Functional Problem Solving (CSC 151 2014F) : Readings
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Summary:
We consider letrec
and “named
let
”, two variants of
let
that permit us to write recursive local procedures.
As you have probably noted, we often find it useful to write helper procedures that accompany our main procedures. For example, we might use a helper to recurse on part of a larger structure or to act as the kernel of a husk-and-kernel procedure.
One issue with this technique is that it often makes little sense for
other procedures to use the helper, so we should restrict access to the
helper procedure. In particular, only the procedure that uses the helper
(unless it's a very generic helper) should be able to access the helper.
We know how to restrict access to variables (using let
and let*
). Can we do the same for procedures?
Yes. As the reading on local bindings suggested, it is possible for a let expression to bind an identifier to a non-recursive procedure:
>
(let ([square (lambda (n) (* n n))]) (square 12))
144
Like any other binding that is introduced in a let expression, this binding is local. Within the body of the let expression, it supersedes any previous binding of the same identifier, but as soon as the value of the let expression has been computed, the local binding evaporates.
However, it is not possible to bind an identifier to a recursively defined procedure in this way. For example, consider the following expression which is intended to find the largest value in a list
>
(let ([largest (lambda (lst) (if (null? (cdr lst)) (car lst) (max (car lst) (largest (cdr lst)))))]) (largest (list 2 1 4 -5 6 13 3)))
largest: undefined; cannot reference an identifier before its definition
The difficulty is that when the lambda expression is
evaluated, the identifier largest
has not yet been bound,
so the value of the lambda expression is a procedure that
includes an unbound identifier. Binding this procedure value to the
identifier largest
creates a new environment, but does not
affect the behavior of procedures that were constructed in the old
environment. So, when the body of the let expression invokes
this procedure, we get the unbound-identifier error.
Changing let
to let*
wouldn't help in this case, since even under let*
the
lambda expression would be completely evaluated before the binding
is established.
letrec
What we need is some variant of let that binds the
identifier to some kind of a place-holder and adds the binding
to the environment first, then computes the value of the
lambda
expression in the new environment, and then finally
substitutes that value for the place-holder. This will work in Scheme,
so long as the procedure is not actually invoked until we get into the
body of the expression.
Fortunately, the designers of Scheme decided to let us write local
procedures using that technique. The keyword associated with this
“recursive binding” variant of let
is named letrec
.
>
(letrec ([largest (lambda (lst) (if (null? (cdr lst)) (car lst) (max (car lst) (largest (cdr lst)))))]) (largest (list 2 1 4 -5 6 13 3)))
13
A letrec expression constructs all of its place-holder bindings simultaneously (in effect), then evaluates all of the lambda expressions simultaneously, and finally replaces all of the place-holders simultaneously. This makes it possible to include binding specifications for mutually recursive procedures (which invoke each other) in the same binding list. Here's a particularly silly example, which takes a list of numbers and alternately adds and subtracts them.
>
(letrec ([up-sum (lambda (ls) (if (null? ls) 0 (+ (car ls) (down-sum (cdr ls)))))] [down-sum (lambda (ls) (if (null? ls) 0 (+ (- (car ls)) (up-sum (cdr ls)))))]) (up-sum (list 1 23 6 12 7)))
-21
; which is 1 - 23 + 6 - 12 + 7.
We can use letrec expressions to separate the husk and the kernel of a recursive procedure without having to define two procedures.
;;; Procedure: ;;; list-index ;;; Parameters: ;;; stuff, a list. ;;; sought, a value. ;;; Purpose: ;;; Find the position of a given value in a given list. ;;; Produces: ;;; pos, a Scheme value ;;; Preconditions: ;;; [No additional.] ;;; Postconditions: ;;; If sought is not in stuff, pos is #f. ;;; If sought is an element of stuff, then ;;; pos is a nonnegative integer. ;;; (list-ref stuff pos) is equal to sought. (define list-index (lambda (stuff sought) (list-index-kernel stuff sought 0))) ;;; Kernel: ;;; list-index-kernel ;;; Parameters: ;;; rest, a list (a sublist of stuff, described above) ;;; sought, a value ;;; bypassed, an integer that counts how many values have ;;; been bypassed in stuff (that is, how many times we've ;;; called cdr since the outside call. ;;; Purpose: ;;; To keep looking for sought in part of the list. (define list-index-kernel (lambda (rest sought bypassed) (cond [(null? rest) #f] [(equal? (car rest) sought) bypassed] [else (list-index-kernel (cdr rest) sought (+ bypassed 1))])))
This technique works, but it's more stylish to construct the kernel procedure inside a letrec expression, so that the extra identifier can be bound to it locally:
(define list-index (lambda (stuff sought) (letrec ([kernel (lambda (rest sought bypassed) (cond [(null? rest) #f] [(equal? (car rest) sought) bypassed] [else (kernel (cdr rest) sought (+ bypassed 1))]))]) (kernel stuff sought 0))))
Of course, now that the recursive kernel procedure is inside the
body of the list-index
procedure, it is not necessary
to pass the value of sought
to the kernel as a
parameter. Instead, the kernel can treat sought
as if it were a constant, since its value doesn't change during any
of the recursive calls.
(define list-index (lambda (stuff sought) (letrec ([kernel (lambda (rest bypassed) (cond [(null? rest) #f] [(equal? (car rest) sought) bypassed] [else (kernel (cdr rest) (+ bypassed 1))]))]) (kernel stuff 0))))
The same approach can be used to perform precondition tests efficiently,
by placing them with the husk in the body of a letrec expression and
omitting them from the kernel. For instance, here's one way to introduce
precondition tests into the drawings-leftmost
procedure from the
reading on verifying preconditions:
(define drawings-leftmost (lambda (drawings) (letrec ([kernel (lambda (drawings) (if (null? (cdr drawings)) (car drawings) (drawing-leftmost (car drawings) (kernel (cdr drawings)))))]) (cond [(null? drawings) (error "drawings-leftmost: Argument must be a *non-empty* list of drawings")] [(not (list? drawings)) (error "drawings-leftmost: Argument must be a *list* of drawings; given" drawings)] [(not (all-drawings? drawings)) (error "drawings-leftmost: Argument list must contain only drawings ; given" drawings)] [else (kernel drawings)]))))
Embedding the kernel inside the definition of
drawings-leftmost
rather than writing a separate
drawings-leftmost-kernel
procedure has another
advantage: It is impossible for an incautious user to invoke
the kernel
procedure directly, bypassing the
precondition tests. The only way to get at the
recursive procedure to which kernel
is bound is
to invoke the procedure within which the binding is established.
We've recycled the name kernel
in this example to drive
home the point that local bindings in separate procedures don't
interfere with one another. Even if both procedures were active at the
same time, the correct kernel
procedure would be used in
each case because the correct local binding would supersede all others.
Hence, even though both list-index
and drawings-leftmost
have a helper named
kernel
, we can safely write
(list-index figure (drawings-leftmost figure))
to find the index of the leftmost drawing in figure
.
Note that drawings-leftmost
effectively has
three local procedures that it relies upon.
Not only does it use kernel
(formerly called
drawings-leftmost-kernel
), but also
all-drawings?
and even
drawing-leftmost
.
While we have used letrec to define a local kernel, we can
also use it to define other local recursive procedures, such
as the all-drawings?
predicate. As we've
seen in the past, we can also use let
or let*
to define non-recursive
procedures. Hence, if we want to define all three helpers within
drawings-leftmost
, we might write something like
the following.
(define drawings-leftmost (let (; drawing-leftmost finds the leftmost of two drawings [drawing-leftmost (lambda (drawing1 drawing2) (if (< (drawing-left drawing1) (drawing-left drawing2)) drawing1 drawing2))]) (lambda (drawings) (letrec (; all-drawings? checks to see if all the values in ; a list are drawings [all-drawings? (lambda (ls) (or (null? ls) (and (drawing? (car ls)) (all-drawings? (cdr ls)))))] ; kernel finds the leftmost drawing w/o verifying preconditions [kernel (lambda (drawings) (if (null? (cdr drawings)) (car drawings) (drawing-leftmost (car drawings) (kernel (cdr drawings)))))]) (cond [(null? drawings) (error "drawings-leftmost: Argument must be a *non-empty* list of drawings")] [(not (list? drawings)) (error "drawings-leftmost: Argument must be a *list* of drawings; given" drawings)] [(not (all-drawings? drawings)) (error "drawings-leftmost: Argument list must contain only drawings ; given" drawings)] [else (kernel drawings)])))))
Many programmers use letrec expressions in writing most of these husk-and-kernel procedures. When there is only one recursive procedure to bind, however, a contemporary Scheme programmer might well use yet another variation of the let expression -- the “named let”.
The named let has the same syntax as a regular let expression, except that there is an identifier between the keyword let and the binding list. The named let binds this extra identifier to a kernel procedure whose parameters are the same as the variables in the binding list and whose body is the same as the body of the let expression. Here's the basic form
(let name ([param1 val1] [param2 val2] ... [paramn valn]) body)
What does this do? In essence, it defines a procedure named
name
with parameters
param1
through
paramn
. Then,
it calls
that procedure on arguments
val1
through
valn
.
You can think of this as a more elegant (and, eventually, more readable)
shorthand for
(letrec ([name (lambda (param1 ... paramn) body)]) (name val1 ... valn))
Alternately, you can think of this as first binding each of
param1
through
paramn
to
the corresponding value. It also creates a procedure with the same
parameters. It then
executes the body. When the body calls the procedure
name
, those bindings get updated, and the body
starts again.
So, for example, one might write the list-index
procedure as
(define list-index (lambda (sought ls) (let kernel ([rest ls] [bypassed 0]) (cond [(null? rest) #f] [(equal? (car rest) sought) bypassed] [else (kernel (cdr rest) (+ bypassed 1))]))))
When we enter the named let, the identifier rest
is bound to the value of ls
and the identifier
bypassed
is bound to 0, just as if we were
entering an ordinary let expression. In addition, however, the
identifier kernel
is bound to a procedure that
has rest
and bypassed
as parameters and the body of the named let as its body. As we
evaluate the cond expression, we may encounter a recursive call
to the kernel
procedure -- in effect, we
re-enter the body of the named let, with rest
now re-bound to the former value of (cdr rest)
and
bypassed
to the former value of (+
bypassed 1)
.
As another example, here's a version of sum
that uses a named let:
(define sum (lambda (numbers) (let kernel ([sum-so-far 0] [remaining numbers]) (if (null? remaining) sum-so-far (kernel (+ sum-so-far (car remaining)) (cdr remaining))))))
Scheme programmers seem to be mixed in their reaction to the named let. Some find it clear and elegant, others find it murky and too special-purpose. Our colleagues prefer it. We'll admit that we first found it murky, but eventually came to prefer it. We hope that you will, too. (That is, we hope that you will come to prefer it, not that we hope you find it murky.)
letrec
?
This question is review: Explain why you cannot name a recursive
procedure locally with let
.
difference
You now have seen a version of list-index
that
uses a letrec
to define its helper locally. Rewrite
newer-difference
so that it similarly uses a
letrec
to define its helper locally.
(define newer-difference (lambda (lst) (new-difference-helper (car lst) (cdr lst)))) (define new-difference-helper (lambda (difference-so-far remaining) (if (null? remaining) difference-so-far (new-difference-helper (- difference-so-far (car remaining)) (cdr remaining)))))
let
us name a local difference
This one might be slightly more difficult, but the best way to
engage the topics is to try them out. You have seen a version of
sum
that uses a named let
to
define its helper locally:
(define new-sum (lambda (numbers) (new-sum-helper 0 numbers))) (define new-sum-helper (lambda (sum-so-far remaining) (if (null? remaining) sum-so-far (new-sum-helper (+ sum-so-far (car remaining)) (cdr remaining))))) (define sum (lambda (numbers) (let kernel ([sum-so-far 0] [remaining numbers]) (if (null? remaining) sum-so-far (kernel (+ sum-so-far (car remaining)) (cdr remaining))))))
Rewrite newer-difference
one more time to
similarly use a named let
to define its helper locally.