A generating function method for impartial games

Transcription

1 A generating function method for impartial games Alex Fink Queen Mary University of London Joint IMU / AMS meeting, Tel Aviv June 2014 Alex Fink Generating functions 1 / 14

2 My aim The purpose of this talk is to ask a question. Question Can we solve any new impartial games via generating functions? Or if not solve them, at least prove more classes periodic, etc.? In this talk I ll show you several games with well-known solutions which a generating function approach does solve. I won t even cheat much! Alex Fink Generating functions 2 / 14

3 Generating functions Generating functions box up combinatorial sequences in a single object. e.g.: i 0 Cat i t i = 1 + t + 2t 2 + 5t t 4 + = 1 1 4t. 2t There is a great toolbox of powerful methods for them. My topic is the generating function for the outcome sequence of a ruleset G, i.e. for the P-positions: p G (t) = δ outcome(i),p t i = t i. i i is P Wanted: an equation that can be solved for p G (t). Alex Fink Generating functions 3 / 14

4 Generating functions Generating functions box up combinatorial sequences in a single object. e.g.: i 0 Cat i t i = 1 + t + 2t 2 + 5t t 4 + = 1 1 4t. 2t There is a great toolbox of powerful methods for them. My topic is the generating function for the outcome sequence of a ruleset G, i.e. for the P-positions: p G (t) = δ outcome(i),p t i = t i. i i is P Wanted: an equation that can be solved for p G (t). Alex Fink Generating functions 3 / 14

5 The functional equation Write the generating function q G (t) for sources of moves to P-positions in terms of p G (t). Examples Subtraction games If the subtraction set is {a, b,...}, then q G (t) = p G (t) (t a + t b + ). Other heap games 2-Mark (Fraenkel): q 2 Mark (t) = tp 2 Mark (t) + (1 + t)p 2 Mark (t 2 ). Then their coefficients are related: { [t i 1 [t i ]q G (t) = 0 ]p G (t) = 0 otherwise Obstacle This isn t an algebraic operation on series! Alex Fink Generating functions 4 / 14

6 The functional equation Write the generating function q G (t) for sources of moves to P-positions in terms of p G (t). Examples Subtraction games If the subtraction set is {a, b,...}, then q G (t) = p G (t) (t a + t b + ). Other heap games 2-Mark (Fraenkel): q 2 Mark (t) = tp 2 Mark (t) + (1 + t)p 2 Mark (t 2 ). Then their coefficients are related: { [t i 1 [t i ]q G (t) = 0 ]p G (t) = 0 otherwise Obstacle This isn t an algebraic operation on series! Alex Fink Generating functions 4 / 14

7 One good move, if any If there is at most one P option from any position, then ( ) p G (t) = t i q G (t) and we can solve for p G (t). Example: Subtraction games with subtraction set i {k, 2k,..., nk} (Any two options have a move between them.) (including n = ). p G = 1 1 t q G = 1 1 t (tk + +t nk )p G = 1 1 t tk t (n+1)k 1 t k p G solves to just as it should. p G = 1 t k (1 t)(1 t (n+1)k ) Alex Fink Generating functions 5 / 14

8 One good move, if any If there is at most one P option from any position, then ( ) p G (t) = t i q G (t) and we can solve for p G (t). Example: Subtraction games with subtraction set i {k, 2k,..., nk} (Any two options have a move between them.) (including n = ). p G = 1 1 t q G = 1 1 t (tk + +t nk )p G = 1 1 t tk t (n+1)k 1 t k p G solves to just as it should. p G = 1 t k (1 t)(1 t (n+1)k ) Alex Fink Generating functions 5 / 14

9 Working mod 2 Few games have only one good move from each N -position. But several exemplary games have an odd number. In this case, ( ) q G (t) t i p G (t) (mod 2) and we can still solve for p G (t) over Z/2Z. Example: Nim, with any number of heaps. i A heap of h has a good move h 1 : iff the leading bit set in the h 2 : nim-sum is also set in h. h 3 : There are oddly many such h. h 4 : (Cheating! But... ) nim-sum: Alex Fink Generating functions 6 / 14

10 Working mod 2 Few games have only one good move from each N -position. But several exemplary games have an odd number. In this case, ( ) q G (t) t i p G (t) (mod 2) and we can still solve for p G (t) over Z/2Z. Example: Nim, with any number of heaps. i A heap of h has a good move h 1 : iff the leading bit set in the h 2 : nim-sum is also set in h. h 3 : There are oddly many such h. h 4 : (Cheating! But... ) nim-sum: Alex Fink Generating functions 6 / 14

11 Solving Nim Our generating function is now multivariate: t (h 1,...,h n) := t h 1 1 thn n So p Nim i = i 1 q Nim 1 t i ( ) 1 t i p Nim 1 t i 1 t i i which solves, modulo 2, to p Nim each i k is 0 or 1 an even number are 1 1 tn in t i 1 1. Alex Fink Generating functions 7 / 14

12 Tips and tricks For misère solutions, add a fake good move from each end position in q G. e.g. q G = (1 + + x k 1 ) +(t k + + t nk )p }{{} G. fake An outcome generating function for G + one-heap Nim is a G-value generating function for G. Negative positions! Generalising beyond monomials! Being indecisive! Alex Fink Generating functions 8 / 14

13 Tips and tricks For misère solutions, add a fake good move from each end position in q G. e.g. q G = (1 + + x k 1 ) +(t k + + t nk )p }{{} G. fake An outcome generating function for G + one-heap Nim is a G-value generating function for G. Negative positions! Generalising beyond monomials! Being indecisive! Alex Fink Generating functions 8 / 14

14 Negative positions Example: 3-heap Lim. (Take a beans from two heaps, add a to the third.) Lim has the oddly many P options property. (Why?!) We take ( ) 1 q Lim = + cyclic p Lim. 1 t 1 t 2 /t 3 But then s contains negative exponents! So we alter the other relation: p Lim = 1 + t 1 + t 2 + t 3 (1 t 1 t 2 /t 3 )(cyclic) q Lim and have to verify that the fake positions have oddly many P options. (Everything works out.) Question Does (2 n 1)-heap Lim have the oddly many P options property? Alex Fink Generating functions 9 / 14

15 Negative positions Example: 3-heap Lim. (Take a beans from two heaps, add a to the third.) Lim has the oddly many P options property. (Why?!) We take ( ) 1 q Lim = + cyclic p Lim. 1 t 1 t 2 /t 3 But then s contains negative exponents! So we alter the other relation: p Lim = 1 + t 1 + t 2 + t 3 (1 t 1 t 2 /t 3 )(cyclic) q Lim and have to verify that the fake positions have oddly many P options. (Everything works out.) Question Does (2 n 1)-heap Lim have the oddly many P options property? Alex Fink Generating functions 9 / 14

16 Welter s game Example: Welter s game has the oddly many P options property. [ONAG, Thm. 82] (still cheating!) Positions in Welter s game can be encoded as Young diagrams of partitions: step up for full cells, right for empty cells. A move is deleting a ribbon. Alex Fink Generating functions 10 / 14

17 Welter s game Example: Welter s game has the oddly many P options property. [ONAG, Thm. 82] (still cheating!) Positions in Welter s game can be encoded as Young diagrams of partitions: step up for full cells, right for empty cells. A move is deleting a ribbon. Alex Fink Generating functions 10 / 14

18 Welter s game is a subtraction game The Schur functions form a basis for the ring of symmetric functions (in ly many vars), indexed by partitions. s λ (t 1,..., t n ) = det(t λ i +n i j )/ det(tj n i ) Proposition (Murnaghan-Nakayama rule) (t n 1 + t n 2 + ) s λ = µ is λ plus a ribbon ±s µ. Using Schurs in the Welter g.f. gives a relation looking exactly like Nim s: ( ) t i q Welter = p Welter. 1 t i But the solution gets messier from there... Question Are there other well-behaved games from invariant theory? i Alex Fink Generating functions 11 / 14

19 Welter s game is a subtraction game The Schur functions form a basis for the ring of symmetric functions (in ly many vars), indexed by partitions. s λ (t 1,..., t n ) = det(t λ i +n i j )/ det(tj n i ) Proposition (Murnaghan-Nakayama rule) (t n 1 + t n 2 + ) s λ = µ is λ plus a ribbon ±s µ. Using Schurs in the Welter g.f. gives a relation looking exactly like Nim s: ( ) t i q Welter = p Welter. 1 t i But the solution gets messier from there... Question Are there other well-behaved games from invariant theory? i Alex Fink Generating functions 11 / 14

20 Welter s game is a subtraction game The Schur functions form a basis for the ring of symmetric functions (in ly many vars), indexed by partitions. s λ (t 1,..., t n ) = det(t λ i +n i j )/ det(tj n i ) Proposition (Murnaghan-Nakayama rule) (t n 1 + t n 2 + ) s λ = µ is λ plus a ribbon ±s µ. Using Schurs in the Welter g.f. gives a relation looking exactly like Nim s: ( ) t i q Welter = p Welter. 1 t i But the solution gets messier from there... Question Are there other well-behaved games from invariant theory? i Alex Fink Generating functions 11 / 14

21 The indecision transform Let G be an impartial ruleset. In indecisive G, I can leave some of my choices to you: positions are nonempty sets of G-positions; the options of {G i } are the subsets of i options(g i). Good news Every N -position in indecisive G has (2 a 1) P options. Bad news The positions don t usually fit in a generating function. Alex Fink Generating functions 12 / 14

22 A case where indecision is manageable Example. Subtraction games with subtraction set an arithmetic progression {k, k + r,..., k + nr}. Which indecisive positions do we care about? Certainly singletons {a}. The indecisive positions {a, a + 2r} and {a, a + r, a + 2r} have the same options, = the same outcome. So if both appear as options in some position, we can cancel them. In fact all options of singletons cancel in pairs, other than {a} and {a, a + r}. And the system of two linear equations for ( ) i : {i} is P p G = i ti : {i, i + r} is P ti is solvable. Alex Fink Generating functions 13 / 14

23 A case where indecision is manageable Example. Subtraction games with subtraction set an arithmetic progression {k, k + r,..., k + nr}. Which indecisive positions do we care about? Certainly singletons {a}. The indecisive positions {a, a + 2r} and {a, a + r, a + 2r} have the same options, = the same outcome. So if both appear as options in some position, we can cancel them. In fact all options of singletons cancel in pairs, other than {a} and {a, a + r}. And the system of two linear equations for ( ) i : {i} is P p G = i ti : {i, i + r} is P ti is solvable. Alex Fink Generating functions 13 / 14

24 A case where indecision is manageable Example. Subtraction games with subtraction set an arithmetic progression {k, k + r,..., k + nr}. Which indecisive positions do we care about? Certainly singletons {a}. The indecisive positions {a, a + 2r} and {a, a + r, a + 2r} have the same options, = the same outcome. So if both appear as options in some position, we can cancel them. In fact all options of singletons cancel in pairs, other than {a} and {a, a + r}. And the system of two linear equations for ( ) i : {i} is P p G = i ti : {i, i + r} is P ti is solvable. Alex Fink Generating functions 13 / 14

25 Closing thoughts If we don t have the oddly many P options property, can we solve for p G with integral equations? E.g. if p(t) = w i t i then w 2 i t i = 1 0 p(te 2πix )p(e 2πix ) dx. Don t get too optimistic: lattice games exhibit universal computation! Question Can we solve any new impartial games via generating functions? And what s up with the oddly many P options property? Thank you! Alex Fink Generating functions 14 / 14

26 Closing thoughts If we don t have the oddly many P options property, can we solve for p G with integral equations? E.g. if p(t) = w i t i then w 2 i t i = 1 0 p(te 2πix )p(e 2πix ) dx. Don t get too optimistic: lattice games exhibit universal computation! Question Can we solve any new impartial games via generating functions? And what s up with the oddly many P options property? Thank you! Alex Fink Generating functions 14 / 14

27 Closing thoughts If we don t have the oddly many P options property, can we solve for p G with integral equations? E.g. if p(t) = w i t i then w 2 i t i = 1 0 p(te 2πix )p(e 2πix ) dx. Don t get too optimistic: lattice games exhibit universal computation! Question Can we solve any new impartial games via generating functions? And what s up with the oddly many P options property? Thank you! Alex Fink Generating functions 14 / 14