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Nim naito@math.nagoya-u.ac.jp,.,.,,,.,,,,,,, Nim,.,,,,. Nim Nim,.,.,.,,.,.,. [1, 3],,, Nim,,., Nim. Date:.

August 10-11, 1999 2 1 Nim.. Pile., Pile.,. normal case.,. reverse case.,.. Pile. N 1, N 2, N 3., N 1, N 2, N 3,., N 1 = N 2 = N 3 =0., N 1 = N 2 = N 3 =0.,., Nim.,,., Nim.. Pile k. N 1,,N k., N 1, N k,., i, N i =0. normal case., i N i =0. reverse case. 1.1 Nim,. Definition 1.1. {N i } k i=1 = {N 1,,N k }, NIM({N 1,,N k })= NIM({N i }). {N i } k i=1 N i, N l i N l 1 i N 1 i N 0 i., N i = Ni l 2 l + Ni l 1 2 l 1 + + Ni 1 2+Ni 0

August 10-11, 1999 3., N l i =0,1., NIM({N i })=(N l 1 + + N l k,n l 1 1 + + N l 1 k,,n 1 1 + + N 1 k,n 0 1 + + N 0 k ) 2 =(N1 l + + Nk)2 l l +(N1 l 1 + + N l 1 k )2 l 1 + +(N1 1 + + N k 1 )2 + (N 1 0 + + N k 0 )., 2. 0 1,, N j i 0, 1. 1 Example 1.2. NIM({N i }). 1. N 1 =3,N 2 =4,N 3 =5. N 1, N 2, N 3, N 1 =3=0 1 1 N 2 =4=1 0 0 N 3 =5=1 0 1 NIM({3, 4, 5}) =0 1 0=2, NIM({3, 4, 5}) =(0, 1, 0) 2 =2. 2. N 1 =2,N 2 =5,N 3 =7. N 1, N 2, N 3, N 1 =2=0 1 0 N 2 =5=1 0 1 N 3 =7=1 1 1 NIM({2, 5, 7}) =0 0 0=0, NIM({2, 5, 7}) =(0, 0, 0) 2 =0. NIM. Proposition 1.3. NIM. 1. {N i } {N i}, NIM({N i }) = NIM({N i}). 2. NIM({M,M}) =0. 3. NIM({N 1,,N k }) = NIM({N 1,,N l }, NIM({N l+1,,n k })). 4. N i N i1, N i2, NIM({N 1,,N k }) = NIM({N 1,,N i1,n i2,,n l }). Proof.. NIM, 1., {N i } m {Ni m }, N1 m + + Nk m mod 2. mod2,.,,.

August 10-11, 1999 4 1. mod2 a + b = b + a. 2. M, 1+1=0,0+0=0 NIM({M,M}) =0. Theorem 1.4. Nim, {N i } NIM({N i })=0, Nim, NIM({N i }) 0., {N i } NIM({N i }) 0, Nim, {N i } NIM({N i })=0. Proof., {N i } NIM({N i })=0., {N i }. N i0, N i 0., 0 N i 0 <N i0., N i0 N i 0, 0 1., NIM({N i }) 1, N i0, NIM({N i }) 0 1.,. {N i }, {N i }, i i 0, N i = N i, N i0 >N i 0., NIM({N i })=0, j =0,,l N j 1 + + N j k =0., N i0 N i 0, j = j 0., N j0 i 0 ((N 1) j 0 + +(N i 0 ) j 0 + +(N k) j 0 ) (N j0 1 (N i 0 ) j0 + + N j0 i 0 + + N j0 j0 k )=Ni 0 (N i 0 ) j 0 =1.,,, N j0 1 + + N j0 i 0 + + N j0 k =0 (N 1) j0 + +(N i 0 ) j0 + +(N k) j0 =1. NIM({N i }) 0., NIM({N i }) 0. M = NIM({N i })., M l +1. {N i } l +1 1, N i0., M l +1 1,., N i0 l +2 l +1 0 N i1, l +1 N i2,, L = NIM(N i2,m). M, N i2 M l +1 1, L l +1 0, N i2 >L., N i0 l +1 N i2 L. {N i2 } i2 {N i } N i0, N i1. Proposition 1.4, NIM({N i } i2,l) = NIM({N i } i2, NIM({N i2,m})) = NIM({N i } i2,n i2,m}) = NIM({N i },M) = NIM(NIM({N i }),M) = NIM(M,M) =0

August 10-11, 1999 5., i 0 (N i+1 L) 2, N i 0, NIM(N i )=0. Example 1.5. 1. N 1 =7,N 2 =6,N 3 =5. N 1 =7=1 1 1 N 2 =6=1 1 0 N 3 =5=1 0 1 NIM({7, 6, 5}) =1 0 0, M =4., N i0, N 1 4, N 1 =3, i 0 =1., L =3. N 1 =3=0 1 1 N 2 =6=1 1 0 N 3 =5=1 0 1 NIM({3, 6, 5}) =0 0 0. 2. N 1 =3,N 2 =4,N 3 =5. N 1 =3=0 1 1 N 2 =4=1 0 0 N 3 =5=1 0 1 NIM({3, 4, 5}) =0 1 0, M =2., i 0 =1., L = NIM(3, 2)=1,, N 1 L =1, N 1 =1=0 0 1 N 2 =4=1 0 0 N 3 =5=1 0 1 NIM({1, 4, 5}) =0 0 0. 1.2 (normal case) normal case., i N i =0, NIM({N i }) = NIM({0,, 0}) =0., (Theorem 1.4), NIM({N i }) 0.,.

August 10-11, 1999 6 Definition 1.6. {N i } = {N 1,,N k }, NIM({N i }) 0. {N i } = {N 1,,N k }, NIM({N i })=0., Theorem 1.4,. Theorem 1.7.,.,.,. Theorem 1.8. Nim normal case,. Proof.,., Theorem 1.7,,. Theorem 1.7,,,.,, {N i }, N i0 0., N i0,. 1.3 (reverse case),.,,,.,,. Definition 1.9. {N i } = {N 1,,N k }, normal case, N i =1 0,, N i =1 0., normal case. Theorem 1.10.,.,. Proof. normal case, reverse case,, N i =1 0, N i =1,., normal case, reverse case,, N i =1 0, N i =1,.,.

August 10-11, 1999 7 1., normal mode, reverse mode, 2.,, reverse mode,, reverse mode, N i =1, 0,, pile 0,, N i =1 0, reverse mode,., pile 1, 0, N i =1, reverse mode,., normal mode, reverse mode, N i =1.,, N i =1, 0., N i > 1 pile,, N i > 1 pile.,, normal mode, reverse mode.,. Theorem 1.11. Nim reverse case,.

August 10-11, 1999 8 2 section., Nim K.. k., k K. N 1,,N k., N 1, N k K,., i, N i =0. normal case, i N i =0. reverse case section Nim Nim K, K =1. 2.1 Nim K Nim. Definition 2.1. {N i } k i=1, NIM K({N 1,,N k }) = NIM K ({N i }). {N i } k i=1, N l i N l 1 i N 1 i N 0 i., N i = Ni l 2 l + Ni l 1 2 l 1 + + Ni 1 2+Ni 0., N l i =0,1., NIM K ({N i })=(N l 1 + + N l k,n l 1 1 + + N l 1 k,,n 1 1 + + N 1 k,n 0 1 + + N 0 k ) 2 =(N1 l + + N k l )(K +1)l +(N1 l 1 + + N l 1 k )(K +1) l 1 + +(N1 1 + + Nk 1 )(K +1)+(N1 0 + + Nk 0 )., K +1. Remark 2.2. NIM({N i }) NIM K ({N i }), 2, K +1., K +1. Example 2.3. NIM K ({N i }). 1. K =2,N 1 =3,N 2 =4,N 3 =5. N 1, N 2, N 3,, NIM 2 ({3, 4, 5}) =(2, 1, 2) 3 =23. N 1 =3=0 1 1 N 2 =4=1 0 0 N 3 =5=1 0 1 NIM 2 ({3, 4, 5}) =2 1 2=2 3 2 +3+2=23

August 10-11, 1999 9 Theorem 2.4. Nim K, {N i } NIM K ({N i })=0, Nim K, NIM K ({N i }) 0., {N i } NIM K ({N i }) 0, Nim K, {N i } NIM K({N i })= 0. Proof., {N i } NIM K ({N i })=0., {N i } K. N 1,,N m,0<m m, N 1,,N m., 0 N s <N s s =1, m., N s N s, 0 1., NIM K ({N i }) 1, N 1,,N m., K, NIM K ({N i }) 0 0.,. {N i }, {N i }, i>m, N i = N i, i m N i >N i., NIM K({N i })=0, j =0,,l N j 1 + + N j k =0., i m N i N i, j = j 0 i i =1,,m., ((N 1) j0 + +(N k) j0 ) (N j0 1 N j0 i (N i )j0 j0 j0 + + Nk ) = N1 (N 1) j0 + Nm j0 (N m) j0 K.,,, N j0 1 + + N j0 k =0 (N 1) j0 + +(N k) j0 0. NIM K ({N i }) 0., NIM K ({N i }) 0. Nim K. M = NIM K ({N i })., M K +1 M l, l +1., {N i } l +1 1 M l. n, M l + n(k +1). N i1,,n im.. l,. N i, i = i 1,,i Ml l +1 0, 1., N i l 2 l 1., NIM K ({N i }) K +1 l +1 0., l, K., NIM K ({N i }) M, K +1 l, M l.,, l <l., N ij l 1,., M l. 1. M l M l., N ij, M l l 0.

August 10-11, 1999 10 2. M l <M l., N ij l 0 NIMK, K +1 l M l M l.,, N i, l 1 M l M l,., NIM K ({N i }), K +1 l 1., K, NIM K ({N i })=0. Example 2.5. NIM K ({N i }) 0. 1. K =2,N 1 =3,N 2 =4,N 3 =5. N 1 =3=0 1 1 N 2 =4=1 0 0 N 3 =5=1 0 1 NIM 2 ({3, 4, 5}) =2 1 2., N i 3 1, N 1, N 2, 011, N 1 =3=0 1 1 N 2 =3=0 1 1 N 3 =3=0 1 1 NIM 2 ({3, 3, 3}) =0 0 0=0. 2. K =2,N 1 =3,N 2 =4,N 3 =5,N 4 =9. N 1 = 3 = 0 0 1 1 N 2 = 4 = 0 1 0 0 N 3 = 5 = 0 1 0 1 N 4 = 9 = 1 0 0 1 NIM 2 ({3, 4, 5, 9}) = 1 2 1 0., N i 4 1, N 4, 0111. N 1 = 3 = 0 0 1 1 N 2 = 4 = 0 1 0 0 N 3 = 5 = 0 1 0 1 N 4 =7=0 1 1 1 NIM 2 ({3, 4, 5, 7}) = 0 0 2 0

August 10-11, 1999 11 2 2, N 4 N 1. 2 01, N 1 = 1 = 0 0 0 1 N 2 = 4 = 0 1 0 0 N 3 = 5 = 0 1 0 1 N 4 = 5 = 0 1 0 1 NIM 2 ({1, 4, 5, 5}) = 0 0 0 0. 2.2 (normal case) normal case. Section,. Definition 2.6. {N i } = {N 1,,N k }, NIM K ({N i }) 0. {N i } = {N 1,,N k }, NIM K ({N i })=0., Theorem 1.7,. Theorem 2.7.,.,.,. Theorem 2.8. Nim K normal case,. 2.3 (reverse case),. K 2, K =1.,, K n,, n 1,,,,.,,. Definition 2.9. {N i } = {N 1,,N k }, normal case, N i =1 K +1, 0, N i =1 K +1, 0., normal case.

August 10-11, 1999 12 Theorem 2.10.,.,. Proof. normal case, reverse case,, N i =1 K +1, 0, N i =1 K +1,., normal case, reverse case,, N i =1 K +1, 0, N i =1 K +1,,.,. 1., normal mode, reverse mode, 2.,, reverse mode,, reverse mode, N i =1 K +1, 0,, pile 0,, 1, pile, N i =1 K +1, 0., reverse mode,., normal mode, reverse mode, N i =1 K +1.,, N i =1 K +1, 0., N i > 1 pile,, N i > 1 pile K.,, normal mode, reverse mode.,. Theorem 2.11. Nim K reverse case,.

August 10-11, 1999 13 3.,., n. 3.1, [2], n Nim K (normal case).,,., Nim n K. n P 1,,P n. k. N 1,,N k. n, N 1, N k,., i, N i =0.,, P i,. P i+1, P i+1,,p i 1,P i,,.. 3.2 Nim K,. Definition 3.1. l-position,, l., n =2, 1-position., Nim K NIMK. Definition 3.2. {N i } k i=1, NIMn K({N 1,,N k }) = NIM n K({N i }), NIM K nk K +1. n =2, nk K +1=K +1, NIM K.,. Theorem 3.3. NIM n K({N i })=0, 1-position. Proof.,. NIM n K ({N i}) δ., δ =0, n 1, δ =0. δ 0, n 1, δ =0.,.

August 10-11, 1999 14 [1] Charles. L. Bouton, Nim, a game with a complete mathematical theorey, Ann. of Math. (2) (1902), no. 3, 35 39. [2] S.-Y. R. Li, N-person Nim and N-person Moore s games, Internat. J. Game Theory 7 (1978), no. 1, 31 36. [3] E.H. Moore, A generalization of the game callednim, Ann. of Math. (2) (1910), no. 11, 93 94.

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