Lecture 2: Self-interpretation in the Lambda-calculus

Transcription

1 Lecture 2: Self-interpretation in the Lambda-calculus H. Geuvers Nijmegen, NL 21st Estonian Winter School in Computer Science Winter 2016 H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 1 / 31

2 Outline Self-interpretation in the Lambda Calculus Self-interpretation in the Lambda Calculus H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 2 / 31

3 More on data types A data type D is a set with some operations (functions) on it. An k-ary operation is a function f : D k D. Thereby a 0-ary operation c : D 0 D is identified with a c D. A datatype is determined by its operations on D: c 1,..., c k0 : D 0 D = D f1 1,..., fk 1 1 : D 1 D f1 2,..., fk 2 2 : D 2 D... Nat has z : Nat, s : Nat Nat. Tree has l : Tree, j : Tree 2 Tree H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 4 / 31

4 Self-interpretation in the Lambda Calculus Defining functions on data types in the lambda-calculus F λ-defines the function f : N N if f (n) = F n for all n: f N N F Λ Λ Can we enode the λ-calculus in itself? F λ-defines the function f : Λ Λ if f (M) = F M M Λ. f Λ Λ for all Λ F Λ Is just the identity? Why is this useful? H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 5 / 31

5 Self-interpretation in the Lambda Calculus Defining functions on codes of lambda-terms Some functions f : Λ Λ can only be defined on codes of terms, not on terms. So we will need to talk about codes. f Λ Λ F Λ Λ Example. There is no λ-term F such that F (M N) = M for all M, N There is a λ-term F such that F M N = M for all M, N. (With a suitable encoding.) We also have an evaluator E: E M = M H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 6 / 31

6 Packing and unpacking λ-terms Given M 1,..., M k, define Define U k i, with 1 i k by M 1,..., M k := λz.z M 1... M k U k i := λx 1... x k.x i Then U k i M 1... M k = M i M 1,..., M k U k i = U k i M 1... M k = M i Note that K = U 2 1 H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 7 / 31

7 Second encoding of data types (Böhm-Piperno-Guerini) Consider the data type D with c : D, f : D D, g : D 2 D The Böhm-Piperno-Guerini coding (also denoted by t ) is c = λe.e U 3 1 e f (t) = λe.e U 3 2 t e g(t 1, t 2 ) = λe.e U 3 3 t 1 t 2 e Proposition. The constructors (c, f, g) can be λ-defined: There are lambda terms F, G such that Proof. Take F t = f (t) G t 1 t 2 = g(t 1, t 2 ) F := λx e.e U 3 2 x e G := λx y e.e U 3 3 x y e. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 8 / 31

8 Recursion Self-interpretation in the Lambda Calculus Theorem. Given A 1, A 2, A 3 Λ there is an H Λ such that Proof. Try H = B 1, B 2, B 3. H c = A 1 H H( f (t) ) = A 2 t H H( g(t 1, t 2 ) ) = A 3 t 1 t 2 H H c = B 1, B 2, B 3 c = c B 1, B 2, B 3 = B 1, B 2, B 3 U 3 1 B 1, B 2, B 3 = B 1 B 1, B 2, B 3 = A 1 B 1, B 2, B 3 if B 1 := λz.a 1 z = A 1 H H f (t) = B 1, B 2, B 3 U 3 2 t B 1, B 2, B 3 = B 2 t B 1, B 2, B 3 = A 2 t H if B 2 := λx z.a 2 x z H g(t 1, t 2 ) = A 3 t 1 t 2 H if B 3 := λx y z.a 3 xy z. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 9 / 31

9 Data type for coding lambda terms To encode λ-terms we consider the data type D with var : app : abs : D D D D D D D This data types is a bit strange: there is no base case. It is a priori unclear how to encode the λ-terms in this data type. How to encode the variable binding? (Later slide.) Like before, we can define the constructors Var, App, Abs: Var := λx e.e U 3 1 x e App := λx y e.e U 3 2 x y e Abs := λx e.e U 3 3 x e H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 10 / 31

10 Recursion for the lambda terms data type Like before, we have a recursion theorem: Theorem I. Given A 1, A 2, A 3 Λ there is an H Λ such that H(Var x) = A 1 x H H(App x y) = A 2 x y H H(Abs x) = A 3 x H Proof. Take H = B 1, B 2, B 3 with B 1 := λx z.a 1 x z B 2 := λx y z.a 2 xy z B 3 := λx z.a 3 x z. Then the equations hold indeed. (Exercise: Check this.). H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 11 / 31

11 Coding of lambda terms Coding lambda terms M M Definition (Mogensen) The coding of λ-terms inside the λ-calculus is defined as follows. x := Var x MN := App M N λx.m := Abs (λx. M ) A variable x is encoded using x itself and abstraction λx. is encoded using the same abstraction λx.. Note: coding is not λ-definable: there is no term C such that C M = M for all M. The reverse operation, evaluation, is λ-definable. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 12 / 31

12 Self-interpretation Theorem. There exists a λ-term E (evaluator) such that for all M Λ E M = M Proof. By recursion we can find an E such that Then E(Var x) = x E(App x y) = E x (E y) E(Abs x) = λz.e (x z) E( x ) = E(Var x) = x E( MN ) = E(App M N ) = E M (E N ) = MN E( λx.m ) = E(Abs(λx. M )) = λx.e M = λx.m. Filling in the details of E one has (writing C := λx y z.x z y) E = K, S, C. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 13 / 31

13 Why is this encoding so cool? There are many encoding of Λ in itself. Why is this one so nice? Older ones all go through a coding of λ-terms as numbers: : Λ N Λ. This one is direct, uses λ for λ-abstraction. E M M, also for terms with free variables. E = K, S, C, the initials of S.C. Kleene, one of the founders of the subject! M 1 α M 2 M 1 α M 2. M 1 α M 2 M 1 β M 2. One can define a term R with R M N, if M has N as normal form. There is a Second Fixed Point Theorem (later) H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 14 / 31

14 Recursion for lambda terms using the encoding We can state the recursion theorem for the encoded lambda terms slightly differently, as follows. Theorem II. Given A 1, A 2, A 3 Λ there is an H Λ such that H x = A 1 x H H M N = A 2 M N H H λx.m = A 3 (λx. M ) H Proof. According to Theorem I, there is an H satisfying H(Var x) = A 1 x H H(App x y) = A 2 x y H H(Abs x) = A 3 x H These equations immediately imply the ones of the statement of Theorem II (check this!), so the same H suffices.. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 15 / 31

15 Application 1 Self-interpretation in the Lambda Calculus If you see someone coming out of arrivals in an airport, you cannot determine where he/she comes from. Similarly, there is no F such that for all X, Y Λ F (X Y ) = X (Given the outcome of applying a function on an argument, there is no way we can determine the function that produced this outcome.) Proposition. There exists an F i Λ, i {1, 2} such that F i X 1 X 2 = X i. Proof. We do this for i = 1. By the Recursion Theorem II, there exists F 1 s.t. F 1 ( X 1 X 2 ) = A 2 X 1 X 2 F 1 = X 1, taking A 2 = U 3 1. This suffices. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 16 / 31

16 Second fixed point theorem Lemma. There exists a term Num Λ such that for all M Λ Num M = β Proof. Use recursion (Theorem I) for the lambda calculus data type with M A 1 x N = App Var (Var x) A 2 m n N = App (App App (N m))(n n) A 3 m N = App Abs (Abs (λx.n(m x))) Second fixed point theorem. For all F there is an X with F X = β X Proof. Let W := λz.f (App z(num z)) and X := W W. Then X = W W = F (App W (Num W ) = F (App W W ) = F W W = F X. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 17 / 31

17 Application 2 Self-interpretation in the Lambda Calculus Self-modifying programs For a given T (the program transformer) there exists a program P such that P c k = c k+1 if k is even, = T P c k otherwise. On even inputs, P performs a standard operation (adding 1) On odd inputs, the program P first modifies its own code using T. To find P, apply the second fixed-point theorem to F := λp x.if (Even x) then (x + 1) else(t p x) H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 18 / 31

18 Lambda-terms themselves are black boxes There is no λ-term F such that F (M N) = M for all M, N There is no λ-term F such that F (M N) = N for all M, N On the other hand, we can compute with the codes of λ-terms There is a λ-term F such that F M N = M for all M, N. There is a λ-term F such that F M N = N for all M, N. We also have an evaluator E: E M = M H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 20 / 31

19 Analogy with programming program text p Text inspect, edit,... compiler executable ˆp Black box input/output ( call ) pass around p : In ˆp Out H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 21 / 31

20 Universal programmable machine Specification of a universal (programmable) machine U: p : U ˆp(x) x Reflection over the programs / programming language: What functions (programs) can we write on program text? H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 22 / 31

21 Programs about programs Examples of programs one can write p : M 1 true if p contains a while false if p contains no while p : M 2 true if p 1000 false if p > 1000 What about this?? p : M 3 true if ˆp halts on all inputs false if ˆp diverges on some input H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 23 / 31

22 The Halting problem is undecidable Theorem It is impossible to write a program H with the following specification p : H true if ˆp halts on x false if ˆp diverges on x x The undecidability of the Halting Problem was first proven by Turing, for his Turing machines. We will prove it for the λ-calculus. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 24 / 31

23 Remember recursion for lambda terms using encoding Theorem II. Given A 1, A 2, A 3 Λ there is an H Λ such that Example There is a λ-term C satisfying H x = A 1 x H H M N = A 2 M N H H λx.m = A 3 (λx. M ) H C x = c 0 C M N = c 1 C λx.m = c 2 Proof Just take A 1 := λxy.c 0, A 2 := λxyz.c 1 and A 3 := λxy.c 2. Exercise: Write a function R that checks whether a code of a term is a redex. (So: R M = T if M is a redex and R M = F if M is not a redex.) H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 25 / 31

24 We need to compute with codes Recall that T = λx y.x and F = λx y.y. The following are impossible to define with λ-terms themselves. There is no term G satisfying G M = T if M has a normal form G M = F if M has no normal form There is no term H (compare the Halting problem) satisfying H M N = T if M N has a normal form H M N = F if M N has no normal form That these are impossible is not surprising: we can t look inside a black box. What if we recast these question with coded λ-terms? H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 26 / 31

25 The Halting problem for λ-calculus The Halting problem is undecidable (λ-calculus version): Theorem There is no term H satisfying H M N = T if M N has a normal form H M N = F if M N has no normal form Proof Suppose H exists, Consider Then Q := λx.h x x Ω T Q Q = H Q Q Ω T { TΩ T = Ω if Q Q has a normal form = FΩ T = T if Q Q has no normal form Contradiction. So H doesn t exist. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 27 / 31.

26 The Blank tape problem for λ-calculus Theorem There is no term B satisfying B M = T if M has a normal form B M = F if M has no normal form Proof Suppose B exists, Consider Q := λx.b x Ω T By the second fixed point theorem, there is a R such that Q R = R, that is B R Ω T = R. Now we have R = B R Ω T = T Ω T = Ω if R has a normal form = F Ω T = T if R has no normal form Contradiction. So B doesn t exist. (NB. There may be a proof by reducing H to B; I didn t see it.) H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 28 / 31

27 Some other terms one can(not) write Example There is a term L satisfying L M = c n if n is the number of λ s inside M Example There is a term V satisfying V M = T if M is of the shape x P 1... P n for some n, P, V M = F if M is not of the shape x P 1... P n. Example There is no term H satisfying H M N = T if M N has a normal form H M N = F if M N has no normal form Proof Reduce B to H. (Show that, if H exists, then we can also define B.) H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 29 / 31

28 Scott s theorem The impossibility (undecidability) results we have seen are all instances of a general theorem due to Scott. We phrase it here purely in terms of λ calculus. We assume a coding function : Λ Λ for which we have λ-terms App and Num satisfying App M N = M N Num M = M Definition Two disjoint subsets of Λ, A, B Λ are separable if there is a λ-term F such that F M = T/F for every M and M A F M = T M B F M = F Theorem (Scott) If A, B are non-empty and closed under = β, then they are not separable. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 30 / 31

29 Proof of Scott s theorem Let A, B Λ closed under = β and say a A and b B. Suppose: F separates A and B, so F M = T/F for every M and We define M A F M = T M B F M = F H := λy.if F (App y (Num y)) then b else a and we consider J := H H : J = H H = if F (App H (Num H )) then b else a = if F ( H H ) then b else a { b iff H H A = a iff H H B This is a contradiction. So F doesn t exist. H. Geuvers - Radboud Univ. EWSCS 2016 Self-interpretation in λ-calculus 31 / 31