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.
Related
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'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 have a requirement to return the last negative number in a list, using a recursive procedure. Right now I have a recursive procedure that returns all negative numbers in the list.
(define returnLastNeg
(lambda (lst)
(if (null? lst)
'()
(if (positive? (car lst))
(returnLastNeg (cdr lst))
(cons (car lst) (returnLastNeg (cdr lst)))))))
calling it with (returnLastNeg'(1 -2 -3 4 -5 6)) returns
Output:
'(-2 -3 -5)
I need it to only return -5 though. I tried to modify my procedure to check to see if the last element in the list is positive. If it is, I want to remove the last element and then call the procedure again. But when I do that I get an error (below)
Modified procedure:
(define returnLastNeg-modified
(lambda (lst)
(if (null? lst)
'()
(if (positive? (last lst))
(remove (last lst) (lst))
(cons (car lst) (returnLastNeg-modified (cdr lst)))))))
ERROR:
application: not a procedure;
expected a procedure that can be applied to arguments
given: '(1 -2 -3 4 -5 6)
arguments...: [none]
>
A simpler approach would be with a helper procedure (called "sub") in this example:
(define returnLastNeg
(lambda (lst)
(define sub
(lambda (lst last-neg)
(if (null? lst)
last-neg
(let ((c (car lst)))
(sub (cdr lst)
(if (negative? c) c last-neg))))))
(sub lst null)))
EDIT
Knowing that
(define <procedureName> (lambda (<params>) ... )
is the same as
(define (<procedureName> <params>) ... )
and reformatting a little, this becomes:
(define (returnLastNeg lst)
(define (sub lst last-neg)
(if (null? lst)
last-neg
(let ((c (car lst)))
(sub (cdr lst) (if (negative? c) c last-neg)))))
(sub lst null))
I hope it's clearer
last-neg gets set to null by the very last expression
the recursive call to sub has 2 parameters (split on 2 lines in the initial version, but newlines don't matter).
This is the same as the even shorter version
(define (returnLastNeg lst)
(let sub ((lst lst) (last-neg null))
(if (null? lst)
last-neg
(let ((c (car lst)))
(sub (cdr lst) (if (negative? c) c last-neg))))))
using a so-called "named let".
I am trying to find the mode of a list
Assuming the list is sorted ascending order
Here is my mode function
(define freq
(lambda (l)
(cond ((null? l)l)
((null? (cdr l))l)
((not(equal? (car l) (car(cdr l))))(freq(cdr(delete l (car l)))))
(else (freq (cdr l)))
)))
(freq '(4 4 4 4 5 7 9 9 9)) => should returns 4 but its returning 9 instead
Here's my solution, which is similar to Óscar's solution but centralises the update of the longest/winning result in one place:
(define (longest-run lst)
(let loop ((result #f)
(cur #f)
(count 0)
(longest 0)
(lst lst))
(cond ((> count longest)
(loop cur cur count count lst))
((null? lst) result)
((eqv? (car lst) cur)
(loop result cur (+ count 1) longest (cdr lst)))
(else
(loop result (car lst) 1 longest (cdr lst))))))
I think my solution is shorter, cleaner, and less repetitive, but Óscar's solution has the advantage of updating the variables fewer times: his solution only updates the variables at the end of a run, whereas mine updates the variables whenever the current length is longer than the longest length seen so far.
What's the logic behind your procedure? how do you expect it to find the mode by removing elements from the input list? you should be counting frequencies instead:
(define (mode lst)
(if (null? lst)
#f ; edge case: an empty list doesn't have a mode
(let loop ((lst lst) ; list to traverse
(current (car lst)) ; current element in sequence
(counter 0) ; number of times current element appears
(max-current (car lst)) ; the mode
(max-counter 0)) ; number of times the mode appears
(cond ((null? lst) ; the list is finished
(if (> counter max-counter) current max-current))
((= (car lst) current) ; current element equal to previous
(loop (cdr lst) ; add 1 to counter and keep iterating
current
(add1 counter)
max-current
max-counter))
(else ; found a different element, a new sequence starts
(loop (cdr lst)
(car lst) ; update current element
1
(if (> counter max-counter) current max-current)
(max counter max-counter)))))))
It works by keeping track of how many times each element appears, returning the element that appears most frequently - because by definition the mode is the value that appears most often in a set of data. We take advantage of the fact that the input list is sorted, because we know that a new sequence of repeated elements starts when the current element is different from the previous one we encountered.
Before I start: YES, this is homework from college. Before I get told that I'm lazy and evil: this part of the homework was to convert two functions we already had, this one is the 6th.
(define (flatten-list a-list)
(cond ((null? a-list) '())
((list? (car a-list))
(append (flatten-list (car a-list)) (flatten-list (cdr a-list))))
(else (cons (car a-list) (flatten-list (cdr a-list))))))
The function, as you can guess, flattens a list even if it's nested. My specific problem with the transformation comes in the (list? (car a-list)) condition, in which I'm doing two recursive calls. I already did fibonacci, which I can do by just having two "acummulators" on the tail recursion. However, my mind is not trained in this yet to know how it should go.
I would appreciate if I was given hints and not the result. Thanks!
Here's my solution:
(define (flatten-iter a-list)
(define (flat-do acc lst-interm lst)
(cond
((null? lst)
(reverse acc))
((and (list? lst-interm) (not (null? lst-interm)))
(flat-do acc (car lst-interm) (append (cdr lst-interm) lst)))
((not (list? lst-interm))
(flat-do (cons lst-interm acc) empty lst))
((list? (car lst))
(flat-do acc (car lst) (cdr lst)))
(else
(flat-do (cons (car lst) acc) empty (cdr lst)))))
(flat-do empty empty a-list))
(flatten-iter (list 1 (list 2 (list 3 4 (list 5 empty 6))) 7 8))
=> (1 2 3 4 5 6 7 8)
Tail-recrusive functions require that they never return, and thus you can't use stack for storing your program's state. Instead, you use function arguments to pass the state between function calls. Therefore, we need to determine how to maintain the state. Because the result of our function is list?, it's meaningful to grow an empty list; we're using acc for this purpose. You can see how it works in else branch above. But we should be able to process nested lists. And while we're going deeper, we should keep the rest elements of the nested list for further processing. Sample list: (list 1 (list 2 3) 4 5)
Until (list 2 3) we have already added 1 to accumulator. Since we can't use stack, we need some other place to store the rest elements of the list. And this place is the lst argument, which contains elements of the original list to be flattened. We can just append the lst to the rest elements (cdr (list 2 3)) which are (list 3), and proceed with the list's head we stumbled upon while flattening, i. e. (car (list 2 3)) which is just 2. Now, (and (list? lst-interm) (not (null? lst-interm))) succeeds because flat-do is called this way:
(flat-do (list 1) (list 2 3) (list 4 5))
and the condition triggers this code:
(flat-do (list 1) (car (list 2 3)) (append (cdr (list 2 3)) (list 4 5)))
flat-do again is called this way: (flat-do (list 1) 2 (list 3 4 5))
The condition (not (list? 2)) now succeeds and the code (flat-do (cons 2 1) empty (list 3 4 5)) is evaluated.
The rest processing is done with else branch until lst is null? and reverse is performed on acc. Function then returns the reversed accumulator.