂. It will be shown that these two games and their misère versions differ slightly only on a subset of positions of Sprague-Grundy values 0 and 1. This result also shows that, as established in [14], the miserability is quite common in impartial games.

This paper continues our investigations of variants of Euclid and related questions; see [3, 15, 16].

2. On the 𝒫-Positions

We show in this section that the two games 𝐺₁ and 𝐺₂ have the same 𝒫-positions. We then present a periodicity property of 𝒫-positions and some special classes of 𝒫-positions.

Lemma 2.1. Let 𝑎≤𝑏≤𝑐. In the game 𝐺₂, if (𝑎,𝑏,𝑐)∈𝒫 then 𝑐<𝑎+𝑏.

Proof. Assume by contradiction that there exists a 𝒫-position (𝑎,𝑏,𝑐) with 𝑎≤𝑏≤𝑐 and 𝑐≥𝑎+𝑏. Then (𝑎,𝑏,𝑐−𝑎)∈𝒩. There must be a move from (𝑎,𝑏,𝑐−𝑎) to some 𝒫-position 𝑝. As 𝑐−𝑎≥𝑏, the position 𝑝 must be of the form (𝑎,𝑏,𝑐−𝑎−𝑖𝑎) for some 𝑖≥1. This is a contradiction as there is then a move from (𝑎,𝑏,𝑐) to 𝑝.

Theorem 2.2. The 𝒫-positions in the game 𝐺₂ are identical to those in the game 𝐺₁.

Proof. Let 𝐴,𝐵 be the sets of 𝒫-positions and 𝒩-positions, respectively, of the game 𝐺₂. We prove that the following two properties hold for the game 𝐺₁:(i)every move from a position in 𝐴 terminates in 𝐵;(ii)from every position in 𝐵, there exists a move that terminates in 𝐴.Since every move in 𝐺₂ is legal in 𝐺₁, (ii) holds. Assume by contradiction that (i) does not hold for 𝐺₁. Then there exists a position (𝑎,𝑏,𝑐)∈𝐴 with 𝑎≤𝑏≤𝑐 and a move in 𝐺₁ leading (𝑎,𝑏,𝑐) to some position in 𝐴. This move cannot be a 1–3 move as (i) holds for 𝐺₂, and so this move must be a 1-2 move. Therefore, there exists 𝑖≥1 such that (𝑎,𝑏−𝑖𝑎,𝑐)∈𝐴. By Lemma 2.1, 𝑐<𝑎+𝑏−𝑖𝑎≤𝑏 giving a contradiction. Therefore, (i) holds for 𝐺₁.

Lemma 2.3. Let 𝑎≤𝑏. For the two games 𝐺₁ and 𝐺₂, there exists exactly one integer 𝑐≥1 in each residue class 𝑟(mod𝑎) such that (𝑎,𝑏,𝑐)∈𝒫.

Proof. We will prove the existence for the game 𝐺₂ and the uniqueness for the game 𝐺₁. The lemma then follows Theorem 2.2. For the existence, let 𝑑 be an integer such that 𝑎≤𝑏≤𝑑 and 𝑑≡𝑟mod𝑎. Consider the position (𝑎,𝑏,𝑑) in the game 𝐺₂. If 𝒢(𝑎,𝑏,𝑑)=0 then we are done. If 𝒢(𝑎,𝑏,𝑑)>0 then there exists a move from (𝑎,𝑏,𝑑) to some 𝒫-position 𝑝. This move must be a 1–3 move, and so 𝑝 is of the form (𝑎,𝑏,𝑑−𝑖𝑎) as required. For the uniqueness, assume by contradiction that there are two positive integers 𝑐₁<𝑐₂ in the residue class 𝑟(mod𝑎) such that both (𝑎,𝑏,𝑐₁) and (𝑎,𝑏,𝑐₂) are 𝒫-positions in the game 𝐺₁. Since 𝑐₁,𝑐₂ are in the same residue class 𝑟mod𝑎, we have 𝑐₂>𝑎. If 𝑐₂<𝑏 then there exists a 1-2 move from (𝑎,𝑐₂,𝑏) to (𝑎,𝑐₁,𝑏). This is a contradiction. If 𝑐₂≥𝑏 then there exists a 1–3 move from (𝑎,𝑏,𝑐₂) to (𝑎,𝑏,𝑐₁). This is a contradiction. Therefore, the uniqueness holds.

We now present a periodicity result for the 𝒫-positions. Note that the following theorem holds for both games by Theorem 2.2.

Theorem 2.4. Let 𝑎≤𝑏≤𝑐. Then (𝑎,𝑏,𝑐)∈𝒫 if and only if (𝑎,𝑏+𝑎,𝑐+𝑎)∈𝒫.

Proof. By Theorem 2.2, it is enough to prove that the theorem holds for 𝐺₂. For the necessary condition, let (𝑎,𝑏,𝑐)∈𝒫, let 𝑚=⌊𝑏/𝑎⌋, the integer part of 𝑏/𝑎, and let 𝑟 denote the remainder, 𝑟=𝑏−𝑚𝑎. Note that 𝑐<𝑎+𝑏 by Lemma 2.1. By Lemma 2.3, there is exactly one integer 𝑛 such that 𝑝=(𝑎,𝑛𝑎+𝑟,𝑐+𝑎)∈𝒫. We show that 𝑛=𝑚+1. First assume that 𝑛≤𝑚. We have 𝑛𝑎+𝑟+𝑎≤𝑚𝑎+𝑟+𝑎=𝑏+𝑎≤𝑐+𝑎. However, as 𝑝∈𝒫, by Lemma 2.1, 𝑎+𝑛𝑎+𝑟>𝑐+𝑎 giving a contradiction. Next, assume that 𝑛≥𝑚+2 and so 𝑛𝑎+𝑟≥𝑏+2𝑎. Note that 𝑎+𝑏>𝑐 by Lemma 2.1 and so 𝑛𝑎+𝑟>𝑐+𝑎. Consider the position 𝑞=(𝑎,𝑏+𝑎,𝑐+𝑎). Since there exists a move from 𝑝=(𝑎,𝑐+𝑎,𝑛𝑎+𝑟) to 𝑞, we have 𝑞∈𝒩 and so (𝑎,𝑏+𝑎,𝑐+𝑎−𝑗𝑎)∈𝒫 for some positive integer 𝑗. By Lemma 2.1, we have 𝑏+𝑎<𝑎+𝑐+𝑎−𝑗𝑎 which implies 𝑗=1, as 𝑏+𝑎>𝑐, and so (𝑎,𝑏+𝑎,𝑐)∈𝒫. Note that 𝑏+𝑎>𝑐 and so there is a move from (𝑎,𝑏+𝑎,𝑐) to (𝑎,𝑏,𝑐) which is also a 𝒫-position. This is a contradiction.
Conversely, assume that (𝑎,𝑏+𝑎,𝑐+𝑎)∈𝒫. By Lemma 2.1, we have 𝑐+𝑎<𝑎+𝑏+𝑎 or 𝑐<𝑎+𝑏. Since there exists a 1–3 move from (𝑎,𝑏+𝑎,𝑐+𝑎) to (𝑎,𝑐,𝑏+𝑎), we have (𝑎,𝑐,𝑏+𝑎)∈𝒩. Then there exists a move from (𝑎,𝑐,𝑏+𝑎) to some 𝒫-position 𝑝^′. Since 𝑏+𝑎>𝑐, 𝑝′ must be of the form (𝑎,𝑏+𝑎−𝑗𝑎,𝑐) for some 𝑗≥1. By Lemma 2.1, we have 𝑎+𝑏+𝑎−𝑗𝑎>𝑐, and so 𝑗=1. Therefore, (𝑎,𝑏,𝑐)∈𝒫.

Collins and Lengyel [11] suggested a similar result for the game 3-Euclid. They claimed that in the game 3-Euclid, if 𝑏,𝑐>𝑎² then (𝑎,𝑏,𝑐)∈𝒫 if and only if (𝑎,𝑏+𝑎,𝑐+𝑎)∈𝒫. To our knowledge, a proof for this claim has not appeared in the literature. In our opinion, a proof for this claim would be much more complicated.

We now solve some special cases for 𝒫-positions.

Corollary 2.5. Let 𝑎≤𝑏. Then (𝑎,𝑏,𝑏)∈𝒫.

Proof. By Theorem 2.2, it is enough to prove that (𝑎,𝑏,𝑏)∈𝒫 in 𝐺₂.
Assume by contradiction that (𝑎,𝑏,𝑏)∈𝒩. Then there exists a positive integer 𝑖 such that (𝑎,𝑏−𝑖𝑎,𝑏)∈𝒫. By Lemma 2.1, 𝑏<𝑎+𝑏−𝑖𝑎≤𝑏 giving a contradiction. Therefore, (𝑎,𝑏,𝑏)∈𝒫.

We state one further result, whose proof we postpone until the end of the next section.

Proposition 2.6. (i)  Let 2≤𝑎≤𝑏. Then (𝑎,𝑏−1,𝑏)∈𝒫 if and only if gcd(𝑎,𝑏)≠1.
(ii)  Let 𝑎<𝑏. Then (𝑎,𝑏−𝑎,𝑏−1)∈𝒫 if and only if gcd(𝑎,𝑏)=1.

3. The Positions of Sprague-Grundy Value 1

In this section, we first give some basic results on the Sprague-Grundy function of the game 𝐺₂ before showing that the two games 𝐺₁ and 𝐺₂ have the same set of positions of Sprague-Grundy value 1. We then show that there is a bridge between the set of 𝒫-positions and the set of positions of Sprague-Grundy value 1. Next, we present two theorems on the periodicity of positions of Sprague-Grundy value 1. Finally, we solve the Sprague-Grundy function for some special cases.

Lemma 3.1. Let 𝑎≤𝑏≤𝑐. In the game 𝐺₂, if 𝒢(𝑎,𝑏,𝑐)=1 then (𝑎,𝑏,𝑐−𝑎)∈𝒫.

Proof. Assume by contradiction that (𝑎,𝑏,𝑐−𝑎)∉𝒫. Then 𝒢(𝑎,𝑏,𝑐−𝑎)>1, and so there exists a move from (𝑎,𝑏,𝑐−𝑎) to some position 𝑝 of Sprague-Grundy value 1. If 𝑐−𝑎≥𝑏, 𝑝 must be of the form (𝑎,𝑏,𝑐−𝑎−𝑖𝑎) for some 𝑖≥1. This is a contradiction as there exists a move from (𝑎,𝑏,𝑐) to 𝑝. Therefore, 𝑐−𝑎<𝑏. Moreover, since 𝒢(𝑎,𝑏,𝑐)=1, there exists a move from (𝑎,𝑏,𝑐) to some 𝒫-position 𝑞. So 𝑞 must be of the form (𝑎,𝑏,𝑐−𝑖𝑎) for some 𝑖≥1. Since (𝑎,𝑏,𝑐−𝑎)∉𝒫, we have 𝑖≥2, and so 𝑎+𝑐−𝑖𝑎≤𝑐−𝑎<𝑏. However, by Lemma 2.1, 𝑎+𝑐−𝑖𝑎>𝑏 giving a contradiction. Therefore, (𝑎,𝑏,𝑐−𝑎)∈𝒫.

Lemma 3.2. Let 𝑎≤𝑏≤𝑐. In the game 𝐺₂, if 𝒢(𝑎,𝑏,𝑐)=1 then 𝑐<2𝑎+𝑏.

Proof. This is immediate from Lemmas 2.1 and 3.1.

Lemma 3.3. Let 𝑎≤𝑏≤𝑐. In the game 𝐺₂,(i)if 𝑐>𝑎+𝑏 then 𝒢(𝑎,𝑏,𝑐)≥2;(ii)if 𝑐<𝑎+𝑏 then 𝒢(𝑎,𝑏,𝑐)≤1.

Proof. The proof is by induction on 𝑐. One can check by hand that the lemma is true for 𝑐≤3. Assume that the lemma is true for 𝑐≤𝑛 for some 𝑛≥3, we will show that the lemma is true for 𝑐=𝑛+1. Note that if 𝑏=𝑐 then the lemma is true by Corollary 2.5. Therefore, it is sufficient to prove the lemma for 𝑎≤𝑏<𝑐=𝑛+1.
For (i), assume by contradiction that there exist 𝑎≤𝑏 such that 𝑎+𝑏<𝑐 and 𝒢(a,𝑏,𝑐)≤1. By Lemma 2.1, we have 𝒢(𝑎,𝑏,𝑐)≠0 and so 𝒢(𝑎,𝑏,𝑐)=1. By Lemma 3.1, we have (𝑎,𝑏,𝑐−𝑎)∈𝒫. Since 𝑐>𝑎+𝑏, we have 𝑐−2𝑎>0. As 𝒢(𝑎,𝑏,𝑐)=1, by Lemma 3.2, we have 𝑐−2𝑎<𝑏. Consider the position (𝑎,𝑐−2𝑎,𝑏). We have 𝑎+𝑐−2𝑎=𝑐−𝑎>𝑏. Note that 𝑏≤𝑛. By the inductive hypothesis on (ii), we have 𝒢(𝑎,𝑐−2𝑎,𝑏)≤1. Note that there exists a move from (𝑎,𝑏,𝑐) whose Sprague-Grundy value is 1 to (𝑎,𝑐−2𝑎,𝑏) and so 𝒢(𝑎,𝑐−2𝑎,𝑏)≠1. Also note that there exists a move from (𝑎,𝑏,𝑐−𝑎)∈𝒫 to (𝑎,𝑐−2𝑎,𝑏) and so 𝒢(𝑎,𝑐−2𝑎,𝑏)≠0, giving a contradiction. Thus, (i) is true for 𝑐=𝑛+1.
For (ii), assume by contradiction that there exist 𝑎,𝑏 such that 𝑎≤𝑏<𝑐, 𝑐<𝑎+𝑏 and 𝒢(𝑎,𝑏,𝑐)≥2. Then there exist integers 𝑖,𝑗≥1 such that 𝒢(𝑎,𝑏,𝑐−𝑖𝑎)=0 and 𝒢(𝑎,𝑏,𝑐−𝑗𝑎)=1. Since 𝒢(𝑎,𝑏,𝑐−𝑖𝑎)=0, by Lemma 2.1, 𝑎+𝑐−𝑖𝑎>𝑏. Note that 𝑐−𝑎<𝑏 and so 𝑖=1. This also implies that 𝑗≥2. We claim that 𝑗=2. Assume by contradiction that 𝑗≥3. Then 2𝑎+𝑐−𝑗𝑎≤𝑐−𝑎<𝑏. If 𝑐−𝑗𝑎<𝑎, since 𝒢(𝑐−𝑗𝑎,𝑎,𝑏)=1, by Lemma 3.2, 2(𝑐−𝑗𝑎)+𝑎>𝑏. However, 𝑐−𝑗𝑎<𝑎 also implies 2(𝑐−𝑗𝑎)+𝑎<2𝑎+𝑐−𝑗𝑎 giving a contradiction. If 𝑎≤𝑐−𝑗𝑎, since 𝒢(𝑎,𝑐−𝑗𝑎,𝑏)=1, by Lemma 3.2, 2𝑎+𝑐−𝑗𝑎>𝑏 giving a contradiction. Therefore, 𝑗=2. Note that 𝑐−2𝑎<𝑏. Now consider the position (𝑎,𝑐−2𝑎,𝑏). Since 𝑎+𝑐−2𝑎=𝑐−𝑎<𝑏 and 𝑏<𝑛+1, by the inductive hypothesis on (i), we have 𝒢(𝑎,𝑐−2𝑎,𝑏)≥2 giving a contradiction. Therefore, (ii) is true for 𝑐=𝑛+1.
Thus, by the inductive principle, the lemma is true.

Question 1. Let 𝑎≤𝑏 be positive integers. What is the relationship between 𝑎,𝑏 so that 𝒢(𝑎,𝑏,𝑎+𝑏)=1?

By the part (i) of Lemma 3.3, we have a result stronger than Lemma 3.2 as follows.

Corollary 3.4. Let 𝑎≤𝑏≤𝑐. In the game 𝐺₂, if 𝒢(𝑎,𝑏,𝑐)=1 then 𝑐≤𝑎+𝑏.

We are now in the position to show that all results early in this section are also true for the game 𝐺₁.

Theorem 3.5. The positions of Sprague-Grundy value 1 in the game 𝐺₂ are identical to those in the game 𝐺₁.

Proof. Let 𝐴 be the set of positions of Sprague-Grundy value 1 in the game 𝐺₂. We show that the following two properties hold for 𝐺₁:(i)there is no move from a position in 𝐴 to a position in 𝐴,(ii)from every position that is not in 𝒫∪𝐴, there is a move to a position in 𝐴.Note that every move in 𝐺₂ is also legal in 𝐺₁ and so (ii) holds for G₁. Assume by contradiction that (i) does not hold for 𝐺₁. Then, there exist a position 𝑝=(𝑎,𝑏,𝑐)∈𝐴 with 𝑎≤𝑏≤𝑐 and a 1-2 move from 𝑝 to some position 𝑞∈𝐴. This implies that 𝑞=(𝑎,𝑏−𝑖𝑎,𝑐) for some 𝑖≥1. By Corollary 3.4, we have 𝑐≤𝑎+𝑏−𝑖a≤𝑏 implying 𝑐=𝑏. But if 𝑐=𝑏 then, by Corollary 2.5, (𝑎,𝑏,𝑐)∈𝒫 giving a contradiction. Therefore (i) holds.

In the next part, we find some connections between the 𝒫-positions and the set of positions of Sprague-Grundy value 1.

Corollary 3.6. Let 𝑎≤𝑏≤𝑐. For the two games 𝐺₁ and 𝐺₂, if 𝒢(𝑎,𝑏,𝑐)=1 then (𝑎,𝑏,𝑐−𝑎)∈𝒫 and (𝑎,𝑐,𝑏+𝑎)∈𝒫.

Proof. We work with the game 𝐺₂. By Lemma 3.1 and Theorem 3.5, (𝑎,𝑏,𝑐−𝑎)∈𝒫. So it remains to show that (𝑎,𝑐,𝑏+𝑎)∈𝒫. Note that 𝑐≤𝑎+𝑏 by Corollary 3.4. Assume by contradiction that (𝑎,𝑐,𝑏+𝑎)∉𝒫. Then there exists a move from (𝑎,𝑐,𝑏+𝑎) to some 𝒫-position 𝑝. So 𝑝 must be of the form (𝑎,𝑐,𝑏+𝑎−𝑖𝑎) for some 𝑖≥1. Since 𝒢(𝑎,𝑏,𝑐)=1, we must have 𝑖≥2, and so 𝑎+𝑏+𝑎−𝑖𝑎≤𝑏≤𝑐. But as (𝑎,𝑐,𝑏+𝑎−𝑖𝑎)∈𝒫, by Lemma 2.1, we have 𝑎+𝑏+𝑎−𝑖𝑎>𝑐 giving a contradiction. Therefore, (𝑎,𝑐,𝑏+𝑎)∈𝒫.

The following corollary is the converse of Corollary 3.6 when 𝑏<𝑐.

Corollary 3.7. Let 𝑎≤𝑏<𝑐. If (𝑎,𝑏,𝑐)∈𝒫, then 𝒢(𝑎,𝑐−𝑎,𝑏)=1 and 𝒢(𝑎,𝑐,𝑏+𝑎)=1.

Proof. By Lemma 2.1, we have 𝑎+𝑏>𝑐. We first prove that 𝒢(𝑎,𝑐−𝑎,𝑏)=1. Assume by contradiction that 𝒢(𝑎,𝑐−𝑎,𝑏)≠1. Then 𝒢(𝑎,𝑐−𝑎,𝑏)>1. By Lemma 3.3, 𝑏≥𝑎+𝑐−𝑎=𝑐 giving a contradiction. Therefore, 𝒢(𝑎,𝑐−𝑎,𝑏)=1.
We now prove that 𝒢(𝑎,𝑐,𝑏+𝑎)=1. Assume by contradiction that 𝒢(𝑎,𝑐,𝑏+𝑎)≠1. Then 𝒢(𝑎,𝑐,𝑏+𝑎)>1. There exists a move from (𝑎,𝑐,𝑏+𝑎) to some position 𝑝 of Sprague-Grundy value 1. So, working in 𝐺₂, 𝑝 must be of the form (𝑎,𝑐,𝑏+𝑎−𝑖𝑎) for some 𝑖≥1. Since 𝑝≠(𝑎,𝑏,𝑐), we have 𝑖≥2 and so 𝑎+𝑏+𝑎−𝑖𝑎≤𝑏<𝑐. But by Lemma 3.3, 𝒢(𝑎,𝑐,𝑏+𝑎−𝑖𝑎)≥2 giving a contradiction. Therefore, 𝒢(𝑎,𝑐,𝑏+𝑎)=1.

Note that Corollary 3.7 does not hold for 𝑏=𝑐. For example, for 𝑎=1,𝑏=3, we have 𝒢(𝑎,𝑏,𝑏)=0 for both games but 𝒢(𝑎,𝑏−𝑎,𝑏)=3 for the game 𝐺₁ and 𝒢(𝑎,𝑏−𝑎,𝑏)=2 for the game 𝐺₂. For 𝑎=1,𝑏=2, we have 𝒢(𝑎,𝑏,𝑏)=0 for both games but 𝒢(𝑎,𝑏,𝑏+𝑎)=3 for the game 𝐺₁ and 𝒢(𝑎,𝑏,𝑏+𝑎)=2 for the game 𝐺₂. An answer for Question 1 would also provide a criteria for the case 𝑏=𝑐 to hold in Corollary 3.7.

We now present two theorems giving periodicity of positions of Sprague-Grundy value 1 of the forms (𝑎,𝑏,𝑐) with 𝑎+𝑏>𝑐 and of the form (𝑎,𝑏,𝑎+𝑏).

Theorem 3.8. Let 𝑎≤𝑏≤𝑐 and 𝑎+𝑏>𝑐. Then 𝒢(𝑎,𝑏,𝑐)=1 if and only if 𝒢(𝑎,𝑏+𝑎,𝑐+𝑎)=1.

Proof. Assume that 𝒢(𝑎,𝑏,𝑐)=1. By Corollary 3.6, we have (𝑎,𝑐,𝑏+𝑎)∈𝒫. By Corollary 3.7, we have 𝒢(𝑎,𝑏+𝑎,𝑐+𝑎)=1. Now assume that 𝒢(𝑎,𝑏+𝑎,𝑐+𝑎)=1. By Corollary 3.6, we have (𝑎,𝑐,𝑏+𝑎)∈𝒫. By Corollary 3.7, we have 𝒢(𝑎,𝑏,𝑐)=1.

The previous theorem does not hold for 𝑐=𝑎+𝑏. For example, 𝒢(1,3,4)=1, but 𝒢(1,4,5)=2 for the game 𝐺₂. Nevertheless, one has the following.

Theorem 3.9. Let 𝑎≤𝑏. Then 𝒢(𝑎,𝑏,𝑎+𝑏)=1 if and only if 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎)=1.

Proof. It is sufficient to prove that the theorem holds for the game 𝐺₂. We first prove the necessary condition. Assume by contradiction that 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎)≠1. By Lemma 2.1, 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎)≠0. Then, 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎)≥2 and so there exists a move from (𝑎,𝑏+2𝑎,𝑏+3𝑎) to some position 𝑝 of Sprague-Grundy value 1. So 𝑝 must be of the form (𝑎,𝑏+2𝑎,𝑏+3𝑎−𝑖𝑎) for some 𝑖≥1. We claim that 𝑖=2. By Corollary 2.5, (𝑎,𝑏+2𝑎,𝑏+2𝑎)∈𝒫, so 𝑖≠1. If 𝑖≥3, then 𝑎+𝑏+3𝑎−𝑖𝑎≤𝑏+𝑎<𝑏+2𝑎 and so 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎−𝑖𝑎)≥2 by Lemma 3.3. So 𝑖=2. But this is impossible as there exists a move from 𝑝 to the position (𝑎,𝑏,𝑎+𝑏). Therefore, 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎)=1.
Conversely, assume that 𝒢(𝑎,𝑏+2𝑎,𝑏+3𝑎)=1. Then 𝒢(𝑎,𝑏+𝑎,𝑏+2𝑎)≠1. Note that (𝑎,𝑏+𝑎,𝑏+𝑎)∈𝒫 by Corollary 2.5. Since there exists a move from (𝑎,𝑏+𝑎,𝑏+2𝑎) to (𝑎,𝑏+𝑎,𝑏+𝑎), 𝒢(𝑎,𝑏+𝑎,𝑏+2𝑎)≥2. It follows that there exists a move from (𝑎,𝑏+𝑎,𝑏+2𝑎) to some position 𝑞 of Sprague-Grundy value 1. So 𝑞 must be of the form (𝑎,𝑏+𝑎,𝑏+2𝑎−𝑖𝑎) for some 𝑖≥1. We claim that 𝑖=2. Note that 𝑖≠1 as (𝑎,𝑏+𝑎,𝑏+𝑎)∈𝒫. If 𝑖≥3 then 𝑎+𝑏+2𝑎−𝑖𝑎≤𝑏<𝑏+𝑎, and so, by Lemma 3.3, we have 𝒢(𝑞)≥2 giving a contradiction. Hence 𝑖=2 and 𝒢(𝑎,𝑏,𝑏+𝑎)=1.

We now present some special classes of positions of Sprague-Grundy value 1. The following corollary follows from the above theorem by induction on 𝑎.

Corollary 3.10. 𝒢(1,𝑎,𝑎+1)=1 if and only if 𝑎 is odd.

Proposition 3.11. (i) Let 2≤𝑎≤𝑏. Then 𝒢(𝑎,𝑏−1,𝑏)=1 if and only if gcd(𝑎,𝑏)=1.
(ii) Let 2≤𝑎<𝑏. Then 𝒢(𝑎,𝑏−𝑎,𝑏−1)=1 if and only if gcd(𝑎,𝑏)≠1.

Proof. Note that (ii) follows from (i) by Corollary 3.6 and Lemma 3.3. Therefore, it is sufficient to prove (i). We prove (i) by induction on 𝑏. It can be checked that (i) holds for 𝑏=2,3. Assume that (i) holds for 𝑏≤𝑛 for some 𝑛≥3. We now show that (i) holds for 𝑏=𝑛+1. If 𝑎=𝑏 then 𝒢(𝑎,𝑏−1,𝑏)=0 by Corollary 2.5. Therefore, (i) holds for 𝑎=𝑏. We claim that (i) also holds for 𝑎=𝑏−1. Note that 𝒢(1,𝑎,𝑎)=0 by Corollary 2.5 and the only move from (𝑎,𝑎,𝑎+1) is to (1,𝑎,𝑎). Therefore, 𝒢(𝑎,𝑎,𝑎+1)=1 by definition. Also note that gcd(𝑎,𝑎+1)=1. Therefore, (i) holds for 𝑎=𝑏−1. Let 2≤𝑎≤𝑏−2. We show that 𝒢(𝑎,𝑏−1,𝑏)=1 if and only if gcd(𝑎,𝑏)=1.
Suppose that 𝒢(𝑎,𝑏−1,𝑏)=1. By Corollary 3.6, 𝒢(𝑎,𝑏−𝑎,𝑏−1)=0. We compare 𝑎 with 𝑏−𝑎. We claim that 𝑎≠𝑏−𝑎. In fact, if 𝑎=𝑏−𝑎 then 𝒢(𝑏−1−𝑎,𝑎,𝑏−𝑎)=0 by Corollary 2.5, but there exists a move from (𝑎,𝑏−𝑎,𝑏−1) to (𝑏−1−𝑎,𝑎,𝑏−𝑎). This is impossible. Hence 𝑎≠𝑏−𝑎. If 𝑎<𝑏−𝑎, then 𝒢(𝑎,𝑏−1−𝑎,𝑏−𝑎)=1 by Corollary 3.7. Since 𝑏−𝑎<𝑛, the inductive hypothesis gives gcd(𝑎,𝑏−𝑎)=1, and so gcd(𝑎,𝑏)=1. If 𝑏−𝑎<𝑎, then 𝒢(𝑏−𝑎,𝑎,𝑏−1−(𝑏−𝑎))=1 by Corollary 3.7; that is, 𝒢(𝑏−𝑎,𝑎−1,𝑎)=1. Since 𝑎<𝑛 and 𝑏−𝑎≥2, the inductive hypothesis gives gcd(𝑏−𝑎,𝑎)=1 and so gcd(𝑎,𝑏)=1.
Conversely, suppose that gcd(𝑎,𝑏)=1. Assume by contradiction that 𝒢(𝑎,𝑏−1,𝑏)≠1. Then 𝒢(𝑎,𝑏−1,𝑏)=0 by Lemma 3.3. By Corollary 3.7, we have 𝒢(𝑎,𝑏−𝑎,𝑏−1)=1. We compare 𝑎 with 𝑏−𝑎. Note that 𝑎≠𝑏−𝑎 as gcd(𝑎,𝑏)=1. If 𝑎<𝑏−𝑎, then 𝒢(𝑎,𝑏−1−𝑎,𝑏−𝑎)=0 by Corollary 3.6. As 𝑏−𝑎<𝑛, by the inductive hypothesis, gcd(𝑎,𝑏−𝑎)≠1 giving a contradiction. If 𝑏−𝑎<𝑎, then 𝒢(𝑏−𝑎,𝑎,𝑏−1−(𝑏−𝑎))=0 by Corollary 3.6; that is, 𝒢(𝑏−𝑎,𝑎−1,𝑎)=0. Since 𝑎<𝑛 and 𝑏−𝑎≥2, by the inductive hypothesis, gcd(𝑏−𝑎,𝑎)≠1, and so gcd(𝑎,𝑏)≠1 giving a contradiction. Therefore, 𝒢(𝑎,𝑏−1,𝑏)=1.

As promised at the end of the previous section, we now give the following.

Proof of Proposition 2.6. For 𝑎≥2, Proposition 2.6 follows immediately from Proposition 3.11 and Lemma 3.3. For 𝑎=1, Proposition 2.6(ii) follows from Corollary 2.5.

4. On the Partition of Sprague-Grundy Values

This section extends a result from Collins and Lengyel’s work on the game 3-Euclid. Let 𝑎≤𝑏, and let 𝑠 be a nonnegative integer. We answer the question as to whether there exists a positive integer 𝑐 such that 𝒢(𝑎,𝑏,𝑐)=𝑠 and whether such an existence is unique.

Theorem 4.1. Let 𝑎≤𝑏, and let 𝑠 be a nonnegative integer. For the game 𝐺₁, there exists exactly one integer 𝑐≥1 in each residue class 𝑟(mod𝑎) such that 𝒢(𝑎,𝑏,𝑐)=𝑠.

Proof. The result holds for 𝑠=0 by Lemma 2.3. So we may assume that 𝑠>0.
For the uniqueness, assume by contradiction that there are two positive integers 𝑐₁<𝑐₂ in the residue class 𝑟(mod𝑎) such that 𝒢(𝑎,𝑏,𝑐₁)=𝒢(𝑎,𝑏,𝑐₂). Since 𝑐₁,𝑐₂ are in the same residue class modulo 𝑎, we have 𝑐₂>𝑎. If 𝑐₂<𝑏 then there exists a 1-2 move from (𝑎,𝑐₂,𝑏) to (𝑎,𝑐₁,𝑏). This is a contradiction. If 𝑐₂≥𝑏 then there exists a 1–3 move from (𝑎,𝑏,𝑐₂) to (𝑎,𝑏,𝑐₁). This is a contradiction. Therefore, the uniqueness holds.
For the existence, assume by contradiction that there is no integer 𝑐 in the residue class 𝑟(mod𝑎) such that 𝒢(𝑎,𝑏,𝑐)=𝑠. Let 𝑚=⌊𝑏/𝑎⌋ and let 𝑑 be an integer such that 𝑑≥𝑏 and 𝑑≡𝑟(mod𝑎). Let 𝑏₀=𝑏−𝑚𝑎. There are two cases for 𝑏₀.
If 𝑏₀=0, we consider the sequence of positions (𝑎,𝑏,𝑑+𝑎),(𝑎,𝑏,𝑑+2𝑎),…,(𝑎,𝑏,𝑑+(𝑠+𝑚)𝑎). (4.1)
Note that each pair of these 𝑠+𝑚 positions has distinct Sprague-Grundy values, and so there are at most 𝑠 positions having Sprague-Grundy values from 0 to 𝑠−1. Consequently, there are, in that sequence, at least 𝑚 positions having Sprague-Grundy values more than 𝑠. Assume that these 𝑚 positions are 𝑎,𝑏,𝑑+𝑖₁𝑎,𝑎,𝑏,𝑑+𝑖₂𝑎,…,𝑎,𝑏,𝑑+𝑖_𝑚𝑎. (4.2) From each position (𝑎,𝑏,𝑑+𝑖_𝑗𝑎), there exists a move to some position 𝑝_𝑗 of Sprague-Grundy value 𝑠. This move must be a 1-2 move, and so 𝑝_𝑗 must be of the form (𝑎,𝑏−𝑘_𝑗𝑎,𝑑+𝑖_𝑗𝑎). Since there are at most 𝑚−1 values for 𝑘_𝑗 (as 𝑏−𝑘_𝑗𝑎>0) while there are 𝑚 positions 𝑝_𝑗, there must be two distinct positions 𝑝_𝑗1=(𝑎,𝑏−𝑘_𝑗1𝑎,𝑑+𝑖_𝑗1𝑎) and 𝑝_𝑗2=(𝑎,𝑏−𝑘_𝑗2𝑎,𝑑+𝑖_𝑗2𝑎) having the same Sprague-Grundy value 𝑠 in which 𝑘_𝑗1=𝑘_𝑗2 and 𝑖_𝑗1≠𝑖_𝑗2. This is a contradiction as there is a 1–3 move (as 𝑏−𝑘_𝑗𝑎≥𝑎) from one of the two positions 𝑝_𝑗1,𝑝_𝑗2 to the other.
If 𝑏₀>0, we consider the sequence of positions 𝑎,𝑏,𝑑+𝑏₀𝑎,𝑎,𝑏,𝑑+2𝑏₀𝑎,…,𝑎,𝑏,𝑑+(𝑠+𝑚+1)𝑏₀𝑎. (4.3)
This sequence contains at least 𝑚+1 positions having Sprague-Grundy values greater than 𝑠. Assume that these 𝑚+1 positions are 𝑎,𝑏,𝑑+𝑖₁𝑏₀𝑎,𝑎,𝑏,𝑑+𝑖₂𝑏₀𝑎,…,𝑎,𝑏,𝑑+𝑖_𝑚+1𝑏₀𝑎. (4.4) From each position (𝑎,𝑏,𝑑+𝑖_𝑗𝑏₀𝑎), there exists a move to some position 𝑞_𝑗 of Sprague-Grundy value 𝑠. This move must be a 1-2 move and so 𝑞_𝑗 must be of the form (𝑎,𝑏−𝑘_𝑗𝑎,𝑑+𝑖_𝑗𝑏₀𝑎). Since there are at most 𝑚 values for 𝑘_𝑗 while there are 𝑚+1 positions 𝑞_𝑗, there must be two distinct positions 𝑞_𝑗1=(𝑎,𝑏−𝑘_𝑗1𝑎,𝑑+𝑖_𝑗1𝑏₀𝑎) and 𝑞_𝑗2=(𝑎,𝑏−𝑘_𝑗2𝑎,𝑑+𝑖_𝑗2𝑏₀𝑎) having the same Sprague-Grundy value 𝑠 in which 𝑘_𝑗1=𝑘_𝑗2 and 𝑖_𝑗1≠𝑖_𝑗2. If 𝑏−𝑘_𝑗1𝑎≥𝑎 then there exists a 1–3 move from one of two positions 𝑞_𝑗1,𝑞_𝑗2 to the other. This is a contradiction. Therefore, 𝑏−𝑘_𝑗1𝑎<𝑎 and so 𝑏−𝑘_𝑗1𝑎=𝑏₀, and the two integers 𝑑+𝑖_𝑗1b₀𝑎 and 𝑑+𝑖_𝑗2𝑏₀𝑎 are in the same residue class 𝑑(mod𝑏₀). Thus, there are two distinct integers 𝑑+𝑖_𝑗1𝑏₀𝑎,𝑑+𝑖_𝑗2𝑏₀𝑎 in the residue class 𝑑(mod𝑏₀) satisfying 𝒢(𝑏₀,𝑎,𝑑+𝑖_𝑗1𝑏₀𝑎)=𝒢(𝑏₀,𝑎,𝑑+𝑖_𝑗2𝑏₀𝑎). This contradicts the uniqueness part of the theorem. Therefore, the existence holds.

Theorem 4.2. Let 𝑎≤𝑏, and let 𝑠 be a nonnegative integer. For the game 𝐺₂, in each residue class 𝑟(mod𝑎), there exists at least one integer 𝑐≥1 such that 𝒢(𝑎,𝑏,𝑐)=𝑠.

Proof. By Lemma 2.3, we may assume that 𝑠>0. Let 𝑑 be an integer such that 𝑑≥𝑏 and 𝑑≡𝑟mod𝑎. Consider the sequence of positions (𝑎,𝑏,𝑑),(𝑎,𝑏,𝑑+𝑎),(𝑎,𝑏,𝑑+2𝑎),…,(𝑎,𝑏,𝑑+𝑠𝑎). (4.5) Note that each pair of these 𝑠+1 positions has distinct Sprague-Grundy values and so there are at most 𝑠 positions having Sprague-Grundy values from 0 to 𝑠−1. Therefore, if none of these positions has Sprague-Grundy value 𝑠 then there exists at least one of these positions having Sprague-Grundy value more than 𝑠, say (𝑎,𝑏,𝑑+𝑖𝑎). Then there exists a 1–3 move from (𝑎,𝑏,𝑑+𝑖𝑎) to some position 𝒢(𝑎,𝑏,𝑑+(𝑖−𝑗)𝑎) of Sprague-Grundy value 𝑠, as required.

In Theorem 4.2, the existence of 𝑐 is not unique. For example, the three positions (4,4,19),(4,8,19), and (4,12,19) have the same Sprague-Grundy value 4. Whenever the existence of the value 𝑐 is not unique, we have the following observation on the values 𝑐 for the game 𝐺₂. We have checked this result as far as 𝑏=100.

Observation 1. Let 𝑎≤𝑏. In the game 𝐺₂, if there are two values 𝑐₁,𝑐₂ in the same residue class modulo 𝑎 such that 𝒢(𝑎,𝑏,𝑐₁)=𝒢(𝑎,𝑏,𝑐₂) then 𝑎≤𝑐₁,𝑐₂<𝑏,𝑏≥3𝑎, and 𝑐₁,𝑐₂ are both multiples of 𝑎.

5. Miserability

In this section, we examine the misère versions of two games 𝐺₁ and 𝐺₂. Recall that an impartial game is under misère convention if the player making the last move loses. A game can be described by a finite directed acyclic graph Γ without multiple edges in which each vertex is a position, and there is a downward edge from 𝑝 to 𝑞 if and only if there is a move from the position 𝑝 to the position 𝑞. Moreover, the graph can be assumed to have only one source. A source is a vertex with no incoming edges. The source of the graph is the original position of the game. The sinks are the vertices with no outgoing edges, so the sinks of the graph are the final (terminal) positions of the game. For convenience, a graph of a game is assumed to have precisely one sink. This is because when the graph has more than one sink, they can be coalesced together into one sink without changing the properties of the game.

Let 𝐺 be an impartial game and Γ the corresponding digraph of the game 𝐺. The misère version of the game 𝐺 can be considered as the graph obtained from Γ by adding an extra vertex 𝑣₀ and a move from the sink of Γ to 𝑣₀ [13]. In [14], a game is said to be miserable if its normal and misère versions are different only on some subset of positions of Sprague-Grundy values 0 and 1. More precisely, a game 𝒢 is miserable if there exist subsets 𝑉₀ of 𝒫-positions and 𝑉₁ of positions of Sprague-Grundy value 1 so that the two functions 𝒢 and 𝒢⁻ swap on positions in 𝑉₀ and 𝑉₁ and are equal on other positions. Here 𝒢 and 𝒢⁻ are the Sprague-Grundy functions for the game 𝐺 and its misère version respectively. If 𝑉₀ is equal to the set of 𝒫-positions and 𝑉₁ is equal to the set of positions of Sprague-Grundy value 1 then the game is said to be strongly miserable. We will show that the two games 𝐺₁ and 𝐺₂ are miserable but not strongly miserable. Before presenting this result, let us discuss some properties of positions of Sprague-Grundy values 0 and 1 of the misère versions. The proofs for the following results are similar to those in previous sections.

Lemma 5.1. Let 𝑎≤𝑏≤𝑐. In the misère versions,(i)if 𝑐>𝑎+𝑏 then 𝒢(𝑎,𝑏,𝑐)≥2;(ii)if 𝑐<𝑎+𝑏 then 𝒢(𝑎,𝑏,𝑐)≤1.

Theorem 5.2. The 𝒫-positions of the misère version of 𝐺₂ are identical to those of the misère version of 𝐺₂.

Theorem 5.3. The positions of Sprague-Grundy value 1 in the misère version of 𝐺₂ are identical to those in the misère version of 𝐺₁.

Some properties of periodicity of 𝒫-positions and positions of Sprague-Grundy value 1 of the two games 𝐺₁ and 𝐺₂ are still true for their misère versions.

Theorem 5.4. Let 𝑎≤𝑏≤𝑐 such that (𝑎,𝑏,𝑐)≠(𝑎,𝑎,2𝑎). In the misère versions of 𝐺₁ and 𝐺₂, if (𝑎,𝑏,𝑐)∈𝒫 then (𝑎,𝑏+𝑎,𝑐+𝑎)∈𝒫.

Theorem 5.5. Let 𝑎≤𝑏≤𝑐 such that 𝑎+𝑏>𝑐 and (𝑎,𝑏,𝑐)≠(𝑎,2𝑎,2𝑎). In the misère versions of 𝐺₁ and 𝐺₂, if (𝑎,𝑏,𝑐) has Sprague-Grundy value 1 then (𝑎,𝑏+𝑎,𝑐+𝑎) has Sprague-Grundy value 1.

Theorem 5.6. Let 𝑎≤𝑏. In the misère versions of 𝐺₁ and 𝐺₂, if (𝑎,𝑏,𝑎+𝑏) has Sprague-Grundy value 1 then (𝑎,𝑏+2𝑎,𝑏+3𝑎) has Sprague-Grundy value 1.

We now come back with the main result of this section. Note that the following theorem is also true for 𝐺₂ by Theorems 5.2 and 5.3.

Theorem 5.7. The game 𝐺₁ is miserable.

Proof. Consider the graph Γ⁻₁ with the sink 𝑣₀ of the misère version. For each vertex (position) 𝑣, the height ℎ(𝑣) of 𝑣 is the length of the longest directed path from 𝑣 to the sink 𝑣₀. We will prove by induction on ℎ(𝑣) that if either 𝒢₁(𝑣)≥2 or 𝒢⁻₁(𝑣)≥2 then 𝒢⁻₁(𝑣)=𝒢₁(𝑣). Note that the claim is true for ℎ(𝑣)=1. Assume that the claim is true for ℎ(𝑣)≤𝑛 for some 𝑛≥1. We show that the claim is true for ℎ(𝑣)=𝑛+1.
We first assume that 𝒢₁(𝑣)≥2. For each 𝑘<𝒢₁(𝑣), there exists 𝑤_𝑘 such that 𝒢₁(𝑤_𝑘)=𝑘 and one can move from 𝑣 to 𝑤_𝑘. By the inductive hypothesis, 𝒢⁻₁(𝑤₀),𝒢⁻₁(𝑤₁)≤1, and if there is a move from 𝑣 to some 𝑤 with 𝒢₁(𝑤)≥2 then 𝒢⁻₁(𝑤)=𝒢₁(𝑤). It remains to show that 𝒢⁻₁(𝑤₀)≠𝒢⁻₁(𝑤₁). We show that there is a move between 𝑤₀,𝑤₁ in the misère version of 𝐺₁. Assume that 𝑣=(𝑎,𝑏,𝑐). There are three following possibilities for moves from 𝑣 to 𝑤₀,𝑤₁:
(i)𝑤₀=(𝑎,𝑏,𝑐−𝑖𝑎),𝑤₁=(𝑎,𝑏,𝑐−𝑗𝑎);(ii)𝑤₀=(𝑎,𝑏−𝑖𝑎,𝑐),𝑤₁=(𝑎,𝑏−𝑗𝑎,𝑐);(iii)either 𝑤₀=(𝑎,𝑏,𝑐−𝑖𝑎),𝑤₁=(𝑎,𝑏−𝑗𝑎,𝑐) or 𝑤₀=(𝑎,𝑏−𝑖𝑎,𝑐),𝑤₁=(𝑎,𝑏,𝑐−𝑗𝑎). One can check that there is a move between 𝑤₀,𝑤₁ for the first two possibilities. We show that the third possibility does not occur. In fact, consider the assumption 𝑤₀=(𝑎,𝑏,𝑐−𝑖𝑎),𝑤₁=(𝑎,𝑏−𝑗𝑎,𝑐) (the other case can be treated similarly). By Lemma 5.1, we have 𝑎+𝑏−𝑗𝑎≤𝑐, and so 𝑗=1,𝑏=𝑐. Note that 𝑎+𝑐−𝑖𝑎≥𝑏 by Lemma 5.1 and so 𝑖=1. Consequently, 𝑤₀=𝑤₁ giving a contradiction. Hence, 𝒢⁻₁(𝑣)=𝒢₁(𝑣). Similarly, we can show that if 𝒢⁻₁(𝑣)≥2 then 𝒢₁(𝑣)=𝒢⁻₁(𝑣). This completes the proof.

Note that the game 𝐺₁ (and so 𝐺₂) is not strongly miserable. A counterexample is the position (1, 3, 3) which is a 𝒫-position in 𝐺₁ and its misère version.

Acknowledgment

The author thanks the referee for his/her helpful comments and references. He thank his supervisor, Dr Grant Cairns, for careful reading of this paper.