I'm beginner in Racket, recently I'm working at trees in racket and i have no idea how to balancing BST ,,, I write a code to insert and build a BST but can't balancing the BST ( my code to insert and build a BST )
(define (insert element tree)
(cond ((null? tree)(list element '() '()))
((= element (car tree)) tre)
((< element (car tree))(list (car tree) (insert element (cadr tree)) (caddr tree)))
(else (list (car tree) (cadr tree) (insert element (caddr tree))))))
(define (build-BST L)
(build-BST-helper L '()))
(define (build-BST-helper L tree)
(cond ((null? L) tree)
(else (build-BST-helper (cdr L) (insert (car L) tree)))))
Thx in advance :) ,,, Sorry for my English
Related
Getting an error for my binary search tree that I created in scheme.
$gosh main.sc
*** ERROR: list required, but got 5
Stack Trace:
_______________________________________
This is my code. I think the error has to do with how I am calling the functions, but I am not sure what exactly is wrong. I am calling the insert function with the two required parameters: tree, and a value of 5.
(define (member? t v)
(cond
((null? t)
#f
)
((< node (car t))
(member? (cadr t) v))
((> node (car t))
(member? (caddr t) v))
(else
#t
)
)
)
(define (insert t v)
(cond
((null? t)
(list v '() '())
)
((< v (car t))
(list (car t) (insert (cadr t) v) (caddr t))
)
((>= v (car t))
(list (car t) (cadr t) (insert (caddr t) v))
)
(else
t
)
)
)
(define (fold func val lst)
(if (null? lst) val (fold func (func val (car lst)) (cdr lst))))
(define (build lst)
(fold (lambda (t v) (insert t v)) '() lst))
(define t (list 10 '() '()))
(insert t 5)
display (member t 5)
display t
You care calling (member t 5) which is the same as (member '(10 '() '()) 5). Now member is not the same as your defined member? since it has a different name. member is the core library that looks like this:
(define (member obj lst)
(cond ((null? lst) #f)
((equal? obj (car lst)) lst)
(else (member obj (cdr lst)))))
Your member? has the two parameters swapped so when you miswrote the name and used the report version member 5 is not null, then it will do (car 5) and that will fail miserably. The error message that 5 is not of the required type list is pretty decent. It might spell the beans that it was member that failed though.
Another thing. If you replace the call to member with a call to member? you hit more problems. You use a variable node that is not defined anywhere.
The indentation and placing of parentheses is not goo lisp style. Your code should be written like this:
;; node doesn't exist in OPs code, but my implementation doesn't like member? without it
(define node 5)
;; possible typo by using the variable node ?
(define (member? t v)
(cond
((null? t)
#f)
((< node (car t))
(member? (cadr t) v))
((> node (car t))
(member? (caddr t) v))
(else
#t)))
(define (insert t v)
(cond
((null? t)
(list v '() '()))
((< v (car t))
(list (car t) (insert (cadr t) v) (caddr t)))
((>= v (car t))
(list (car t) (cadr t) (insert (caddr t) v)))
(else
t)))
(define (fold func val lst)
(if (null? lst) val (fold func (func val (car lst)) (cdr lst))))
(define (build lst)
(fold (lambda (t v) (insert t v)) '() lst))
(define t (list 10 '() '()))
(insert t 5)
;; NB doesn't call a procedure, just evaluates it.
display
;; Here the arguments are the wrong order and you don't use memeber?
(member t 5)
;; NB doesn't call a procedure, just evaluates it.
display
t
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)
Can someone show me the error in this code please?
I want to generalize the member function to support nested lists. I need to search thing inside the nested list and return the rest of the list when I found thing. I don't really understand whats wrong with the code below.
(define (memberk thing lis)
(cond
((null? lis) #f)
((list? (car lis))
(cons (memberk thing (car lis))
(memberk thing (cdr lis))))
(else
(if (equal? (car lis) thing)
lis
(memberk thing (cdr lis))))))
Expexted output: (memberk 3 '(1 4 (3 1) 2)) = '((3 1) 2)
Actual output from the code above: '((3 1) . #f)
So how I see this you would like the top level cons that has the key found somewhere in car. I'm thinking something like:
(define (memberk needle lst)
(define (found? haystack)
(or (equal? needle haystack)
(and (pair? haystack)
(or (found? (car haystack))
(found? (cdr haystack))))))
(let loop ((lst lst))
(cond ((null? lst) #f)
((found? (car lst)) lst)
(else (loop (cdr lst))))))
(memberk '(a) '(a b (b (a) c) c d)) ; ==> ((b (a) c) c d)
Something like this?
It is a bit unclear what you want - since there is only one test case.
(define (memberk thing lis)
(cond
[(null? lis)
#f]
[(and (cons? (car lis)) (memberk thing (car lis)))
=> (λ (found) (cons found (cdr lis)))]
[(equal? (car lis) thing)
lis]
[else
(memberk thing (cdr lis))]))
I've been working on a vector multiply function in scheme and have found myself in rut. I dont want to use any looping and I dont want to use any scheme built in functions other than the ones I've already included. I've created a helper function called rotate and dotproduct. I can get the correct values if I do this in racket (vectormult '(1 2 -1) (rotate '((0 2 3) (1 2 0) (1 0 3)))). How can I rotate the initial parameter without re-rotating after every recursive call? NOTE: I dont want to introduce additional paramaters. If my logic/approach to this is all wrong please help me get on the right track.
Code
(define dotproduct
(lambda (l1 l2)
(if (or (null? l1) (null? l2))
0
(+ (* (car l1) (car l2)) (dotproduct (cdr l1) (cdr l2))))))
(define getFirsts
(lambda (l)
(cond
((null? l) `())
(else (cons (first* l) (getFirsts (cdr l)))))))
(define removeFirsts
(lambda (l)
(cond
((null? l) `())
((null? (car l)) `())
(else (cons (cdr (car l)) (removeFirsts (cdr l)))))))
(define rotate
(lambda (l)
(cond
((null? l) `())
((null? (first* l)) `())
(else (cons (getFirsts l) (rotate (removeFirsts l)))))))
(define vectormult
(lambda (l1 l2)
(cond
((null? l2) `())
(else (cons (dotproduct l1 (car l2)) (vectormult l1 (cdr l2)))))))
If I understand the question, you could rename your current vectormult to, say, rotatedvectormult (and change its recursive call accordingly), and then have vectormult just rotate the parameter before calling rotatedvectormult. This way, rotatedvectormult would know the parameter was already rotated, but vectormult could still take an unrotated vector.
I ended up ditching the rotate function in favor of adding 2 functions getFirsts and removeFirsts.
Code
(define getFirsts
(lambda (l)
(cond
((null? l) `())
(else (cons (first* l) (getFirsts (cdr l)))))))
(define removeFirsts
(lambda (l)
(cond
((null? l) `())
((null? (car l)) `())
(else (cons (cdr (car l)) (removeFirsts (cdr l)))))))
;(define rotate
; (lambda (l)
; (cond
; ((null? l) `())
; ((null? (first* l)) `())
; (else (cons (getFirsts l) (rotate (removeFirsts l)))))))
(define vectormult
(lambda (l1 l2)
(cond
((null? (first* l2)) `())
(else (cons (dotproduct l1 (getFirsts l2)) (vectormult l1 (removeFirsts l2)))))))
Create the function AllEqual?(L) which will be given a list of (at least 2) elements and will return #t if all of them are the same, otherwise #f
(define (allequal? l)
(if (< (length l) 2)
#t
(equal? (car l) (allequal? (cdr l)))))
(equal? (car l) (allequal? (cdr l)))
tests if (car l) is equal? to (allequal? ...). Since allequal? returns a boolean (#t or #f), you test if (car l) is #t or #f, which is not what you want.
You need to compare the first two elements, if available, and then recurse on the cdr of your list:
(define (allequal? l)
(if (< (length l) 2)
#t
(and (equal? (car l) (cadr l))
(allequal? (cdr l)))))
or, easier,
(define (allequal? l)
(or (< (length l) 2)
(and (equal? (car l) (cadr l))
(allequal? (cdr l)))))
or even
(define (allequal? l)
(or (null? l)
(null? (cdr l))
(and (equal? (car l) (cadr l))
(allequal? (cdr l)))))