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tail-recursive function appending element to list
(5 answers)
Closed 2 years ago.
For a long time I'm trying to implement the append procedure in Racket using tail recursion(iterative).
The best solution I did so far is this:
(define (my-reverse lst)
(define (iter lst reversed)
(if (null? lst)
reversed
(iter (cdr lst) (cons (car lst) reversed))))
(iter lst '()))
(define (append-iter el lst)
(my-reverse (cons el (my-reverse lst))))
(append-iter 4 '(1 2 3)) ; (1 2 3 4)
My question is whether there is a better a way of implementation?
Well, it depends on your definition of "better" :) . Your solution is simple, and you won't find a more straightforward way to write your own procedure to append an element at the end of a list, using tail recursion.
My only comment is that my-reverse is doing the same as the built-in reverse procedure, which most certainly will be tail recursive, so you can simply write it as:
(define (append-iter el lst)
(reverse (cons el (reverse lst))))
If you're ok with using continuation passing style, the following solution is also tail-recursive and it only depends on the most basic primitive procedures:
(define (append-iter el lst)
(append-cps lst (cons el '()) (lambda (x) x)))
(define (append-cps lst1 lst2 k)
(if (null? lst1)
(k lst2)
(append-cps
(cdr lst1)
lst2
(lambda (appended-cdr)
(k (cons (car lst1) appended-cdr))))))
Either way, it works as expected:
(append-iter 4 '(1 2 3))
=> '(1 2 3 4)
If you're curious, the CPS solution will evaluate to an expression like the one shown below, which naturally leads to the answer:
((lambda (append-cdr)
((lambda (appended-cdr)
((lambda (appended-cdr)
((lambda (x) x)
(cons 1 appended-cdr)))
(cons 2 appended-cdr)))
(cons 3 append-cdr)))
'(4))
=> '(1 2 3 4)
Yes,
(define (rev-append lst reversed)
(if (null? lst)
reversed
(rev-append (cdr lst)
(cons (car lst) reversed))))
(define (append-iter el lst)
(rev-append (rev-append lst '())
(cons el '())))
is marginally "better".
rev-append is your my-reverse's iter. It deserves its own top level binding.
Related
I'm new to racket and trying to write a function that checks if a list is in strictly ascending order.
'( 1 2 3) would return true
'(1 1 2) would return false (repeats)
'(3 2 4) would return false
My code so far is:
Image of code
(define (ascending? 'list)
(if (or (empty? list) (= (length 'list) 1)) true
(if (> first (first (rest list))) false
(ascending? (rest list)))))
I'm trying to call ascending? recursively where my base case is that the list is empty or has only 1 element (then trivially ascending).
I keep getting an error message when I use check-expect that says "application: not a procedure."
I guess you want to implement a procedure from scratch, and Alexander's answer is spot-on. But in true functional programming style, you should try to reuse existing procedures to write the solution. This is what I mean:
(define (ascending? lst)
(apply < lst))
It's shorter, simpler and easier to understand. And it works as expected!
(ascending? '(1 2 3))
=> #t
(ascending? '(1 1 2))
=> #f
Some things to consider when writing functions:
Avoid using built in functions as variable names. For example, list is a built in procedure that returns a newly allocated list, so don't use it as an argument to your function, or as a variable. A common convention/alternative is to use lst as a variable name for lists, so you could have (define (ascending? lst) ...).
Don't quote your variable names. For example, you would have (define lst '(1 2 3 ...)) and not (define 'lst '(1 2 3 ...)).
If you have multiple conditions to test (ie. more than 2), it may be cleaner to use cond rather than nesting multiple if statements.
To fix your implementation of ascending? (after replacing 'list), note on line 3 where you have (> first (first (rest list))). Here you are comparing first with (first (rest list)), but what you really want is to compare (first lst) with (first (rest lst)), so it should be (>= (first lst) (first (rest lst))).
Here is a sample implementation:
(define (ascending? lst)
(cond
[(null? lst) #t]
[(null? (cdr lst)) #t]
[(>= (car lst) (cadr lst)) #f]
[else
(ascending? (cdr lst))]))
or if you want to use first/rest and true/false you can do:
(define (ascending? lst)
(cond
[(empty? lst) true]
[(empty? (rest lst)) true]
[(>= (first lst) (first (rest lst))) false]
[else
(ascending? (rest lst))]))
For example,
> (ascending? '(1 2 3))
#t
> (ascending? '(1 1 2))
#f
> (ascending? '(3 2 4))
#f
If you write down the properties of an ascending list in bullet form;
An ascending list is either
the empty list, or
a one-element list, or
a list where
the first element is smaller than the second element, and
the tail of the list is ascending
you can wind up with a pretty straight translation:
(define (ascending? ls)
(or (null? ls)
(null? (rest ls))
(and (< (first ls) (first (rest ls)))
(ascending? (rest ls)))))
This Scheme solution uses an explicitly recursive named let and memoization:
(define (ascending? xs)
(if (null? xs) #t ; Edge case: empty list
(let asc? ((x (car xs)) ; Named `let`
(xs' (cdr xs)) )
(if (null? xs') #t
(let ((x' (car xs'))) ; Memoization of `(car xs)`
(if (< x x')
(asc? x' (cdr xs')) ; Tail recursion
#f)))))) ; Short-circuit termination
(display
(ascending?
(list 1 1 2) )) ; `#f`
I'm trying to reverse a list, here's my code:
(define (reverse list)
(if (null? list)
list
(list (reverse (cdr list)) (car list))))
so if i enter (reverse '(1 2 3 4)), I want it to come out as (4 3 2 1), but right now it's not giving me that. What am I doing wrong and how can I fix it?
The natural way to recur over a list is not the best way to solve this problem. Using append, as suggested in the accepted answer pointed by #lancery, is not a good idea either - and anyway if you're learning your way in Scheme it's best if you try to implement the solution yourself, I'll show you what to do, but first a tip - don't use list as a parameter name, that's a built-in procedure and you'd be overwriting it. Use other name, say, lst.
It's simpler to reverse a list by means of a helper procedure that accumulates the result of consing each element at the head of the result, this will have the effect of reversing the list - incidentally, the helper procedure is tail-recursive. Here's the general idea, fill-in the blanks:
(define (reverse lst)
(<???> lst '())) ; call the helper procedure
(define (reverse-aux lst acc)
(if <???> ; if the list is empty
<???> ; return the accumulator
(reverse-aux <???> ; advance the recursion over the list
(cons <???> <???>)))) ; cons current element with accumulator
Of course, in real-life you wouldn't implement reverse from scratch, there's a built-in procedure for that.
Here is a recursive procedure that describes an iterative process (tail recursive) of reversing a list in Scheme
(define (reverse lst)
(define (go lst tail)
(if (null? lst) tail
(go (cdr lst) (cons (car lst) tail))))
(go lst ())))
Using substitution model for (reverse (list 1 2 3 4))
;; (reverse (list 1 2 3 4))
;; (go (list 1 2 3 4) ())
;; (go (list 2 3 4) (list 1))
;; (go (list 3 4) (list 2 1))
;; (go (list 4) (list 3 2 1))
;; (go () (list 4 3 2 1))
;; (list 4 3 2 1)
Here is a recursive procedure that describes a recursive process (not tail recursive) of reversing a list in Scheme
(define (reverse2 lst)
(if (null? lst) ()
(append (reverse2 (cdr lst)) (list (car lst)))))
(define (append l1 l2)
(if (null? l1) l2
(cons (car l1) (append (cdr l1) l2))))
Using substitution model for (reverse2 (list 1 2 3 4))
;; (reverse2 (list 1 2 3 4))
;; (append (reverse2 (list 2 3 4)) (list 1))
;; (append (append (reverse2 (list 3 4)) (list 2)) (list 1))
;; (append (append (append (reverse2 (list 4)) (list 3)) (list 2)) (list 1))
;; (append (append (append (append (reverse2 ()) (list 4)) (list 3)) (list 2)) (list 1))
;; (append (append (append (append () (list 4)) (list 3)) (list 2)) (list 1))
;; (append (append (append (list 4) (list 3)) (list 2)) (list 1))
;; (append (append (list 4 3) (list 2)) (list 1))
;; (append (list 4 3 2) (list 1))
;; (list 4 3 2 1)
Tail recursive approach using a named let:
(define (reverse lst)
(let loop ([lst lst] [lst-reversed '()])
(if (empty? lst)
lst-reversed
(loop (rest lst) (cons (first lst) lst-reversed)))))
This is basically the same approach as having a helper function with an accumulator argument as in Oscar's answer, where the loop binding after let makes the let into an inner function you can call.
Here's a solution using build-list procedure:
(define reverse
(lambda (l)
(let ((len (length l)))
(build-list len
(lambda (i)
(list-ref l (- len i 1)))))))
This one works but it is not a tail recursive procedure:
(define (rev lst)
(if (null? lst)
'()
(append (rev (cdr lst)) (car lst))))
Tail recursive solution:
(define (reverse oldlist)
(define (t-reverse oldlist newlist)
(if (null? oldlist)
newlist
(t-reverse (cdr oldlist) (cons (car oldlist) newest))))
(t-reverse oldlist '()))
Just left fold the list using cons:
(define (reverse list) (foldl cons null list))
This is also efficient because foldl is tail recursive and there is no need for append. This can also be done point-free (using curry from racket):
(define reverse (curry foldl cons null))
(define reverse?
(lambda (l)
(define reverse-aux?
(lambda (l col)
(cond
((null? l) (col ))
(else
(reverse-aux? (cdr l)
(lambda ()
(cons (car l) (col))))))))
(reverse-aux? l (lambda () (quote ())))))
(reverse? '(1 2 3 4) )
One more answer similar to Oscar's. I have just started learning scheme, so excuse me in case you find issues :).
There's actually no need for appending or filling the body with a bunch of lambdas.
(define (reverse items)
(if (null? items)
'()
(cons (reverse (cdr items)) (car items))))
I think it would be better to use append instead of cons
(define (myrev l)
(if (null? l)
'()
(append (myrev (cdr l)) (list (car l)))
)
)
this another version with tail recursion
(define (myrev2 l)
(define (loop l acc)
(if (null? l)
acc
(loop (cdr l) (append (list (car l)) acc ))
)
)
(loop l '())
)
I am watching SICP video lectures and i came to a section where tutors are showing procedures to work with lists, so, here is one of them:
(define (map p l)
(if (null? l)
(list)
(cons (p (car l))
(map p (cdr l)))))
What i want to ask is: is there a way to define map in iterative way, or that cons requires lazy evaluation to be executed right?
You original code is almost tail recursive.. the only thing that makes it not is the cons part. If Scheme had equal requirement for having TRMC optimization as it has TCO requirement you could leave your code as is and the implementation would have made it tail recursive for you.
Since it isn't a requirement we need to do our own TRMC optimization. Usually when iterating a list in a loop and having it tail recursive by using an accumulator you get the result in the opposite order, thus you can do linear update reverse:
(define (map proc lst)
(let loop ((lst lst) (acc '()))
(cond ((null? lst) (reverse! acc) acc)
(else (loop (cdr lst)
(cons (proc (car lst)) acc))))))
Or you can do it all in one pass:
(define (map proc lst)
(define head (list 1))
(let loop ((tail head) (lst lst))
(cond ((null? lst) (cdr head))
(else (set-cdr! tail (list (proc (car lst))))
(loop (cdr tail) (cdr lst))))))
Now in both cases you mutate only the structure the procedure has itself created, thus for the user it might as well be implemented in the same manner as your example.
When you use higher order procedures like map from your implementation it could happen it has been implemented like this. It's easy to find out by comparing performance on the supplied map with the different implementations with a very long list. The difference between the executions would tell you if it's TRMCO or how the supplied map probably has been implemented.
You need to embrace recursion in order to appreciate SICP and Scheme in general, so try to get used to it, you will appreciate it later, promised.
But yes, you can:
(define (iterative-map f lst)
(define res null)
(do ((i (- (length lst) 1) (- i 1))) ((= i -1))
(set! res (cons (f (list-ref lst i)) res)))
res)
(iterative-map (lambda (x) (+ x 1)) '(1 3 5))
=> '(2 4 6)
but using set! is considered bad style if avoidable.
In Racket you have a different set of loops that are more elegant:
(define (for-map f lst)
(for/list ((i lst))
(f i)))
(for-map add1 '(1 3 5))
=> '(2 4 6)
First off, this is homework, but I am simply looking for a hint or pseudocode on how to do this.
I need to sum all the items in the list, using recursion. However, it needs to return the empty set if it encounters something in the list that is not a number. Here is my attempt:
(DEFINE sum-list
(LAMBDA (lst)
(IF (OR (NULL? lst) (NOT (NUMBER? (CAR lst))))
'()
(+
(CAR lst)
(sum-list (CDR lst))
)
)
)
)
This fails because it can't add the empty set to something else. Normally I would just return 0 if its not a number and keep processing the list.
I suggest you use and return an accumulator for storing the sum; if you find a non-number while traversing the list you can return the empty list immediately, otherwise the recursion continues until the list is exhausted.
Something along these lines (fill in the blanks!):
(define sum-list
(lambda (lst acc)
(cond ((null? lst) ???)
((not (number? (car lst))) ???)
(else (sum-list (cdr lst) ???)))))
(sum-list '(1 2 3 4 5) 0)
> 15
(sum-list '(1 2 x 4 5) 0)
> ()
I'd go for this:
(define (mysum lst)
(let loop ((lst lst) (accum 0))
(cond
((empty? lst) accum)
((not (number? (car lst))) '())
(else (loop (cdr lst) (+ accum (car lst)))))))
Your issue is that you need to use cond, not if - there are three possible branches that you need to consider. The first is if you run into a non-number, the second is when you run into the end of the list, and the third is when you need to recurse to the next element of the list. The first issue is that you are combining the non-number case and the empty-list case, which need to return different values. The recursive case is mostly correct, but you will have to check the return value, since the recursive call can return an empty list.
Because I'm not smart enough to figure out how to do this in one function, let's be painfully explicit:
#lang racket
; This checks the entire list for numericness
(define is-numeric-list?
(lambda (lst)
(cond
((null? lst) true)
((not (number? (car lst))) false)
(else (is-numeric-list? (cdr lst))))))
; This naively sums the list, and will fail if there are problems
(define sum-list-naive
(lambda (lst)
(cond
((null? lst) 0)
(else (+ (car lst) (sum-list-naive (cdr lst)))))))
; This is a smarter sum-list that first checks numericness, and then
; calls the naive version. Note that this is inefficient, because the
; entire list is traversed twice: once for the check, and a second time
; for the sum. Oscar's accumulator version is better!
(define sum-list
(lambda (lst)
(cond
((is-numeric-list? lst) (sum-list-naive lst))
(else '()))))
(is-numeric-list? '(1 2 3 4 5))
(is-numeric-list? '(1 2 x 4 5))
(sum-list '(1 2 3 4 5))
(sum-list '(1 2 x 4 5))
Output:
Welcome to DrRacket, version 5.2 [3m].
Language: racket; memory limit: 128 MB.
#t
#f
15
'()
>
I suspect your homework is expecting something more academic though.
Try making a "is-any-nonnumeric" function (using recursion); then you just (or (is-any-numeric list) (sum list)) tomfoolery.
Given a list of numbers, say, (1 3 6 10 0), how do you compute differences (xi - xi-1), provided that you have x-1 = 0 ?
(the result in this example should be (1 2 3 4 -10))
I've found this solution to be correct:
(define (pairwise-2 f init l)
(first
(foldl
(λ (x acc-data)
(let ([result-list (first acc-data)]
[prev-x (second acc-data)])
(list
(append result-list (list(f x prev-x)))
x)))
(list empty 0)
l)))
(pairwise-2 - 0 '(1 3 6 10 0))
;; => (1 2 3 4 -10)
However, I think there should be more elegant though no less flexible solution. It's just ugly.
I'm new to functional programming and would like to hear any suggestions on the code.
Thanks.
map takes multiple arguments. So I would just do
(define (butlast l)
(reverse (cdr (reverse l))))
(let ((l '(0 1 3 6 10)))
(map - l (cons 0 (butlast l)))
If you want to wrap it up in a function, say
(define (pairwise-call f init l)
(map f l (cons init (butlast l))))
This is of course not the Little Schemer Way, but the way that avoids writing recursion yourself. Choose the way you like the best.
I haven't done scheme in dog's years, but this strikes me as a typical little lisper type problem.
I started with a base definition (please ignore misplacement of parens - I don't have a Scheme interpreter handy:
(define pairwise-diff
(lambda (list)
(cond
((null? list) '())
((atom? list) list)
(t (pairwise-helper 0 list)))))
This handles the crap cases of null and atom and then delegates the meat case to a helper:
(define pairwise-helper
(lambda (n list)
(cond
((null? list) '())
(t
(let ([one (car list)])
(cons (- one n) (pairwise-helper one (cdr list))))
))))
You could rewrite this using "if", but I'm hardwired to use cond.
There are two cases here: null list - which is easy and everything else.
For everything else, I grab the head of the list and cons this diff onto the recursive case. I don't think it gets much simpler.
After refining and adapting to PLT Scheme plinth's code, I think nearly-perfect solution would be:
(define (pairwise-apply f l0 l)
(if (empty? l)
'()
(let ([l1 (first l)])
(cons (f l1 l0) (pairwise-apply f l1 (rest l))))))
Haskell tells me to use zip ;)
(define (zip-with f xs ys)
(cond ((or (null? xs) (null? ys)) null)
(else (cons (f (car xs) (car ys))
(zip-with f (cdr xs) (cdr ys))))))
(define (pairwise-diff lst) (zip-with - (cdr lst) lst))
(pairwise-diff (list 1 3 6 10 0))
; gives (2 3 4 -10)
Doesn't map finish as soon as the shortest argument list is exhausted, anyway?
(define (pairwise-call fun init-element lst)
(map fun lst (cons init-element lst)))
edit: jleedev informs me that this is not the case in at least one Scheme implementation. This is a bit annoying, since there is no O(1) operation to chop off the end of a list.
Perhaps we can use reduce:
(define (pairwise-call fun init-element lst)
(reverse (cdr (reduce (lambda (a b)
(append (list b (- b (car a))) (cdr a)))
(cons (list init-element) lst)))))
(Disclaimer: quick hack, untested)
This is the simplest way:
(define (solution ls)
(let loop ((ls (cons 0 ls)))
(let ((x (cadr ls)) (x_1 (car ls)))
(if (null? (cddr ls)) (list (- x x_1))
(cons (- x x_1) (loop (cdr ls)))))))
(display (equal? (solution '(1)) '(1))) (newline)
(display (equal? (solution '(1 5)) '(1 4))) (newline)
(display (equal? (solution '(1 3 6 10 0)) '(1 2 3 4 -10))) (newline)
Write out the code expansion for each of the example to see how it works.
If you are interested in getting started with FP, be sure to check out How To Design Program. Sure it is written for people brand new to programming, but it has tons of good FP idioms within.
(define (f l res cur)
(if (null? l)
res
(let ((next (car l)))
(f (cdr l) (cons (- next cur) res) next))))
(define (do-work l)
(reverse (f l '() 0)))
(do-work '(1 3 6 10 0))
==> (1 2 3 4 -10)