Algorithmic Tools for the Asymptotics of Diagonals

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1 Algorithmic Tools for the Asymptotics of Diagonals Bruno Salvy Inria & ENS de Lyon Lattice walks at the Interface of Algebra, Analysis and Combinatorics September 19, 2017

2 Asymptotics & Univariate Generating Functions counts the number of objects of size n (a n ) 7! A(z) := X n 0 a n z n captures some structure (a n ) P-recursive A(z) D-finite p 0 (n)a n+k + + p k (n)a n =0 q 0 (z)a (`) (z)+ + q`(z)a(z) =0 1. Possible exponential growth (a n n, 6= 0) root of the characteristic polynomial of the leading coeff of the recurrence wrt n q 0 (1/ ) =0 2. Possible sub-exponential growth (a n c n (n), (n + 1) (n)! 1) a basis can be computed and then c approximated Qu.: How can we get,,c? and how fast? 1/25

3 Univariate Generating Functions D-FINITE Aim: asymptotics of the coefficients, automatically. DIAGONAL ALGEBRAIC RATIONAL Def diagonal: later. 2/25

4 I. A Quick Review of Analytic Combinatorics in One Variable Principle: Dominant singularity exponential behaviour local behaviour subexponential terms

5 Coefficients of Rational Functions F 1 =1= 1 2 i I 1 dz 1 z z 2 z 2 a n = 1 2 i Z f(z) z n+1 dz = = As n increases, the smallest singularities dominate. F n = n n /25

6 Conway s sequence 1,11,21,1211,111221, Generating function for lengths: f(z)=p(z)/q(z) with deg Q=72. Smallest singularity: δ(f) ρ=1/δ(f) l n ρ n Fast univariate resolution: Sagraloff-Mehlhorn16 ρ Res(f,δ(f)) remainder exponentially small 4/25

7 Singularity Analysis Ex: Rational Functions A 3-Step Method: 1. Locate dominant singularities a. singularities; b. dominant ones 2. Compute local behaviour 3. Translate into asymptotics (1 z) log k 1 1 z 7! n 1 ( ) logk n, ( 62 N? ) 17z 3 9z 2 7z Numerical resolution with sufficient precision + algebraic manipulations 2. Local expansion (easy). 3. Easy. Useful property [Pringsheim Borel] a n 0 for all n real positive dominant singularity. 5/25

8 Algebraic Generating Functions P (z, y(z)) = 0 1a. Location of possible singularities Implicit Function P (z, y(z)) = (z, y(z)) = Numerical resolution with sufficient precision + algebraic manipulations 1b. Analytic continuation finds the dominant ones: not so easy [FlSe NoteVII.36]. 2. Local behaviour (Puiseux): (1 z), ( 2 Q) 3. Translation: easy. 6/25

9 Differentially-Finite Generating Functions a n (z)y (n) (z)+ + a 0 (z)y(z) =0, a i polynomials 1a. Location of possible singularities. Cauchy-Lipshitz Theorem: 1b. Analytic continuation finds the dominant ones: only numerical in general. Sage code exists [Mezzarobba2016]. 2. Local behaviour at regular singular points: 3. Translation: easy. a n (z) =0 (1 z) log k 1, ( 2 Q,k 2 N) 1 z Numerical resolution with sufficient precision + algebraic manipulations 7/25

10 Example: Apéry s Sequences a n = nx k=0 2 2 n n + k X n, b n = a n k k k=1 1 n k 3 + X k=1 kx m=1 ( 1) m n 2m 3 k n m 2 n+k k n+m m 2 and c n = b n (3)a n have generating functions that satisfy vanishes at 0, = p = p 2 ' 0.03, 2 ' 34. z 2 (z 2 34z + 1)y (z 5)y =0 In the neighborhood of, all solutions behave like analytic µ p z(1 + ( z)analytic). Mezzarobba s code gives µ a ' 4.55, µ b ' 5.46, µ c ' 0. Slightly more work gives µ c =0, then c n and eventually, a proof that (3) is irrational. n [Apéry1978] 8/25

11 II. Diagonals

12 If F (z) = G(z) H(z) its diagonal is Definition is a multivariate rational function with Taylor expansion F (z) = X i2n n c i z i, F (t) = X k2n c k,k,...,k t k. in this talk 1 k +1 2k k 2k k : : 1 1 x y =1+x + y +2xy + x2 + y x 2 y x (1 x y)(1 x) =1+y+1xy x2 +y x 2 y 2 + Apéry s a k : 1 1 t(1 + x)(1 + y)(1 + z)(1 + y + z + yz + xyz) =1+ +5xyzt + 9/25

13 Diagonals & Multiple Binomial Sums Ex. S n = X r 0 X n ( 1) n+r+s r s 0 n s n + s n + r 2n r s s r n Thm. Diagonals binomial sums with 1 free index. > BinomSums[sumtores](S,u): ( ) defined properly 1 1 t(1 + u 1 )(1 + u 2 )(1 u 1 u 3 )(1 u 2 u 3 ) has for diagonal the generating function of S n [Bostan-Lairez-S.17] 10/25

14 Multiple Binomial Sums over a field K Sequences constructed from - Kronecker s : n 7! n ; - geometric sequences n 7! C n, C 2 K; - the binomial sequence (n, k) 7! n k ; using algebra operations and - affine changes of indices (u n ) 7! (u (n) ); - indefinite summation (u n,k ) 7! mx k=0 u n,k!. 10 /25

15 Diagonals are Differentially Finite [Christol84,Lipshitz88] a n (z)y (n) (z)+ + a 0 (z)y(z) =0, Thm. If F has degree d in n variables, ΔF satisfies a LDE with order d n, coeffs of degree d O(n). + algo in ops. Õ(d 8n ) ad-finite i diag. alg. rat. Compares well with creative telescoping when both apply. asymptotics from that LDE Christol s conjecture: All differentially finite power series with integer coefficients and radius of convergence in (0, ) are diagonals. [Bostan-Lairez-S.13,Lairez16] 11/25

16 Asymptotics Thm. [Katz70,André00,Garoufalidis09] a 0 +a 1 z+ D-finite, a i in z, radius in (0, ), then its singular points are regular with rational exponents X a n n n log k 1 (n) f,,k. n (,,k)2 finite set in Q Q N Ex. The number a n of walks from the origin taking n steps {N,S,E,W,NW} and staying in the first quadrant behaves like C n n with 62 Q not D-finite. = 1+ [Bostan-Raschel-S.14] arccos(u), 8u3 8u 2 +6u 1=0, u > 0. 12/25

17 Bivariate Diagonals are Algebraic [Pólya21,Furstenberg67] D-finite diag. alg. = diag 2 [Bostan-Dumont-S.17] rat. Thm. F=A(x,y)/B(x,y), deg d in x and y, then ΔF cancels a polynomial of degree 4 d in y and t. x satisfies 1 x 2 y t y t y t t y t y 6 t t t y t t y 4 54 t t y t t y 2 t 3 50 t t t y t t =0 + quasi-optimal algorithm. the differential equation is often better. 13/25

18 III. Analytic Combinatorics in Several Variables, with Computer Algebra Here, we restrict to rational diagonals and simple cases

19 Starting Point: Cauchy s Formula If f = X i 1,...,i n 0 c i1,...,i z i 1 n 1 z i n n neighborhood of 0, then is convergent in the c i1,...,i n = 1 2 i n Z T f(z 1,...,z n ) dz 1 dz n z i 1+1 z i n+1 1 n for any sufficiently small torus T ( z j = re i j ) around 0. Asymptotics: deform the torus to pass where the integral concentrates asymptotically. 14/25

20 Coefficients of Diagonals F (z) = G(z) H(z) c k,...,k = 1 2 i n Z T G(z) H(z) dz 1 dz n (z 1 z n ) k+1 rank Critical points: minimize z 1 z n on V = {z @z 1 z n 1 z n n Minimal ones: on the boundary of the domain of convergence of F (z).! apple 1 i.e. 1 = = n A 3-step method 1a. locate the critical points (algebraic condition); 1b. find the minimal ones (semi-algebraic condition); 2. translate (easy in simple cases). 15/25

21 Ex.: Central Binomial Coefficients 2k k : 1 1 x y =1+x + y +2xy + x2 + y x 2 y 2 + (1). Critical points: 1 x y =0,x= y =) x = y =1/2. (2). Minimal ones. Easy. In general, this is the difficult step. (3). Analysis close to the minimal critical point: a k = 1 ZZ (2 i) 2 Z 4k+1 2 i 1 1 x y dx dy (xy) k i exp(4(k + 1)(x 1/2) 2 ) dx 4k p k. Z dx (x(1 x)) k+1 residue saddle-point approx 16/25

22 Kronecker Representation for the Critical Points Algebraic part: ``compute the solutions of the system H(z) =0 1 = = n If Under genericity assumptions, a probabilistic algorithm running in bit ops finds: History and Background: see Castro, Pardo, Hägele, and Morais (2001) [Giusti-Lecerf-S.01;Schost02;SafeySchost16] System reduced to a univariate polynomial. 17/25

23 Example (Lattice Path Model) The number of walks from the origin taking steps {NW,NE,SE,SW} and staying in the first quadrant is F, F (x, y, t) = (1 + x)(1 + y) 1 t(1 + x 2 + y 2 + x 2 y 2 ) Kronecker representation of the critical points: P (u) =4u u u u Q x (u) = 336u u Q y (u) = 160u u Q t (u) =4u u u/ /2 ie, they are given by: P (u) =0, x = Q x(u) P 0 (u), y = Q y(u) P 0 (u), t = Q t(u) P 0 (u) Which one of these 4 is minimal? 18/25

24 Testing Minimality Def. F(z 1,,z n ) is combinatorial if every coefficient is 0. Prop. [PemantleWilson] In the combinatorial case, one of the minimal critical points has positive real coordinates. Thus, we add the equation for a new variable and select the positive real point(s) z with no from a new Kronecker representation: P (v) =0 P 0 (v)z 1 Q 1 (v) =0 P 0 (v)z n Q n (v) =0 P 0 (v)t Q t (v) =0.. This is done numerically, with enough precision. 19/25

25 Example F = 1 H = 1 (1 x y)(20 x 40y) 1 Critical point : x(2x + 41y 21) = y(41x + 80y 60) 4 critical points, 2 of which are real: (x 1,y 1 )=(0.2528, ), (x 2,y 2 )=( , ) Add H(tx, ty) =0and compute a Kronecker representation: P (u) =0, x = Q x(u) P 0 (u), y = Q y(u) P 0 (u), Solve numerically and keep the real positive sols: t = Q t(u) P 0 (u) (0.31, 0.55, 0.99), (0.31, 0.55, 1.71), (0.25, 9.99, 0.09), (0.25, 0.99, 0.99) (x 1,y 1 ) is not minimal, (x 2,y 2 ) is. 20/25

26 Algorithm and Complexity Thm. If F (z) is combinatorial, then under regularity conditions, the points contributing to dominant diagonal asymptotics can be determined in Õ(hd 5 D 4 ) bit operations. Each contribution has the form A k = T k k (1 n)/2 (2 ) (1 n)/2 (C + O(1/k)) T, C can be found to 2 apple precision in Õ(h(dD) 3 + Dapple) bit ops. explicit algebraic number This result covers the easiest cases. All conditions hold generically and can be checked within the same complexity, except combinatoriality. [Melczer-S.16] 21/25

27 Example: Apéry's sequence 1 1 t(1 + x)(1 + y)(1 + z)(1 + y + z + yz + xyz) =1+ +5xyzt + Kronecker representation of the critical points: x = 2u 1006 P 0 (u) P (u) =u 2 366u u , y = z = P 0, t = (u) P 0 (u) There are two real critical points, and one is positive. After testing minimality, one has proven asymptotics 22/25

28 Example: Restricted Words in Factors F (x, y) = 1 x 3 y 6 + x 3 y 4 + x 2 y 4 + x 2 y 3 1 x y + x 2 y 3 x 3 y 3 x 4 y 4 x 3 y 6 + x 4 y 6 words over {0,1} without or /25

29 Minimal Critical Points in the Noncombinatorial Case Then we use even more variables and equations: H(z) =0 1 = = n H(u) =0 u 1 2 = t z 1 2,..., u n 2 = t z n 2 + critical point equations for the projection on the t axis And check that there is no solution with t in (0,1). Prop. Under regularity assumptions, this can be done in Õ(hd 4 2 3n D 9 ) bit operations. 24/25

30 Summary & Conclusion D-finite diag. alg. rat. Diagonals are a nice and important class of generating functions for which we now have many good algorithms. ACSV can be made effective (at least in simple cases). Requires nice semi-numerical Computer Algebra algorithms. Without computer algebra, these computations are difficult. Complexity issues become clearer. Work in progress: extend beyond some of the assumptions (see Melczer s thesis). The End