J. Math. Tokushima Univ. 83 Vol. 54 (2020), 83−91
Transfinite Version of Welter’s Game By Tomoaki Abuku Graduate school of pure and applied sciences, University of Tsukuba, 1-1-1 Tennodai, Tsukuba, Ibaraki 305-8577 Japan. e-mail address : [email protected] (Received October 10, 2019. Revised September 28, 2020.) Abstract We study the transfinite version of Welter’s Game, a combinatorial game played on a belt divided into squares numbered with general ordinal numbers. In particular, we give a solution for the game, based on those of the transfinite version of Nim and the original version of Welter’s Game using Cantor normal form. 2010 Mathematics Subject Classifications. 91A46, 03E10 Introduction We assume that the reader is familiar with basic terminology on combinatorial game theory, in particular about impartial games, for example, -position, - N P position, nim-sum, and -values (see [2], [7]). G Welter’s Game is an impartial game investigated by Welter in 1954 [9]. Since it was also investigated by Mikio Sato, it is often called Sato’s Maya Game in Japan. The rules of Welter’s games are as follows:
It is played with several coins placed on a belt divided into squares numbered with the nonnegative integers 0, 1, 2,... from the left as shown in Fig. 1. The legal move is to move any one coin from its present square to any unoc- cupied square with a smaller number. The game terminates when a player is unable to move any coin, namely, the coins are jammed in the squares with the smallest numbers as shown in Fig. 2.
84 Tomoaki Abuku 2 3
We express the binary nim-sum by symbol , and finitely many summation by ○ ○ ○ ○ ⊕ . 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
⊕ Figure 2: End position of Welter’s Game Definition 0.1 (Welter function). Let (a1,...,an) be a Welter’s Game position. Then we define the value [a a ] of Welter function at (a ,...,a ) as follows: 1|···| n 1 n Theorem 1.1 (Cantor Normal Form theorem [4]). Every α (α>0) can be ∈ ON expressed as [a1 an]=a1 an (ai aj), |···| ⊕···⊕ ⊕ | γk γ1 γ0 1 i
Definition 1.4 ( -value). Let G and G′ be game positions. The value (G) called G G the -value of G is defined as follows: 1. Transfinite Game G (G) = mex (G′) G G′ . G {G | → } We denote by Z the set of all integers and by N0 the set of all nonnegative integers. Theorem 1.5. Let G be a game position. Then
Let us denote by the class of all ordinal numbers. Later we see that the (G) = 0 if and only if G is an -position, ON G ̸ N nim-sum operation can be extended naturally on . (G) = 0 if and only if G is a -position. ON { G P The following is known about general ordinal numbers.
2. Main Results ○ ○ ○ ○ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2.1. Transfinite Nim
Figure 1: Welter’s Game We extend Nim into its transfinite version (Transfinite Nim) by allowing the size of the heaps of tokens to be a general ordinal number. The legal moves are to replace an arbitrary ordinal number α by a smaller ordinal number β. Transfinite Version of Welter’s Game 85 2 3
We express the binary nim-sum by symbol , and finitely many summation by ○ ○ ○ ○ ⊕ . 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
⊕ Figure 2: End position of Welter’s Game Definition 0.1 (Welter function). Let (a1,...,an) be a Welter’s Game position. Then we define the value [a a ] of Welter function at (a ,...,a ) as follows: 1|···| n 1 n Theorem 1.1 (Cantor Normal Form theorem [4]). Every α (α>0) can be ∈ ON expressed as [a1 an]=a1 an (ai aj), |···| ⊕···⊕ ⊕ | γk γ1 γ0 1 i
Definition 1.4 ( -value). Let G and G′ be game positions. The value (G) called G G the -value of G is defined as follows: 1. Transfinite Game G (G) = mex (G′) G G′ . G {G | → } We denote by Z the set of all integers and by N0 the set of all nonnegative integers. Theorem 1.5. Let G be a game position. Then
Let us denote by the class of all ordinal numbers. Later we see that the (G) = 0 if and only if G is an -position, ON G ̸ N nim-sum operation can be extended naturally on . (G) = 0 if and only if G is a -position. ON { G P The following is known about general ordinal numbers.
2. Main Results ○ ○ ○ ○ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2.1. Transfinite Nim
Figure 1: Welter’s Game We extend Nim into its transfinite version (Transfinite Nim) by allowing the size of the heaps of tokens to be a general ordinal number. The legal moves are to replace an arbitrary ordinal number α by a smaller ordinal number β. 86 Tomoaki Abuku 4 5
Definition 2.1. For ordinal numbers α ,...,α , we define their nim-sum If we put α′ = β , α = β (j =i). Then, α >β and we have 1 n ∈ ON i i j j ̸ i i as follows: β1 βi 1 βi βi+1 βn = β ⊕··· − ⊕ ⊕ ⊕··· γk α1 αn = ω (m1k mnk). ⊕···⊕ ⊕···⊕ Therefore, for each β (<α), there is a position (β1,...,βn) reached by a single ∑k move from (α1,...,αn). We give a proof for the -value of the Transfinite Nim using this form. G Example 2.3. In the case of position (1,ω 2+4,ω2 3+9,ω2 2+ω 4 + 16,ω2 + n · · · · Theorem 2.2. For Transfinite Nim position (α1,...,αn) ，we have the ω 5 + 25): ⊆ ON · following: Let us calculate the value of α α α α α . 1 ⊕ 2 ⊕ 3 ⊕ 4 ⊕ 5 We get (α ,...,α )=α α . G 1 n 1 ⊕···⊕ n β2 β1 2 α1 = ω m12 + ω m11 + m10 = ω 0+ω 0+1 Proof. The proof is by induction. Let α1 αn = α (α ). We have to · · · · ⊕···⊕ ∈ ON α = ωβ2 m + ωβ1 m + m = ω2 0+ω 2+4 show that, for each β (<α), there exists a position with -value β reached by a 2 · 22 · 21 20 · · G β2 β1 2 single move from (α1,...,αn). α = ω m + ω m + m = ω 3+ω 0+9 3 · 32 · 31 30 · · Let (α1,...,αn) (β1,...,βn), by induction hypothesis we have α = ωβ2 m + ωβ1 m + m = ω2 2+ω 4 + 16 → 4 · 42 · 41 40 · · β2 β1 2 (β ,...,β )=β β . α5 = ω m52 + ω m51 + m50 = ω 1+ω 5 + 25. G 1 n 1 ⊕···⊕ n · · · · If α = 0, no ordinal β (β<α) exists. We can assume α>0. So, we have We can write α and β as m12 m22 m32 m42 m52 =0 0 3 2 1 α = ωγk a + + ωγk a + a ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ · k ··· · 1 0 β = ωγk b + + ωγk b + b , =0 · k ··· · 1 0 m11 m21 m31 m41 m51 =0 2 0 4 5 where a0,...,ak,b0,...,bk N0. By definition, ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ∈ =3 as = m1s mns, for s =1,...,k. m m m m m =1 4 9 16 25 ⊕···⊕ 10 ⊕ 20 ⊕ 30 ⊕ 40 ⊕ 50 ⊕ ⊕ ⊕ ⊕ Since α>β, there exsists s such that =5.
as >bs, at = bt for all t (
mit′ = mit as bs for all t ( s) ⊕ ⊕ ≤ 2.2. Transfinite Welter’s Game and In transfinite version (Transfinite Welter’s Game), the size of the belt of Welter’s γ γ +1 γ α′ = ω k m + ω s m + ω s m′ Game is extended into general ordinal numbers, but played with finite number of i · ik ··· · is+1 · is γs 1 γ0 coins. The legal moves are to move one coin toward the left (jumping is allowed), and + ω − mis′ 1 + + ω mi′0, · − ··· · one cannot place two or more coins on the same square as in the original Welter’s where m a b = m′ . Game (see Fig. 3). We will define Welter function of a position of Transfinite is ⊕ s ⊕ s is Welter’s Game. Transfinite Version of Welter’s Game 87 4 5
Definition 2.1. For ordinal numbers α ,...,α , we define their nim-sum If we put α′ = β , α = β (j =i). Then, α >β and we have 1 n ∈ ON i i j j ̸ i i as follows: β1 βi 1 βi βi+1 βn = β ⊕··· − ⊕ ⊕ ⊕··· γk α1 αn = ω (m1k mnk). ⊕···⊕ ⊕···⊕ Therefore, for each β (<α), there is a position (β1,...,βn) reached by a single ∑k move from (α1,...,αn). We give a proof for the -value of the Transfinite Nim using this form. G Example 2.3. In the case of position (1,ω 2+4,ω2 3+9,ω2 2+ω 4 + 16,ω2 + n · · · · Theorem 2.2. For Transfinite Nim position (α1,...,αn) ，we have the ω 5 + 25): ⊆ ON · following: Let us calculate the value of α α α α α . 1 ⊕ 2 ⊕ 3 ⊕ 4 ⊕ 5 We get (α ,...,α )=α α . G 1 n 1 ⊕···⊕ n β2 β1 2 α1 = ω m12 + ω m11 + m10 = ω 0+ω 0+1 Proof. The proof is by induction. Let α1 αn = α (α ). We have to · · · · ⊕···⊕ ∈ ON α = ωβ2 m + ωβ1 m + m = ω2 0+ω 2+4 show that, for each β (<α), there exists a position with -value β reached by a 2 · 22 · 21 20 · · G β2 β1 2 single move from (α1,...,αn). α = ω m + ω m + m = ω 3+ω 0+9 3 · 32 · 31 30 · · Let (α1,...,αn) (β1,...,βn), by induction hypothesis we have α = ωβ2 m + ωβ1 m + m = ω2 2+ω 4 + 16 → 4 · 42 · 41 40 · · β2 β1 2 (β ,...,β )=β β . α5 = ω m52 + ω m51 + m50 = ω 1+ω 5 + 25. G 1 n 1 ⊕···⊕ n · · · · If α = 0, no ordinal β (β<α) exists. We can assume α>0. So, we have We can write α and β as m12 m22 m32 m42 m52 =0 0 3 2 1 α = ωγk a + + ωγk a + a ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ · k ··· · 1 0 β = ωγk b + + ωγk b + b , =0 · k ··· · 1 0 m11 m21 m31 m41 m51 =0 2 0 4 5 where a0,...,ak,b0,...,bk N0. By definition, ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ∈ =3 as = m1s mns, for s =1,...,k. m m m m m =1 4 9 16 25 ⊕···⊕ 10 ⊕ 20 ⊕ 30 ⊕ 40 ⊕ 50 ⊕ ⊕ ⊕ ⊕ Since α>β, there exsists s such that =5. as >bs, at = bt for all t (
○ ○ ○ ○ In the former case, as in Transfinite Nim, there is an index i and λ′ (<λ) such ··· ··· ··· i 0 1 2 3 ω ω +2 ω2 that (λ1,...,λi 1,λi′ ,λi+1,...,λn) has -value λ′ and we can adjust the finite part − G of αi so that the resulting Welter function to be β. Figure 3: Transfinite Version of Welter’s Game Therefore, for each β (<α), there is a position reached by a single move from (α ,...,α ) and its -value is β. 1 n G Definition 2.4. Let α1,...,αn . Each αi can be expressed as αi = ω λi +mi, Corollary 2.6. A position in Transfinite Welter’s Game is a -position if and only ∈ ON · P where λi and mi N0. Welter function in general ordinal numbers is defined if it satisfies the following conditions: ∈ ON ∈ as follows: ω (λ λ )=0 · 1 ⊕···⊕ n [α α ]=ω (λ λ )+ [S ], [Sλ]=0. 1|···| n · 1 ⊕···⊕ n λ λ λ ⊕∈ON ⊕∈ON By this corollary, we can easily calculate a winning move in Transfinite Welter’s where [Sλ] is Welter function, and Sλ = mn λn = λ . { | } Game by its Welter function. The following is our main result. Example 2.7. In the case of position (1,ω 2+4,ω 2+9,ω2+ω 4+16,ω2+ω 5+25): · · · · Theorem 2.5. Let α ,...,α . The -value of general position (α ,...,α ) Let us calculate the value of [α1 α2 α3 α4 α5]. We get 1 n ∈ ON G 1 n | | | | in Transfinite Welter’s Game is equal to its Welter function. Namely, α = ωβ2 m + ωβ1 m + m = ω2 0+ω 0+1 1 · 12 · 11 10 · · β2 β1 2 (α ,...,α )=[α α ]. α2 = ω m22 + ω m21 + m20 = ω 0+ω 2+4 G 1 n 1|···| n · · · · β2 β1 2 α3 = ω m32 + ω m31 + m30 = ω 0+ω 2+9 Proof. Let [α1 αn]=α. We have to show that, for each β (<α), there exists a · · · · β2 β1 2 |···| α4 = ω m42 + ω m41 + m40 = ω 1+ω 4 + 16 position with -value β reached by a single move from (α1,...,αn). · · · · G β2 β1 2 Let (α ,...,α ) (β ,...,β ). Then by the assumption of induction we have α5 = ω m52 + ω m51 + m50 = ω 1+ω 5 + 25. 1 n → 1 n · · · · (β ,...,β )=[β β ]. So, we have G 1 n 1|···| n m12 m22 m32 m42 m52 =0 0 0 1 1 If α = 0, there exist no β (<α). We can assume α>0 and ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ =0 α = ω λ + a0 and β = ω λ′ + b0, m m m m m =0 2 2 4 5 · · 11 ⊕ 21 ⊕ 31 ⊕ 41 ⊕ 51 ⊕ ⊕ ⊕ ⊕ where λ, λ′ , a0, b0 N0. Since α>β，we have =1 ∈ ON ∈ [m ] [m m ] [m ] m ]=[1] [4 9] [16] [25] (λ>λ′)or(λ = λ′ and a0 >b0). 10 ⊕ 20 | 30 ⊕ 40 ⊕ 50 ⊕ | ⊕ ⊕ [ =1 (4 9 1) 16 25 In the latter case, since a = [S ] >b, from the theory of Nim ([1], [3], ⊕ ⊕ − ⊕ ⊕ 0 λ 0 =4. λ [8]) there exists some λ and nonnegative⊕∈ON integer c (< [S ]) such that 0 0 λ0 Therefore, by the definition of Welter function for general ordinal number
a0 [Sλ ] c0 = b0. [α α α α α ]=ω + 4. ⊕ 0 ⊕ 1| 2| 3| 4| 5 Since, this shows that we are in an -position, we will calculate a winning move. Next since [Sλ0 ] >c0, from the theory of Welter function ([2]), there is an index N i and m ( ○ ○ ○ ○ In the former case, as in Transfinite Nim, there is an index i and λ′ (<λ) such ··· ··· ··· i 0 1 2 3 ω ω +2 ω2 that (λ1,...,λi 1,λi′ ,λi+1,...,λn) has -value λ′ and we can adjust the finite part − G of αi so that the resulting Welter function to be β. Figure 3: Transfinite Version of Welter’s Game Therefore, for each β (<α), there is a position reached by a single move from (α ,...,α ) and its -value is β. 1 n G Definition 2.4. Let α1,...,αn . Each αi can be expressed as αi = ω λi +mi, Corollary 2.6. A position in Transfinite Welter’s Game is a -position if and only ∈ ON · P where λi and mi N0. Welter function in general ordinal numbers is defined if it satisfies the following conditions: ∈ ON ∈ as follows: ω (λ λ )=0 · 1 ⊕···⊕ n [α α ]=ω (λ λ )+ [S ], [Sλ]=0. 1|···| n · 1 ⊕···⊕ n λ λ λ ⊕∈ON ⊕∈ON By this corollary, we can easily calculate a winning move in Transfinite Welter’s where [Sλ] is Welter function, and Sλ = mn λn = λ . { | } Game by its Welter function. The following is our main result. Example 2.7. In the case of position (1,ω 2+4,ω 2+9,ω2+ω 4+16,ω2+ω 5+25): · · · · Theorem 2.5. Let α ,...,α . The -value of general position (α ,...,α ) Let us calculate the value of [α1 α2 α3 α4 α5]. We get 1 n ∈ ON G 1 n | | | | in Transfinite Welter’s Game is equal to its Welter function. Namely, α = ωβ2 m + ωβ1 m + m = ω2 0+ω 0+1 1 · 12 · 11 10 · · β2 β1 2 (α ,...,α )=[α α ]. α2 = ω m22 + ω m21 + m20 = ω 0+ω 2+4 G 1 n 1|···| n · · · · β2 β1 2 α3 = ω m32 + ω m31 + m30 = ω 0+ω 2+9 Proof. Let [α1 αn]=α. We have to show that, for each β (<α), there exists a · · · · β2 β1 2 |···| α4 = ω m42 + ω m41 + m40 = ω 1+ω 4 + 16 position with -value β reached by a single move from (α1,...,αn). · · · · G β2 β1 2 Let (α ,...,α ) (β ,...,β ). Then by the assumption of induction we have α5 = ω m52 + ω m51 + m50 = ω 1+ω 5 + 25. 1 n → 1 n · · · · (β ,...,β )=[β β ]. So, we have G 1 n 1|···| n m12 m22 m32 m42 m52 =0 0 0 1 1 If α = 0, there exist no β (<α). We can assume α>0 and ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ =0 α = ω λ + a0 and β = ω λ′ + b0, m m m m m =0 2 2 4 5 · · 11 ⊕ 21 ⊕ 31 ⊕ 41 ⊕ 51 ⊕ ⊕ ⊕ ⊕ where λ, λ′ , a0, b0 N0. Since α>β，we have =1 ∈ ON ∈ [m ] [m m ] [m ] m ]=[1] [4 9] [16] [25] (λ>λ′)or(λ = λ′ and a0 >b0). 10 ⊕ 20 | 30 ⊕ 40 ⊕ 50 ⊕ | ⊕ ⊕ [ =1 (4 9 1) 16 25 In the latter case, since a = [S ] >b, from the theory of Nim ([1], [3], ⊕ ⊕ − ⊕ ⊕ 0 λ 0 =4. λ [8]) there exists some λ and nonnegative⊕∈ON integer c (< [S ]) such that 0 0 λ0 Therefore, by the definition of Welter function for general ordinal number a0 [Sλ ] c0 = b0. [α α α α α ]=ω + 4. ⊕ 0 ⊕ 1| 2| 3| 4| 5 Since, this shows that we are in an -position, we will calculate a winning move. Next since [Sλ0 ] >c0, from the theory of Welter function ([2]), there is an index N i and m ( (2 2 4 5) 1=1 1 = 0. [5] T. Kayada, Generalizations of Sato-Welter game (in Japanese). RIMS ⊕ ⊕ ⊕ ⊕ ⊕ Kokyuroku, 1738 (2011), 24-33. Thus, the only legal move is 5 5 1 = 4. So, our good move is in ω 5 + 25. → ⊕ · Then，in such moves，we will search for a move that satisfy the second condition． [6] R. J. Nowakowski, ... Welter’s Game, Sylver Coinage, Dots-and-Boxes, ..., It is obtained from the knowledge of Welter function. Combinatorial Games, Proc. Symp. Appl. Math. 43, Amer. Math. Soc. Provi- The finite part should satisfy dence, RI, 1991 (R. K. Guy ed.), pp. 155-182. 1 [4 9] [x 16] = 0. [7] A. N. Siegel, Combinatorial Game Theory, American Mathematical Society, ⊕ | ⊕ | 2013． So we have [8] R. P. Sprague, Uber¨ mathematische Kampfspiele, Tˆohoku Math. J., 41 (1935- x = 30. 36), 438-444. Therefore, the only good move is ω 5 + 25 ω 4 + 30. [9] C. P. Welter, The theory of a class of games on a sequence of squares, in terms · → · In fact， of the advancing operation in a special abelian group, Indagationes Math., 16 (1954), 194-200. m m m m m =0 0 0 1 1 12 ⊕ 22 ⊕ 32 ⊕ 42 ⊕ 52 ⊕ ⊕ ⊕ ⊕ =0 m m m m m =0 2 2 4 4 11 ⊕ 21 ⊕ 31 ⊕ 41 ⊕ 51 ⊕ ⊕ ⊕ ⊕ =0 [m ] [m m ] [m ] [m ]=[1] [4 9] [30 16] 10 ⊕ 20 | 30 ⊕ 40 ⊕ 50 ⊕ | ⊕ | =1 (4 9 1) (30 16 1) ⊕ ⊕ − ⊕ ⊕ − =1 12 13 ⊕ ⊕ =0. Thus, this position is a -position. P 3. Acknowledgements The author would like to thank Dr. KˆoSakai for useful comments and discussions. References [1] C. L. Bouton, Nim, a game with a complete mathmatical theory, Ann. of math. 3 (1902), 35-39． [2] J. H. Conway, On Numbers and Games (second edition)，A. K. Peters, 2001． [3] P. M. Grundy, Mathematics and games, Eureka, 2 (1939), 6-8. [4] T. Jech, Set Theory (third edition), Springer, 2002. Transfinite Version of Welter’s Game 91 8 9 (2 2 4 5) 1=1 1 = 0. [5] T. Kayada, Generalizations of Sato-Welter game (in Japanese). RIMS ⊕ ⊕ ⊕ ⊕ ⊕ Kokyuroku, 1738 (2011), 24-33. Thus, the only legal move is 5 5 1 = 4. So, our good move is in ω 5 + 25. → ⊕ · Then，in such moves，we will search for a move that satisfy the second condition． [6] R. J. Nowakowski, ... Welter’s Game, Sylver Coinage, Dots-and-Boxes, ..., It is obtained from the knowledge of Welter function. Combinatorial Games, Proc. Symp. Appl. Math. 43, Amer. Math. Soc. Provi- The finite part should satisfy dence, RI, 1991 (R. K. Guy ed.), pp. 155-182. 1 [4 9] [x 16] = 0. [7] A. N. Siegel, Combinatorial Game Theory, American Mathematical Society, ⊕ | ⊕ | 2013． So we have [8] R. P. Sprague, Uber¨ mathematische Kampfspiele, Tˆohoku Math. J., 41 (1935- x = 30. 36), 438-444. Therefore, the only good move is ω 5 + 25 ω 4 + 30. [9] C. P. Welter, The theory of a class of games on a sequence of squares, in terms · → · In fact， of the advancing operation in a special abelian group, Indagationes Math., 16 (1954), 194-200. m m m m m =0 0 0 1 1 12 ⊕ 22 ⊕ 32 ⊕ 42 ⊕ 52 ⊕ ⊕ ⊕ ⊕ =0 m m m m m =0 2 2 4 4 11 ⊕ 21 ⊕ 31 ⊕ 41 ⊕ 51 ⊕ ⊕ ⊕ ⊕ =0 [m ] [m m ] [m ] [m ]=[1] [4 9] [30 16] 10 ⊕ 20 | 30 ⊕ 40 ⊕ 50 ⊕ | ⊕ | =1 (4 9 1) (30 16 1) ⊕ ⊕ − ⊕ ⊕ − =1 12 13 ⊕ ⊕ =0. Thus, this position is a -position. P 3. Acknowledgements The author would like to thank Dr. KˆoSakai for useful comments and discussions. References [1] C. L. Bouton, Nim, a game with a complete mathmatical theory, Ann. of math. 3 (1902), 35-39． [2] J. H. Conway, On Numbers and Games (second edition)，A. K. Peters, 2001． [3] P. M. Grundy, Mathematics and games, Eureka, 2 (1939), 6-8. [4] T. Jech, Set Theory (third edition), Springer, 2002.