Recursion Theorem. Ziv Scully Ziv Scully Recursion Theorem / 28

Transcription

1 Recursion Theorem Ziv Scully Ziv Scully Recursion Theorem / 28

2 A very λ-calculus appetizer P-p-p-plot twist! λ Ziv Scully Recursion Theorem / 28

3 A very λ-calculus appetizer The master plan 1 A very λ-calculus appetizer 2 Main theorem 3 Applications 4 Fixed points and diagonalization Ziv Scully Recursion Theorem / 28

4 A very λ-calculus appetizer The story so far 1 A very λ-calculus appetizer 2 Main theorem 3 Applications 4 Fixed points and diagonalization Ziv Scully Recursion Theorem / 28

5 A very λ-calculus appetizer What is computation? Depending on who you ask, computation is following clear procedures step by step (Turing machines). Ziv Scully Recursion Theorem / 28

6 A very λ-calculus appetizer What is computation? Depending on who you ask, computation is following clear procedures step by step (Turing machines).... a composition of primitive functions on (µ-recursive functions). Ziv Scully Recursion Theorem / 28

7 A very λ-calculus appetizer What is computation? Depending on who you ask, computation is following clear procedures step by step (Turing machines).... a composition of primitive functions on (µ-recursive functions).... complexity that emerges from simple rules (cellular automata). Ziv Scully Recursion Theorem / 28

8 A very λ-calculus appetizer What is computation? Depending on who you ask, computation is following clear procedures step by step (Turing machines).... a composition of primitive functions on (µ-recursive functions).... complexity that emerges from simple rules (cellular automata).... applying functions to functions to get more functions (λ-calculus). Ziv Scully Recursion Theorem / 28

9 A very λ-calculus appetizer Definition of λ-calculus A λ-calculus term is one of three things. A variable, such as x, y, or z. Ziv Scully Recursion Theorem / 28

10 A very λ-calculus appetizer Definition of λ-calculus A λ-calculus term is one of three things. A variable, such as x, y, or z. A function application, one term applied to another, such as f x, f (g x), or (f x) y = f x y. Ziv Scully Recursion Theorem / 28

11 A very λ-calculus appetizer Definition of λ-calculus A λ-calculus term is one of three things. A variable, such as x, y, or z. A function application, one term applied to another, such as f x, f (g x), or (f x) y = f x y. A function abstraction, making a term a one-argument function, such as λx. x, λx. (λy. x) = λx. λy. x, or λf. λg. λx. f (g x). Ziv Scully Recursion Theorem / 28

12 A very λ-calculus appetizer Definition of λ-calculus A λ-calculus term is one of three things. A variable, such as x, y, or z. A function application, one term applied to another, such as f x, f (g x), or (f x) y = f x y. A function abstraction, making a term a one-argument function, such as λx. x, λx. (λy. x) = λx. λy. x, or λf. λg. λx. f (g x). Additionally, we require that, at the top level, no variable appear outside the scope of a function abstraction that declares it. Ziv Scully Recursion Theorem / 28

13 A very λ-calculus appetizer Evaluating λ-calculus terms There are two rules for evaluating terms. α-renaming: change variable names, such as λx. λy. x λf. λg. f. Ziv Scully Recursion Theorem / 28

14 A very λ-calculus appetizer Evaluating λ-calculus terms There are two rules for evaluating terms. α-renaming: change variable names, such as λx. λy. x λf. λg. f. β-reduction: apply a function to an argument by substitution, such as (λx. λy. x)(λz. z) λy. λz. z. Ziv Scully Recursion Theorem / 28

15 A very λ-calculus appetizer Evaluating λ-calculus terms There are two rules for evaluating terms. α-renaming: change variable names, such as λx. λy. x λf. λg. f. β-reduction: apply a function to an argument by substitution, such as (λx. λy. x)(λz. z) λy. λz. z. These rules are powerful enough to simulate a Turing machine. Ziv Scully Recursion Theorem / 28

16 A very λ-calculus appetizer Hello, factorial! Let s define the factorial function. 1 x = 0 F = λx. x (F(x 1)) otherwise. Ziv Scully Recursion Theorem / 28

17 A very λ-calculus appetizer Hello, factorial! Let s define the factorial function. 1 x = 0 F = λx. x (F(x 1)) otherwise. Problem: we can t have F refer to itself in λ-calculus. Ziv Scully Recursion Theorem / 28

18 A very λ-calculus appetizer Sneakier self-reference What if we can t use self-reference? F = G G, where 1 x = 0 G = λg. λx. x (g g (x 1)) otherwise. Ziv Scully Recursion Theorem / 28

19 A very λ-calculus appetizer Sneakier self-reference What if we can t use self-reference? F = G G, where 1 x = 0 G = λg. λx. x (g g (x 1)) otherwise. We use an open definition: G refers to whatever function we want it to use next. Passing G to itself gives G a reference to itself. Ziv Scully Recursion Theorem / 28

20 A very λ-calculus appetizer Sneakier self-reference What if we can t use self-reference? F = G G, where 1 x = 0 G = λg. λx. x (g g (x 1)) otherwise. We use an open definition: G refers to whatever function we want it to use next. Passing G to itself gives G a reference to itself. 1 x = 0 G G = λg. λx. G x (g g (x 1)) otherwise 1 x = 0 = λx. x (G G (x 1)) otherwise. Ziv Scully Recursion Theorem / 28

21 A very λ-calculus appetizer Laziness is a virtue Using a self-application g g for every recursive call is cumbersome. What if we re lazy and want to write f instead of g g? 1 x = 0 F = λf. λx. x (f (x 1)) otherwise. Ziv Scully Recursion Theorem / 28

22 A very λ-calculus appetizer Laziness is a virtue Using a self-application g g for every recursive call is cumbersome. What if we re lazy and want to write f instead of g g? 1 x = 0 F = λf. λx. x (f (x 1)) otherwise. F is a shell that needs to be filled in by the true factorial function, f. We can think of F as having type ( ) ( ). Ziv Scully Recursion Theorem / 28

23 A very λ-calculus appetizer Laziness is a virtue Using a self-application g g for every recursive call is cumbersome. What if we re lazy and want to write f instead of g g? 1 x = 0 F = λf. λx. x (f (x 1)) otherwise. F is a shell that needs to be filled in by the true factorial function, f. We can think of F as having type ( ) ( ). If f computes factorial, then so does F f, in which case f = F f. That is, we necessarily want a fixed point of F. Ziv Scully Recursion Theorem / 28

24 A very λ-calculus appetizer Wishful thinking Let s wish for a function Y that finds a fixed point of its input. Namely, Y should satisfy Y F = F (Y F). Ziv Scully Recursion Theorem / 28

25 A very λ-calculus appetizer Wishful thinking Let s wish for a function Y that finds a fixed point of its input. Namely, Y should satisfy Y F = F (Y F). When F is our factorial shell, we get 1 x = 0 Y F = F (Y F) = λx. x (Y F (x 1)) otherwise. Ziv Scully Recursion Theorem / 28

26 A very λ-calculus appetizer Wishful thinking Let s wish for a function Y that finds a fixed point of its input. Namely, Y should satisfy Y F = F (Y F). When F is our factorial shell, we get 1 x = 0 Y F = F (Y F) = λx. x (Y F (x 1)) otherwise. That is, to do recursion, it suffices to find a fixed point of F. Ziv Scully Recursion Theorem / 28

27 A very λ-calculus appetizer Wish really, really sneakily We know how to do self-reference: make a function accept any reference as an argument, then feed the function to itself. With that inspiration, we wish for some G F such that Y F = F (Y F) = G F G F. Ziv Scully Recursion Theorem / 28

28 A very λ-calculus appetizer Y combinator? Because functions! Y F = F (Y F) = G F G F Ziv Scully Recursion Theorem / 28

29 A very λ-calculus appetizer Y combinator? Because functions! Y F = F (Y F) = G F G F G F G F = F (G F G F ) Ziv Scully Recursion Theorem / 28

30 A very λ-calculus appetizer Y combinator? Because functions! Y F = F (Y F) = G F G F G F G F = F (G F G F ) = (λg. F (g g)) G F Ziv Scully Recursion Theorem / 28

31 A very λ-calculus appetizer Y combinator? Because functions! Y F = F (Y F) = G F G F G F G F = F (G F G F ) = (λg. F (g g)) G F G F = λg. F (g g). Ziv Scully Recursion Theorem / 28

32 A very λ-calculus appetizer Y combinator? Because functions! Y F = F (Y F) = G F G F G F G F = F (G F G F ) = (λg. F (g g)) G F G F = λg. F (g g). Theorem (Existence of the Y combinator) The combinator Y = λf. G f G f = λf. (λg. f (g g)) (λg. f (g g)) satisfies Y F = F (Y F) for all F. Ziv Scully Recursion Theorem / 28

33 Main theorem The story so far 1 A very λ-calculus appetizer 2 Main theorem 3 Applications 4 Fixed points and diagonalization Ziv Scully Recursion Theorem / 28

34 Main theorem Turing machines are computers, too! Notation: [n] is the Turing machine n encodes. Theorem (Recursion theorem) There exists a total (always halting) computable function Y such that for all F, if [F] is total, then [Y(F)] = [[F](Y(F))]. Ziv Scully Recursion Theorem / 28

35 Main theorem Turing machines are computers, too! Notation: [n] is the Turing machine n encodes. Theorem (Recursion theorem) There exists a total (always halting) computable function Y such that for all F, if [F] is total, then [Y(F)] = [[F](Y(F))]. Put another way: if we consider numbers equivalent if they encode equivalent Turing machines, then [F] has a fixed point for all F, and that fixed point is computable from F. Ziv Scully Recursion Theorem / 28

36 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) Ziv Scully Recursion Theorem / 28

37 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) Ziv Scully Recursion Theorem / 28

38 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) Ziv Scully Recursion Theorem / 28

39 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) G F = g [F]([g](g)). Ziv Scully Recursion Theorem / 28

40 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. We can compute G F, so we win! Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) G F = g [F]([g](g)). Ziv Scully Recursion Theorem / 28

41 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. We can compute G F, so we win! What happens if [F](x) = x + 1? Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) G F = g [F]([g](g)). Ziv Scully Recursion Theorem / 28

42 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) G F = g [F]([g](g)). We can compute G F, so we win! What happens if [F](x) = x + 1? Y(F) = Y(F) + 1, so the universe explodes. Ziv Scully Recursion Theorem / 28

43 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) G F = g [F]([g](g)). We can compute G F, so we win! What happens if [F](x) = x + 1? Y(F) = Y(F) + 1, so computing Y(F) = [G F ](G F ) must not halt. Ziv Scully Recursion Theorem / 28

44 Main theorem Proof of recursion theorem Notation: [n] is the Turing machine n encodes, x E(x) is the procedure that maps x to expression E(x), and P is the encoding of procedure P. Y(F) = [F](Y(F)) = [G F ](G F ) [G F ](G F ) = [F]([G F ](G F )) = (g [F]([g](g)))(G F ) G F = g [F]([g](g)). We can compute G F, so we win! What happens if [F](x) = x + 1? Y(F) = Y(F) + 1, so computing Y(F) = [G F ](G F ) must not halt. We tried to find a fixed point, but we only need a fixed point modulo equivalence of encoded Turing machines. Ziv Scully Recursion Theorem / 28

45 Main theorem Proof of recursion theorem 2: electric boogaloo We try again with a slightly weaker goal. Y(F) = [G F ](G F ) [Y(F)] = [[F](Y(F))] Ziv Scully Recursion Theorem / 28

46 Main theorem Proof of recursion theorem 2: electric boogaloo We try again with a slightly weaker goal. Y(F) = [G F ](G F ) [Y(F)] = [[F](Y(F))] [G F ](G F ) = x [[F]([G F ](G F ))](x) Ziv Scully Recursion Theorem / 28

47 Main theorem Proof of recursion theorem 2: electric boogaloo We try again with a slightly weaker goal. Y(F) = [G F ](G F ) [Y(F)] = [[F](Y(F))] [G F ](G F ) = x [[F]([G F ](G F ))](x) = (g x [[F]([g](g))](x) )(G F ) Ziv Scully Recursion Theorem / 28

48 Main theorem Proof of recursion theorem 2: electric boogaloo We try again with a slightly weaker goal. Y(F) = [G F ](G F ) [Y(F)] = [[F](Y(F))] [G F ](G F ) = x [[F]([G F ](G F ))](x) = (g x [[F]([g](g))](x) )(G F ) G F = g x [[F]([g](g))](x). Ziv Scully Recursion Theorem / 28

49 Main theorem Proof of recursion theorem 2: electric boogaloo We try again with a slightly weaker goal. Y(F) = [G F ](G F ) [Y(F)] = [[F](Y(F))] [G F ](G F ) = x [[F]([G F ](G F ))](x) = (g x [[F]([g](g))](x) )(G F ) G F = g x [[F]([g](g))](x). This time, not only can we compute G F, but [G F ] is a total function, so computing Y(F) = [G F ](G F ) always halts! Ziv Scully Recursion Theorem / 28

50 Main theorem For the skeptical Just to make sure we have this right, if Y(F) = [G F ](G F ), then [Y(F)](X) = [[G F ](G F )](X) = [[ g x [[F]([g](g))](x) ](G F )](X) = [ x [[F]([G F ](G F ))](x) ](X) = [[F]([G F ](G F ))](X) = [[F](Y(F))](X). As desired, Y(F) and [F](Y(F)) encode equivalent Turing machines. Ziv Scully Recursion Theorem / 28

51 Applications The story so far 1 A very λ-calculus appetizer 2 Main theorem 3 Applications 4 Fixed points and diagonalization Ziv Scully Recursion Theorem / 28

52 Applications Recursion (duh) Letting [F](f) =... [f]... gives so [Y(F)] can make recursive calls. [Y(F)] = [[F](Y(F))] =... [Y(F)]..., Ziv Scully Recursion Theorem / 28

53 Applications Recursion (duh) Letting [F](f) =... [f]... gives so [Y(F)] can make recursive calls. In general, [F](f) =... f... gives [Y(F)] = [[F](Y(F))] =... [Y(F)]..., [Y(F)] = [[F](Y(F))] =... Y(F)..., which gives us something stronger: [Y(F)] can access its own source code. Practical application is that, when defining a procedure P, we can use P in the definition. Ziv Scully Recursion Theorem / 28

54 Applications Quines A Quine is a program that prints its own source code. This sounds easy at first, but after some trial and error it starts to seem impossible. Ziv Scully Recursion Theorem / 28

55 Applications Quines A Quine is a program that prints its own source code. This sounds easy at first, but after some trial and error it starts to seem impossible. But we just said that defining P(x) = P is okay! Ziv Scully Recursion Theorem / 28

56 Applications Quines A Quine is a program that prints its own source code. This sounds easy at first, but after some trial and error it starts to seem impossible. But we just said that defining P(x) = P is okay! Letting [F](f) = x f gives [Y(F)](x) = [[F](Y(F))](x) = Y(F). That is, [Y(F)] always returns its encoding. Ziv Scully Recursion Theorem / 28

57 Applications Quines A Quine is a program that prints its own source code. This sounds easy at first, but after some trial and error it starts to seem impossible. But we just said that defining P(x) = P is okay! Letting [F](f) = x f gives [Y(F)](x) = [[F](Y(F))](x) = Y(F). That is, [Y(F)] always returns its encoding. This example is simple enough that we can examine G F directly. G F = g x [[F]([g](g))](x) = g x [g](g). Ziv Scully Recursion Theorem / 28

58 Applications Quines A Quine is a program that prints its own source code. This sounds easy at first, but after some trial and error it starts to seem impossible. But we just said that defining P(x) = P is okay! Letting [F](f) = x f gives [Y(F)](x) = [[F](Y(F))](x) = Y(F). That is, [Y(F)] always returns its encoding. This example is simple enough that we can examine G F directly. G F = g x [[F]([g](g))](x) = g x [g](g). G F says apply my first argument, decoded, to itself, still encoded. Y(F) = [G F ](G F ) decodes G F and applies it to a still encoded G F. Ziv Scully Recursion Theorem / 28

59 Applications Quines A Quine is a program that prints its own source code. This sounds easy at first, but after some trial and error it starts to seem impossible. But we just said that defining P(x) = P is okay! Letting [F](f) = x f gives [Y(F)](x) = [[F](Y(F))](x) = Y(F). That is, [Y(F)] always returns its encoding. This example is simple enough that we can examine G F directly. G F = g x [[F]([g](g))](x) = g x [g](g). G F says apply my first argument, decoded, to itself, still encoded. Y(F) = [G F ](G F ) decodes G F and applies it to a still encoded G F. This echoes the famous natural-language Quine: quoted and followed by itself is a Quine quoted and followed by itself is a Quine. Ziv Scully Recursion Theorem / 28

60 Applications Impossibility results Theorem (Rice s theorem) No nontrivial predicate of Turing machines is decidable. Ziv Scully Recursion Theorem / 28

61 Applications Impossibility results Theorem (Rice s theorem) No nontrivial predicate of Turing machines is decidable. Proof. Fix some nontrivial predicate decidable by P, and let [A] and [B] satisfy and not satisfy the predicate, respectively. Ziv Scully Recursion Theorem / 28

62 Applications Impossibility results Theorem (Rice s theorem) No nontrivial predicate of Turing machines is decidable. Proof. Fix some nontrivial predicate decidable by P, and let [A] and [B] satisfy and not satisfy the predicate, respectively. The procedure B P( Q ) Q(x) = A otherwise gives a contradiction. Ziv Scully Recursion Theorem / 28

63 Fixed points and diagonalization The story so far 1 A very λ-calculus appetizer 2 Main theorem 3 Applications 4 Fixed points and diagonalization Ziv Scully Recursion Theorem / 28

64 Fixed points and diagonalization Yet another proof of Cantor s theorem A surjection â : 2 gives a function a : 2 such that for any b : 2, there is some n such that b = a(n, ), in which case we say that b is representable by a. Ziv Scully Recursion Theorem / 28

65 Fixed points and diagonalization Yet another proof of Cantor s theorem A surjection â : 2 gives a function a : 2 such that for any b : 2, there is some n such that b = a(n, ), in which case we say that b is representable by a. Let (n) = (n, n) and σ(p) = 1 p. Ziv Scully Recursion Theorem / 28

66 Fixed points and diagonalization Yet another proof of Cantor s theorem A surjection â : 2 gives a function a : 2 such that for any b : 2, there is some n such that b = a(n, ), in which case we say that b is representable by a. Let (n) = (n, n) and σ(p) = 1 p. a 2 b 2 σ Ziv Scully Recursion Theorem / 28

67 Fixed points and diagonalization Yet another proof of Cantor s theorem A surjection â : 2 gives a function a : 2 such that for any b : 2, there is some n such that b = a(n, ), in which case we say that b is representable by a. Let (n) = (n, n) and σ(p) = 1 p. a 2 b 2 σ By construction, because σ has no fixed points, b(n) a(n, n) for all n, which means b a(n, ) for all n. Therefore, â is not a surjection. Ziv Scully Recursion Theorem / 28

68 Fixed points and diagonalization Yet another proof the halting problem is undecidable Suppose a : 2 decides the halting problem. That is, a(f, X) is 1 if and only if [F](X) halts. Ziv Scully Recursion Theorem / 28

69 Fixed points and diagonalization Yet another proof the halting problem is undecidable Suppose a : 2 decides the halting problem. That is, a(f, X) is 1 if and only if [F](X) halts. Let σ(0) = 1 and σ(1) never terminate. Ziv Scully Recursion Theorem / 28

70 Fixed points and diagonalization Yet another proof the halting problem is undecidable Suppose a : 2 decides the halting problem. That is, a(f, X) is 1 if and only if [F](X) halts. Let σ(0) = 1 and σ(1) never terminate. a 2 b 2 σ Ziv Scully Recursion Theorem / 28

71 Fixed points and diagonalization Yet another proof the halting problem is undecidable Suppose a : 2 decides the halting problem. That is, a(f, X) is 1 if and only if [F](X) halts. Let σ(0) = 1 and σ(1) never terminate. a 2 b 2 σ By construction, because σ has no fixed points, b( b ) a( b, b ). This manifests itself as the usual contradiction: a( b, b ) = 0 b( b ) doesn t halt a( b, b ) = 1 Ziv Scully Recursion Theorem / 28

72 Fixed points and diagonalization Yet another cookie-cutter diagonalization proof The general picture, due originally to Lawvere and nicely explained by Yanofsky ( X X a Y X b Y σ X = domain = diagonal a = application Y = truth values σ = shuffle b = bad, where bad means not representable as a(x, ) for some x. Ziv Scully Recursion Theorem / 28

73 Fixed points and diagonalization Cookies contrapositively cut X X a Y X b Y σ Contrapositively, if b is representable as a(x, ) for some x, then σ must have a fixed point. Ziv Scully Recursion Theorem / 28

74 Fixed points and diagonalization A sketchier, boxier recursion theorem Fix total computable P. Let a(f, x) = [f](x) and σ([f]) = σ([p(f)]). (We strategically choose not to worry about how σ is well-defined as part of the composition but isn t on its own.) a TM b σ TM Ziv Scully Recursion Theorem / 28

75 Fixed points and diagonalization A sketchier, boxier recursion theorem Fix total computable P. Let a(f, x) = [f](x) and σ([f]) = σ([p(f)]). (We strategically choose not to worry about how σ is well-defined as part of the composition but isn t on its own.) a TM b σ TM All of, a, and σ do straightforward computations, so b is computable and therefore representable as a( b, ). Ziv Scully Recursion Theorem / 28

76 Fixed points and diagonalization A sketchier, boxier recursion theorem Fix total computable P. Let a(f, x) = [f](x) and σ([f]) = σ([p(f)]). (We strategically choose not to worry about how σ is well-defined as part of the composition but isn t on its own.) a TM b σ TM All of, a, and σ do straightforward computations, so b is computable and therefore representable as a( b, ). This means σ has a fixed point, or, equivalently, P has a fixed point modulo equivalence of encoded Turing machines. Ziv Scully Recursion Theorem / 28