I'm new to Racket and I'm trying to define a function that takes a list of divisors and a list of numbers to test, and applies direct recursion and a function 'drop-divisible' for each element in the list of divisors.
I defined a function drop-divisible which takes a number and a list of numbers, and returns a new list containing only those numbers not "non-trivially divisible" by the number. This function is not the problem, it works. So what I have trouble with is the function who is gonna call this function, and itself.
Here is what I've come up with. I can imagine that this is far from correct, but I have no idea what to do.
(define (sieve-with divisors testlist)
(if (null? divisors)
'()
(begin
(drop-divisible (first divisors) testlist)
(sieve-with (rest divisors) testlist))))
You need to use tail recursion:
(define (sieve-with divisors list)
(cond [(empty? divisors) list]
[else (sieve-with (rest divisors)
(drop-divisible (first divisors) list))]))
Also, stay away from begin as much as possible. Stick to the functional paradigm.
Related
I wonder how do you, experienced lispers / functional programmers usually make decision what to use. Compare:
(define (my-map1 f lst)
(reverse
(let loop ([lst lst] [acc '()])
(if (empty? lst)
acc
(loop (cdr lst) (cons (f (car lst)) acc))))))
and
(define (my-map2 f lst)
(if (empty? lst)
'()
(cons (f (car lst)) (my-map2 f (cdr lst)))))
The problem can be described in the following way: whenever we have to traverse a list, should we collect results in accumulator, which preserves tail recursion, but requires list reversion in the end? Or should we use unoptimized recursion, but then we don't have to reverse anything?
It seems to me the first solution is always better. Indeed, there's additional complexity (O(n)) there. However, it uses much less memory, let alone calling a function isn't done instantly.
Yet I've seen different examples where the second approach was used. Either I'm missing something or these examples were only educational. Are there situations where unoptimized recursion is better?
When possible, I use higher-order functions like map which build a list under the hood. In Common Lisp I also tend to use loop a lot, which has a collect keyword for building list in a forward way (I also use the series library which also implements it transparently).
I sometimes use recursive functions that are not tail-recursive because they better express what I want and because the size of the list is going to be relatively small; in particular, when writing a macro, the code being manipulated is not usually very large.
For more complex problems I don't collect into lists, I generally accept a callback function that is being called for each solution. This ensures that the work is more clearly separated between how the data is produced and how it is used.
This approach is to me the most flexible of all, because no assumption is made about how the data should be processed or collected. But it also means that the callback function is likely to perform side-effects or non-local returns (see example below). I don't think it is particularly a problem as long the the scope of the side-effects is small (local to a function).
For example, if I want to have a function that generates all natural numbers between 0 and N-1, I write:
(defun range (n f)
(dotimes (i n)
(funcall f i)))
The implementation here iterates over all values from 0 below N and calls F with the value I.
If I wanted to collect them in a list, I'd write:
(defun range-list (N)
(let ((list nil))
(range N (lambda (v) (push v list)))
(nreverse list)))
But, I can also avoid the whole push/nreverse idiom by using a queue. A queue in Lisp can be implemented as a pair (first . last) that keeps track of the first and last cons cells of the underlying linked-list collection. This allows to append elements in constant time to the end, because there is no need to iterate over the list (see Implementing queues in Lisp by P. Norvig, 1991).
(defun queue ()
(let ((list (list nil)))
(cons list list)))
(defun qpush (queue element)
(setf (cdr queue)
(setf (cddr queue)
(list element))))
(defun qlist (queue)
(cdar queue))
And so, the alternative version of the function would be:
(defun range-list (n)
(let ((q (queue)))
(range N (lambda (v) (qpush q v)))
(qlist q)))
The generator/callback approach is also useful when you don't want to build all the elements; it is a bit like the lazy model of evaluation (e.g. Haskell) where you only use the items you need.
Imagine you want to use range to find the first empty slot in a vector, you could do this:
(defun empty-index (vector)
(block nil
(range (length vector)
(lambda (d)
(when (null (aref vector d))
(return d))))))
Here, the block of lexical name nil allows the anonymous function to call return to exit the block with a return value.
In other languages, the same behaviour is often reversed inside-out: we use iterator objects with a cursor and next operations. I tend to think it is simpler to write the iteration plainly and call a callback function, but this would be another interesting approach too.
Tail recursion with accumulator
Traverses the list twice
Constructs two lists
Constant stack space
Can crash with malloc errors
Naive recursion
Traverses list twice (once building up the stack, once tearing down the stack).
Constructs one list
Linear stack space
Can crash with stack overflow (unlikely in racket), or malloc errors
It seems to me the first solution is always better
Allocations are generally more time-expensive than extra stack frames, so I think the latter one will be faster (you'll have to benchmark it to know for sure though).
Are there situations where unoptimized recursion is better?
Yes, if you are creating a lazily evaluated structure, in haskell, you need the cons-cell as the evaluation boundary, and you can't lazily evaluate a tail recursive call.
Benchmarking is the only way to know for sure, racket has deep stack frames, so you should be able to get away with both versions.
The stdlib version is quite horrific, which shows that you can usually squeeze out some performance if you're willing to sacrifice readability.
Given two implementations of the same function, with the same O notation, I will choose the simpler version 95% of the time.
There are many ways to make recursion preserving iterative process.
I usually do continuation passing style directly. This is my "natural" way to do it.
One takes into account the type of the function. Sometimes you need to connect your function with the functions around it and depending on their type you can choose another way to do recursion.
You should start by solving "the little schemer" to gain a strong foundation about it. In the "little typer" you can discover another type of doing recursion, founded on other computational philosophy, used in languages like agda, coq.
In scheme you can write code that is actually haskell sometimes (you can write monadic code that would be generated by a haskell compiler as intermediate language). In that case the way to do recursion is also different that "usual" way, etc.
false dichotomy
You have other options available to you. Here we can preserve tail-recursion and map over the list with a single traversal. The technique used here is called continuation-passing style -
(define (map f lst (return identity))
(if (null? lst)
(return null)
(map f
(cdr lst)
(lambda (r) (return (cons (f (car lst)) r))))))
(define (square x)
(* x x))
(map square '(1 2 3 4))
'(1 4 9 16)
This question is tagged with racket, which has built-in support for delimited continuations. We can accomplish map using a single traversal, but this time without using recursion. Enjoy -
(require racket/control)
(define (yield x)
(shift return (cons x (return (void)))))
(define (map f lst)
(reset (begin
(for ((x lst))
(yield (f x)))
null)))
(define (square x)
(* x x))
(map square '(1 2 3 4))
'(1 4 9 16)
It's my intention that this post will show you the detriment of pigeonholing your mind into a particular construct. The beauty of Scheme/Racket, I have come to learn, is that any implementation you can dream of is available to you.
I would highly recommend Beautiful Racket by Matthew Butterick. This easy-to-approach and freely-available ebook shatters the glass ceiling in your mind and shows you how to think about your solutions in a language-oriented way.
I'm trying to reverse a list in scheme and I came up with to the following solution:
(define l (list 1 2 3 4))
(define (reverse lista)
(car (cons (reverse (cdr (cons 0 lista))) 0)))
(display (reverse l))
Although it works I don't really understand why it works.
In my head, it would evaluate to a series of nested cons until cons of () (which the cdr of a list with one element).
I guess I am not understanding the substitution model, could someone explain me why it works?
Obs:
It is supposed to work only in not nested lists.
Taken form SICP, exercise 2.18.
I know there are many similar questions, but as far as I saw, none presented
this solution.
Thank you
[As this happens quite often, I write the answer anyway]
Scheme implementations do have their builtin versions of reverse, map, append etc. as they are specified in RxRS (e.g. https://www.cs.indiana.edu/scheme-repository/R4RS/r4rs_8.html).
In the course of learning scheme (and actually any lisp dialect) it's really valuable to implement them anyway. The danger is, one's definition can collide with the built-in one (although e.g. scheme's define or lisp's label should shadow them). Therefore it's always worth to call this hand-made implementation with some other name, like "my-reverse", "my-append" etc. This way you will save yourself much confusion, like in the following:
(let ([append
(lambda (xs ys)
(if (null? xs)
ys
(cons (car xs) (append (cdr xs) ys))))])
(append '(hello) '(there!)))
-- this one seems to work, creating a false impression that "let" works the same as "letrec". But just change the name to "my-append" and it breaks, because at the moment of evaluating the lambda form, the symbol "my-append" is not yet bound to anything (unlike "append" which was defined as a builtin procedure).
Of course such let form will work in a language with dynamic scoping, but scheme is lexical (with the exception of "define"s), and the reason is referential transparency (but that's so far offtopic that I can only refer interested reader to one of the lambda papers http://repository.readscheme.org/ftp/papers/ai-lab-pubs/AIM-453.pdf).
This reads pretty much the same as the solutions in other languages:
if the list is empty, return an empty list. Otherwise ...
chop off the first element (CAR)
reverse the remainder of the list (CDR)
append (CONS) the first element to that reversal
return the result
Now ... given my understanding from LISP days, the code would look more like this:
(append (reverse (cdr lista)) (list (car lista)))
... which matches my description above.
There are several ways to do it. Here is another:
(define my-reverse
(lambda (lst)
(define helper
(lambda (lst result)
(if (null? lst)
result
(helper (cdr lst) (cons (car lst) result)))))
(helper lst '())))
NOTE: I would like to do this without rackets built in exceptions if possible.
I have many functions which call other functions and may recursively make a call back to the original function. Under certain conditions along the way I want to stop any further recursive steps, and no longer call any other functions and simply return some value/string (the stack can be ignored if the condition is met).. here is a contrived example that hopefully will show what I'm trying to accomplish:
(define (add expr0 expr1)
(cond
[(list? expr0) (add (cadr expr0) (cadr (cdr expr0)))]
[(list? expr1) (add (cadr expr1) (cadr (cdr expr1)))]
[else (if (or (equal? expr0 '0) (equal? expr1 '0))
'(Adding Zero)
(+ expr0 expr1))]
))
If this were my function and I called it with (add (add 2 0) 3), Then the goal would be to simply return the entire string '(Adding Zero) ANYTIME that a zero is one of the expressions, instead of making the recursive call to (add '(Adding Zero) 3)
Is there a way to essentially "break" out of recursion? My problem is that if i'm already deep inside then it will eventually try to evaluate '(Adding Zero) which it doesn't know how to do and I feel like I should be able to do this without making an explicit check to each expr..
Any guidance would be great.
In your specific case, there's no need to "escape" from normal processing. Simply having '(Adding Zero) in tail position will cause your add function to return (Adding Zero).
To create a situation where you might need to escape, you need something a
little more complicated:
(define (recursive-find/collect collect? tree (result null))
(cond ((null? tree) (reverse result))
((collect? tree) (reverse (cons tree result)))
((not (pair? tree)) (reverse result))
(else
(let ((hd (car tree))
(tl (cdr tree)))
(cond ((collect? hd)
(recursive-find/collect collect? tl (cons hd result)))
((pair? hd)
(recursive-find/collect collect? tl
(append (reverse (recursive-find/collect collect? hd)) result)))
(else (recursive-find/collect collect? tl result)))))))
Suppose you wanted to abort processing and just return 'Hahaha! if any node in the tree had the value 'Joker. Just evaluating 'Hahaha! in tail position
wouldn't be enough because recursive-find/collect isn't always used in
tail position.
Scheme provides continuations for this purpose. The easiest way to do it in my particular example would be to use the continuation from the predicate function, like this:
(call/cc
(lambda (continuation)
(recursive-find/collect
(lambda (node)
(cond ((eq? node 'Joker)
(continuation 'Hahaha!)) ;; Processing ends here
;; Otherwise find all the symbols
;; in the tree
(else (symbol? node))))
'(Just 1 arbitrary (tree (stucture) ((((that "has" a Joker in it)))))))))
A continuation represents "the rest of the computation" that is going to happen after the call/cc block finishes. In this case, it just gives you a way to escape from the call/cc block from anywhere in the stack.
But continuations also have other strange properties, such as allowing you to jump back to whatever block of code this call/cc appears in even after execution has left this part of the program. For example:
(define-values a b (call/cc
(lambda (cc)
(values 1 cc))))
(cc 'one 'see-see)
In this case, calling cc jumps back to the define-values form and redefines a and b to one and see-see, respectively.
Racket also has "escape continuations" (call/ec or let/ec) which can escape from their form, but can't jump back into it. In exchange for this limitation you get better performance.
I have some code to find the maximum height and replace it with the associated name. There are separate lists for height and names, each the same length and non-empty.
I can solve this using structural recursion but have to change it into accumulative, and I am unsure how to do that. All the examples I have seen are confusing me. Is anybody able to turn the code into one using accumulative recursion?
(define (tallest names heights)
(cond
[(empty? names) heights]
[(> (first heights) (first (rest heights)))
(cons (first names) (tallest (rest (rest names)) (rest (rest heights))))]
[else (tallest (rest names) (rest heights))]))
First of all, your provided tallest function doesn't actually work (calling (tallest '(Bernie Raj Amy) (list 1.5 1.6 1.7)) fails with a contract error), but I see what you're getting at. What's the difference between structural recursion and accumulative recursion?
Well, structural recursion works by building a structure as a return value, in which one of the values inside that structure is the result of a recursive call to the same function. Take the recursive calculation of a factorial, for example. You might define it like this:
(define (factorial n)
(if (zero? n) 1
(* n (factorial (sub1 n)))))
Visualize how this program would execute for an input of, say, 4. Each call leaves a "hole" in the multiplication expression to be filled in by the result of a recursive subcall. Here's what that would look like, visualized, using _ to represent one of those "holes".
(* 4 _)
(* 3 _)
(* 2 _)
(* 1 _)
1
Notice how a large portion of the work is done only after the final case is reached. Much of the work must be done in the process of popping the calls off the stack as they return because each call depends on performing some additional operation on its subcall's result.
How is accumulative recursion different? Well, in accumulative recursion, we use an extra argument to the function called an accumulator. Rewriting the above factorial function to use an accumulator would make it look like this:
(define (factorial n acc)
(if (zero? n) acc
(factorial (sub1 n) (* acc n))))
Now if we wanted to find the factorial of 4, we'd have to call (factorial 4 1), providing a starting value for the accumulator (I'll address how to avoid that in a moment). If you think about the call stack now, it would look quite different.
Notice how there are no "holes" to be filled in—the result of the factorial function is either the accumulator or a direct call to itself. This is referred to as a tail call, and the recursive call to factorial is referred to as being in tail position.
This turns out to be helpful for a few reasons. First of all, some functions are just easier to express with accumulative recursion, though factorial probably isn't one of them. More importantly, however, Scheme requires that implementations provide proper tails calls (sometimes also called "tail call optimization"), which means that the call stack will not grow in depth when tail calls are made.
There is plenty of existing information about how tail calls work and why they're useful, so I won't repeat that here. What's important to understand is that accumulative recursion involves an accumulator argument, which usually causes the resulting function to be implemented with a tail call.
But what do we do about the extra parameter? Well, we can actually just make a "helper" function that will do the accumulative recursion, but we will provide a function that automatically fills in the base case.
(define (factorial n)
(define (factorial-helper n acc)
(if (zero? n) acc
(factorial-helper (sub1 n) (* acc n))))
(factorial-helper n 1))
This sort of idiom is common enough that Racket provides a "named let" form, which simplifies the above function to this:
(define (factorial n)
(let helper ([n n] [acc 1])
(if (zero? n) acc
(helper (sub1 n) (* acc n)))))
But that's just some syntactic sugar for the same idea.
Okay, so: how does any of this apply to your question? Well, actually, using accumulative recursion makes implementing your problem quite easy. Here's a breakdown of how you'd structure the algorithm:
Just like in your original example, you'd iterate through the list until you get empty. This will form your "base case".
Your accumulator will be simple—it will be the current maximum element you've found.
Upon each iteration, if you find an element greater than the current maximum, that becomes the new accumulator. Otherwise, the accumulator remains the same.
Putting these all together, and here's a simple implementation:
(define (tallest-helper names heights current-tallest)
(cond
[(empty? names)
(car current-tallest)]
[(> (first heights) (cdr current-tallest))
(tallest-helper (rest names) (rest heights)
(cons (first names) (first heights)))]
[else
(tallest-helper (rest names) (rest heights)
current-tallest)]))
This can be further improved in plenty of ways—wrapping it in a function that provides the accumulator's starting value, using named let, removing some of the repetition, etc.—but I'll leave that as an exercise for you.
Just remember: the accumulator is effectively your "working sum". It's your "running total". Understand that, and things should make sense.
I am attempting to write a small recursive program that tests a list and returns t if every element is an atom. The problem I am having is that when the function receives an empty list, it returns t instead of the desired result of nil. I cannot come up with a way to have it return nil for an initially empty list and still function properly in a recursive manner.
(defun only-atoms (in)
(if (null in)
t
(and (atom (first in)) (only-atoms (cdr in)) )
)
)
The function can be implemented without recursion using e.g. every, as in:
(defun only-atoms (list)
(and list (every #'atom list)))
When it comes to your stated problem that the function returns T instead of the desired result of NIL when the function is called with an empty list:
Your recursive implementation explicitly returns T if (null in) is true, which explains your finding. Simply change it to the desired value NIL. Consider changing the if construct to and.
Only make the recursive call when the list has more than one item. A well placed test for (rest in) will do. Provide a true value instead of making the recursive call if the list is at its last item.
Carefully locate the only-atoms call to ensure that the function can be tail-recursive.
For example:
(defun only-atoms (list)
(and list
(atom (first list))
(or (null (rest list))
(only-atoms (rest list)))))
Use COND, which allows you to test for several cases:
empty list -> nil
one element list -> atom? ; hint: don't use LENGTH
else -> atom? for first element and recursive call for rest elements
The empty list does fulfill the condition that every element is an atom! Your requirement that it should contain at least one element is an additional requirement.
The simplest way to express "every element of the list is an atom" is (every #'atom list). You can combine it with your additional requirement with and.
If you insist on doing it recursively in SICP-style, separate your requirements:
(defun not-empty-and-all-atoms (list)
(and list
(all-atoms list)))
(defun all-atoms (list)
(if (endp list)
t
(and (atom (first list))
(all-atoms (rest list)))))
This solution works correctly:
(defun lat-p (lst)
(labels ((lat-p* (lst)
(cond
((null lst) t)
((atom (car lst)) (lat-p* (cdr lst)))
(t nil))))
(if lst
(lat-p* lst)
nil)))
However a much more elegant solution(with no recursion) would be:
(defun lat-p (lst)
(and lst (every #'atom lst)))
You can split your function in two, and provide the initial nil screening before you enter recursion. The following code is one way to do so (I tried to keep it as close to provided code as possible):
(defun only-atoms (in)
(defun only-atoms-iter (in)
(if (null in)
t
(and (atom (first in)) (only-atoms-iter (cdr in)))))
(unless (null in)
(only-atoms-iter in)))
This is also a good opportunity to make your function tail recursive:
(defun only-atoms (in)
(defun only-atoms-iter (in state)
(if (null in)
state
(only-atoms-iter (cdr in) (and state (atom (first in))))))
(unless (null in)
(only-atoms-iter in t)))