COMP4630: λ-calculus

Transcription

1 COMP4630: λ-calculus 4. Standardsaton Mcael Norrs Canberra Researc Lab., NICTA Semester 2, 2015

2 Last Tme Confluence Te property tat dvergent evaluatons can rejon one anoter Proof Damond propertes Uses parallel reducton (= β ); and Many nductons Consequences Soundness of λ (Wt an analogous proof) Soundness of λη Incompleteness of λ (See end of lecture 2)

3 Today Introducton Head Reducton Weak Head Reducton Te Proof Falng Approaces Te Rgt Approac Consequences Concluson

4 Objectve From last tme, we know tat a term M as at most one normal form. Unfortunately, we also know tat not all evaluaton strateges wll lead to tat normal form. Ts s not nconsstent wt confluence. Wy? Introducton 4/27

5 Evaluaton Strateges An evaluaton strategy s bascally a way of answerng te queston: Were (.e., n wc sub-term) sould I do my next reducton? Or: Were sould I do te next bt of work? Some languages (e.g., Java) do not allow for coces to be made at all. Tey specfy a precse evaluaton order. Wy would tey do tat? Introducton 5/27

6 Evaluaton Strategy #1: Applcatve Order Evaluate everytng from te bottom up. I.e., n (λv. M) N work wll start wt M, passng to N and performng te top-level β-reducton last A functon s arguments (and te functon tself) wll be evaluated before te argument s passed to te functon. Also known as strct evaluaton. (Used n Fortran, C, Pascal, Ada, SML, Java... ) Causes (λx. y) Ω to go nto an nfnte loop. Introducton 6/27

7 Evaluaton Strategy #2: Normal Order Evaluate top-down, left-to-rgt. Wt (λv. M) N, start by performng te β-reducton, producng M[v := N] Fnd te top-most, left-most β-redex and reduce t. Keep gong Ts strategy s bend te lazy evaluaton of languages lke Haskell. Normal order evaluaton wll always termnate wt a normal form f a term as one. (Proof to come... ) Introducton 7/27

8 Evaluaton Strategy Trade-offs An evaluaton strategy mgt 1. be guaranteed to fnd normal forms; or 2. am to perform te least number of β-reductons Naïvely, normal order reducton does 1; applcatve order sort of aceves 2, but gves up on 1 (In fact, optmal reducton s very dffcult to get rgt.) Introducton 8/27

9 Provng Normal Order Evaluaton Our focus s n sowng tat normal order evaluaton s guaranteed to fnd normal forms. (Tat s wy t s called normal order... ) Here are te rules: (λv. M) N n M[v := N] M n M (λv. M) n (λv. M ) M n M M not an abstracton M N n M N N n N M not an abstracton M n β-nf M N n M N Introducton 9/27

10 Head Reducton We can dvde normal order reducton nto two dfferent sorts of reducton. Frst, normal order reducton: (λv. M) N n M[v := N] M n M (λv. M) n (λv. M ) M n M M not an abstracton M N n M N N n N M not an abstracton M n β-nf M N n M N Head Reducton 10/27

11 Head Reducton We can dvde normal order reducton nto two dfferent sorts of reducton. Ten, ead reducton: (λv. M) N M[v := N] M M (λv. M) (λv. M ) M M M not an abstracton M N M N N N M not an abstracton M n β-nf M N M N Wen ead reducng, you never reduce to te rgt of an applcaton Head Reducton 10/27

12 Hence, Head Normal Forms Head Normal Form s a reasonable stoppng place. Rules agan: (λv. M) N M[v := N] M M (λv. M) (λv. M ) Examples: v s n nf (λv. v) s n nf M M M not an abstracton M N M N (λv w. v (λu. M) N) s n nf (λu w z. v ((λu. M) N)) s n nf Head Reducton 11/27

13 Head Normal Forms, Generally If term M s n nf, ten t wll look lke: (λ v. u M 1 M n ) Te vector v may be empty, u may be free or bound, and te number of extra arguments, n, may be 0. Once a term s n nf, ts top-level structure can t cange. Head Reducton 12/27

14 After Head Reductons Once a term s n nf, ts top-level structure can t cange. Any furter reductons (of any sort) nsde wll be reductons wtn an M. (λ v. u M 1 M n ) Eac argument wll evolve ndependently, and te number of arguments can t cange. Tese are nternal reductons. If te term s (λ v. (λu. M) N 1 N 2... ) (not n nf) and M reduces, ten tat s an nternal reducton too All reductons are eter ead or nternal Head Reducton 13/27

15 Normal Order Reducton Splts n Two Te last rule of normal order reducton (wc we deleted to get ead reducton): N n N M not an abstracton M n β-nf M N n M N If M n N, and te reductons aren t all ead, ten tere must be a frst ead normal form P, suc tat M P N We want to sow te same sort of splt for arbtrary β-reducton ( β ) Head Reducton 14/27

16 Interlude: Weak Head Reducton Tanks to: M M (λv. M) (λv. M ) ead reducton proceeds nsde functon bodes. If you take ts rule out, you get weak ead reducton. Weak ead normal forms are ead normal forms, or abstractons. Weak ead reducton s used n mplementatons of functonal programmng languages. Head Reducton Weak Head Reducton 15/27

17 Basc Strategy We want to know tat, f by some pat: M β N wt N a normal form, ten normal order reducton wll take M to N too. Wll do ts by sowng a more general result. Tat for any N, f M β N, ten tere exsts a P suc tat M P N Te Proof 16/27

18 We Want to Commute Steps If we ad M N P Te Proof Falng Approaces 17/27

19 We Want to Commute Steps If we ad M N N P we d lke to know tat tere was a N tat we could get to va ead reducton, and from wc we could make nternal reductons to get to P Te Proof Falng Approaces 17/27

20 We Want to Commute Steps If we ad M N N P we d lke to know tat tere was a N tat we could get to va ead reducton, and from wc we could make nternal reductons to get to P (maybe wt multple steps?). Te Proof Falng Approaces 17/27

21 We Want to Commute Steps If we ad M N N P we d lke to know tat tere was a N tat we could get to va ead reducton, and from wc we could make nternal reductons to get to P (maybe wt multple steps?). Maybe ts would allow ead and nternal steps to be bubble-sorted so tat all ead steps came frst. Te Proof Falng Approaces 17/27

22 But Drect Commutng s Hard Commutng does requre multple steps: (λx. f x x) ((λy. y z) u) (λx. f x x) (u z) f ((λy. y z) u) ((λy. y z) u) f (u z) (u z) f (u z) ((λy. y z) u) f ((λy. y z) u) (u z) f (u z) (u z) f (u z) (u z) Ts example requres multple (2) nternal reductons. Te Proof Falng Approaces 18/27

23 But Drect Commutng s Hard Commutng does requre multple steps: (λu. (λv. v u z) f) N (λu. f u z) N (λv. v N z) f f N z f N z Ts example requres multple ead reductons (and no nternals). Te Proof Falng Approaces 18/27

24 Commutng wt Multple Steps Isn t Good Enoug Ts s a teorem: But t s not good enoug. M N P = N. M N P Our examples sow us tat we can ave 2 2 Te Proof Falng Approaces 19/27

25 Te Bubble-Sort Tat Never Ends Our examples sow us tat we can ave 2 2 Start wt a reducton sequence : Ts s not progress: we stll ave two ead reductons tat aven t been sorted to te start of te sequence. Te Proof Falng Approaces 20/27

26 A Better Lemma Toug commutng nternal and ead reductons can result n multple nternal reductons, te latter are all parallel (wrte So, prove nstead: = ). M P N If a nternal parallel reducton s followed by a ead reducton, Te Proof Te Rgt Approac 21/27

27 A Better Lemma Toug commutng nternal and ead reductons can result n multple nternal reductons, te latter are all parallel (wrte So, prove nstead: = ). M P P N If a nternal parallel reducton s followed by a ead reducton, tere s an alternatve route were ead reductons come frst, and tere s one nternal parallel reducton afterwards. Te Proof Te Rgt Approac 21/27

28 Proof n More Detal Have M = P N As P ead-reduces t s (λ v. (λu. P 0 ) P 1 P 2 P n ), wt n 1 And M s of form (λ v. (λu. M 0 ) M 1 M 2 M n ), wt M = β P Te Proof Te Rgt Approac 22/27

29 Proof n More Detal Have M = P N As P ead-reduces t s (λ v. (λu. P 0 ) P 1 P 2 P n ), wt n 1 And M s of form (λ v. (λu. M 0 ) M 1 M 2 M n ), wt M = β P So M (λ v. M 0 [u := M 1 ] M 2 M n ) Te Proof Te Rgt Approac 22/27

30 Proof n More Detal Have M = P N As P ead-reduces t s (λ v. (λu. P 0 ) P 1 P 2 P n ), wt n 1 And M s of form (λ v. (λu. M 0 ) M 1 M 2 M n ), wt M = β P So M Last transton s = β, not at top level, makng t ead. (λ v. M 0 [u := M 1 ] M 2 M n ) = β N = because M 0 s reducton may be A lttle more work s stll requred (decomposng = β nto ead and nternal parts). Te Proof Te Rgt Approac 22/27

31 Te Last Bg Lemma If M = β N, ten tere are M suc tat M M 1 M 2 M n = N and eac M = β N Ts gves: P M β N Te Proof Te Rgt Approac 23/27

32 Puttng te Lemmas Togeter Provng: M β N = P. M P Have M β N, and so also M = β N. (Base case of zero steps trval.) N Te Proof Te Rgt Approac 24/27

33 Puttng te Lemmas Togeter Provng: M β N = P. M P Have M β N, and so also M = β N. So, assume M = β M = β N. N Te Proof Te Rgt Approac 24/27

34 Puttng te Lemmas Togeter Provng: M β N = P. M P Have M β N, and so also M = β N. So, assume M = β M = β N. By Last Bg Lemma, also ave P 1 s.t. M N P 1 = M Te Proof Te Rgt Approac 24/27

35 Puttng te Lemmas Togeter Provng: M β N = P. M P Have M β N, and so also M = β N. So, assume M = β M = β N. By Last Bg Lemma, also ave P 1 s.t. M By nductve ypotess, ave P 2 s.t. M I.e., M P 1 = M P 2 N P 1 = M P 2 N N Te Proof Te Rgt Approac 24/27

36 Puttng te Lemmas Togeter Provng: M β N = P. M P Have M β N, and so also M = β N. So, assume M = β M = β N. By Last Bg Lemma, also ave P 1 s.t. M By nductve ypotess, ave P 2 s.t. M I.e., M P 1 = M P 2 N P 1 = M P 2 N N Now, we can bubble ead reductons after M up over te usng te Better Lemma. =, Te Proof Te Rgt Approac 24/27

37 Consequences: Standardsaton Have sown: M β N = P. M P N It s obvously possble to order te nternal reductons so tat tey occur left-to-rgt. By nducton. Te nternal terms wtn N are all smaller tan N tself, so te nternal reductons wtn eac N can temselves be sorted approprately. Gves Standardsaton: If M β N s possble, ten N can be reaced from M n a standard way (dong reductons n left to rgt order) Te Proof Consequences 25/27

38 Consequences: Normal Order Evaluaton Works Recall tat normal order evaluaton s a standard evaluaton strategy tat does all possble reductons. Te Proof Consequences 26/27

39 Consequences: Normal Order Evaluaton Works Recall tat normal order evaluaton s a standard evaluaton strategy tat does all possble reductons. If M can reduce to N, a β-normal form, ten tere s a standard reducton tat does te same. Te Proof Consequences 26/27

40 Consequences: Normal Order Evaluaton Works Recall tat normal order evaluaton s a standard evaluaton strategy tat does all possble reductons. If M can reduce to N, a β-normal form, ten tere s a standard reducton tat does te same. If a standard reducton termnates n a β-normal form, t as done all possble reductons. Te Proof Consequences 26/27

41 Consequences: Normal Order Evaluaton Works Recall tat normal order evaluaton s a standard evaluaton strategy tat does all possble reductons. If M can reduce to N, a β-normal form, ten tere s a standard reducton tat does te same. If a standard reducton termnates n a β-normal form, t as done all possble reductons. And so tat standard reducton was a normal order reducton. Te Proof Consequences 26/27

42 Consequences: Normal Order Evaluaton Works Recall tat normal order evaluaton s a standard evaluaton strategy tat does all possble reductons. If M can reduce to N, a β-normal form, ten tere s a standard reducton tat does te same. If a standard reducton termnates n a β-normal form, t as done all possble reductons. And so tat standard reducton was a normal order reducton. So, normal order evaluaton fnds normal forms f tey exst. Te Proof Consequences 26/27

43 Summary An nvolved proof. Lesson #1: Te λ-calculus s a plausble programmng language tere s an algortm for turnng λ-terms nto values (wen tose terms ave values at all) Lesson #2: Evaluaton Orders are a Desgn Queston Do you want to guarantee as muc termnaton as possble? Use normal order Or, do you want more speed, and unnecessary non-termnatons? Use applcatve order (lke C, Java etc) Next tme: addng plausblty. Numbers, Pars and Lsts for te λ-calculus. Concluson 27/27