Lecture 7: Recursive Definitions and Environments |
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Consider the function
B = λ f . λ x . f (x f x)
If we evaluate B f B with call-by-name semantics, we get:
B f B |-> f (B f B)
We define Y as λ f . B f B. Observe that:
Y f |-> f (Y f)
The Y combinator takes f and finds its fixed point. Suppose we want to find the fixed point of the function:
λ fact λ n.if n = 1 then 1 else n*(fact n-1)
Now we can apply the Y combinator to this function to get its fixed point (assuming call-by-name):
Y λ fact λ n.(if n = 1 then 1 else n*(fact n-1)) |-> λ n.(if n = 1 then 1 else n*((Y λ fact λ x.if x = 1 then 1 else x*(fact x-1)) x-1)
Note that we are assuming call-by-name semantics here. If we are using call-by-value, then
B f B |-> f(B f B) |-> f(f(B f B)) ...
which diverges. In call-by-value, the evaluation of B must be suspended by wrapping it in a lambda:
B = λ f . λ x . f (λ n.(x f x) n)
so that
B f B |-> f (λ n.(B f B) n)
Which does not diverge.
You don't always want to use the Y combinator to actually implement recursion. Instead, we add a fixed point operator to Lambda Calculus. The new expression type is:
type debruijn =
Cn of int
| Vr of int
| Lm of debruijn
| Ap of debruijn * debruijn
| Fi of debruijn
With substitution semantics for variable bindings, the semantics of this new language are:
exception Error of string;;
let rec plus e m =
match e with
Cn i -> Cn i
| Vr n -> Vr (if n<m then n else n+1)
| Ap (e1,e2) -> Ap (plus e1 m, plus e2 m)
| Lm (e) -> Lm (plus e (m+1))
| Fi (e) -> Fi (plus e (m+1))
let rec subst i v e =
match e with
Cn i -> Cn i
| Vr n -> (if n==i then v
else if n>i then Vr (n-1)
else Vr n)
| Ap (e1,e2) -> Ap (subst i v e1, subst i v e2)
| Lm (e) -> Lm (subst (i+1) (plus v 0) e)
| Fi (e) -> Fi (subst (i+1) (plus v 0) e)
(* Big-step semantics *)
let rec ev5 e =
match e with
Cn i -> Cn i
| Vr x ->
raise (Error "Unbound variable")
| Lm e' -> e
| Ap (e1,e2) ->
(match (ev5 e1) with
Lm e ->
let v = ev5 e2 in
ev5 (subst 0 v e)
| _ -> raise (Error "Applying non-function"))
| Fi e' -> ev5 (subst 0 (Fi e') e')
The expression Fi e evaluates the contained expression e with e bound to a variable. The substitution replaces occurrences of the newly bound variable with the expression Fi e, delaying evaluation until it is needed.
With environment semantics, we define the fixed point operator as follows:
type debValue =
V_Cn of int
| V_Lm of (debValue list) * debruijn
| V_Fi of (debValue list) * debruijn
let rec ev6 env e =
match e with
Cn i -> V_Cn i
| Vr 0 ->
(match env with
v::vs ->
(match v with
V_Fi (env',e') -> ev6 (v::env') e'
| _ -> v)
| _ -> raise (Error "Unbound variable"))
| Vr n ->
(match env with
v::vs -> ev6 vs (Vr (n-1))
| _ -> raise (Error "Unbound variable"))
| Lm e' -> V_Lm (env, e')
| Ap (e1,e2) ->
(match (ev6 env e1) with
(V_Lm (env',e)) ->
let v = ev6 env e2 in
ev6 (v::env') e
| _ -> raise (Error "Applying non-function"))
| Fi e' -> ev6 ((V_Fi (env, e'))::env) e'
The fixed point expression Fi e evaluates to the evaluation of e with Fi e bound in the environment. Again, the expression e, and not the result of evaluating this expression, is stored in the environment; evaluation is delayed until it is needed.
In the rest of these notes, we consider in Scheme some additional ways to implement recursive definitions.
Slogan: Definitions build environments.
Let us add a recursive binding mechanism to our language where we restrict right-hand sides to lambda expressions.
(define-struct rec-let (lhs ; variable
rhs ; required to be a lambda-expression
body))
where lhs is the new local variable, rhs is the lambda-expression defining the value of the new variable, and body is an expression that can use the new local variable. The new variable lhs is visible in both rhs and body. The code for it in the interpreter might look like
((rec-let? M) ... (MEval (rec-let-body M)
(extend env
(rec-let-lhs M)
(make-closure (rec-let-rhs M) E))))
To turn the rhs expression of M into a recursive closure, we desire that E be exactly like the environment that we are in the process of constructing. In other words, we would like a special kind of environment, env*, with the following property:
env* = (extend env (rec-let-lhs M) (make-closure (rec-let-rhs M) env*))
Using the following procedure
(define FENV
(lambda (env*)
(extend env (rec-let-lhs M) (make-closure (rec-let-rhs M) env*))))
the equation for env* can be rewritten as
env* = (FENV env*)
which shows that we want the environment to be the fixed-point of the function FENV (from environments to environments).
If environments are implemented as functions (symbol -> Value), the definition of fix-env, the function that finds the fixed-point of FENV (and any other function mapping environments to enviroments), is simple:
(define fix-env
(lambda (F)
(local
[(define renv (F (lambda (id) (lookup id renv))))]
renv)))
If Scheme were a call-by-name functional language, the code would be even simpler:
(define fix-env
(lambda (F)
(local
[(define renv (F renv))]
renv)))
In a call-by-name language where local is a recursive binding form (as in Scheme),
(local [(define renv (F renv))] renv)))
binds renv to a suspension (closure of no arguments). When renv is evaluated in the body of the local, the suspension evaluates (F renv) in the environment constructed by local, which includes the new binding of renv. Hence,
(fix-env FENV)
= (local
[(define renv (FENV renv))]
renv)
= (local
[(define renv
(extend env (rec-let-lhs M) (make-closure (rec-let-rhs M) renv)))]
renv)
If
M = (make-rec-let x proc ast),
then
(local
[(define renv
(extend env (rec-let-lhs M) (make-closure (rec-let-rhs M) renv)))]
renv)
= (local
[(define renv
(extend env x (make-closure proc renv)))]
renv)
= (define renv' (extend env (make-closure proc renv')))
renv'
which is exactly the recursively defined environment that we need. The (call-by-value) Scheme solution requires the expression
(lambda (id) (lookup id renv))
instead of renv because call-by-value evaluation requires a lambda form to suspend evaluation. On inputs of type environment (symbol -> Value), the function
(lambda (env) (lambda (id) (lookup id env))
is the identity function except on improper values of env (undefined, divergence, or errors). On improper values for env,
(lambda (id) (lookup id env))
is still a legal environment (which will blow up only if it is applied to an Id).
In the absence of this suspension trick, the evaluation of the right hand side of the definition
(define renv
(extend env (rec-let-lhs M) (make-closure (rec-let-rhs M) renv)))
in Scheme, blows up because renv has no value (is undefined) until the evaluation of the definition is complete. The suspension trick wraps the argument renv inside a lambda-expresssion without changing its meaning for proper values of env.
We have used the recursive binding mechanism local in Scheme (and call-by-name Scheme) to implement rec-let. Note that we have placed a strong requirement on the rhs of rec-let in LC. Therefore, this interpreter does not completely explain how local works in Scheme. We will give a more complete explanation later in the course.
| Challenge: Do we need Scheme's local or letrec to interpret rec-let from LC? |
| Hint: Take a look at the sample programs in1 and in2 for assignment 2. The book The Seasoned Schemer by Matthias Felleisen and Daniel Friedman contains an extensive discussion of the Y operator. |
If we represent environments as structures, how do we write fix-env? We need to construct an environment that contains references to copies of itself. In a "lazy" functional language like Haskell or Miranda, we could easily build the required structures using "lazy" constructors that defer the evaluation of their arguments. For example if we added lazy-cons to LC, the code would
(rec-let x (lazy-cons 0 x) ...)
define x as a list consisting of 0 followed by x.
Instead of storing a value in each field of the constructed structure, a lazy constructor stores a suspension specifying how to evaluate the field when it is needed. This is exactly the same mechanism used to support call-by-name argument passing to program defined functions. Note that this approach formulates self referential environments as infinite data structures rather data structures with cycles.
In functional Scheme, we cannot use ordinary data structures (non-functions) to solve the recursion equation
env* = (FENV env*)
because these structures cannot be infinite. However, in imperative Scheme, we can create the the necessary self-references using cycles created by brute-force. We can overwrite the environment field inside the new closure to point to the extended environment. If we adhere to the list representation of environments introduced in Lecture 4, we must abandon the explicit definition the fix-env operation because finding the references to env* that we have to patch in (FENV env*) is either impossible (if F uses constructors that we have not anticipated) or inefficient (because we have to traverse all of the data structure (F env*)).
Fortunately, we don't need a a general function fix-env for our interpreter; all we need is a function (rec-extend M env) that builds the fixed-point for FENV:
(FENV (rec-extend M env)) = (rec-extend M env)
We can define rec-extend as follows
(define rec-extend
(lambda (M env)
(local
[(define var (rec-let-lhs M))
(define rhs (rec-let-rhs M))
(define clsr (make-closure rhs null))
(define renv (extend env var clsr))]
; Is renv the appropriate environment yet? No.
(set-closure-env! clsr renv))))
Given the environment env and the rec-let expression M of a function, (rec-extend env M) extends the environment with the new function binding in M such that the environment embedded in the function binding is the extended environment. The rec-let clause of our interpreter can now be written
((rec-let? M) ... (MEval (rec-let-body M) (rec-extend env M)))
| Exercise: Devise a minor change to the representation of environments as structures that makes it easy to write a general fix-env function. |
| Hint: Scheme boxes. |
In LC we restricted the right hand sides of recursive bindings to lambda-expressions. What happens if we allow arbitrary expressions as right hand sides. To evaluate such a right hand side rhs, we must call (MEval rhs env*) where env* is our self-referential environment. If this evaluation actually tries to lookup the value of var, the computation should abort because the value of var is still being computed. (DrScheme follows this convention, but unfortunately, it is left unspecified in the Scheme standard.) For LC, we can rewrite rec-extend to support this generalization as follows:
(define rec-extend
(lambda (M env)
(local
[(define var (rec-let-lhs M))
(define rhs (rec-let-rhs M))
(define renv (extend env var (void)))]
; (void) is a dummy value for var
; Is renv the appropriate environment yet? No.
(set-first-pair-val! renv (MEval rhs renv)))))
(define set-first-pair-val!
(lambda (env val)
(local
[(define first-pair (first env))]
(set-pair-val! first-pair val))))
To make MEval abort on the inspection of (void) we also need to add the clause
[(void? M) (error 'MEval
"variable referenced before definition in rec-let")]
to the definition of MEval.
| Exercise: Generalize the code from the previous exercise to allow an arbitrary expression as the right hand side of a rec-let. |
| Hint: It is much simpler than the code above. |
Say we have functions from R to R, where R denotes the set of real numbers. We might have a function like
f(x) = 2x + 1
The fixed-point of f is the value x such that x = f(x). Hence, x = 2 x + 1, implying that the fixed-point x = -1.
Does every function from R to R have a fixed-point? No: consider g(x) = x + 1. Substituting x and reducing, we get 0 = 1, which is a contradiction. On the other hand, h(x) = x has an infinite number of solutions. Hence, a function from R to R could have zero, one, several, or an infinite number of fixed-points. However, we will show that for every function from environments to environments, we can construct a fixed-point.
We can use fixed-points to assign a mathematical meaning to any recursive definition expressed in a computational framework. For example, we can assign a mathematical meaning to any recursive definition of a function in terms of already defined functions.
Consider the equation
fact(n) = if zero?(n) then 1 else n * fact(n - 1)
over the natural numbers N = {0, 1, 2, ...} where the primitive functions zero?, *, -, and if-then-else have their usual meaning. The defined function
fact
obviously must satisfy the equation
fact = map n to if zero?(n) then 1 else n * fact(n - 1)
which is mathematically equivalent to being a fixed-point of the functional FACT defined by the non-recursive equation
FACT(f) = map n to if zero?(n) then 1 else n * f(n - 1)
How do we find the fixed-point of such an equation? We interpret F as an inductive definition of the graph of the partial function fix(F).
A partial function mapping N to N is any a subset G of N x N such that (a,b) in G and (a,c) in G imply that b = c. A (total) function mapping N to N is a partial function F such that for every a in N, there exists b such that (a,b) in F. Computed functions correspond to partial functions rather than total functions because they may diverge on some inputs.
We use the term graph of a function to indicate that we are interpreting the function simply as a set of ordered pairs.
Let us restate the definition of the fixed-point fact of FACT as an inductive definition of the graph of a function.
Given a, b in N, the pair (a,b) belongs to the graph of fact iff either:
All of the inductive data definitions presented in Comp 210 can be interpreted in exactly the same way. Each data definition specifies a function D mapping sets to sets. The set specfied by the definition is the least set d such that D(d) = d.
How is the fixed-point of a function F mapping sets over N to sets over N constructed? By repeatedly applying the function F to the empty set. The fixed-point f of F is
Union {Fk(empty) | k = 0,1,...}
where Fk(empty) is k applications of F to empty: F(F(F( ... F(empty)...))) and F0(empty) = empty.
Consider the FACT example above.
FACT1(empty) = {(0,1)}
FACT2(empty) = {(0,1), (1,1)}
FACT3(empty) = {(0,1), (1,1), (2,2)}
...
FACTN+1(empty) = {(0,1), (1,1), (2,2), ..., (N,N!)}
...
Each additional application of FACT permits the recursive evaluation of our equation defining fact to go one level deeper. FACT1 is the function computed by the equation for fact if the first recursive call diverges. FACTN is the function computed by the equation if the Nth recursive call (in depth) diverges.
Any recursive definition of a function from N to N can be interpreted in this way. How do we apply this idea to solving our recursion equation over environments? If we represent envivronments as functions, then exactly the same technique works. The only difference is that the function FENV maps environments to environments instead of N to N.
If environments are represented some other way, then the representation must permit us to find a non-empty least fixed-point for FENV. Otherwise, the representation is incorrect.
Consider the function FADD on sets of N defined by
FADD(S) = {s+1 | s in S}
What is its least fixed-point? The emtpy set! In a computational setting every computable function mapping a data domain D (e.g., N, sets over N, functions over N) into D has a least fixed-point. But the fixed-point may be the trivial "empty" element of D. By definition, a computational data domain must be a partial order with a least element that contains the limits of infinite ascending chains of elements. Similarly, any computable function over such a domain must preserve limits of ascending chains (continuity). It is not difficult to prove that every continuous function f on a domain D has a least fixed point: it the limit (least upper bound) of the ascending chain empty, f(empty), f(f(empty)), ... Consider the recursive definition
x = x + 1
over N. What is the "least" element of N satisfying this equation? There is no conventional natural number that suffices. But the computational analog of the set N includes one additional element, called bottom, which is the meaning of a non-terminating computation. Does the equation
bottom = bottom + 1
hold? Yes! Both sides of the equation represent non-termination.
At this point in the course, we have explained several language constructs, such as procedures and applications, in terms of their Scheme counterparts. A notable exception is environments, which we have represented explicitly. Indeed, since we explicitly manage variable-value associations in our interpreters, we can analyze some of the lower-level behavior of the language such as its memory management.
Consider the evaluation of
(let (x 1)
(let (y 2)
(let (z 3)
...)))
We begin with the empty environment; as we encounter each let, we add a new level to our environment; as we leave the let body, we remove that level. At the end of the entire expression, our environment will be empty again. Hence, our environment will have behaved like a stack. Do environments always behave in this manner?
Consider this expression:
((lambda (x) (lambda (y) (y x))) 10)
This evaluates to an ``apply to 10'' function. This expression
((lambda (z) (lambda (w) (+ z w))) 20)
evaluates to a procedure that adds 20 to its argument. If we apply the former to the latter, what happens to the environment at each stage?
We begin with the empty environment; the first application is performed and the environment is extended with the binding [x = 10]. The closure is now created, and the environment is emptied. Then we evaluate the second expression, which adds the binding [z = 20]. The second closure (call it C2) is produced, and the environment is emptied again.
At this point, we are ready to perform the desired application. When we do, the environment has the bindings [x = 10] and [y = C2]. Now C2 is applied, which has the effect of replacing the current environment with its own, which contains the bindings [z = 20]. The application adds the binding [w = 10], at which point the addition is performed, 30 is returned, and the environment is emptied again.
The moral of this example is that environments for LC programs, no matter how we choose to represent them, branch out like trees, and a simple stack discipline is insufficient for maintaining them. In many programming languages (such as Fortran, Pascal, and C/C++), it is often the programmer's job to manage memory and the collection of environments in existence at any point in the program must be representable by a stack. Such languages prohibit language constructions that can create a collection of environments that cannot represented by a stack. In our interpreters for LC, we have left the task of managing the storage for LC environments to Scheme's run-time system. Otherwise, our interpreters would have been much more complex programs.
| Exercise: Impose restrictions on LC so that it remains a procedural language yet does not require a tree-based management of environments |
| Hint: Follow the example of Pascal. |