Atomic Weight Calculus of Subversion

Transcription

1 , West Chester University, with N. McKay, R. J. Nowakowski, P. Ottaway, and C. Santos January 26, 2017

2 Table of contents The game

3 The game Subversion is a partizan self-referential subtraction game played on pairs of non-negative integers, (a, b), dictated by the following rules:

4 The game Subversion is a partizan self-referential subtraction game played on pairs of non-negative integers, (a, b), dictated by the following rules: If a = 0 or b = 0, then the position (a, b) is terminal.

5 The game Subversion is a partizan self-referential subtraction game played on pairs of non-negative integers, (a, b), dictated by the following rules: If a = 0 or b = 0, then the position (a, b) is terminal. If a > 0 and b > 0, then

6 The game Subversion is a partizan self-referential subtraction game played on pairs of non-negative integers, (a, b), dictated by the following rules: If a = 0 or b = 0, then the position (a, b) is terminal. If a > 0 and b > 0, then if a b, then Left can move to (a, 0); if a < b, then Left can move to (a, b a);

7 The game Subversion is a partizan self-referential subtraction game played on pairs of non-negative integers, (a, b), dictated by the following rules: If a = 0 or b = 0, then the position (a, b) is terminal. If a > 0 and b > 0, then if a b, then Left can move to (a, 0); if a < b, then Left can move to (a, b a); if b a, then Right can move to (0, b); if a > b, then Right can move to (a b, b).

8 Consider (25, 9): (25, 9) (25, 0) (16, 9) (16, 0) (7, 9) (7, 2) (0, 9) (7, 0) (5, 2) (5, 0) (3, 2) (3, 0) (1, 2) (1, 1) (0, 2) (1, 0) (0, 1)

9 Notice that 25 9 has continued fraction representation Using the short-hand notation, we write [2, 1, 3, 2].

10 Notice that 25 9 has continued fraction representation Using the short-hand notation, we write [2, 1, 3, 2]. (25, 9) (25, 0) (16, 9) (16, 0) (7, 9) (7, 2) (0, 9) (7, 0) (5, 2) (5, 0) (3, 2) (3, 0) (1, 2) (1, 1) (0, 2) (1, 0) (0, 1)

11 Adding negatives for left-hand branching, we utilize the more descriptive notation: (25, 9) = [2, 1, 3, 2]. (25, 9) (25, 0) (16, 9) (16, 0) (7, 9) (7, 2) (0, 9) (7, 0) (5, 2) (5, 0) (3, 2) (3, 0) (1, 2) (1, 1) (0, 2) (1, 0) (0, 1)

12 The Subversion position (25, 9) has value {0 0 0}. (25, 9) (25, 0) (16, 9) (16, 0) (7, 9) (7, 2) (0, 9) (7, 0) (5, 2) (5, 0) (3, 2) (3, 0) (1, 2) (1, 1) (0, 2) (1, 0) (0, 1)

13 Hackenbush variation The Subversion position (25, 9) = [2, 1, 3, 2] can be thought of as the following Hackenbush variation.

14 The class of Subversion forms is constructed as follows: 0 = { } is the only Subversion game born on day 0; The Subversion games born on day n + 1 have the possible forms {0 G R } or {H L 0} where G R and H L are Subversion games born on day n.

15 An impartial self-referential subtraction game is a subtraction game where the sizes of the (non-empty) piles of tokens in a given position form the subtraction set. The subtraction set is dynamic, changing with every move.

16 An impartial self-referential subtraction game is a subtraction game where the sizes of the (non-empty) piles of tokens in a given position form the subtraction set. The subtraction set is dynamic, changing with every move. Impartial Susen is an example of such a game; from a position of heaps with sizes 1, 3, and 4, one can remove 1, 3, or 4 tokens. For instance, removing one token from heap two leaves the position {1, 2, 4}. The new subtraction set is now {1, 2, 4}.

17 In Partizan Susen, we are given two sets S L = {a 1, a 2,... } and S R = {b 1, b 2,... }. For any i, Left can remove a i tokens from any b j provided that a i b j. Right s moves are similar.

18 In Partizan Susen, we are given two sets S L = {a 1, a 2,... } and S R = {b 1, b 2,... }. For any i, Left can remove a i tokens from any b j provided that a i b j. Right s moves are similar. R. J. Nowakowski proposed the ruleset Subversion as a variant of Partizan Susen while he was trying to generate many self-referential games.

19 Recall that if G is an infinitesimal, then n < G < n for some sufficiently large n. Hence, one can use the sequence,, 3,... as a scale to calibrate the size of any infinitesimal. The atomic weight of G describes precisely where G should be placed on this scale.

20 Remote stars Consider the following flower garden.

21 Remote stars In the presence of a sufficiently remote nimber ( 5 is the smallest such nimber here), the position becomes a Left win. This position has atomic weight 1.

22 Remote stars As it turns out, if k is remote for a given position, then so is any m, where m k. It is convenient to adopt the symbol as shorthand for a sufficiently remote nimber. We generally refer to as a remote star (or far star).

23 criterion aw(g) = g if and only if < G g <.

24 Two ahead rule 1. If aw(g) 2, then G > If aw(g) 1, then G If aw(g) 2, then G < If aw(g) 1, then G 0.

25 calculus Let G be dicotic. Consider the form {aw(g L ) 2 aw(g R ) + 2} and let w(g) be w(g) = {aw(g L ) 2 aw(g R ) + 2} if the form is not a integer. min(i ) if G < w(g) = max(i ) if G > if the form is a integer. 0 otherwise where I = {x Z aw(g L ) 2 x aw(g R ) + 2}. Then, aw(g) = w(g).

26 Why not just work out canonical forms for Subversion positions?

27 Why not just work out canonical forms for Subversion positions? Consider a disjunctive sum like (211, 155) + (5, 4) + (13, 17).

28 Why not just work out canonical forms for Subversion positions? Consider a disjunctive sum like (211, 155) + (5, 4) + (13, 17). After simplifying this position with help of CGSuite, we get the following interesting form:

29 Why not just work out canonical forms for Subversion positions? Consider a disjunctive sum like (211, 155) + (5, 4) + (13, 17). After simplifying this position with help of CGSuite, we get the following interesting form: {{ , {,, 3 3 6} 3}, { { { , { }} }} {{ { , { }} }, {,, 3 3 6} 3 {,, {,, 3 3 6} 3 3, {, {,, 3 3 6} 3} 3} 0 2 }, {, {,, 3 3 6} 3, {,, 3 3 6} 3 {,, {,, 3 3 6} 3 3, {, {,, 3 3 6} 3} 3} 0 2 }}, where {0 k G} means {0 {0 k 1 G}}.

30 Before we present our main result, we need just a tiny bit more preparation.

31 Before we present our main result, we need just a tiny bit more preparation. We say that G has adorned atomic weight 0 L (written as aw(g) = 0 L ) if aw(g) = 0, aw({0 G}) = 1, and {G 0} =.

32 Before we present our main result, we need just a tiny bit more preparation. We say that G has adorned atomic weight 0 L (written as aw(g) = 0 L ) if aw(g) = 0, aw({0 G}) = 1, and {G 0} =. Similarly, we say that G has adorned atomic weight 0 R (written as aw(g) = 0 R ) if aw(g) = 0, aw({g 0}) = 1, and {0 G} =.

33 The Subversion position (9, 4) is an example of a position with atomic weight 0 L

34 Main result Theorem If G is a Subversion position, then the following table can be used to compute the atomic weight of {0 G}. aw(g) k { 2 k} 1 0 R If G = then {0 G} = aw({0 G}) { 2 k + 2} L 0 L If G = {G L 0} then 1 {0 G} = If G = {0 G R } then 0 L aw(g) G = 0 G = 0 L {k 2} k 4 aw({0 G}) {0 G} = {0 G} = k + 2 k + 1

35 Main result Theorem (Mirror image version) If G is a Subversion position, then the following table can be used to compute the atomic weight of {G 0}. aw(g) k { 2 k} 1 0 R aw({g 0}) k k aw(g) G = 0 G = 0 L 1 If G = then {G 0} = aw({g 0}) {G 0} = If G = {G L 0} then 0 R 0 R 1 If G = {0 G R } then {k 2} k R {k 2 2}

36 Returning to the Subversion position (9, 4) we see how to apply the theorem 0 L 0 3/

37 Algorithm for computing aw(a, b) Consider E, the possible entries of the table shown in theorem. The classification of the turning points motivates the partitions E = L 1 L 2 L 3 = R 1 R 2 R 3, where L 1 = {..., 3, 2, 1}, L 2 = { 3 2, { 2 k},, 0, 0R }, L 3 = {0 L,, 1, 3 2, 2,..., {k 2}}, and R 1 = {1, 2, 3,...}, R 2 = { 3 2, {k 2},, 0, 0L }, R 3 = {{ 2 k},..., 2, 3 2, 1,, 0R }.

38 Algorithm for computing aw(a, b) Two functions which we ll need are the ceiling function g = min n Z {n g} and the floor function g = max n Z {n g}. Note that for the special case when g =, we define = = 0. We will also need the following two functions when n 2: ψ L (n, g) = adorning 0 L if the result is 0 and n 2 if g L 1 n 1 if g L 2, n + g if g L 3

39 Algorithm for computing aw(a, b) ψ R (n, g) = adorning 0 R if the result is 0. 2 n if g R 1 1 n if g R 2, g n if g R 3

40 Algorithm for computing aw(a, b) Finally, consider the class of objects Υ = { [n 1, n 2, n 3, n 4,...] g n i Z + and g E } and the function ξ Υ Υ defined by the following rules 1. ξ([ ] g) = [ ] g ξ([n 1, n 2,..., n k ] g) = [n 1,..., n k 1 ] aw({0 G}) if n k = 1 [n 1,..., n k 1 ] if n k = 2 and (g = 0 or g = 0 R or g = ) [n 1,..., n k 1 ] ψ L (n k, g) otherwise [n 1,..., n k 1 ] aw({g 0}) if n k = 1 ξ([n 1, n 2,..., n k ] g) = [n 1,..., n k 1 ] if n k = 2 and (g = 0 or g = 0 L or g = ) [n 1,..., n k 1 ] ψ R (n k, g) otherwise (Note that means to use the table in the main theorem.)

41 Example Let s first compute the atomic weight of (25, 9). Recall that the signed continued fraction representation is [2, 1, 3, 2]. Using the algorithm presented in the preceding slides, we perform the following calculations [2, 1, 3, 2] 0 ξ [2, 1, 3] ξ [2, 1] ψ L (3, ) = [2, 1] 2 ξ [2] 1 ξ [ ] ψ L (2, 1) = [ ] 3

42 Example Let s first compute the atomic weight of (25, 9). Recall that the signed continued fraction representation is [2, 1, 3, 2]. Using the algorithm presented in the preceding slides, we perform the following calculations [2, 1, 3, 2] 0 ξ [2, 1, 3] ξ [2, 1] ψ L (3, ) = [2, 1] 2 Hence, aw(25, 9) = 3. ξ [2] 1 ξ [ ] ψ L (2, 1) = [ ] 3

43 Example Next, we compute the atomic weight of (37, 24). The signed continued fraction representation is [1, 1, 1, 5, 2]. Using the algorithm presented in the preceding slides, we perform the following calculations [1, 1, 1, 5, 2] 0 ξ [1, 1, 1, 5] ξ [1, 1, 1] ψ R (5, ) = [1, 1, 1] 4 ξ [1, 1] 2 ξ [1] 4 ξ [ ] 2

44 Example Next, we compute the atomic weight of (37, 24). The signed continued fraction representation is [1, 1, 1, 5, 2]. Using the algorithm presented in the preceding slides, we perform the following calculations [1, 1, 1, 5, 2] 0 ξ [1, 1, 1, 5] ξ [1, 1, 1] ψ R (5, ) = [1, 1, 1] 4 ξ [1, 1] 2 Hence, aw(37, 24) = 2. ξ [1] 4 ξ [ ] 2

45 Example Lastly, we compute the atomic weight of (2017, 26). The signed continued fraction representation is [77, 1, 1, 2, 1, 3]. Using the algorithm presented in the preceding slides, we perform the following calculations [77, 1, 1, 2, 1, 3] 0 ξ [77, 1, 1, 2, 1] ψ R (3, 0) = [77, 1, 1, 2, 1] 2 ξ [77, 1, 1, 2] 1 ξ [77, 1, 1] ψ R (2, 1) = [77, 1, 1] 3 ξ [77, 1] 3/2 = [ ] 75 ξ [77] 3 ξ [ ] ψ L (77, 3)

46 Example Lastly, we compute the atomic weight of (2017, 26). The signed continued fraction representation is [77, 1, 1, 2, 1, 3]. Using the algorithm presented in the preceding slides, we perform the following calculations [77, 1, 1, 2, 1, 3] 0 ξ [77, 1, 1, 2, 1] ψ R (3, 0) = [77, 1, 1, 2, 1] 2 ξ [77, 1, 1, 2] 1 ξ [77, 1, 1] ψ R (2, 1) = [77, 1, 1] 3 ξ [77, 1] 3/2 ξ [77] 3 = [ ] 75 Hence, aw(2017, 26) = 75. ξ [ ] ψ L (77, 3)

47 Given an all-small ruleset, is there any relationship between the number of eccentric cases in the atomic weight calculus being finite and the nim dimension of the ruleset?

48 Given an all-small ruleset, is there any relationship between the number of eccentric cases in the atomic weight calculus being finite and the nim dimension of the ruleset? Analyze the full version of Subversion. Positions in this generalization are ordered (n + m)-tuples, (p 1, p 2,..., p n ; q 1, q 2,..., q m ). A Left move is to choose some p i and subtract it from some q j, etc. Right s moves are similar.

49 The game The End