On space efficiency of Krivine s abstract machine and Hyland-Ong games

Transcription

1 13/02/08, LIPN p.1/28 On space efficiency of Krivine s abstract machine and Hyland-Ong games Kazushige Terui terui@nii.ac.jp National Institute of Informatics, Tokyo Laboratoire d Informatics de Paris Nord, CNRS

2 13/02/08, LIPN p.2/28 ICC Intensional Computational Complexity (ICC) algorithms (eg. terms) evaluation mechanisms (eg. fi-reduction)

3 13/02/08, LIPN p.3/28 Space sensitive Turing machines Input tape (read only) Work tapes (read/write) Output tape (write only) Finite control Space = the num of cells used on worktapes. The input and output can be significantly larger than the space: M If works in f (n) space then the output is of O(2 size ). Given two Turing machines M, N, we want to compose them. f(n) How?

4 13/02/08, LIPN p.4/28 Space sensitive Turing machines Functional composition is not good: M;N If work in f space (n);g(n), M ; N then works in space O(f (n) + g(2 f(n) ) + 2 f(n) ). In particular, logspace functions do not compose. Composition must be interactive: M fl N works in O(f (n) + g(2 f(n) )). From functional to interactive computation!

5 13/02/08, LIPN p.5/28 Towards a logical theory of computational complexity Shortcomings of Turing Machines: Poor data structures (only tapes) Poor control mechanisms (only transitions) Ad hoc constructions: no canonical way of composition Results in lack of beautiful mathematical theory Logic and lambda calculus Rich data structures and controls Canonical composition T ffi U = x:t (U (x)) Hope to make complexity theory more mathematical

6 13/02/08, LIPN p.6/28 Space in lambda calculus Composition does not preserve good space bounds when using fi-reduction, because fi-reduction is not interactive! Interactive evaluation mechanisms Geometry of Interaction Krivine s abstract machine (KAM) Game semantics

7 13/02/08, LIPN p.7/28 KAM KAM implements weak head linear reduction: ( x :x ~ U ) T =) ( x :T ~ U ) T GoI with jumps under A ) B =!(A ffi B) (Danos-Regnier 94) Syntax: Terms t ::= x j x:t1 j t1t2 Environments ρ ::= [x1 7! t1ρ1; :::; x n 7! t n ρ n ] Closures Stacks ß ::= t1ρ1 : : t n ρ n tρ States S ::= (tρ; ß)

8 ((tu)ρ; ß)! (tρ; uρ : ß) 13/02/08, LIPN p.8/28 KAM Initial state: (t[]; nil) Transitions (xρ; ß)! (ρ(x); ß) if x 2 Dom(ρ) (( x:t)ρ; uρ 0 : ß)! (tρ[x 7! uρ 0 ]; ß) Terminations ß) x Dom(ρ) (xρ; 62 x nil) (( x:t)ρ; with Success with output Failure Not space efficient: ρ contains a lot of redundant bindings.

9 13/02/08, LIPN p.9/28 Optimized KAM Optimized transitions (xρ; ß)! (ρ(x); ß) if x 2 Dom(ρ) ß)! (t(ρj t ); ρ(x) : ß) if x 2 Dom(ρ) ((tx)ρ; ß)! (t(ρj t ); u(ρj u ) : ß) otherwise ((tu)ρ; (( x:t)ρ; uρ 0 : ß)! (tρ[x 7! uρ 0 ]; ß) ρj t is a restriction of ρ to the free variables of t.

10 (2Kc; nil) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

11 (2Kc; nil) (2K; c) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

12 (2Kc; nil) (2K; c) (2; K : c) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

13 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

14 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

15 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) (f [f 7! K]; fx[f 7! K; x 7! c]) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

16 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) (f [f 7! K]; fx[f 7! K; x 7! c]) (K; fx[f 7! K; x 7! c]) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

17 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) (f [f 7! K]; fx[f 7! K; x 7! c]) (K; fx[f 7! K; x 7! c]) ( z:z; fx[f 7! K; x 7! c]) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

18 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) (f [f 7! K]; fx[f 7! K; x 7! c]) (K; fx[f 7! K; x 7! c]) ( z:z; fx[f 7! K; x 7! c]) (fx[f 7! K; x 7! c]; nil) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

19 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) (f [f 7! K]; fx[f 7! K; x 7! c]) (K; fx[f 7! K; x 7! c]) ( z:z; fx[f 7! K; x 7! c]) (fx[f 7! K; x 7! c]; nil) (f [f 7! K]; c) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

20 (2Kc; nil) (2K; c) (2; K : c) ( x:f (fx)[f 7! K]; c) (f (fx)[f 7! K; x 7! c]; nil) (f [f 7! K]; fx[f 7! K; x 7! c]) (K; fx[f 7! K; x 7! c]) ( z:z; fx[f 7! K; x 7! c]) (fx[f 7! K; x 7! c]; nil) (f [f 7! K]; c) (K; c) 13/02/08, LIPN p.10/28 Example Suppose (K[]; nil)! Λ ( z:z; nil), 2 = fx:f (fx).

21 ]ρ = ]ρ1 + + ]ρ n + n if ρ = [x1 7! t1ρ1;:::;x n 7! t n ρ n ]ß = ]ρ1 + + ]ρ n + n if t1ρ1 : : t n ρ n ](tρ; ß) = ]ρ + ]ß /02/08, LIPN p.11/28 Size of a state Subterm property: (t[]; nil)! if S, all terms occurring in S are subterms of t. Λ Allows us to think of each state as made of pointers rather than actual terms. = the number of pointers in S: ]S In particular, ](t[]; nil) = 1. KAM evaluates t with n pointers if for any intermediate state (t[]; nil)! Λ S, ]S» n.

22 13/02/08, LIPN p.12/28 TM space-efficiently simulates KAM Naively, TM requires of log n overhead to simulate KAM: if KAM t evaluates n with pointers, M (t) then works in O(n log space jtj). We are interested in terms td, where t (program) is fixed while d (input) is varying. Consider a huge alphabet ± = fu j u is a subterm occ. of tg [ f0; 1g

23 13/02/08, LIPN p.13/28 TM space-efficiently simulates KAM t satisfies the input condition if in any intermediate state nil) S, the number of pointers pointing subterms of d (td[];! is constant (independent Λ of d). Analogy: the num of heads on input tape is fixed. Theorem: There is a M TM which simulates KAM linearly: if KAM td evaluates f (jdj) with t pointers, satisfies the input condition f (x) log and x, M (t) then works in O(f space (jdj)).

24 " = x " x0x1:x " 13/02/08, LIPN p.14/28 Data structures Booleans: t = xy:x, f = xy:y Scott numerals: w for w 2 f0; 1g Λ 0 w = x " x0x1:x0w 1 w = x " x0x1:x1w Purely linear. Successor, predecessor and conditional are also linear. Useful for worktapes. What are read-only/write-only tapes? They should be interactive entities (analogy: Book vs Dialogue)

25 13/02/08, LIPN p.15/28 Data structures Databases: For d 2 f0; 1g 2n, define a term d : Word! Bool s.t. Given w 2 f0; 1g n, dw returns the wth bit of d. d = w:wiωω if d = i 2 f0; 1g = w:wωd0d1 if d1 = i2 n i2 n 1 +1, d0 = i2 n 1 i 1, d = d1d0

26 13/02/08, LIPN p.16/28 KAM space-efficiently simulates TM Theorem: For any TM M, there is a term t M s.t. d 2 f0; 1g Given, t M d Λ0 Λ1 returns Λ0 or Λ1 according to the output of Λ (d). M When M works in space f (n) log n, KAM evaluates t M d Λ0 Λ1 with O(f (n)) pointers. Now the space hierarchy theorem implies There is no TM M that evaluates t M d in space O(f (n) ffl ) with ffl < 1. Corollary: There is no evaluator that is uniformly (i.e. for all terms) and significantly (i.e. super-linearly) more space-efficient than KAM.

27 t M : (Word! Bool)! (Word! Bool) 13/02/08, LIPN p.17/28 Composition M;N Let be transducer TMs which work f (n);g(n) in space and produce outputs 2 of size 2 and. g(n) f(n) N : (Word! Bool)! (Word! Bool) t M ffi t N : (Word! Bool)! (Word! Bool) t (t N ffi t M )dw Λ0 Λ1 returns the wth bit of N ffi M (d). KAM evaluates it with O(f (jdj) + g(jd 0 j)) pointers, where d 0 is the output of M (d). KAM + canonical composition simulates the interactive composition of TMs!

28 13/02/08, LIPN p.18/28 Moral Turing Machines -calculus Composition non canonical canonical Evaluation canonical non canonical In -calculus, TIME-SPACE tradeoff shows up at the stage of evaluation: KAM is space-efficient CBV seems to be time-efficient (cf. Dal Lago-Martini). Work to be done: identify a subclass of -terms (via a type system) on which space bounds are well preserved by KAM composition.

29 13/02/08, LIPN p.19/28 Warning There is no uniformly more efficient evaluator than KAM. KAM is like a student Bob with score Math Science Grammer Latin Music Although there is no uniformly better student than Bob, this does not mean Bob is the best student.

30 u n = ( x1 x n :x n ) Λ1 Λ n 13/02/08, LIPN p.20/28 From KAM to HO games Empirically, KAM is good at tall terms: n = ( fx:f n (x))( y:y)λ t KAM t evaluates n O(1) with pointers. KAM is not good at wide terms: KAM evaluates u n with O(n) pointers. For simply typed terms with unbounded width and fixed rank, HO games seem to be more efficient.

31 13/02/08, LIPN p.21/28 HO games A fully abstract model of PCF (Hyland-Ong 00). Successfully applied to various programming languages. Useful for compositional and higher-order model checking (Ghica-Mckusker 00, Ong 04, etc.) Types = Arenas Terms = Strategies Computation = play Play = interactive composition of two strategies Composition is effective (cf. Ong 04). Corresponds to Pointer Abstract Machine (Danos-Herbelin-Regnier 96).

32 ( X:X (0) ^ X (1)) (bool!bool)!bool x:f (x) 13/02/08, LIPN p.22/28 Ranks and complexity The rank of a type: rank( ) = 0, rank(a! B) = max(rank(a) + 1; rank(b)) A term is at rank n if all of its subterms have rank n types. Rank Quantative Qualitative 1 boolean formulas (open) addition = xy:b!! yx :b 2 boolean circuits multiplication (x; : : : ;x)) bool!bool G(y) n ffi m ( x:f 3 8x:F (x) = QBFs exponentiation (n) int!int m

33 13/02/08, LIPN p.23/28 Ranks and complexity Normalization problem: Given two t; u terms u where is normal, t does reduce to u? (Schubert 2001) shows that this problem is 1. DTIME(n log n) in for rank 1 terms 2. P-complete for rank 2 terms 3. PSPACE-complete for rank 3 terms Hardness is easy. Membership is difficult, since fi-reduction does not work. Schubert uses graph rewriting. We use HO games.

34 13/02/08, LIPN p.24/28 HO games for simple types Arena for type A: A Moves = occurrences of atomic types in P-moves = negative occurrences O-moves = positive occurrences Initial move = tail(a) ` tail(c) where C is an immediate subformula of B tail(b) Legal play: a PO-alternating pointing sequence like s = m0m1 m j m i m0 is the initial O-move each m i (1» i) is justified by a preceding move m j (m j ` m i, 0» j < i).

35 dse m 2 fi 13/02/08, LIPN p.25/28 Let s play the game! P-view: a legal play in which any O-move is justfied by the move immediately before. Given an arbitrary legal play s, a dse P-view can be extracted. Innocent strategy (as view function): a P-deterministic tree made of P-ending P-views. Every closed normal t : A term can be interpreted by an innocent strategy. HO-dialogue: Given innocent ff : A! B strategies fi : and A, 1. Start with the initial m0 = tail(b) move 2. Expand an odd length s play s m to such dse m 2 ff that 3. Expand an even length m0 s play m0 s m to such that

36 13/02/08, LIPN p.26/28 A lemma in game semantics Difficulty: HO game is history sensitive. But recording the whole history leads to TIME = SPACE. Lemma (taught by P. Boude): For any legal play satisfying P- and O-visibility s = m1 m2 if m2 justifies no moves, it never happens that = m1 m2 s So one can safely s contract to s 0 = m1m2 and continue the play.

37 13/02/08, LIPN p.27/28 Complexity of Rank 3 games Theorem: Given a t term at rank 3 (which is an applicative combination of closed normal terms), the length of play can be kepth within O(jtj). Corollary: One can determine the head variable nf (t) of in polynomial space via HO games. Work to be done: What about rank 4? Dowehavea hierarchy result as in Kristiansen s talk?

38 13/02/08, LIPN p.28/28 Conclusion Lambda calculus admits a canonical M ffi N composition KAM is optimal for simulating TM. Composition works nicely with KAM. (KAM allows natural programming eg. boolean matrix closure) HO games seem to be good at least for rank bounded terms. Good understanding of game semantics leads to clever space management. Game semantics may implicitly explain complexity.