So, you're telling us that the universe is written in continuation-passing style?
As you proceed with this assignment, you may find the following resources helpful.
1. Define and test a procedure
binary-to-decimal-cps that is a CPSed version of the following
(define binary-to-decimal (lambda (n) (cond [(null? n) 0] [else (+ (car n) (* 2 (binary-to-decimal (cdr n))))])))
binary-to-decimal uses little-endian binary numbers; you should consider binary sequences with one or more trailing
0s to be ill-formed binary numbers (bad data). Here are a few sample calls to make the meaning clear.
> (binary-to-decimal '()) 0 > (binary-to-decimal '(1)) 1 > (binary-to-decimal '(0 1)) 2 > (binary-to-decimal '(1 1 0 1)) 11
2. Define and test a procedure
rember*1-cps that is a CPSed version of the following
rember*1 procedure, which removes the first
? in the arbitrarily nested list
(define rember*1 (lambda (ls) (cond [(null? ls) '()] [(pair? (car ls)) (cond [(equal? (car ls) (rember*1 (car ls))) (cons (car ls) (rember*1 (cdr ls)))] [else (cons (rember*1 (car ls)) (cdr ls))])] [(eqv? (car ls) '?) (cdr ls)] [else (cons (car ls) (rember*1 (cdr ls)))])))
This part of the assignment will be completed in several stages. You should do each of these steps in a separate file. Turn in a file containing the last step, and call it a7.rkt. But it is imperative that you keep each of the previous steps. We will ask to see them at some point in the future.
You should begin with the interpreter below. The interpreter below uses a slightly different language than the interpreters we have written before. Also, instead of Racket's
match, we are using
union-case. The function
union-case takes an element of a union, a union, which is a collection of patterns for special forms, and then a line for each pattern in the union. You can treat it like you would
match. To begin with, you can download the a7-0 and parenthec files. Leave
parenthec in the same directory as your a7 files.
#lang racket (require "parenthec.rkt") (define-union exp (const expr) (mult x1 x2) (sub1 x) (zero x) (if test conseq alt) (capture body) (return k-exp v-exp) (let e body) (var expr) (lambda body) (app rator rand)) (define value-of (lambda (expr env) (union-case expr exp [(const expr) expr] [(mult x1 x2) (* (value-of x1 env) (value-of x2 env))] [(sub1 x) (sub1 (value-of x env))] [(zero x) (zero? (value-of x env))] [(if test conseq alt) (if (value-of test env) (value-of conseq env) (value-of alt env))] [(capture body) (call/cc (lambda (k) (value-of body (lambda (y) (if (zero? y) k (env (sub1 y)))))))] [(return k-exp v-exp) ((value-of k-exp env) (value-of v-exp env))] [(let e body) (let ((a (value-of e env))) (value-of body (lambda (y) (if (zero? y) a (env (sub1 y))))))] [(var expr) (env expr)] [(lambda body) (lambda (a) (value-of body (lambda (y) (if (zero? y) a (env (sub1 y))))))] [(app rator rand) ((value-of rator env) (value-of rand env))]))) (value-of (exp_app (exp_lambda (exp_var 0)) (exp_const 5)) (lambda (y) (error 'value-of "unbound variable ~s" y)))
The a7-0 file contains the
value-of and test program, so you can see how to invoke it. It's not essential that you keep this test program in subsequent files. When you write your own test programs, you will probably prefer not to write them directly in the language of this interpreter. We provide the file parse.rkt, which contains a function
parse that will convert normal Scheme-like syntax of our usual interpreter into the language we are using on this assignment. For example,
> (require "parse.rkt") > (parse '((lambda (a b c) (* a c)) 5 6 7)) '(exp_app (exp_app (exp_app (exp_lambda (exp_lambda (exp_lambda (exp_mult (exp_var 2) (exp_var 0))))) (exp_const 5)) (exp_const 6)) (exp_const 7))
Use this to generate tests for your program. Notice that the output is curried for you. Note also that our parser produces quoted output, but your
value-of will not take quoted expressions as input.
value-of-cps. Use the “let trick” to eliminate the let binding when you CPS the
letline. You might consider always using
k^as the additional continuation variable to your extended environments. Do not apply
k^to the call to
empty-env. This is similar to the behavior of
times-cps-shortcutfrom Assignment 6. Scheme's
call/ccmay not be used in your CPSed interpreter.
apply-k, and add all the calls to those functions. Notice your
apply-envnow take three arguments.
extend-envand replace all 3 of your explicitly higher-order representations of environments with calls to
k) are the same as the formal parameters to the inner functions in
apply-env. Remember, if you add
(else (env y k))as the last line of your
matchexpression, you can test each transformation one at a time.
closureand replace your explicitly higher-order representation of a closure with a call to
k) are the same as the formal parameters to the inner function in
^to each of the formal parameters of your continuation helpers, and change the body to match (that is, do an alpha substitution). Then, ensure that the inner function in each of your continuation helpers uses the same formal parameter as the second argument to
(else (k v))as the last line of your match, you can test each transformation one at a time. If you're good, you can do almost all of this step with a keyboard macro.
(else (k v))line to ensure that you've properly removed all higher-order function representations of continuations.
Place your final version in a file named
a7.rkt and submit it via Oncourse.
For the brainteaser this week, you'll get to learn about
streams, a data-structure
that enables us to process infinite lists of items. Its a lazily-evaluated, memoized
(also termed delayed) list.
To play around, we'll first need to implement a few tools.
(define-syntax cons$ (syntax-rules () ((cons$ x y) (cons x (delay y))))) (define car$ car) (define cdr$ (lambda ($) (force (cdr $))))
We'll get back to those helpers. For now, its enough that they're tweaked
The first question to ask is: how do you build an infinite list? The only reasonable answer is: one item at a time, as needed.
Here, we're going to define an infinite stream of ones.
(define inf-1s (cons$ 1 inf-1s))
It looks like that can't possibly work. We're definining the stream in terms of itself. That's a circular definition. But, in fact, that's precisely what we're after. We're defining a list that has a 1 in front and whose cdr - well, whatever that thing is, it has a 1 in the front of it. And thus its 1s all the way down.
So we can build a procedure
(define take$ (lambda (n $) (cond ((zero? n) '()) (else (cons (car$ $) (take$ (sub1 n) (cdr$ $)))))))
that pulls n items from the stream, and returns them in a list (the $ stands for $tream, by the way).
> (take$ 5 inf-1s) (1 1 1 1 1) > (take$ 10 inf-1s) (1 1 1 1 1 1 1 1 1 1)
So how is this all working? There's nothing to
car$. That's just
car with fancy window-dressing.
cons$ is the first macro definition we've looked at in class. But there's nothing much to it, either. You can think of it as performing a textual transformation – every time we see something of the form (
<thing2>), that code is actually transformed as (
<thing2> isn't evaluated in the process. But do take note of the
delay form, and the
force form in
cdr$. As we've already seen, there are times in which we might like to evaluate an expression only once, and thereafter just return that already computed value. And we might like to hold off on doing that evaluation, instead of doing it right away.
delay does both of those – it creates a promise, of which
force can then force the evaluation. From there on out, every time we force that promise, we get the same value.
> (define worst-random (delay (random 4))) > (force worst-random) 2 > (force worst-random) 2 > (force worst-random) 2
So, to put it all together, we define
inf-1s to be a pair whose
car is 1 and whose
cdr is a promise. When we finally get around to evaluating that promise, we find that its value is in fact
inf-1s – that is, a pair whose
car is 1 and whose
cdr is a promise.
Hopefully that all makes sense. Your task this week is to implement the tribonacci stream. Its the sequence from the exam: 0 is the first tribonacci number, 1 is the second, 1 is the third, and the values thereafter are each the sum of the three previous values in the sequence. Call it
> (car$ trib$) 0 > (car$ (cdr$ trib$)) 1 > (take$ 7 trib$) (0 1 1 2 4 7 13)
By now, you probably have a pretty strong intuition as to the mechanical process by which you CPS programs. As mentioned in class, it is possible to write a program that automatically performs CPS transformations. Write a procedure that takes an expression and returns a CPSed version of that expression. You may find it exceedingly helpful to consult Danvy and Nielsen's "A First-Order One-Pass CPS Transformation", specifically Figure 2 on page 244 and the surrounding discussion. Take note of the four specific cases in which they treat applications.
This is a little more open-ended than most of our problems; you can get it to work on the lambda-calculus, or add more forms if you want (this will probably make it easier to test and to use it). Add a comment in your assignment to tell us how to call and to use your CPSer.