(define (long a b)
(cond ((null? a) '())
((null? b) '())
(else
(if (equal? (car a) (car b))
(cons a (long (cdr a) (cdr b)))
(cons b (long (cdr a) (cdr b)))))))
This function is suppose to the find the longest common prefix between 2 lists. So when I input '(1 2) and '(1 2 3), I should get '(1 2). However, when I run (long '(1 2) '(1 2 3)), I get ((1 2) (2)). When I retraced my steps through my code, I don't seem to find the issue. Can someone please help me find the issue?
Thank you.
A simple solution is the following:
(define (long a b)
(cond ((null? a) '())
((null? b) '())
((equal? (car a) (car b)) (cons (car a) (long (cdr a) (cdr b))))
(else '())))
When a list terminates the prefix is the empty list; if both the cars are equal, the result is the list with the car and the longest prefix of the cdrs of the lists. Otherwise, the lists differ for the first element, so the result is the empty list.
One can use 'for/list' loop with conditions here:
(define (long a b)
(for/list ((i a)
(j b)
#:break (not(= i j)))
i))
(long '(1 2 3) '(1 2))
Output:
'(1 2)
Related
I'm finishing up a Scheme assignment and I'm having some trouble with the recursive cases for two functions.
The first function is a running-sums function which takes in a list and returns a list of the running sums i.e (summer '(1 2 3)) ---> (1 3 6) Now I believe I'm very close but can't quite figure out how to fix my case. Currently I have
(define (summer L)
(cond ((null? L) '())
((null? (cdr L)) '())
(else (cons (car L) (+ (car L) (cadr L))))))
I know I need to recursively call summer, but I'm confused on how to put the recursive call in there.
Secondly, I'm writing a function which counts the occurrences of an element in a list. This function works fine through using a helper function but it creates duplicate pairs.
(define (counts L)
(cond ((null? L) '())
(else (cons (cons (car L) (countEle L (car L))) (counts (cdr L))))))
(define (countEle L x)
(if (null? L) 0
(if (eq? x (car L)) (+ 1 (countEle (cdr L) x)) (countEle (cdr L) x))))
The expected output is:
(counts '(a b c c b b)) --> '((a 1) (b 3) ( c 2))
But it's currently returning '((a . 1) (b . 3) (c . 2) (c . 1) (b . 2) (b . 1)). So it's close; I'm just not sure how to handle checking if I've already counted the element.
Any help is appreciated, thank you!
To have a running sum, you need in some way to keep track of the last sum. So some procedure should have two arguments: the rest of the list to sum (which may be the whole list) and the sum so far.
(define (running-sum L)
(define (rs l s)
...)
(rs L 0))
For the second procedure you want to do something like
(define (count-elems L)
(define (remove-elem e L) ...)
(define (count-single e L) ...)
(if (null? L)
'()
(let ((this-element (car L)))
(cons (list this-element (count-single this-element L))
(count-elems (remove-elem this-element (cdr L)))))))
Be sure to remove the elements you've counted before continuing! I think you can fill in the rest.
To your first problem:
The mistake in your procedure is, that there is no recursive call of "summer". Have a look at the last line.
(else (cons (car L) (+ (car L) (cadr L))))))
Here is the complete solution:
(define (summer LL)
(define (loop sum LL)
(if (null? LL)
'()
(cons (+ sum (car LL)) (loop (+ sum (car ll)) (cdr LL)))))
(loop 0 LL))
The problem is to:
Write a function (encode L) that takes a list of atoms L and run-length encodes the list such that the output is a list of pairs of the form (value length) where the first element is a value and the second is the number of times that value occurs in the list being encoded.
For example:
(encode '(1 1 2 4 4 8 8 8)) ---> '((1 2) (2 1) (4 2) (8 3))
This is the code I have so far:
(define (encode lst)
(cond
((null? lst) '())
(else ((append (list (car lst) (count lst 1))
(encode (cdr lst)))))))
(define (count lst n)
(cond
((null? lst) n)
((equal? (car lst) (car (cdr lst)))
(count (cdr lst) (+ n 1)))
(else (n)))))
So I know this won't work because I can't really think of a way to count the number of a specific atom in a list effectively as I would iterate down the list. Also, Saving the previous (value length) pair before moving on to counting the next unique atom in the list. Basically, my main problem is coming up with a way to keep a count of the amount of atoms I see in the list to create my (value length) pairs.
You need a helper function that has the count as additional argument. You check the first two elements against each other and recurse by increasing the count on the rest if it's a match or by consing a match and resetting count to 1 in the recursive call.
Here is a sketch where you need to implement the <??> parts:
(define (encode lst)
(define (helper lst count)
(cond ((null? lst) <??>)
((null? (cdr lst)) <??>))
((equal? (car lst) (cadr lst)) <??>)
(else (helper <??> <??>))))
(helper lst 1))
;; tests
(encode '()) ; ==> ()
(encode '(1)) ; ==> ((1 1))
(encode '(1 1)) ; ==> ((1 2))
(encode '(1 2 2 3 3 3 3)) ; ==> ((1 1) (2 2) (3 4))
Using a named let expression
This technique of using a recursive helper procedure with state variables is so common in Scheme that there's a special let form which allows you to express the pattern a bit nicer
(define (encode lst)
(let helper ((lst lst) (count 1))
(cond ((null? lst) <??>)
((null? (cdr lst)) <??>))
((equal? (car lst) (cadr lst)) <??>)
(else (helper <??> <??>)))))
Comments on the code in your question: It has excess parentheses..
((append ....)) means call (append ....) then call that result as if it is a function. Since append makes lists that will fail miserably like ERROR: application: expected a function, got a list.
(n) means call n as a function.. Remember + is just a variable, like n. No difference between function and other values in Scheme and when you put an expression like (if (< v 3) + -) it needs to evaluate to a function if you wrap it with parentheses to call it ((if (< v 3) + -) 5 3); ==> 8 or 2
i need to create a list such that the min is always at the outside in a list.
Example
input (1 2 3)
output (1 (2 3))
Here is my code, assuming that the numbers are in descending order, which i wish to extent later to a general case.
I am getting an unexpected output of (3 2 1 0 -1 -2 -3 ()).
How do I achieve this in scheme any ideas?'
(define (find-min-rest L)
(if (null? (cdr L)) (let ( (x (car L))) (cons x '( ())))
(let* ((ret-ans (find-min-rest (cdr L))) (cur-elem (car L)) (mini (car ret-ans)) (rem-list (cdr ret-ans)))
(cond ((> cur-elem mini) (cons cur-elem (cons mini rem-list)))))))
It'll be simpler if you use built-in procedures, and split the problem in parts. Notice that the following assumes that there's a single minimum, adjust as necessary:
(define (find-min-rest L)
(let* ((the-min (apply min L))
(the-rest (remove the-min L)))
(list the-min the-rest)))
(find-min-rest '(1 2 3))
=> '(1 (2 3))
The code
(define (find-min-rest L)
(if (null? (cdr L)) (let ( (x (car L))) (cons x '( ())))
(let* ((ret-ans (find-min-rest (cdr L))) (cur-elem (car L)) (mini (car ret-ans)) (rem-list (cdr ret-ans)))
(cond ((> cur-elem mini) (cons cur-elem (cons mini rem-list)))))))
I am trying to reverse a list in Scheme using DrRacket.
Code:
(define rev
(lambda(l)
(if (null? l)
'()
(append (rev (cdr l)) (list (car l))))))
If I input (rev '(a((b)(c d)(((e)))))), the output is (((b) (c d) (((e)))) a).
I want it to be (((((e)))(d c)(b))a). I looked here: How to Reverse a List? but I get an even worse output. What am I doing wrong? Any help would be appreciated!
This is trickier than it looks, you're trying to do a "deep reverse" on a list of lists, not only the elements are reversed, but also the structure … here, try this:
(define (rev l)
(let loop ((lst l)
(acc '()))
(cond ((null? lst) acc)
((not (pair? lst)) lst)
(else (loop (cdr lst)
(cons (rev (car lst))
acc))))))
It works as expected:
(rev '(a ((b) (c d) (((e))))))
=> '(((((e))) (d c) (b)) a)
This code will do it:
(define (rev-list lst)
(if (null? lst)
null
(if (list? lst)
(append (rev-list (cdr lst)
(list (rev-list (car lst))))
lst)))
And the result is:
>>> (display (rev-list '((1 7) 5 (2 4 (5 9))) ))
(((9 5) 4 2) 5 (7 1))
The idea is simple: Return the arg if it's not a list, return rev-list(arg) otherwise.
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)