In class we talked about two functions in racket i-e letcc and throw. The instructor said that let/cc has something to do with calling with the current continuation and throw just applies a to b e.g.
(throw a b)
I haven't been able to find much about these online. I did find a question asked for letcc on stackoverflow but I have not been able to understand the functioning of letcc from it completely. Could someone explain these two in simple words along with a simple example?
Edit1: Also in my practice mid exam we are given two question related to it.
For each of the following expressions that contain uses of let/cc, what is the value of each expression?
(let/cc k (throw (throw k 5) 6))
(let/cc k (throw k ((( lambda (x) x) k) (∗ 5 5))))
The answers to these are 5 and 25 respectively. I just wanna understand the two concepts so that I can work with questions like these in my midterm exam.
Let's look at let/cc first.
The expression (let/cc k e) will 1) capture the current continuation (represented as a function) then 2) bind the variable k to the captured continuation and finally 3) evaluate the expression e.
A few examples are in order.
If during evaluation of the expression e, the captured contination k is not called, then the value of the let/cc expression is simply the value(s) to which the expression e evaluated to.
> (+ 10 (let/cc k 32))
42
If on the other hand k is called with a value v then the value of the entire let\cc expression becomes v.
> (+ 10 (let/cc k (+ 1 (k 2))))
12
Notice that the part (+ _) around the call (k 2) is skipped.
The value is return to the continuation of (let/cc ...) immediately.
The most common use of let/cc is to mimic the control struct return known from many statement based languages. Here is the classical is-it-a-leap-year problem:
(define (divisible-by? y k)
(zero? (remainder y k)))
(define (leap-year? y)
(let/cc return
(when (not (divisible-by? y 4))
(return #f))
(when (not (divisible-by? y 100))
(return #t))
(when (not (divisible-by? y 400))
(return #f))
#t))
(for/list ([y (in-range 1898 1906)])
(list y (leap-year? y)))
Now what about throw ? That depends on which throw we are talking about.
Is it the one from misc1/throw ?
Or perhaps the one from Might's article? http://matt.might.net/articles/programming-with-continuations--exceptions-backtracking-search-threads-generators-coroutines/
Or perhaps you are using the definition
(define (throw k v)
(k v))
If the latter, then you can replace (k v) in my examples with (throw k v).
UPDATE
Note that the continuation bound to k can be used more than once - it can also be used outside the let/cc expression. Consider this example:
(define n 0)
(let ([K (let/cc k k)])
(when (< n 10)
(displayln n)
(set! n (+ n 1))
(K K)))
Can you figure out what it does without running it?
Here is a "step by step" evaluation of the nested let/cc example.
(let/cc k0
((let/cc k1
(k0 (sub1 (let/cc k2
(k1 k2)))))
1))
(let/cc k0
(k2 1))
(let/cc k0
((let/cc k1
(k0 (sub1 1)))
1))
(let/cc k0
((let/cc k1
(k0 0))
1))
0
Related
im currently writing a compiler in OCaml for a subset of scheme and am having trouble understanding how to compile with continuations. I found some great resources, namely:
The cps slides of the cmsu compiler course:
https://www.cs.umd.edu/class/fall2017/cmsc430/
This explanation of another cs course:
https://www.cs.utah.edu/~mflatt/past-courses/cs6520/public_html/s02/cps.pdf
Matt Mights posts on a-normal form and cps:
http://matt.might.net/articles/a-normalization/ and
http://matt.might.net/articles/cps-conversion/
Using the anormal transformation introduced in the anormal-paper, I now have code where function calls are either bound to a variable or returned.
Example:
(define (fib n)
(if (<= n 1)
n
(+ (fib (- n 1))
(fib (- n 2)))))
becomes:
(define (fib n)
(let ([c (<= n 1)])
(if c
n
(let ([n-1 (- n 1)])
(let ([v0 (fib n-1)])
(let ([n-2 (- n 2)])
(let ([v1 (fib n-2)])
(+ v0 v1)))))))
In order to cps-transform, I now have to:
add cont-parameters to all non-primitive functions
call the cont-parameter on tail-positions
transform all non-primitive function calls, so that they escape the let-binding and become an extra lambda with the previous let-bound variable as sole argument and the previous let-body
as the body
The result would look like:
(define (fib n k)
(let ([c (<= n 1)])
(if c
(k n)
(let ([n-1 (- n 1)])
(fib n-1
(lambda (v0)
(let ([n-2 (- n 2)])
(fib n-2
(lambda (v1)
(k (+ v0 v1))))))))))
Is this correct?
The csmu-course also talks about how Programs in CPS require no stack and never return. Is that because we don't need to to save the adresses to return to and closures as well as other datatypes are stored on the heap and references are kept alive by using the closures?
The csmu also talks about desugaring of call/cc:
(call/cc) => ((lambda (k f) (f k k)))
when using such desugaring, how does:
(+ 2 (call/cc (lambda (k) (k 2))))
in MIT-Scheme return 4, since the current continuation would probably be something like display?
is this correct?
(define (fib n k)
(let ([c (<= n 1)])
(if c
(k n)
(let ([n-1 (- n 1)])
(fib n-1
(lambda (v0)
(let ([n-2 (- n 2)])
(fib n-2
(lambda (v1)
(k (+ v0 v1))))))))))
you get an A+ 💯
The csmu-course also talks about how Programs in CPS require no stack and never return. Is that because we don't need to to save the addresses to return to and closures as well as other datatypes are stored on the heap and references are kept alive by using the closures?
Exactly! See Chicken Complilation Process for an in-depth read about such a technique.
The csmu also talks about desugaring of call/cc:
(call/cc) => ((lambda (k f) (f k k)))
That doesn't look quite right. Here's a desugar of call/cc from Matt Might -
call/cc => (lambda (f cc) (f (lambda (x k) (cc x)) cc))
The essence of the idea of compiling with continuations is that you want to put an order on the evaluation of arguments passed to each function and after you evaluate that argument you send its value to the continuation passed.
It is required for the language in which you rewrite the code in CPS form to be tail recursive, otherwise it will stack empty frames, followed only by a return. If the implementation language does not impose tail-recursion you need to apply more sophisticated methods to get non-growing stack for cps code.
Take care, if you do it, you also need to change the signature of the primitives. The primitives will also be passed a continuation but they return immediately the answer in the passed continuation, they do not create other continuations.
The best reference about understanding how to compile with continuations remains the book of Andrew W. Appel and you need nothing more.
Im trying to created the function "apply N times" in Dr Racket but Don't know where I'm going wrong. My code seems to be correct but clearly I'm missing something. Printed below is the code and the error I'm getting.
(define (applyNtimes F n)
(lambda (x)
(if (= n 0) x
(F (applyNtimes F (- n 1))))))
(define cdr3 (applyNtimes cdr 3))
(cdr3 '(1 2 3 4 4 5))
and This is the error I'm getting:
cdr: contract violation
expected: pair?
given: #
the expected output should be
(4 4 5)
Here's the problem, you are trying to apply F to the return of applyNtimes but think for a second, what does that return? a lambda.
This means in your case, we're trying to apply cdr to a lambda, whoops.
In order to get the value of this lambda to actually use it, we have to invoke it. Doing this is pretty easy, just changing
(F (applyNtimes F (- n 1))))))
to
(F ( (applyNtimes F (- n 1)) ;;Get the Lambda
x))))) ;; and apply it to x
To understand this let's break it apart, first we apply F to something. Since in our case F is really cdr we'd better give it some form of pair.
In this code the 'Something' is the result of applying (applyNTimes F (- n 1)) to x.
Well this leads us to some recursion, where what we're really doing is
(F (F (F ... ((applyNTimes F (- n whatever)) x))))
Ok so how does this recursion end? Well when (- n whatever) is 0, we return the lambda
(lambda (x) x)
Which really turns this whole thing into
(F (F (F (F (F (F x))))))
Which is our desired applyNtimes function.
I'm in a Scheme class and I was curious about writing a recursive function without using define. The main problem, of course, is that you cannot call a function within itself if it doesn't have a name.
I did find this example: It's a factorial generator using only lambda.
((lambda (x) (x x))
(lambda (fact-gen)
(lambda (n)
(if (zero? n)
1
(* n ((fact-gen fact-gen) (sub1 n)))))))
But I can't even make sense of the first call, (lambda (x) (x x)): What exactly does that do? And where do you input the value you want to get the factorial of?
This is not for the class, this is just out of curiosity.
(lambda (x) (x x)) is a function that calls an argument, x, on itself.
The whole block of code you posted results in a function of one argument. You could call it like this:
(((lambda (x) (x x))
(lambda (fact-gen)
(lambda (n)
(if (zero? n)
1
(* n ((fact-gen fact-gen) (sub1 n)))))))
5)
That calls it with 5, and returns 120.
The easiest way to think about this at a high level is that the first function, (lambda (x) (x x)), is giving x a reference to itself so now x can refer to itself, and hence recurse.
The expression (lambda (x) (x x)) creates a function that, when evaluated with one argument (which must be a function), applies that function with itself as an argument.
Your given expression evaluates to a function that takes one numeric argument and returns the factorial of that argument. To try it:
(let ((factorial ((lambda (x) (x x))
(lambda (fact-gen)
(lambda (n)
(if (zero? n)
1
(* n ((fact-gen fact-gen) (sub1 n)))))))))
(display (factorial 5)))
There are several layers in your example, it's worthwhile to work through step by step and carefully examine what each does.
Basically what you have is a form similar to the Y combinator. If you refactored out the factorial specific code so that any recursive function could be implemented, then the remaining code would be the Y combinator.
I have gone through these steps myself for better understanding.
https://gist.github.com/z5h/238891
If you don't like what I've written, just do some googleing for Y Combinator (the function).
(lambda (x) (x x)) takes a function object, then invokes that object using one argument, the function object itself.
This is then called with another function, which takes that function object under the parameter name fact-gen. It returns a lambda that takes the actual argument, n. This is how the ((fact-gen fact-gen) (sub1 n)) works.
You should read the sample chapter (Chapter 9) from The Little Schemer if you can follow it. It discusses how to build functions of this type, and ultimately extracting this pattern out into the Y combinator (which can be used to provide recursion in general).
You define it like this:
(let ((fact #f))
(set! fact
(lambda (n) (if (< n 2) 1
(* n (fact (- n 1))))))
(fact 5))
which is how letrec really works. See LiSP by Christian Queinnec.
In the example you're asking about, the self-application combinator is called "U combinator",
(let ((U (lambda (x) (x x)))
(h (lambda (g)
(lambda (n)
(if (zero? n)
1
(* n ((g g) (sub1 n))))))))
((U h) 5))
The subtlety here is that, because of let's scoping rules, the lambda expressions can not refer to the names being defined.
When ((U h) 5) is called, it is reduced to ((h h) 5) application, inside the environment frame created by the let form.
Now the application of h to h creates new environment frame in which g points to h in the environment above it:
(let ((U (lambda (x) (x x)))
(h (lambda (g)
(lambda (n)
(if (zero? n)
1
(* n ((g g) (sub1 n))))))))
( (let ((g h))
(lambda (n)
(if (zero? n)
1
(* n ((g g) (sub1 n))))))
5))
The (lambda (n) ...) expression here is returned from inside that environment frame in which g points to h above it - as a closure object. I.e. a function of one argument, n, which also remembers the bindings for g, h, and U.
So when this closure is called, n gets assigned 5, and the if form is entered:
(let ((U (lambda (x) (x x)))
(h (lambda (g)
(lambda (n)
(if (zero? n)
1
(* n ((g g) (sub1 n))))))))
(let ((g h))
(let ((n 5))
(if (zero? n)
1
(* n ((g g) (sub1 n)))))))
The (g g) application gets reduced into (h h) application because g points to h defined in the environment frame above the environment in which the closure object was created. Which is to say, up there, in the top let form. But we've already seen the reduction of (h h) call, which created the closure i.e. the function of one argument n, serving as our factorial function, which on the next iteration will be called with 4, then 3 etc.
Whether it will be a new closure object or same closure object will be reused, depends on a compiler. This can have an impact on performance, but not on semantics of the recursion.
I like this question. 'The scheme programming language' is a good book. My idea is from Chapter 2 of that book.
First, we know this:
(letrec ((fact (lambda (n) (if (= n 1) 1 (* (fact (- n 1)) n))))) (fact 5))
With letrec we can make functions recursively. And we see when we call (fact 5), fact is already bound to a function. If we have another function, we can call it this way (another fact 5), and now another is called binary function (my English is not good, sorry). We can define another as this:
(let ((another (lambda (f x) .... (f x) ...))) (another fact 5))
Why not we define fact this way?
(let ((fact (lambda (f n) (if (= n 1) 1 (* n (f f (- n 1))))))) (fact fact 5))
If fact is a binary function, then it can be called with a function f and integer n, in which case function f happens to be fact itself.
If you got all the above, you could write Y combinator now, making a substitution of let with lambda.
With a single lambda it's not possible. But using two or more lambda's it is possible. As, all other solutions are using three lambdas or let/letrec, I'm going to explain the method using two lambdas:
((lambda (f x)
(f f x))
(lambda (self n)
(if (= n 0)
1
(* n (self self (- n 1)))))
5)
And the output is 120.
Here,
(lambda (f x) (f f x)) produces a lambda that takes two arguments, the first one is a lambda(lets call it f) and the second is the parameter(let's call it x). Notice, in its body it calls the provided lambda f with f and x.
Now, lambda f(from point 1) i.e. self is what we want to recurse. See, when calling self recursively, we also pass self as the first argument and (- n 1) as the second argument.
I was curious about writing a recursive function without using define.
The main problem, of course, is that you cannot call a function within
itself if it doesn't have a name.
A little off-topic here, but seeing the above statements I just wanted to let you know that "without using define" does not mean "doesn't have a name". It is possible to give something a name and use it recursively in Scheme without define.
(letrec
((fact
(lambda (n)
(if (zero? n)
1
(* n (fact (sub1 n)))))))
(fact 5))
It would be more clear if your question specifically says "anonymous recursion".
I found this question because I needed a recursive helper function inside a macro, where one can't use define.
One wants to understand (lambda (x) (x x)) and the Y-combinator, but named let gets the job done without scaring off tourists:
((lambda (n)
(let sub ((i n) (z 1))
(if (zero? i)
z
(sub (- i 1) (* z i)) )))
5 )
One can also put off understanding (lambda (x) (x x)) and the Y-combinator, if code like this suffices. Scheme, like Haskell and the Milky Way, harbors a massive black hole at its center. Many a formerly productive programmer gets entranced by the mathematical beauty of these black holes, and is never seen again.
I'm unsure of how to turn count-forwards into a tail-recursive program. It takes a non-negative number, n, and returns the list of integers from 0 to n (including n).
Edit: Okay, I finally got this one to work. The problem wasn't that my current program was recursive and I needed to make it tail-recursive- It was just plain wrong. The actual answer is really short and clean. So if anyone else is stuck on this and is also a total programming noob, here's a few hints that might help:
1) Your helper program is designed to keep track of the list so far.
2) Its base case is.. If x = 0.. what do you do? add 0 onto.. something.
3) Recur on x - 1, and then add x onto your list so far.
4) When you get to your actual program, count-forwards, all you need is the helper. But remember that it takes two arguments!
The only recursive function here is list-reverse. It is tail-recursive, because the call to itself is the last operation in the function body.
Your function for generating a nondecreasing sequence from zero to m, which contains the successive results of adding 1 to the previous element, would look something like:
(define (my-reverse lst)
(define (rev-do xs ys)
(if (empty? xs)
ys
(rev-do (cdr xs) (cons (car xs) ys))))
(rev-do lst empty))
(define (seq m n)
(seq-do m n (list m)))
(define (seq-do m n xs)
(if (= m n)
(my-reverse xs)
(let ((next (add1 m)))
(seq-do next n (cons next xs)))))
(define (seq-from-zero m)
(seq 0 m))
Test:
> (seq-from-zero 10)
(0 1 2 3 4 5 6 7 8 9 10)
seq-do is a general function for generating nondecreasing sequences from m to n; it is tail-recursive, because the last operation is the call to itself.
I've also implemented reverse from scratch, so that you can use it in your homework problems.
My solution to exercise 1.11 of SICP is:
(define (f n)
(if (< n 3)
n
(+ (f (- n 1)) (* 2 (f (- n 2))) (* 3 (f (- n 3))))
))
As expected, a evaluation such as (f 100) takes a long time. I was wondering if there was a way to improve this code (without foregoing the recursion), and/or take advantage of multi-core box. I am using 'mit-scheme'.
The exercise tells you to write two functions, one that computes f "by means of a recursive process", and another that computes f "by means of an iterative process". You did the recursive one. Since this function is very similar to the fib function given in the examples of the section you linked to, you should be able to figure this out by looking at the recursive and iterative examples of the fib function:
; Recursive
(define (fib n)
(cond ((= n 0) 0)
((= n 1) 1)
(else (+ (fib (- n 1))
(fib (- n 2))))))
; Iterative
(define (fib n)
(fib-iter 1 0 n))
(define (fib-iter a b count)
(if (= count 0)
b
(fib-iter (+ a b) a (- count 1))))
In this case you would define an f-iter function which would take a, b, and c arguments as well as a count argument.
Here is the f-iter function. Notice the similarity to fib-iter:
(define (f-iter a b c count)
(if (= count 0)
c
(f-iter (+ a (* 2 b) (* 3 c)) a b (- count 1))))
And through a little trial and error, I found that a, b, and c should be initialized to 2, 1, and 0 respectively, which also follows the pattern of the fib function initializing a and b to 1 and 0. So f looks like this:
(define (f n)
(f-iter 2 1 0 n))
Note: f-iter is still a recursive function but because of the way Scheme works, it runs as an iterative process and runs in O(n) time and O(1) space, unlike your code which is not only a recursive function but a recursive process. I believe this is what the author of Exercise 1.1 was looking for.
I'm not sure how best to code it in Scheme, but a common technique to improve speed on something like this would be to use memoization. In a nutshell, the idea is to cache the result of f(p) (possibly for every p seen, or possibly the last n values) so that next time you call f(p), the saved result is returned, rather than being recalculated. In general, the cache would be a map from a tuple (representing the input arguments) to the return type.
Well, if you ask me, think like a mathematician. I can't read scheme, but if you're coding a Fibonacci function, instead of defining it recursively, solve the recurrence and define it with a closed form. For the Fibonacci sequence, the closed form can be found here for example. That'll be MUCH faster.
edit: oops, didn't see that you said forgoing getting rid of the recursion. In that case, your options are much more limited.
See this article for a good tutorial on developing a fast Fibonacci function with functional programming. It uses Common LISP, which is slightly different from Scheme in some aspects, but you should be able to get by with it. Your implementation is equivalent to the bogo-fig function near the top of the file.
To put it another way:
To get tail recursion, the recursive call has to be the very last thing the procedure does.
Your recursive calls are embedded within the * and + expressions, so they are not tail calls (since the * and + are evaluated after the recursive call.)
Jeremy Ruten's version of f-iter is tail-recursive rather than iterative (i.e. it looks like a recursive procedure but is as efficient as the iterative equivalent.)
However you can make the iteration explicit:
(define (f n)
(let iter
((a 2) (b 1) (c 0) (count n))
(if (<= count 0)
c
(iter (+ a (* 2 b) (* 3 c)) a b (- count 1)))))
or
(define (f n)
(do
((a 2 (+ a (* 2 b) (* 3 c)))
(b 1 a)
(c 0 b)
(count n (- count 1)))
((<= count 0) c)))
That particular exercise can be solved by using tail recursion - instead of waiting for each recursive call to return (as is the case in the straightforward solution you present), you can accumulate the answer in a parameter, in such a way that the recursion behaves exactly the same as an iteration in terms of the space it consumes. For instance:
(define (f n)
(define (iter a b c count)
(if (zero? count)
c
(iter (+ a (* 2 b) (* 3 c))
a
b
(- count 1))))
(if (< n 3)
n
(iter 2 1 0 n)))