Type Systems. Today. 1. What is the Lambda Calculus. 1. What is the Lambda Calculus. Lecture 2 Oct. 27th, 2004 Sebastian Maneth

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1 Today 1. What is the Lambda Calculus? Type Systems 2. Its Syntax and Semantics 3. Church Booleans and Church Numerals 4. Lazy vs. Eager Evaluation (call-by-name vs. call-by-value) Lecture 2 Oct. 27th, 2004 Sebastian Maneth 1. What is the Lambda Calculus introduced in late 1930 s by Alonzo Church and Stephen Kleene 1. What is the Lambda Calculus introduced in late 1930 s by Alonzo Church and Stephen Kleene used in 1936 by Church to prove the undecidability of the Entscheidungsproblem can compute the same as Turing Machines, which is everything we can (intuitively) compute (Church-Turing Thesis). is a formal system designed to investigate function definition function plication recursion is a formal system designed to investigate function definition function plication recursion

2 1. What is the Lambda Calculus 2. Syntax of the Lambda Calculus what do we want? a small core language, into which other language constructs can be translated. There are many such languages: why do we pick out the Lambda Calulus? Turing Machines µ Recursive Functions Chomsky s Type-0 Grammars Cellular Automata etc. Let V be a countable set of variable names. The set of lambda terms (over V) is the smallest set T such that 1. if x V, then x T variable 2. if x V and t 1 T, then x. t 1 T abstraction 3. if t 1, t 2 T, then t 1 t 2 T plication Function abstraction: instead of f(x) = x + 5 write f = x.x + 5 because types are about values of program variables. a lambda term (i.e., T) representing a nameless function, which adds 5 to its parameter 2. Syntax of the Lambda Calculus 2. Syntax of the Lambda Calculus x Function plication: instead of f(x) write f x Example: (x. x + 5) a ply x x + 5 surface syntax Conventions (to save parenthesis) Abstract Syntax Tree a (AST) abstract syntax Example: x.y.z. x z (y z) = x.(y.z. x z (y z)) = x.(y.(z. (x z (y z)) y z ply = x. (y. (z. ((x z) (y z)))) ply ply x z y z surface syntax abstract syntax Conventions (to save parenthesis) plication is left associative: x y z = (x y) z x (y z) scope of abstraction extends as far to the right as possible: plication is left associative: x y z = (x y) z x (y z) scope of abstraction extends as far to the right as possible: x. x y = x. (x y) ( x. x) y x. x y = x. (x y) ( x. x) y

3 2. Semantics of the Lambda Calculus (x.x + 5) a SUBSTITUTE a for x in x Semantics of the Lambda Calculus Example: ( x.y. f (y x) ) 5 ( x. x ) ply redex (REDucible EXpression): ( x. t ) t 1 x x + 5 a can be reduced to (evaluates to): β reduction [ x a ] x + 5 = a + 5 To compute in Lambda Calculus, ALL you do is SUBSTITUTE!! 2. Semantics of the Lambda Calculus Example: ( x.y. f (y x) ) 5 ( x. x ) 2. Semantics of the Lambda Calculus Example: ( x.y. f (y x) ) 5 ( x. x ) = (( x.y. f (y x) ) 5) ( x. x ) because App binds to the left! = (( x.y. f (y x) ) 5) ( x. x ) [ x 5 ]( y. f (y x) ) ( x. x ) because App binds to the left! = ( y. f (y 5) ) ( x. x )

4 2. Semantics of the Lambda Calculus Example: ( x.y. f (y x) ) 5 ( x. x ) 2. Semantics of the Lambda Calculus Example: ( x.y. f (y x) ) 5 ( x. x ) = (( x.y. f (y x) ) 5) ( x. x ) [ x 5 ]( y. f (y x) ) ( x. x ) because App binds to the left! = (( x.y. f (y x) ) 5) ( x. x ) [ x 5 ]( y. f (y x) ) ( x. x ) because App binds to the left! = ( y. f (y 5) ) ( x. x ) = ( y. f (y 5) ) ( x. x ) [ y x. x ]( f (y 5) ) [ y x. x ]( f (y 5) ) = f (x. x 5) = f (x. x 5) f 5 (normal form = cannot be reduced further) 2. Semantics of the Lambda Calculus 2. Semantics of the Lambda Calculus Example: Does every -term have a normal form? Example: Does every -term have a normal form? NO!!! NO!!! ( x. x x ) ( x. x x ) ( x. x x ) ( x. x x ) [ x ( x. x x ) ] ( x x ) [ x ( x. x x ) ] ( x x ) = ( x. x x ) ( x. x x )

5 2. Semantics of the Lambda Calculus 2. Semantics of the Lambda Calculus Example: Does every -term have a normal form? Example: Does every -term have a normal form? NO!!! NO!!! ( x. x x ) ( x. x x ) [ x ( x. x x ) ] ( x x ) ( x. x x ) ( x. x x ) is called the omega combinator =: omega = ( x. x x ) ( x. x x ) ( x. x x ) ( x. x x ) ( x. x x ) ( x. x x ) combinator = closed lambda term = lambda term with no free variables The simplest combinator, identity: id := x. x 2. Semantics of the Lambda Calculus Free vs. Bound Variables: bound x. x y = x. (x y) free scope of x x is bound in its scope Define the set of free variables of a term t, FV(t), as if t = x V, then FV(t) = { x } if t = x. t 1, then FV(t) = FV(t 1 ) \ { x } 3. Church Booleans and Numerals How to encode BOOLEANS into the lambda calculus? tru takes two arguments, selects the FIRST fls takes two arguments, selects the SECOND THEN: if-then-else can be defined as: test x u w = ply x to u w = (k. m. n. k m n) x u w tru := m. n. m fls := m. n. n =: test if t = t 1 t 2, then FV(t) = FV(t 1 ) FV(t 2 ) test tru u w u

6 3. Church Booleans and Numerals How to encode BOOLEANS into the lambda calculus? tru takes two arguments, selects the FIRST fls takes two arguments, selects the SECOND tru := m. n. m fls := m. n. n test := k. m. n. k m n How to do and on these BOOLEANS? and u w = ply u to w fls := (m. n. m n fls) u w =: and 3. Church Booleans and Numerals How to encode BOOLEANS into the lambda calculus? tru takes two arguments, selects the FIRST fls takes two arguments, selects the SECOND tru := m. n. m fls := m. n. n test := k. m. n. k m n How to do and on these BOOLEANS? and u w = ply u to w fls := (m. n. m n fls) u w =: and Define the or and not functions! 3. Church Booleans and Numerals How to encode NUMBERS into the lambda calculus? c0 := s. z. z c1 := s. z. s z c2 := s. z. s (s z) c3 := s. z. s (s (s z)) etc. THEN, the successor function can be defined as scc := n. s. z. s (n s z) 3. Church Booleans and Numerals How to encode NUMBERS into the lambda calculus? c0 := s. z. z c1 := s. z. s z scc := n. s. z. s (n s z) c2 := s. z. s (s z) c3 := s. z. s (s (s z)) How to do plus and times on these Church Numerals? plus := m. n. s. z. m s (n s z) scc c0 s. z. s (c0 s z) s. z. s z = c1 ply m times the successor to n just like fls! Select the second argument.

7 3. Church Booleans and Numerals How to encode NUMBERS into the lambda calculus? 3. Church Booleans and Numerals How to encode NUMBERS into the lambda calculus? c0 := s. z. z c1 := s. z. s z c2 := s. z. s (s z) c3 := s. z. s (s (s z)) scc := n. s. z. s (n s z) c0 := s. z. z c1 := s. z. s z c2 := s. z. s (s z) c3 := s. z. s (s (s z)) scc := n. s. z. s (n s z) plus := m. n. s. z. m s (n s z) How to do plus and times on these Church Numerals? plus := m. n. s. z. m s (n s z) ply m times the successor to n times := m. n. m (plus n) c0 Questions: 1. Write a function subt for subtraction on Church Numerals. 2. How can other datatypes be encoded into the lambda calculus, like, e.g., lists, trees, arrays, and variant records? ply m times (plus n) to c0 4. Lazy vs. Eager Evaluation What does this lambda term evaluate to?? tru id omega 4. Lazy vs. Eager Evaluation What does this lambda term evaluate to?? tru id omega (m. n. m) (x. x) ((x. x x) (x. x x)) where to start evaluating? which redex?? ply ply ply m id x x n ply ply m x x x x

8 4. Lazy vs. Eager Evaluation What does this lambda term evaluate to?? tru id omega (m. n. m) (x. x) ((x. x x) (x. x x)) where to start evaluating? which redex?? 4. Lazy vs. Eager Evaluation What does this lambda term evaluate to?? tru id omega (m. n. m) (x. x) ((x. x x) (x. x x)) where to start evaluating? which redex?? redex1 ply redex2 ply ply m id x x n ply ply m x x x x redex1 ply redex2 ply ply m n id x ply x ply m x x x x if we always reduce redex2 then this lambda term has NO semantics. 4. Lazy vs. Eager Evaluation A redex if outermost, if in the AST it has no ancestor that is a redex. 4. Lazy vs. Eager Evaluation A redex if outermost, if in the AST it has no ancestor that is a redex. outermost redexes outermost redexes A redex if leftmost, if in the AST it has no redex to the left of it.

9 4. Lazy vs. Eager Evaluation A redex if outermost, if in the AST it has no ancestor that is a redex. outermost redexes A redex if leftmost, if in the AST it has no redex to the left of it. 4. Lazy vs. Eager Evaluation A redex if outermost, if in the AST it has no ancestor that is a redex. outermost redexes A redex if leftmost, if in the AST it has no redex to the left of it. 4. Lazy vs. Eager Evaluation A redex if outermost, if in the AST it has no ancestor that is a redex. leftmost outermost redexes A redex if leftmost, if in the AST it has no redex to the left of it. 4. Lazy vs. Eager Evaluation leftmost Evaluation Strategies: normal order outermost redexes always reduce leftmost outermost redex first lazy eager call-by-name call-by-need call-by-value like normal order, but NOT inside abstractions like call-by-name but with sharing reduce only value-redexes (= argument is a value) and do this leftmost right branch of

10 4. Lazy vs. Eager Evaluation Lazy seems better than eager, because more terms can be evaluated! can you define an infinite list consisting of all prime numbers? (with lazy evaluation you can fetch the first n numbers of this list!) > fetch c3 primelist should compute the list [ 2, 3, 5 ] 4. Lazy vs. Eager Evaluation Lazy seems better than eager, because more terms can be evaluated! can you define an infinite list consisting of all prime numbers? (with lazy evaluation you can fetch the first n numbers of this list!) > fetch c3 primelist should compute the list [ 2, 3, 5 ] If a term evaluates to a normal form n using eager evaluation, then it also evaluates to n using lazy evaluation. can you prove this?!? what about the number of eval. steps needed by eager vs. lazy? If a term evaluates to a normal form n using eager evaluation, then it also evaluates to n using lazy evaluation. can you prove this?!? what about the number of eval. steps needed by eager vs. lazy? Lazy is hard to implement efficiently because copies of unevaluated lambda terms must be shared in order not to have duplicate reductions Lazy is hard to implement it efficiently because lots of duplicate reductions might be done. Most FL s use call-by-value. Also the TaPL book! fct = n.if eq n c0 then c1 else (times n (fct (prd n))) e.g. fct c3 needs to unroll 4 times the definition (expand) recursion fct c3 = if eq c3 c0 then c1 else (times c3 ( if eq c2 c0 then c1 else (times c2 ( if eq c1 c0 then c1 else (times c1 ( if eq c0 c0 then c1 else (..)..) ( evaluates to c6 ) fct = n.if eq n c0 then c1 else (times n (fct (prd n))) e.g. fct c3 needs to unroll 4 times the definition (expand) recursion fct c3 = if eq c3 c0 then c1 else (times c3 ( if eq c2 c0 then c1 else (times c2 ( if eq c1 c0 then c1 else (times c1 ( if eq c0 c0 then c1 else (..)..) ( evaluates to c6 ) Is there a combinator doing the unrolling, when plied to fct?

11 fct = n.if eq n c0 then c1 else (times n (fct (prd n))) recursion such a combinator is similar to omega! (x. x x) (x. x x) First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 additionally to copying itself it should each time split of one plication of the definition of fct Is there a combinator doing the unrolling, when plied to fct? First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 lazy! n.if eq n c0 then c1 else (times n (h h (prd n))) c3

12 First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 eager! g (g (h h)) c3 g(g(g(h h) c3 lazy! n.if eq n c0 then c1 else (times n (h h (prd n))) c3 First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 lazy! n.if eq n c0 then c1 else (times n (h h (prd n))) c3 if eq c3 c0 then c1 else (times c3 (h h (prd c3))) First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 lazy! n.if eq n c0 then c1 else (times n (h h (prd n))) c3 if eq c3 c0 then c1 else (times c3 (h h (prd c3))) times c3 (h h (prd c3)) First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 lazy! n.if eq n c0 then c1 else (times n (h h (prd n))) c3 if eq c3 c0 then c1 else (times c3 (h h (prd c3))) times c3 (h h (prd c3)) times c3 (g (h h) (prd c3))

13 First, under call-by-name (lazy) evaluation. (cbn) fixed-point combinator Y := f. (x. f (x x)) (x. f (x x)) g := fct. n.if eq n c0 then c1 else (times n (fct (prd n))) Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 Y g c3 (x. g (x x)) (x. g (x x)) c3 g (h h) c3 lazy! n.if eq n c0 then c1 else (times n (h h (prd n))) c3 if eq c3 c0 then c1 else (times c3 (h h (prd c3))) times c3 (h h (prd c3)) times c3 (g (h h) (prd c3)) times c3 c2 c1 c1 Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 g (y. h h y) c3 -guard

14 Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 g (y. h h y) c3 -guard n.if eq n c0 then c1 else (times n ((y. h h y)(prd n))) c3 Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 g (y. h h y) c3 -guard n.if eq n c0 then c1 else (times n ((y. h h y)(prd n))) c3 if eq c3 c0 then c1 else (times c3 ((y. h h y)(prd c3))) Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 g (y. h h y) c3 -guard n.if eq n c0 then c1 else (times n ((y. h h y)(prd n))) c3 if eq c3 c0 then c1 else (times c3 ((y. h h y)(prd c3))) times c3 ((y. h h y)(prd c3)) Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 g (y. h h y) c3 -guard n.if eq n c0 then c1 else (times n ((y. h h y)(prd n))) c3 if eq c3 c0 then c1 else (times c3 ((y. h h y)(prd c3))) unguard times c3 ((y. h h y)(prd c3)) times c3 h h (prd c3)

15 Now, under eager (call-by-value) evaluation. (cbv) fixed-point combinator fix := f. (x. f (y. x x y)) (x. f (y. x x y)) fix g c3 (x. g (y. x x y)) (x. g (y. x x y)) c3 Question: Can you feel why the lambda calculus is Turing complete? Can you prove it? What does it take to be Turing complete? g (y. h h y) c3 -guard n.if eq n c0 then c1 else (times n ((y. h h y)(prd n))) c3 if eq c3 c0 then c1 else (times c3 ((y. h h y)(prd c3))) unguard times c3 ((y. h h y)(prd c3)) times c3 h h (prd c3) times c3 g (y. h h y) (prd c3) times c3 c2 c1 c1 redex (REDucible EXpression): ( x. t ) s β reduction: ( x. t ) s := [ x s ] t substitution A. only replace the FREE occurrences of x in t!! [ x s ] : B. if replacing within ( y.u ) then y should NOT be FREE in s!! DEFINE [ x s ] t, by induction on the structure of t: redex (REDucible EXpression): ( x. t ) s β reduction: ( x. t ) s := [ x s ] t substitution A. only replace the FREE occurrences of x in t!! [ x s ] : B. if replacing within ( y.u ) then y should NOT be FREE in s!! DEFINE [ x s ] t, by induction on the structure of t: 1. [ x s ] y = 2. [ x s ] y. t 1 = 3. [ x s ] t 1 t 2 =

16 redex (REDucible EXpression): ( x. t ) s β reduction: ( x. t ) s := [ x s ] t substitution A. only replace the FREE occurrences of x in t!! [ x s ] : B. if replacing within ( y.u ) then y should NOT be FREE in s!! DEFINE [ x s ] t, by induction on the structure of t: redex (REDucible EXpression): ( x. t ) s β reduction: ( x. t ) s := [ x s ] t substitution A. only replace the FREE occurrences of x in t!! [ x s ] : B. if replacing within ( y.u ) then y should NOT be FREE in s!! DEFINE [ x s ] t, by induction on the structure of t: 1. [ x s ] y = s if y=x, and y otherwise 2. [ x s ] y. t 1 = 3. [ x s ] t 1 t 2 = 1. [ x s ] y = s if y=x, and y otherwise 2. [ x s ] y. t 1 = y. [ x s ] t 1 if y x and y FV(s) A,B 3. [ x s ] t 1 t 2 = redex (REDucible EXpression): ( x. t ) s β reduction: ( x. t ) s := [ x s ] t substitution A. only replace the FREE occurrences of x in t!! [ x s ] : B. if replacing within ( y.u ) then y should NOT be FREE in s!! DEFINE [ x s ] t, by induction on the structure of t: redex (REDucible EXpression): ( x. t ) s β reduction: ( x. t ) s := [ x s ] t substitution A. only replace the FREE occurrences of x in t!! [ x s ] : B. if replacing within ( y.u ) then y should NOT be FREE in s!! DEFINE [ x s ] t, by induction on the structure of t: 1. [ x s ] y = s if y=x, and y otherwise 2. [ x s ] y. t 1 = y. [ x s ] t 1 if y x and y FV(s) A,B 3. [ x s ] t 1 t 2 = ([ x s ] t 1 ) ([ x s ] t 2 ) 1. [ x s ] y = s if y=x, and y otherwise 2. [ x s ] y. t 1 = y. [ x s ] t 1 if y x and y FV(s) A,B 3. [ x s ] t 1 t 2 = ([ x s ] t 1 ) ([ x s ] t 2 ) to py 2., renaming of BOUND y s in t 1 might be necessary!!! = alpha-conversion

17 Idea: let variable occurrences directly point to their binders, rather than referring to them by name. use natural numbers k, meaning the k-th enclosing e.g. x. y. x ( y x ) BECOMES.. 1 ( 0 1 ) Idea: let variable occurrences directly point to their binders, rather than referring to them by name. use natural numbers k, meaning the k-th enclosing distance: 0 e.g. x. y. x ( y x ) BECOMES.. 1 ( 0 1 ) distance: 1 Idea: let variable occurrences directly point to their binders, rather than referring to them by name. use natural numbers k, meaning the k-th enclosing distance: 0 e.g. x. y. x ( y x ) BECOMES.. 1 ( 0 1 ) distance: 1 Then, every CLOSED term has a unique debruijn representation! Idea: let variable occurrences directly point to their binders, rather than referring to them by name. use natural numbers k, meaning the k-th enclosing distance: 0 e.g. x. y. x ( y x ) BECOMES.. 1 ( 0 1 ) distance: 1 Then, every CLOSED term has a unique debruijn representation! what to do with free variables?? use naming context Γ V *. E.g., bca means b 2, c 1, a 0

18 fix a naming context Γ V *. lambda term (Γ = xu) y. u y removenames Γ restorenames Γ remov Γ resto Γ nameless lambda term. 1 0 (Γ = xuy) shift(d, s) := shiftb(d, 0, s) DON T shift vars with index <0!! substitution [ 1 s ](. 2 ) Γ=xu Γ =Γy increment all free vars in s by one! substitution [ 1 s ](. 2 ) Γ=xu Γ =Γy increment all free vars in s by one! [ j s ](. t 1 ) =. [ j+1 shift(1, s) ] t 1 [ j s ](. t 1 ) =. [ j+1 shift(1, s) ] t 1 shift function must keep track of BOUND vars in order to ONLY shift the FREE vars. shift function must keep track of BOUND vars in order to ONLY shift the FREE vars. shift(d, s) := shiftb(d, 0, s) substitution [ 1 s ](. 2 ) Γ=xu Γ =Γy [ j s ](. t 1 ) =. [ j+1 shift(1, s) ] t 1 DON T shift vars with index <0!! shiftb(d, b, k) = k if k<b, and k+d otherwise shiftb(d, b,. t 1 ) =. shiftb(d, b+1, t 1 ) shiftb(d, b, t 1 t 2 ) = shiftb(d, b, t 1 ) shiftb(d, b, t 2 ) increment all free vars in s by one! fix a naming context Γ V *. removenames(γ, x) = index of rightmost x in Γ removenames(γ, x. t 1 ) =. removenames(γx, t 1 ) removenames(γ, t 1 t 2 ) = removenames(γ, t 1 ) removenames(γ, t 2 ) restorenames(γ, k) = k-th name in Γ restorenames(γ,. t) = x. restorenames(γx, t 1 ) x is the first name not in Γ restorenames(γ, t 1 t 2 ) = restorenames(γ, t 1 ) restorenames(γ, t 2 ) shift function must keep track of BOUND vars in order to ONLY shift the FREE vars.