1. Algebraic Structure in a Family of Nim-like Arrays Lowell Abrams The George Washington University Dena Cowen-Morton Xavier University TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AA A AAAA CanaDaM 2009

2. Combinatorial Games full information no probability winning strategy (“1 st player win” vs. “2 nd player win”) Basics

3. Nim even piles --- who wins? Combinatorial Games

4. Nim (this is a “sum” of three individual single-pile games) Combinatorial Games

5. Nimbers and Nim addition Nim pile with n stones has nimber n nimber 0 means the second player wins two side-by-side Nim piles, both with nimber n, have sum 0: n+n = 0 if (r+s)+n = 0 (i.e. is a second player win) then r+s=n

6. the Nimbers table (G+H) 01234567 10325476 23016745 32107654 456701 23 547610 32 67452301 76543210 Rule: Seed with 0. Rule: Enter smallest non-negative integer appearing neither above nor to left.

7. Another way to combine games Ullman and Stromquist: sequential compound G → H misère play: G → 1 misère nim addition: G+H → 1 something else: G+H → s for integer s ¸ 2

8. the Nimbers table (G+H → 2) 2013456789 01243658710 12056349107 3450127698 436107251112 563270141211 654721031314 7896543012 87109111213103 91078121114230 Rule: Seed with 2. Proceed with same algorithm.

9. An algebraic approach... view array as defining an operation ¤ on N ¸ 0 20134560124365120563434501274361072563270165472102013456012436512056343450127436107256327016547210 3 ¤ 3 = 0 4 ¤ 5 = 7

10. Basic algebraic structure view array as defining an operation ¤ on N ¸ 0 20134560124365120563434501274361072563270165472102013456012436512056343450127436107256327016547210 ¤ is commutative 2 is the ¤ -identity ¤ is not associative e.g. 1 ¤ (1 ¤ 4) = 1 ¤ 6 = 4 (1 ¤ 1) ¤ 4 = 0 ¤ 4 = 3 write: A 2 := ( N ¸ 0, ¤ ) have A s, by analogy, for each seed s

11. “(Q, ¤ ) is a quasigroup” means: for every i,j 2 Q there exist unique p,q 2 Q such that i ¤ p = j and q ¤ i = j “(Q, ¤ ) is a loop” means: (Q, ¤ ) is a quasigroup with a two-sided ¤ -identity Basic algebraic structure, continued...

12. Quasigroups all groups are quasigroups x123411234224133314244321x123411234224133314244321 (units in Z/5Z, under multiplication) but not every quasigroup is a group (units in Z/5Z, under division) /1234 11234 23142 32413 44321 note: 2/(3/2) = 2/4 = 2 but (2/3)/2 = 4/2 = 3

13. “(Q, ¤ ) is a quasigroup” means: for every i,j 2 Q there exist unique p,q 2 Q such that i ¤ p = j and q ¤ i = j “(Q, ¤ ) is a loop” means: (Q, ¤ ) is a quasigroup with a two-sided ¤ -identity Basic algebraic structure, continued... observe: A s is a loop

14. Algebraic results provide a way to encode combinatorial properties Take-Home Point:

15. Main Results (in brief) Theorem For each seed s ≥ 2, A s is monogenic. Theorem There are no nontrivial homomorphisms A s → A t if s ≥ 2 or t ≥ 2. Otherwise, there are a lot of them.

16. Monogenicity Notation: « x; ◊ » is the free unital groupoid on generator x with operation ◊ Note, e.g. : (x◊x) ◊ (x◊x) ≠ x ◊ (x ◊ (x◊x) ) Write x n for x ◊ (… ◊ (x◊x) ) n times

17. Monogenicity define L is monogenic:there is n ∈ L such that φ n is surjective loop L, element n ∈ L define φ n : « x; ◊ » → L operation-preserving φ n (e ◊ ) = e L φ n (x) = n note: this differs a little from the standard definition...

18. Monogenicity Theorem (A. and Cowen-Morton) A s is monogenic iff s ≥ 2 For s=2, every element n>2 is a generator. For s>2, every element n ≠ s is a generator. apparently, a novelty in the literature

19. Homomorphisms Theorem (A. and Cowen-Morton) The only loop homomorphism f:A s → A t for s ≠ t and either s ≥ 2 or t ≥ 2 (or both) is the trivial map A s → {t}. For s=t ≥ 2, homomorphism f is either the trivial map A s → {s} or the identity map.

20. Homomorphisms Terri Evans (1953): description of homomorphisms of finitely presented monogenic loops Theorem (A. and Cowen-Morton) For any seed s, the loop A s is not finitely presented.

21. Homomorphisms Essence of proof ● monogenicity ● commutativity of this diagram: « x; ◊ » AsAs AtAt φnφn ψ f(n) f ψ is the appropriate evaluation map

22. Homomorphisms case: s = t = 2 and f(3) > 6 set α= (x 2 ) 2 ◊ [ x 2 ◊ ((x 2 ) 2 ◊ x) ] β = ( (x 2 ) 2 ◊ x ) ◊ ( x ◊ [ x 2 ◊ ((x 2 ) 2 ◊ x) ] ) αβ 4 « x; ◊ » A2A2 A2A2 φ3φ3 φ f(3) f φ f(3) (α) ≠ φ f(3) (β) ?

23. Homomorphisms case: s = t = 2 and f(3) = 4, 5, or 6 set α= (x 2 ◊ x) ◊ [ (x 2 ) 2 ◊ [ x 2 ◊ ( x ◊ [ (x 2 ) 2 ◊ (x 2 ◊ x) ] )] ] β = [ (x 2 ) 2 ◊ (x 2 ◊ x) ] ◊ [ x 2 ◊ ( x ◊ [ (x 2 ) 2 ◊ (x 2 ◊ x) ] ) ] αβ 13 « x; ◊ » A2A2 A2A2 φ3φ3 φ f(3) f φ f(3) (α) ≠ φ f(3) (β) ?

24. Homomorphisms case: s = 2, t = 0 for δ ∈ « x; ◊ » define |δ| = number of x’s in δ « x; ◊ » A2A2 A0A0 φnφn ψ f(n) f for δ ∈ « x; ◊ », f○φ n (δ) = ψ f(n) (x |δ| ) = f(n) if |δ| ≡ 1 (mod 2) 0 otherwise A 0 is asociative in A 0, m 2 =0 for all m

25. Homomorphisms case: s = 2, t = 0 « x; ◊ » A2A2 A0A0 φ3φ3 ψ f(3) f set α= (x 2 ) 2 ◊ [ x ◊ ( x 3 ◊ (x 2 ) 2 ) ] β = x ◊ ( x 2 ◊ [ x ◊ ( x 3 ◊ (x 2 ) 2 ) ] ) then we have 0 = f○φ 3 (α) = f○φ 3 (β) = f(3) |α| = 12 φ 3 (α) = 9 = φ 3 (β) |β| = 11 since 3 generates A 2 and 0 is the identity in A 0, f is trivial

26. Homomorphisms Hom(A 0,A 0 ) = ∏ ≥0 A 0 Hom(A 0,A 1 ) = ∏ ≥0 Ζ/2Z Hom(A 1,A 0 ) = ∏ ≥0 A 0 Hom(A 1,A 1 ) = ∏ ≥1 Ζ/2Z [ [ Inj(A 0,A 0 ) £ {0,1} N ] Theorem (A. and Cowen-Morton)

27. Homomorphisms Each element 2 i in A 0 (i≥0) generates a subgroup H i isomorphic to Z/2Z. A 0 is the weak product of the H i since its operation is bitwise XOR. Each element 2 i in A 1 (i≥1) generates a subgroup G i = {2 i, 0, 2 i +1, 1} isomorphic to Z/4Z. A 1 is not the weak product of the G i but the G i stay out of each other’s way. behind the proof...

28. Homomorphisms Let Q1 denote the loop quotient of A 1 by the relation 0 ≡ 1. Let Q2 denote the loop quotient of A 1 by the relations {2k ≡ 2k+1 | k = 1, 2,...}. Let Q3 denote the loop quotient of A 1 by all relations enforcing associativity. Then each of these quotients is isomorphic to A 0 under an isom’m sending G i to H i-1 for each i, for which all three quotient maps are the same. Theorem (A. and Cowen-Morton) behind the proof...

29. Homomorphisms Hom(A 0,A 0 ) = ∏ ≥0 A 0 Hom(A 0,A 1 ) = ∏ ≥0 Ζ/2Z Hom(A 1,A 0 ) = ∏ ≥0 A 0 Hom(A 1,A 1 ) = ∏ ≥1 Ζ/2Z [ [ Inj(A 0,A 0 ) £ {0,1} N ] Theorem (A. and Cowen-Morton) ◄