I need to write the function (quick-sort pred lst)
lst is the list of numbers to be sorted
pred is the predicate by which the list is ordered, the signature of this predicate is: (lambda (x y) …)
- (quick-sort < lst) will sort ascending (small to large)
- (quick-sort > lst) will sort descending (large to small)
- (quick-sort (lambda (x y) (< (car x) (car y))) lst) will sort a list
with inner lists according to the first element of the inner list, ascending.
I started with regular quick-sort:
(define (quick-sort lst)
(cond
((null? lst) '())
((= (length lst) 1) lst)
(else (append (quick-sort (filter (lambda (n) (< n (car lst))) lst))
(list (car lst))
(quick-sort (filter (lambda (n) (> n (car lst))) lst))))))
And now I'm trying to do this with pred:
(define (quick-sort pred lst)
(define (quick-sort-help lst)
(cond ((null? lst) ())
((= (length lst) 1) lst)
(else
(append (quick-sort-help (filter (lambda (n) (pred n (car lst))) lst))
(list (car lst))
(quick-sort-help (filter (lambda (n) (not(pred n (car lst)))) lst)))))) (quick-sort-help lst))
And I get an infinite recursion or something.
Can you help me solve this problem please?
Thanks!
First of you don't need the helper function quick-sort-help.
It recurs infinitely because you apply your helper function to lst instead cdr lst. In your regular quicksort you have (filter (lambda (n) (< n (car lst))) and (filter (lambda (n) (> n (car lst))). But then in the one with the predicate you have the problem that (not (pred ...) would cover the cases for <= and not < if the predicate is > and vice versa. So it gets stuck because the first element in the list is always equal with itself.
Here a correct quicksort:
(define (qsort f lst)
(if (null? lst)
null
(let ([pivot (car lst)])
(append (qsort f (filter (λ (n) (f n pivot)) (cdr lst)))
(list pivot)
(qsort f (filter (λ (n) (not (f n pivot))) (cdr lst)))))))
Related
I've been self-teaching myself Scheme R5RS for the past few months and have just started learning about mutable functions. I've did a couple of functions like this, but seem to find my mistake for this one.
(define (lst-functions)
(let ((lst '()))
(define (sum lst)
(cond ((null? lst) 0)
(else
(+ (car lst) (sum (cdr lst))))))
(define (length? lst)
(cond ((null? lst) 0)
(else
(+ 1 (length? (cdr lst))))))
(define (average)
(/ (sum lst) (length? lst)))
(define (insert x)
(set! lst (cons x lst)))
(lambda (function)
(cond ((eq? function 'sum) sum)
((eq? function 'length) length?)
((eq? function 'average) average)
((eq? function 'insert) insert)
(else
'undefined)))))
(define func (lst-functions))
((func 'insert) 2)
((func 'average))
You're not declaring the lst parameter in the procedures that use it, but you're passing it when invoking them. I marked the lines that were modified, try this:
(define (lst-functions)
(let ((lst '()))
(define (sum lst) ; modified
(cond ((null? lst) 0)
(else
(+ (car lst) (sum (cdr lst))))))
(define (length? lst) ; modified
(cond ((null? lst) 0)
(else
(+ 1 (length? (cdr lst))))))
(define (average)
(/ (sum lst) (length? lst)))
(define (insert x)
(set! lst (cons x lst)))
(lambda (function)
(cond ((eq? function 'sum) (lambda () (sum lst))) ; modified
((eq? function 'length) (lambda () (length? lst))) ; modified
((eq? function 'average) average)
((eq? function 'insert) insert)
(else
'undefined)))))
Now it works as expected:
(define func (lst-functions))
((func 'insert) 2)
((func 'average))
=> 2
((func 'sum))
=> 2
((func 'length))
=> 1
Some of your functions are recursive but defined without argument. Thus (sum (cdr lst)) shouldn't work since sum uses lst. You could do it by defining a helper:
(define (sum-rec lst)
(if (null? lst)
0
(+ (car lst) (sum-rec (cdr lst)))))
Or perhaps with an accumulator:
(define (sum-iter lst acc)
(if (null? lst)
acc
(sum-iter (cdr lst) (+ (car lst) acc)))
Your sum would of course use it passing the lst:
(define (sum)
(sum-iter lst 0))
Or you can just have the driver partial apply them like this:
(lambda (function)
(cond ((eq? function 'sum) (lambda () (sum-iter lst))
...))
A side note. length? is a strangely named function. A question mark in the end of a name is usually reserved for functions that return a true or a false value and this clearly returns a number.
I'm new to Scheme, and I've hit a wall. I have my sort and average functions, and I'm trying to change a median function I found on this site. However, no matter what I try, I keep getting errors where I have more than one expression in the median function, or when I try to use sort in the median function it's "undefined".
(define (sort1 L)
(if (or (null? L) (<= (length L) 1)) L
(let loop ((l null) (r null)
(pivot (car L)) (rest (cdr L)))
(if (null? rest)
(append (append (sort1 l) (list pivot)) (sort1 r))
(if (<= (car rest) pivot)
(loop (append l (list (car rest))) r pivot (cdr rest))
(loop l (append r (list (car rest))) pivot (cdr rest)))))))
(define (avg lst)
(let loop ((count 0) (sum 0) (args lst))
(if (not (null? args))
(loop (add1 count) (+ sum (car args)) (cdr args))
(/ sum count))))
(define (median L)
(if (null? L) (error "The list is empty")
(let loop ((L1 L) (L2 L))
(cond ((null? (cdr L2)) (car L1))
((null? (cddr L2)) (list (car L1) (cadr L1)))
(else (loop (cdr L1) (cddr L2)))))))
I'm trying to edit the median function to first sort the list, and if there are an even number of elements, I need to take the average of the list, and use the element closest to the average.
Any help would be appreciated, thank you in advance.
Like I said in a comment, what you want isn't a let, it's function composition.
Your current median function is this:
(define (median L)
(if (null? L)
(error "The list is empty")
(let loop ((L1 L) (L2 L))
(cond ((null? (cdr L2)) (car L1))
((null? (cddr L2)) (list (car L1) (cadr L1)))
(else (loop (cdr L1) (cddr L2)))))))
But as Oscar Lopez pointed out, this doesn't properly compute the median. However, it does some of the work, so keep it. Rename it to median-helper or something.
(define (median-helper L)
(if (null? L)
(error "The list is empty")
(let loop ((L1 L) (L2 L))
(cond ((null? (cdr L2)) (car L1))
((null? (cddr L2)) (list (car L1) (cadr L1)))
(else (loop (cdr L1) (cddr L2)))))))
Then you can use function composition to define the "real" median function:
(define (median lst)
(median-helper (sort1 lst)))
This returns the middle element for odd-length lists, and the middle-two elements for even length lists. If this is want you wanted, great. If not, then you can fix median-helper by returning the average in the second case of the cond. So instead of (list (car L1) (cadr L1)) there, you would have (avg (list (car L1) (cadr L1))).
;; median-helper : (Listof Number) -> Number
(define (median-helper L)
(if (null? L)
(error "The list is empty")
(let loop ((L1 L) (L2 L))
(cond ((null? (cdr L2)) (car L1))
((null? (cddr L2)) (avg (list (car L1) (cadr L1))))
(else (loop (cdr L1) (cddr L2)))))))
;; median : (Listof Number) -> Number
(define (median lst)
(median-helper (sort1 lst)))
I think you're misunderstanding the definition of a median. A very simple (if not particularly efficient) implementation follows:
(define (my-sort L)
(sort L <))
(define (average x y)
(exact->inexact (/ (+ x y) 2)))
(define (median L)
(if (null? L)
(error "The list is empty")
(let* ((n (length L))
(sorted (my-sort L))
(half (quotient n 2)))
(if (odd? n)
(list-ref sorted half)
(average (list-ref sorted half)
(list-ref sorted (sub1 half)))))))
It works as defined:
(median '())
=> The list is empty
(median '(3 2 1 5 4))
=> 3
(median '(6 4 3 1 2 5))
=> 3.5
I'm trying to learn how to do this, and I know it involves stacks, but I can't wrap my head around it unless I see a function in action. We've been given this example of a function to create and I need some help. Here it is:
;leafpile takes a list and returns the result of pushing all
;occurrences of the symbol 'leaf to the end of the list
> (leafpile '(rock grass leaf leaf acorn leaf))
(list 'rock 'grass 'acorn 'leaf 'leaf 'leaf)
We can use a helper function but the function needs to be written in a way to minimize recursive passes
update (heres what I got so far)
(define (leafpile/help ls pile)
(local
[
(define (helper 2ls leafpile)
(cond
[(empty? 2ls) (filter ls 'leaf)]
[(equal? (first 2ls) 'leaf)
(cons (first 2ls) (helper (rest 2ls) leafpile))]
[else (helper (rest 2ls) leafpile)]))]
(helper ls pile)))
OK snow I have this:
(define (helper lsh)
(cond
[(empty? lsh) '()]
[(not(equal? (first lsh) 'leaf))
(cons (first lsh) (helper (rest lsh)))]
[else (helper (rest lsh))]))
(define (count-leaf ls)
(cond
[(empty? ls) 0]
[(not (equal? 'leaf (first ls))) (count-leaf (rest ls))]
[else (add1 (count-leaf (rest ls)))]))
(define (leafpile ls)
(append (helper ls) (make-list (count-leaf ls) 'leaf)))
but I need it in one simple function with the least recursive passes possible.
Here is the solution I came up with:
(define (leafpile lst)
(for/fold ([pile (filter (lambda (leaf?) (not (equal? leaf? 'leaf))) lst)])
([i (build-list (for/fold ([leaves 0])
([leaf? lst])
(if (equal? leaf? 'leaf)
(add1 leaves)
leaves)) values)])
(append pile '(leaf))))
How it works:
The main for/fold loop iterates over a list with a length of the number of leaves there are, and the 'collection value' is a list of all the elements in lst that aren't the symbol 'leaf (achieved by filter).
Sample input/output:
> (leaf-pile '(rock grass leaf leaf acorn leaf))
'(rock grass acorn leaf leaf leaf)
Really simple way to do this:
(define (leaf? v)
(eq? v 'leaf))
(define (leafpile lst)
(append (filter (compose not leaf?) lst)
(filter leaf? lst)))
It really doesn't need to be more to it unless you experience performance issues and I usually don't for small lists. I tend to think of lists with fewer than a million elements as small. The obvious recursive one that might not be faster:
(define (leafpile lst)
(local [(define (leafpile lst n) ; screaming for a named let here!
(cond
((null? lst) (make-list n 'leaf))
((leaf? (car lst)) (leafpile (cdr lst) (add1 n)))
(else (cons (car lst) (leafpile (cdr lst) n)))))]
(leafpile lst 0)))
A tail recursive one that accumulates non leaf values, counts leaf values and uses srfi/1 append-reverse! to produce the end result:
(require srfi/1)
(define (leafpile lst)
(local [(define (leafpile lst acc n) ; I'm still screaming
(cond
((null? lst) (append-reverse! acc (make-list n 'leaf)))
((leaf? (car lst)) (leafpile (cdr lst) acc (add1 n)))
(else (leafpile (cdr lst) (cons (car lst) acc) n))))]
(leafpile lst '() 0)))
I'm trying to teach myself functional language thinking and have written a procedure that takes a list and returns a list with duplicates filtered out. This works, but the output list is sorted in the order in which the last instance of each duplicate item is found in the input list.
(define (inlist L n)
(cond
((null? L) #f)
((= (car L) n) #t)
(else (inlist (cdr L) n))
))
(define (uniquelist L)
(cond
((null? L) '())
((= 1 (length L)) L)
((inlist (cdr L) (car L)) (uniquelist (cdr L)))
(else (cons (car L) (uniquelist (cdr L))))
))
So..
(uniquelist '(1 1 2 3)) => (1 2 3)
...but...
(uniquelist '(1 2 3 1)) => (2 3 1)
Is there a simple alternative that maintains the order of the first instance of each duplicate?
The best way to solve this problem would be to use Racket's built-in remove-duplicates procedure. But of course, you want to implement the solution from scratch. Here's a way using idiomatic Racket, and notice that we can use member (another built-in function) in place of inlist:
(define (uniquelist L)
(let loop ([lst (reverse L)] [acc empty])
(cond [(empty? lst)
acc]
[(member (first lst) (rest lst))
(loop (rest lst) acc)]
[else
(loop (rest lst) (cons (first lst) acc))])))
Or we can write the same procedure using standard Scheme, as shown in SICP:
(define (uniquelist L)
(let loop ((lst (reverse L)) (acc '()))
(cond ((null? lst)
acc)
((member (car lst) (cdr lst))
(loop (cdr lst) acc))
(else
(loop (cdr lst) (cons (car lst) acc))))))
The above makes use of a named let for iteration, and shows how to write a tail-recursive implementation. It works as expected:
(uniquelist '(1 1 2 3))
=> '(1 2 3)
(uniquelist '(1 2 3 1))
=> '(1 2 3)
(define filter-in
(lambda (predicate list)
(let((f
(lambda (l)
(filter-in-sexpr predicate l))))
(map f list))))
(define filter-in-aux
(lambda (pred lst)
(if (null? lst) '()
(cons (filter-in-sexpr pred (car lst))
(filter-in-aux pred (cdr lst))))))
(define filter-in-sexpr
(lambda (pred sexpr)
(if (equal? (pred sexpr) #t)
sexpr
'())))
Calling (filter-in number? ’(a 2 (1 3) b 7)) produces ( () 2 () () 7).
How I can skip null elements from the generated list to get final outcome of (2 7) ?
The problem is that you're mapping filter-in-sxpr over the list. You can either run another filter pass to remove the nulls, or use a modified filter-in-aux like this:
(define filter-in-aux
(lambda (pred lst)
(if (null? lst) '()
(let ((h (filter-in-sexpr pred (car lst)))
(t (filter-in-aux pred (cdr lst))))
(if (null? h) t
(cons h t))))))