Algorithmic Tools for the Asymptotics of Linear Recurrences

Transcription

1 Algorithmic Tools for the Asymptotics of Linear Recurrences Bruno Salvy Inria & ENS de Lyon

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sha1_base64="bidnl+9cqlsvii6tjsd6rvkccse=">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</latexit> Problem p 0 (n)a n+k + + p k (n)a n =0, 0 62 p 0 (N), p i 2 Z[n] Aim: given a 0,...,a k 1, predict the behaviour of a n as n!1. Simplified version: ``compute, when they exist,,, m, c 6= 0 such that a n c n n log m n local (at ) 1/30

3 Singularity Analysis sequence (a n ) 7! A(z) := X n 0 a n z n Generating function captures some structure A 3-Step Method: 1. Locate dominant singularities a. singularities; b. dominant ones 2. Compute local behaviour 3. Translate into asymptotics a n = 1 2 i I A(z) dz zn+1 A(z) z! c [Flajolet-Odlyzko1990] 1 z log m 1 1 z a n full asymptotic expansion available Next: computation of,, m, c c n n 1 n!1 ( ) logm n ( 62 N) 3/30

4 P-recursivity & D-finiteness (a n ) 7! A(z) := X n 0 a n z n (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 Classical properties of LDEs: 1. singularities satisfy q 0 ( ) =0; 2. one can compute a basis of formal solutions at (regular) singular points, of the form convergent 1 z log m 1 1 z (1 + ), 2 Q, m 2 N. local (at ρ) More recently (M. Mezzarobba s ore_algebra_analytic): certified analytic continuation ( c numerically). Semi-decision 4/30

5 Asymptotics of P-recursive Combinatorial Sequences Thm. [Katz70,Chudnovsky85,André00] a 0 +a 1 z+ D-finite, a i integers, 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. 8/30

6 Ex: Pólya s 3D Random Walk Start from the origin in Z d ; move one step along one of the axes; repeat. What is the probability p d of returning to 0? Numerical approximation by analytic continuation: 1. u n := P(3D-walk returns to 0 in 2n steps) satisfies (2n+3)(2n+1)(n+1)u n 2(2n+3)(10n 2 +30n+23)u n+1 +36(n+2) 3 u n+2 =0 nx 2. a n := u k! c := 1 converges slowly (1 is a singularity) 1 p 3 3. k=0 Given a 0,a 1,a 2, MM s code produces 100 digits of 1 1 A(z) c 1 z + + c 2 p + + c 3 (1 + ) 1 z c = p [Watson39;Glasser-Zucker77;Koutschan et alii 13,16] c, c 2,c not accessible to the algorithms presented here. s.t. in 0.4 sec. 6/30

7 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] 7/30

8 Univariate Generating Functions Aim: asymptotics of the D-FINITE DIAGONAL coefficients, automatically. ALGEBRAIC RATIONAL More structure more complete algorithm (analytic continuation Def diagonal: later. easier) Christol s conjecture: All differentially finite power series with integer coefficients and radius of convergence in (0, ) are diagonals. 9/30

9 Pringsheim s Theorem (a n ) 7! A(z) := X n 0 a n z n Useful property [Pringsheim Borel] a n 0 for all n real positive dominant singularity. A couple of years after the publication of my Thesis, Edmund Landau wrote to me to say that the German mathematician Pringsheim thought he had discovered the theorem [ ] ``But, he added, ``of course, this result is yours. [ ] what would have happened if I had not been alive to receive his letter? He would not have been undeceived and the discovery would have been thought to be mine until other readers would have restored it, not to Pringsheim but to E. Borel. J. Hadamard (1954) 10/30

10 I. Rational Generating Functions (Linear Recurrences with Constant Coefficients)

11 Fibonacci Numbers 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 /30

12 Conway s sequence 1,11,21,1211,111221, Generating function for lengths: f(z)=p(z)/q(z) with deg Q=72. Smallest singularity: ρ l n ρ -n Fast univariate resolution: Sagraloff-Mehlhorn16 c=ρ -1 Res(f,ρ) algebraic [Conway 1987] remainder exponentially small 12/30

13 Singularity Analysis for Rational Functions Ex: 17z 3 9z 2 7z +8 dist 10-5 A 3-Step Method: 1. Locate dominant singularities a. singularities; b. dominant ones 2. Compute local behaviour 3. Translate into asymptotics? Prop. [Mahler64] Cor. Numerical resolution with sufficient precision finds the dominant roots in polynomial complexity Known to be tight only for n=2 & 3 13/30

14 II. Algebraic Generating Functions P (z,f(z)) = 0 with P (z,y) 2 Z[z,y] \{0}

15 Algebraic Generating Functions 1a. Location of possible singularities Implicit Function Theorem: P (z,y(z)) = 0 1b. Analytic continuation finds the dominant ones 2. Local behaviour (Puiseux): 3. Translation: easy: a n P (z,y(z)) = 0 c n n 1 n!1 ( ) (discriminant) with c, algebraic, rational. (1 z/ ), ( 2 Q) Numerical resolution with sufficient precision + algebraic manipulations 14/30

16 3-regular 2-connected Planar Graphs U =2G 3 +T +2U 2 = T (1 U) 3,T = z(1+b)3,b = G 3 + B 2 1+B +z B + 12 B2 define power series U(z),G 3 (z),t(z),b(z). The aim is to compute the asymptotic behaviour of [z n ]B(z). 1. Eliminating U,T,G 3 gives P = 16B 6 z z 2 (z z 1). 2. The discriminant has degree 20, but only one root in (0,1]:.102 root of 54z z z At z =, P has only 1 (double) real positive root: B( ) 4. Computing more terms gives B(z) =B( )+c Conclusion: z ± c 1 [z n ]B(z) 3c 3/2 z + 4 p n 5/2 n. with an explicit c Analytic continuation exploiting the combinatorial origin. 15/30

17 Singularity Analysis of Algebraic Series Prop. [Abel1827;Cockle1861;Harley1862;Tannery1875] Algebraic series are D-finite. Exact analytic continuation for singularity analysis via LDE: A. Compute a LDE starting from P; B. For all roots of disc(p), sorted by increasing modulus, 1. compute exactly the local branches; 2. match with numerical continuation (MM s code); 3. if a singular behaviour is encountered, return it. 16/30

18 III. Diagonals

19 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 + 17/30

20 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] 18/30

21 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!. 18 /30

22 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, ad-finite i diag. alg. rat. ΔF satisfies a LDE with order d n, coeffs of degree d O(n). + algo in ops. Õ(d 8n ) asymptotics from that LDE Starting point: F (z) = G(z) H(z) ) F = n 1 2 i 1 I F z 1,...,z n 1, t z 1 z n 1 dz1 dz n 1 z 1 z n 1. [Bostan-Lairez-S.13,Lairez16] 19/30

23 LDE for Integrals: Griffiths-Dwork Method Basic idea: I(t) = 1. While m>1, reduce modulo and integrate by parts P Q m I P(t, x) Q m (t, x) dx J := h@ 1 Q,...,@ n Qi Q square-free Int. over a cycle where Q 0. = r+v 1@ 1 Q+ +v n Q Q m = r Q m + P Q m 1 + derivatives 2. Apply to I,I,I, until a linear dependency is found. Thm. If P/Q has degree d in n variables, I(t) satisfies a LDE with order d n, coeffs of degree d O(n). +Algo in Õ(d 8n ) Diagonals: J becomes hz 1 H z n H,..., z n n 1 H z n Hi. [Griffiths70;Christol84;Bostan-Lairez-S.13;Lairez16] 20/30

24 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. 21/30

25 <latexit sha1_base64="lzscpwjyul+titgoutrnnrnkddo=">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</latexit> <latexit sha1_base64="hwqrw/oenos0hh+s6qwbunadkbs=">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</latexit> <latexit sha1_base64="hwqrw/oenos0hh+s6qwbunadkbs=">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</latexit> <latexit sha1_base64="oe8d4sasspmtm81kyz+etoeybgg=">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</latexit> Summary of this part Prop. Algebraic series are the diagonals of bivariate rational functions. Diagonals are D-finite; they are closed under sum, product, Hadamard product; their coefficients are multiple binomial sums (and conversely). D-finite diag. alg. rat. Christol s conjecture: All D- finite power series with integer coefficients and radius of convergence in (0, ) are diagonals F, 4 9, , z? = (?) All these properties are effective, with good bounds and complexity. asymptotics from the LDE [Pólya21,Furstenberg67,Christol84,BostanLairezS.13,Lairez16,BostanDumontS.17] 21 /30

26 IV. Analytic Combinatorics in Several Variables, with Computer Algebra Wanted: complete algorithms, good complexity, more cases with `explicit c. Solution: 1. restrict to simplest class; 2. avoid amoebas and deal only with polynomial systems; 3. control all degrees & sizes.

27 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 Critical points: minimize z 1 z n on V = {z H(z) @z 1 z n 1 z n n! apple 1 1 = = n Minimal ones: on the boundary of the domain of convergence. i.e. J from G-D method 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). Analytic continuation from the rational function 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. 22/30

28 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 23/30

29 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. 24/30

30 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? 25/30

31 Testing Minimality 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. 26/30

32 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 numbers [Melczer-S.16] half-integer This result covers the easiest cases. All conditions hold generically and can be checked within the same complexity, except combinatoriality. 27/30

33 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 proved asymptotics 28/30

34 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 /30

35 Summary & Conclusion D-finite diag. alg. rat. In many cases, LDE + certified analytic continuation works. 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) and recovers explicit constants. Complexity issues become clearer. Work in progress: extend beyond some of the assumptions The End 30/30