I'm having trouble using my member? function. I need to recurse on my set? function until the last element in my list 'lst' is reached. I believe I have the navigation down correctly, but maybe my inputs syntax is wrong. I know there are three cases:
1) What happens if the list is empty? that means that there aren't any duplicates in it
2) What happens if the current element of the list exists somewhere in the rest of the list? then it means that there's a duplicate in the list (hint: the member procedure might be useful)
3) If none of the above are true, continue with the next element.
Here is my code.
(define (member? e lst)
(if (null? lst) #f
(if (equal? e (car lst)) #t
(member? e (cdr lst)))))
(define (set? lst)
(if (null? lst) #t ;Case1
(if (member? (car lst) lst) #f ;Case2
(set? (cdr lst))))) ;Case3
;Example tests for the set? function
(set? '(x y z))
(set? '(a 1 b 2 c 3))
(set? '())
(set? '(6 2 2))
(set? '(x y z x))
There's a small mistake with your code, look how it gets fixed:
(define (set? lst)
(if (null? lst)
#t
(if (member? (car lst) (cdr lst)) ; in here
#f
(set? (cdr lst)))))
In particular, notice what this line is doing:
(member? (car lst) lst)
That won't work: the test is checking whether the first element in lst is a member of lst - and that'll always be true. The solution is simple, just check to see if the current element is in the rest of the list, if it's there, then we know that we've found a duplicate:
(member? (car lst) (cdr lst))
And by the way, the above code would look much nicer using cond, which is great when you have nested ifs:
(define (set? lst)
(cond ((null? lst) #t)
((member? (car lst) (cdr lst)) #f)
(else (set? (cdr lst)))))
Related
I am trying to write a code in scheme -last_two(lst) where it should output the last two elements in the list.
(define (last_two lst)
(if (null? lst)
'() )
(if (null? (cdr lst))
lst)
(if(null? (cddr lst)))
lst)
(last_two (cdr lst)))
last_two(1 2 3 4) -> (3 4)
A procedure returns the value of the last expression that's executed in the body. In your procedure, when one of the first if conditions is true, you don't return the value from the procedure, you just go on to test the next condition. Eventually you get to the end, which is the recursive call -- you never stop at the base case. Eventually you'll get an error when you try to call (cdr lst) on an empty list.
You can use cond to test a series of conditions and return the value from the first one that's true.
Also, you should be returning lst when the conditions match; your code will always return (), since you're returning the same thing that matched the null? condition.
(define (last_two lst)
(cond ((null? lst) '())
((null? (cdr lst)) lst)
((null? (cddr lst)) lst)
(else (last_two (cdr lst)))))
(last_two '(1 2 3 4))
You might notice that the first 3 conditions all return lst, so you could combine them using or.
(define (last_two lst)
(if (or (null? lst) (null? (cdr lst)) (null? (cddr lst)))
lst
(last_two (cdr lst))))
I am writing a program in Scheme and having difficulty with this one part. Below is an example to make my question clear
(endsmatch lst) should return #t if the first element in the list is the same as the last element in the list and return #f otherwise.
For example:
(endsmatch '(s t u v w x y z)) should return: #f
and
(endsmatch (LIST 'j 'k 'l 'm 'n 'o 'j)) should return: #t
Here is what I have so far (just error handling). The main issue I am having is solving this recursively. I understand there are easier solutions that are not recursive but I need to solve this using recursion.
My code so far:
(define (endsmatch lst)
(if (not(list? lst))
"USAGE: (endsmatch [list])"
(if (or (null? lst)
(= (length lst) 1))
#t
(equal? ((car lst)) (endsmatch(car lst)))
)))
I believe my code starting at "(equal? " is where it is broken and doesn't work. This is also where I believe recursion will take place. Any help is appreciated!
Easiest way is to use a (recursive) helper function to do the looping:
(define (endsmatch lst)
(define (helper no1 lst)
(if (null? (cdr lst))
(equal? no1 (car lst))
(helper no1 (cdr lst))))
(if (or (not (list? lst)) (null? lst))
"USAGE: (endsmatch [list])"
(helper (car lst) lst)))
The reason I pass lst and not (cdr lst) as the second argument in the last line is so that it also works for 1-element lists.
I tend to use KISS when programming. aka. "Keep it simple, stupid!"
With that regard I would have oped for:
(define (ends-match? lst)
(or (null? lst)
(equal? (car lst)
(last lst))))
Now last we can define like this:
(define (last lst)
(foldl (lambda (e a) e) last lst))
It's not perfect. It should signal an error if you pass an empty list, but in the ends-match? you check for this and thus it's not a problem.
Consider:
(define (nested-reverse lst)
(cond ((null? lst) '())
((list? (car lst)) (nested-reverse (car lst)))
(else
(cons (nested-reverse (cdr lst))
(list (car lst))))))
When I input,
(nested-reverse '((a b c) 42))
it gives me ((() 42) (a b c)). It's supposed to give me (42 (c b a)). How I would change my code so that the nested lists also get reversed?
Keep in mind that a list (1 2 3) is (cons 1 (cons 2 (cons 3 '()))). Using append is a very poor choice on how to reverse a list since append is implemented like this:
(define (append lst1 lst2)
(if (null? lst1)
lst2
(cons (car lst1) (append (cdr lst1) lst2))))
A list can be iterated from the first element towards the end while it can only be made in reverse. Thus the obvious none recursive reverse would look like something like this:
(define (simple-reverse lst)
(let loop ((lst lst) (result '()))
(if (null? lst)
result
(loop (cdr lst) (cons (car lst) result)))))
To make it work for nested list you check if you need to reverse (car lst) by checking of it's a list or not and use the same procedure as you are creating to do the reverse on the element as well. Other than that it's very similar.
I wrote a small function that takes a list and returns a list composed of only positive numbers. This all works fine, but for some reason, it is reversing the order. More information below on that. Could someone please tell me if this is normal, or if I miss-wrote something? Thank you in advance.
(define (positive-nums-only lst)
(if (empty? lst)
'()
(append (positive-nums-only (cdr lst))
(if (>= (car lst) 0)
(list (car lst))
'()))))
(positive-nums-only '(1 2 -4 90 -4))
The above test case returns '(90 2 1)
You did not make a mistake, the program is making what you asked.
See, the program finishes the recursion calls first, before going into resolving the if statement. This causes the (list ... ) to start listing from the last element that is positive, in this example it will be 90.
Changing the code order will produce the result you want.
(define (positive-nums-only lst)
(if (empty? lst) '()
(append (if (>= (car lst) 0 )
(list (car lst))
'())
(positive-nums-only (cdr lst)))
)
)
On the other hand, this kind of recursion could be expensive to the computer. I'd use tail recursion, like this:
(define positive-nums-only-tail
(λ (lst r)
(cond ((empty? lst) (reverse r))
((positive? (car lst))
(positive-nums-only-tail (cdr lst)
(cons (car lst) r)))
(else (positive-nums-only-tail (cdr lst) r))
)
)
)
Have you tried reversing the append?
(define (positive-nums-only lst)
(if (empty? lst)
'()
(append (if (>= (car lst) 0) (list (car lst)) '())
(positive-nums-only (cdr lst)))))
Personally I find it more natural to write it like this:
(define (positive-nums-only lst)
(if (empty? lst)
'()
(let ((rest (positive-nums-only (cdr lst))))
(if (>= (car lst) 0)
(cons (car lst) rest)
rest))))
So i have these two functions that work fine alone. I am trying to write one function to accomplish both but i keep getting a car error. Any guidance on the best way to solve this?
(define (countNumbers lst)
(cond
((null? lst) 0)
((number? (car lst))(+ 1 (countNumbers (cdr lst))))
(else (countNumbers (cdr lst)))))
(define (flatten x)
(cond ((null? x) '())
((pair? x) (append (flatten (car x)) (flatten (cdr x))))
(else (list x))))
I tried something like this im rather new to functional programming in general so im still trying to wrap my mind around it it says the problem is after number?(car lst)
(define (flatten lst)
(cond ((null? lst) '())
((pair? lst) (append (flatten (car lst)) (flatten (cdr lst))))
(else (list(cond
((null? lst) 0)
((number? (car lst))(+ 1 (flatten (cdr lst))))
(else (flatten (cdr lst))))))))
As I mentioned in my comment, I don't think it's a good idea to stick everything in a single function. Anyway, you were kinda on the right track, but we have to remember that if we're going to return a number as the final result, then our base case should reflect this and also return a number (not an empty list), and the combining step should add numbers, not append them. This is what I mean:
(define (count-flatten lst)
(cond ((null? lst) 0)
((pair? lst)
(+ (count-flatten (car lst))
(count-flatten (cdr lst))))
((number? lst) 1)
(else 0)))
But I'd rather do this:
(define (count-flatten lst)
(countNumbers (flatten lst)))
We can even write an idiomatic solution using only built-in procedures, check your interpreter's documentation, but in Racket we can do this:
(define (count-flatten lst)
(count number? (flatten lst)))
Anyway, it works as expected:
(count-flatten '(1 x (x 2) x (3 (4 x (5) 6) 7)))
=> 7