In this paper, we shall introduce a game called Corner the Knight Game, which is a game changed the chess queen of Corner the Queen Game to the chess knight. Although this game is very simple, we could not find any literature on this game. Hence it should be of some meaning to investigate this elementary game

Katayama, Shin-ichi

Koyama, Yuya

English

Tokushima University Institutional Repository

83J. Math. Tokushima Univ. Vol. 55 (2021), 83−88Corner the Knight GameByShin-ichi Katayama and Yuya KoyamaShin-ichi KatayamaDepartment of Mathematical Sciences, Graduate School of Science and Technology,Tokushima University, Minamijosanjima-cho 2-1, Tokushima 770-8506, JAPANe-mail address : shinkatayama@tokushima-u.ac.jpandYuya KoyamaGraduate School of Integrated Arts and Sciences, Tokushima University,Minamijosanjima-cho 1-1, Tokushima 770-8502, JAPANe-mail address : lfc.8.oct17@gmail.com(Received December 25, 2020. Revised February 26, 2021)AbstractIn this paper, we shall introduce a game called Corner the KnightGame, which is a game changed the chess queen of Corner the QueenGame to the chess knight. Although this game is very simple, we couldnot find any literature on this game. Hence it should be of some meaningto investigate this elementary game.2010 Mathematics Subject Classification. Primary 91A46; Secondary91A05)1 IntroductionWythoff’s Nim game is a two player’s game which was introduced by W. A. Wythoffin his paper [8](1907). Later in 1960’s R. P. Isaacs gave an equivalent description ofWythoff’s nim game, which is now called Corner the Queen Game. In corner thequeen game, a single chess queen is placed somewhere on a large grid of squares(or an184 Shin-ichi Katayama and Yuya Koyamainfinite chess board) and the queen is restricted to move horizontally left, verticallydownward and diagonally lower left. Each player can move the queen toward thelower left(or south west) corner of the grid. Let (m,n) be the position of the queen.Then the winner of this game is the player who moves the queen to the lower leftcorner (0, 0).In this short note, we shall change the queen to the knight and call this new gameCorner the Knight Game. In the following , we shall study the elementary propertiesof the grundy numbers of this game. It is a natural problem to change the queento other chess pieces and the properties are very simple. We wonder this new gameCorner the Knight Game has been known to the specialists, but we could not find anyliterature referred to this game. Hence it should be of some interest to write downthese elementary results explicitly. In corner the knight game, the movement of theknight is also restricted as the queen of the corner the queen game as follows.The Rule of Corner the Knight Game:1) It is a two player game and each player can move the knight on the (generalized)chess board as follows.2) Movement.The knight at the position (m,n) is restricted to move to the places (m−2, n−1),(m− 1, n− 2), (m− 2, n+ 1) or (m+1, n-2), where m− i, n− j must be ≥ 0.3) WinnerThe winner of this game is the player who moves the queen to the lower left corners(0, 0), (1, 0), (0, 1) or (1, 1).One can easily obtain the following table of grundy numbers for small (m,n) bychecking the mex, i.e., minimum excluded value inductively:Table 1 Table of the grundy numbers gS(m,n) of corner the knight game forsmall m and n.2　...................................................11 1 1 2 3 1 1 2 3 1 1 2 1 4 3 2 3 . . .10 1 1 2 2 1 1 2 2 1 2 2 2 3 2 2 2 . . .9 0 0 2 3 0 0 2 3 0 0 2 1 0 0 1 1 . . .8 0 0 3 3 0 0 3 4 0 0 1 1 0 0 1 1 . . .7 1 1 2 3 1 1 2 1 4 3 2 3 3 3 2 3 . . .6 1 1 2 2 1 2 2 2 3 2 2 2 3 2 2 2 . . .5 0 0 2 3 0 0 2 1 0 0 1 1 0 0 1 1 . . .4 0 0 3 4 0 0 1 1 0 0 1 1 0 0 1 1 . . .3 1 1 2 1 4 3 2 3 3 3 2 3 3 3 2 3 . . .2 1 2 2 2 3 2 2 2 3 2 2 2 3 2 2 2 . . .1 0 0 2 1 0 0 1 1 0 0 1 1 0 0 1 1 . . .0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 . . .n/m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . . .2 Proof of the Main TheoremIn the first place, we shall verify the following preparatory proposition.Proposition 2.1 With the above notation, we havegS(m, 0) =[m2](mod 2),gS(m, 1) ={2 for the case m = 2,[m2](mod 2) otherwise,gS(m, 2) =1 for the case m = 1,2 for the case m = 2,3 for the cases ≡ 0 (mod 4) and m ≥ 4,2 otherwise,gS(m, 3) =1 for the case m = 0, 1, 3,4 for the case m = 4,2 for the cases m ≡ 2 (mod 4),3 otherwise.Moreover gS(m, 4) = gS(m− 4, 0) and gS(m, 5) = gS(m− 4, 1) for the cases m ≥ 4.Proof. One can verify the prposition’s claims are true for 0 ≤ m ≤ 15 as in Table1. Inthe first place, we shall concentrate on proving the periodicity of gS(m, 0), gS(m, 1)for m ≥ 3. Assume gS(4k, 0) = gS(4k, 1) = gS(4k + 1, 0) = gS(4k + 1, 1) = 0385Corner the Knight Gameinfinite chess board) and the queen is restricted to move horizontally left, verticallydownward and diagonally lower left. Each player can move the queen toward thelower left(or south west) corner of the grid. Let (m,n) be the position of the queen.Then the winner of this game is the player who moves the queen to the lower leftcorner (0, 0).In this short note, we shall change the queen to the knight and call this new gameCorner the Knight Game. In the following , we shall study the elementary propertiesof the grundy numbers of this game. It is a natural problem to change the queento other chess pieces and the properties are very simple. We wonder this new gameCorner the Knight Game has been known to the specialists, but we could not find anyliterature referred to this game. Hence it should be of some interest to write downthese elementary results explicitly. In corner the knight game, the movement of theknight is also restricted as the queen of the corner the queen game as follows.The Rule of Corner the Knight Game:1) It is a two player game and each player can move the knight on the (generalized)chess board as follows.2) Movement.The knight at the position (m,n) is restricted to move to the places (m−2, n−1),(m− 1, n− 2), (m− 2, n+ 1) or (m+1, n-2), where m− i, n− j must be ≥ 0.3) WinnerThe winner of this game is the player who moves the queen to the lower left corners(0, 0), (1, 0), (0, 1) or (1, 1).One can easily obtain the following table of grundy numbers for small (m,n) bychecking the mex, i.e., minimum excluded value inductively:Table 1 Table of the grundy numbers gS(m,n) of corner the knight game forsmall m and n.2　...................................................11 1 1 2 3 1 1 2 3 1 1 2 1 4 3 2 3 . . .10 1 1 2 2 1 1 2 2 1 2 2 2 3 2 2 2 . . .9 0 0 2 3 0 0 2 3 0 0 2 1 0 0 1 1 . . .8 0 0 3 3 0 0 3 4 0 0 1 1 0 0 1 1 . . .7 1 1 2 3 1 1 2 1 4 3 2 3 3 3 2 3 . . .6 1 1 2 2 1 2 2 2 3 2 2 2 3 2 2 2 . . .5 0 0 2 3 0 0 2 1 0 0 1 1 0 0 1 1 . . .4 0 0 3 4 0 0 1 1 0 0 1 1 0 0 1 1 . . .3 1 1 2 1 4 3 2 3 3 3 2 3 3 3 2 3 . . .2 1 2 2 2 3 2 2 2 3 2 2 2 3 2 2 2 . . .1 0 0 2 1 0 0 1 1 0 0 1 1 0 0 1 1 . . .0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 . . .n/m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . . .2 Proof of the Main TheoremIn the first place, we shall verify the following preparatory proposition.Proposition 2.1 With the above notation, we havegS(m, 0) =[m2](mod 2),gS(m, 1) ={2 for the case m = 2,[m2](mod 2) otherwise,gS(m, 2) =1 for the case m = 1,2 for the case m = 2,3 for the cases ≡ 0 (mod 4) and m ≥ 4,2 otherwise,gS(m, 3) =1 for the case m = 0, 1, 3,4 for the case m = 4,2 for the cases m ≡ 2 (mod 4),3 otherwise.Moreover gS(m, 4) = gS(m− 4, 0) and gS(m, 5) = gS(m− 4, 1) for the cases m ≥ 4.Proof. One can verify the prposition’s claims are true for 0 ≤ m ≤ 15 as in Table1. Inthe first place, we shall concentrate on proving the periodicity of gS(m, 0), gS(m, 1)for m ≥ 3. Assume gS(4k, 0) = gS(4k, 1) = gS(4k + 1, 0) = gS(4k + 1, 1) = 0386 Shin-ichi Katayama and Yuya Koyamaand gS(4k + 2, 0) = gS(4k + 2, 1) = gS(4k + 3, 0) = gS(4k + 3, 1) = 1. Then theset of grundy numbers of the next places of the each knight at (4k + 1, 2), (4k +1, 3), (4K + 2, 2), (4k + 2, 3) contains the set {0, 1}. Hence each grundy number ofgS(4k+1, 2), gS(4k+1, 3), gS(4K+2, 2), gS(4k+2, 3) must be ≥ 2. gS(4k+2, 2) ≥ 2and gS(4k + 2, 0) = 1 imply gS(4k + 4, 1) = 0 and gS(4k + 2, 1) = 1 implies gS(4k +4, 0) = 0. gS(4k + 2, 0) = gS(4k + 2, 1) = 1 and gS(4k + 4, 0) = gS(4k + 4) = 0 implygS(4k+ 3, 2), gS(4k+ 3, 3) are also ≥ 2. Hence we know gS(m, 0) = gS(m, 1) ∈ {0, 1}have the period 4 for m ≥ 3 inductively. At the same time we have also verifiedgS(m, 2), gS(m, 3) ≥ 2 for m ≥ 4.In the next place, we shall prove the facts gS(m, 2), gS(m, 3) ≥ 2 for m ≥ 4 implygS(m, 4) = gS(m−4, 0) and gS(m, 5) = gS(m−4, 1). and hence gS(m, 4), gS(m, 5) arealso periodic with the period 4. In the casesm = 4, 5, 6, 7, 8, 9, gS(m, 4) = gS(m−4, 0)and gS(m, 5) = gS(m − 4, 1), gS(m, 6) = gS(m − 4, 2) can be verified case by case.In the case m ≥ 10 and n = 4, 5, 6, we shall assume gS(m′, n) = gS(m′ − 4, n − 4)for any m′ (4 ≥ m′ < m). By definition, gS(m, 4) = mex {gS(m − 2, 5), gS(m −2, 3), gS(m − 1, 2), gS(m + 1, 4)}. gS(m − 2, 3), gS(m − 1, 2), gS(m + 1, 4) ≥ 2 implygS(m, 4)( ̸= gS(m − 2, 5) ∈ {0, 1}. By assumption gS(m − 2, 5) = gS(m − 6, 1) andgS(m−4, 0)( ̸= gS(m−6, 1) ∈ {0, 1}. Hence gS(m, 4) = gS(m−4, 0). Similarly the factsgS(m, 2), gS(m, 3) ≥ 2 for m ≥ 4 and the assumptions on gS(m′, 4) = gS(m′ − 4, 0)and gS(m′, 6) = gS(m′−4, 2)for any m′ (4 ≥ m′ < m) imply gS(m, 5) = gS(m−4, 1).In the next place we shall show the periodicty of gS(m, 2) and gS(m, 3). One canverify the grundy numbers for 0 ≤ m ≤ 6. Since gS(m′, 0), gS(m′, 1), gS(m′, 4) ∈{0, 1} for 4 ≤ m′, we see gS(m, 2)+gS(m+2, 3) = gS(m, 3)+gS(m+2, 2) = 5. Hencewe have verified the properties of gS(m, 2) and gS(m, 3), which completes the proofof tis proposition.Now we shall prove the following main theorem using the induction on the valuep = m+ n.Theorem 2.2 With the above notation, we have gS(i, j) = gS(j, i) and gS(i+ 4, j +44) = gS(i, j) for any i, j ≥ 0. More precisely we havegS(m, 4n) =[m2](mod 2),gS(m, 4n+ 1) ={2 for the case m = 4n+ 2,[m2](mod 2) otherwise,gS(m, 4n+ 2) ={3 for the case m ≡ 0 (mod 4),2 otherwise,gS(m, 4n+ 3) =1 for the case m = 4n+ 3,4 for the case m = 4n+ 4,2 for the case m ≡ 2 (mod 4),3 otherwise,where each gS(i, j) is restricted for i ≥ j, because of the symmetry of grundy numbersgS(i, j) = gS(j, i).Proof. From the proposition, we have already shown the periodicity gS(i,+4, j+4) =gS(i, j) is true for the cases i = 0, 1 or j = 0, 1. Hence we have to verify i, j ≥ 2,i.e., m,n ≥ 6. Put p = m + n. Then we can verify the periodicity gS(m′, n′) =gS(m′− 4, n′− 4) for any 4 ≤ m′.n′ and p ≤ 12 as in Table 1. Assume the periodicitygS(m′, n′) = gS(m′ − 4, n′ − 4) for any 4 ≤ m′.n′ ≤ 6 and p. Then by definition ofthe grundy numbergS(m,n) = mex {gS(m−2, n−1), gS(m−2, n+1), gS(m−1, n−2), gS(m+1, n−2)}for any m ≥ 6, n ≥ 6 with m+ n = p+ 1. Then p′ = m′ + n′ = p− 3 or p− 1 for thecases (m′, n′) = (m−2, n−1), (m−2, n+1), (m−1, n−2) and (m+1, n−2). From theassumption on the periodicity of gS(m′, n′), we obtain gS(m′, n′) = gS(m′− 4, n′− 4)for above m′, n′. ThusgS(m,n) = mex {gS(m−2, n−1), gS(m−2, n+1), gS(m−1, n−2), gS(m+1, n−2)}= mex {gS(m − 6, n − 5), gS(m − 6, n − 3), gS(m − 5, n − 6), gS(m − 3, n − 6)} =gS(m− 4, n− 4).Hence the periodicity holds for m+n = p+1, which completes the induction. Finallywe give the following modified table with shadows, the periodicity with the period 4was visualized as the L-shaped hook structure of the grundy numbers.Table 4 Table of the grundy numbers gS(m,n) of Corner the Knight game for smallm and n.587Corner the Knight Gameand gS(4k + 2, 0) = gS(4k + 2, 1) = gS(4k + 3, 0) = gS(4k + 3, 1) = 1. Then theset of grundy numbers of the next places of the each knight at (4k + 1, 2), (4k +1, 3), (4K + 2, 2), (4k + 2, 3) contains the set {0, 1}. Hence each grundy number ofgS(4k+1, 2), gS(4k+1, 3), gS(4K+2, 2), gS(4k+2, 3) must be ≥ 2. gS(4k+2, 2) ≥ 2and gS(4k + 2, 0) = 1 imply gS(4k + 4, 1) = 0 and gS(4k + 2, 1) = 1 implies gS(4k +4, 0) = 0. gS(4k + 2, 0) = gS(4k + 2, 1) = 1 and gS(4k + 4, 0) = gS(4k + 4) = 0 implygS(4k+ 3, 2), gS(4k+ 3, 3) are also ≥ 2. Hence we know gS(m, 0) = gS(m, 1) ∈ {0, 1}have the period 4 for m ≥ 3 inductively. At the same time we have also verifiedgS(m, 2), gS(m, 3) ≥ 2 for m ≥ 4.In the next place, we shall prove the facts gS(m, 2), gS(m, 3) ≥ 2 for m ≥ 4 implygS(m, 4) = gS(m−4, 0) and gS(m, 5) = gS(m−4, 1). and hence gS(m, 4), gS(m, 5) arealso periodic with the period 4. In the casesm = 4, 5, 6, 7, 8, 9, gS(m, 4) = gS(m−4, 0)and gS(m, 5) = gS(m − 4, 1), gS(m, 6) = gS(m − 4, 2) can be verified case by case.In the case m ≥ 10 and n = 4, 5, 6, we shall assume gS(m′, n) = gS(m′ − 4, n − 4)for any m′ (4 ≥ m′ < m). By definition, gS(m, 4) = mex {gS(m − 2, 5), gS(m −2, 3), gS(m − 1, 2), gS(m + 1, 4)}. gS(m − 2, 3), gS(m − 1, 2), gS(m + 1, 4) ≥ 2 implygS(m, 4)( ̸= gS(m − 2, 5) ∈ {0, 1}. By assumption gS(m − 2, 5) = gS(m − 6, 1) andgS(m−4, 0)( ̸= gS(m−6, 1) ∈ {0, 1}. Hence gS(m, 4) = gS(m−4, 0). Similarly the factsgS(m, 2), gS(m, 3) ≥ 2 for m ≥ 4 and the assumptions on gS(m′, 4) = gS(m′ − 4, 0)and gS(m′, 6) = gS(m′−4, 2)for any m′ (4 ≥ m′ < m) imply gS(m, 5) = gS(m−4, 1).In the next place we shall show the periodicty of gS(m, 2) and gS(m, 3). One canverify the grundy numbers for 0 ≤ m ≤ 6. Since gS(m′, 0), gS(m′, 1), gS(m′, 4) ∈{0, 1} for 4 ≤ m′, we see gS(m, 2)+gS(m+2, 3) = gS(m, 3)+gS(m+2, 2) = 5. Hencewe have verified the properties of gS(m, 2) and gS(m, 3), which completes the proofof tis proposition.Now we shall prove the following main theorem using the induction on the valuep = m+ n.Theorem 2.2 With the above notation, we have gS(i, j) = gS(j, i) and gS(i+ 4, j +44) = gS(i, j) for any i, j ≥ 0. More precisely we havegS(m, 4n) =[m2](mod 2),gS(m, 4n+ 1) ={2 for the case m = 4n+ 2,[m2](mod 2) otherwise,gS(m, 4n+ 2) ={3 for the case m ≡ 0 (mod 4),2 otherwise,gS(m, 4n+ 3) =1 for the case m = 4n+ 3,4 for the case m = 4n+ 4,2 for the case m ≡ 2 (mod 4),3 otherwise,where each gS(i, j) is restricted for i ≥ j, because of the symmetry of grundy numbersgS(i, j) = gS(j, i).Proof. From the proposition, we have already shown the periodicity gS(i,+4, j+4) =gS(i, j) is true for the cases i = 0, 1 or j = 0, 1. Hence we have to verify i, j ≥ 2,i.e., m,n ≥ 6. Put p = m + n. Then we can verify the periodicity gS(m′, n′) =gS(m′− 4, n′− 4) for any 4 ≤ m′.n′ and p ≤ 12 as in Table 1. Assume the periodicitygS(m′, n′) = gS(m′ − 4, n′ − 4) for any 4 ≤ m′.n′ ≤ 6 and p. Then by definition ofthe grundy numbergS(m,n) = mex {gS(m−2, n−1), gS(m−2, n+1), gS(m−1, n−2), gS(m+1, n−2)}for any m ≥ 6, n ≥ 6 with m+ n = p+ 1. Then p′ = m′ + n′ = p− 3 or p− 1 for thecases (m′, n′) = (m−2, n−1), (m−2, n+1), (m−1, n−2) and (m+1, n−2). From theassumption on the periodicity of gS(m′, n′), we obtain gS(m′, n′) = gS(m′− 4, n′− 4)for above m′, n′. ThusgS(m,n) = mex {gS(m−2, n−1), gS(m−2, n+1), gS(m−1, n−2), gS(m+1, n−2)}= mex {gS(m − 6, n − 5), gS(m − 6, n − 3), gS(m − 5, n − 6), gS(m − 3, n − 6)} =gS(m− 4, n− 4).Hence the periodicity holds for m+n = p+1, which completes the induction. Finallywe give the following modified table with shadows, the periodicity with the period 4was visualized as the L-shaped hook structure of the grundy numbers.Table 4 Table of the grundy numbers gS(m,n) of Corner the Knight game for smallm and n.588 Shin-ichi Katayama and Yuya Koyama　...................................................11 1 1 2 3 1 1 2 3 1 1 2 1 4 3 2 3 . . .10 1 1 2 2 1 1 2 2 1 2 2 2 3 2 2 2 . . .9 0 0 2 3 0 0 2 3 0 0 2 1 0 0 1 1 . . .8 0 0 3 3 0 0 3 4 0 0 1 1 0 0 1 1 . . .7 1 1 2 3 1 1 2 1 4 3 2 3 3 3 2 3 . . .6 1 1 2 2 1 2 2 2 3 2 2 2 3 2 2 2 . . .5 0 0 2 3 0 0 2 1 0 0 1 1 0 0 1 1 . . .4 0 0 3 4 0 0 1 1 0 0 1 1 0 0 1 1 . . .3 1 1 2 1 4 3 2 3 3 3 2 3 3 3 2 3 . . .2 1 2 2 2 3 2 2 2 3 2 2 2 3 2 2 2 . . .1 0 0 2 1 0 0 1 1 0 0 1 1 0 0 1 1 . . .0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 . . .n/m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . . .References[ 1 ] C. L. Bouton, Nim, A Game with a Complete Mathematical Theory, Ann. ofMath., 3 (1901–1902), 35–39.[ 2 ] S. Katayama and T. Kubo, On Restricted Wythoff’s Nim, Journal of theMathematics, Tokushima University 52, (2018), 53–57.[ 3 ] S. Katayama and Y. Koyama, On Square and Rectangular Nim Games, Journalof the Mathematics, Tokushima University 54, (2020) 93–104.[ 4 ] R. Miyadera, M. Fukui, T. Inoue, Y, Nakaya and Y. Tokuni, A Variant ofCorner the Queen Game, IPSJ Technical Report. 2016-GI-3506.[ 5 ] R. Miyadera, M. Fukui, Y, Nakaya and Y. Tokuni, A Generalized Ryuoh-Nim:A Variant of the Classical Game of Wythoff Nim, IPSJ Technical Report.2016-GI-3604.[ 6 ] F. Sato, On the Mathematics of Extracting Games of Stones-Wonderful Rela-tions between Games and Algebra, Sugaku-Shobou, 2014 (in Japanese).[ 7 ] K. Suetsugu and M. Fukui, Reseach on a Variant of Wythoff Nim WhichRestricts the Moves of Tokens, IPSJ Tecnical Report 2017-GI-3803(in Japanese).[ 8 ] W. A. Wythoff, A Modification of the Game of Nim, Nieuw Arch, Wisk. 7(1907), 199–202.6

Corner the Knight Game

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Corner the Knight Game

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