How can I write a function using abstract list functions (foldr, map, and filter) without recursion that consumes a list of numbers (list a1 a2 a3 ...) and produces a new list removing the minimum number from the original list?
The recursion code is:
(define (find-min lst)
(cond
[(empty? (rest lst)) (first lst)]
[else
(local [(define min-rest (find-min (rest lst)))]
(cond
[(< (first lst) min-rest) (first lst)]
[else min-rest]))]))
A fold applies a 2-argument function against a given value and the car of a list uses the result against the successive cars or the cdrs or the list. this is what we want.
Whereas map returns a new list by doing something with each element of a list.
And filter returns a smaller or equal list based on some predicate.
Now just to formulate a function that can choose the lessor of two arguments
(define (the-lessor x y)
(if (< x y)
x
y))
From there implementation is straightforward.
(define (min L) (fold the-lessor (car L) (cdr L)))
Since this looks like a homework question, I'm not going to provide all the code, but hopefully push you in the right direction.
From the HTDP book, we see that "The Intermediate Student language adds local bindings and higher-order functions." The trick here is probably going to using "local bindings".
Some assumptions:
(remove-min-from-list '()) => not allowed: input list must be non-empty
(remove-min-from-list '(1)) => '()
(remove-min-from-list '(1 2 3 1 2 3)) => '(2 3 2 3) ; all instances of 1 were removed
Somehow, we need to find the minimum value of the list. Call this function min-of-list. What are its inputs and outputs? It's input is a list of numbers and its output is a number. Of the abstract list functions, which ones allow us to turn a list of numbers into a number? (And not another list.) This looks like foldl/foldr to me.
(define (min-of-list lst)
(foldr some-function some-base lst))
Since you already showed that you could write min-of-list recursively, let's move on. See #WorBlux's answer for hints there.
How would we use this in our next function remove-min-from-list? What are the inputs and outputs of remove-min-from-list? It takes a list of numbers and returns a list of numbers. Okay, that looks like map or filter. However, the input list is potentially shorter than that output list, so filter and not map.
(define (remove-min-from-list lst)
....
(filter some-predicate list))
What does some-predicate look like? It needs to return #f for the minimum value of the list.
Let's pull this all together and use local to write one function:
(define (remove-min-from-list lst)
(local [(define min-value ...)
(define (should-stay-in-list? number) ...min-value ...)]
(filter should-stay-in-list? lst)))
The key here, is that the definition for should-stay-in-list? can refer to min-value because min-value came before it in the local definitions block and that the filter later on can use should-stay-in-list? because it is in the body of the local.
(define (comparator n) (local [(define (compare v) (not (equal? v n)))] compare))
(define (without-min list) (filter (comparator (foldr min (foldr max 0 list) list)) list))
Related
New to scheme but trying to learn the basics.
Let's say I passed a list in as a parameter and I wanted to multiply each element by -1. Right now I have this:
(define (negative b)
(* (car b) -1 )))
Which returns the first element as -1 * that element
So in this case giving it (negative '(5 1 2 3)) returns -5.
But lets say I want it to return
-5 -1 -2 -3
How would I go about making the rest of the list negative? Using cdr recursively?
Do it recursively.
(define (negative l)
(if (null? l)
'()
(cons (* (car l) -1)
(negative (cdr l)))))
If the list is empty, this just returns an empty list, as the base case.
Otherwise, it calculates -1 * the first element, the negative of the rest of the list, and combines them to produce the result.
The purpose of your exercise may be for you to code up your own map procedure, in which case that's fine. But if not, use scheme's built in 'map' procedure which is intended for just this kind of purpose.
'map' has been available at least since R4RS (that is, a long time ago) and possibly earlier.
by using map. If you want it returned as list.
It would be like this
(define negative
(lambda (b)
(map - b)))
Map is going through list b, and apply procedure "-" to each number in list
If you want to return as single numbers not in list you apply values on the list.
(define negative1
(lambda (b)
(apply values (map - b))))
Edit: I saw that you are asking for recursive solution, which would go like this
(define negative1
(lambda (b)
(if (null? b)
'()
(cons (- (car b)) (negative1 (cdr b))))))
So currently I'm working on a problem of implementing the shortest-seek-time disk scheduling algorithm. Our task is, being passed a list of tracks, to return a list of length 3 containing the Avg Movement, Total movement, and then a list of the movement amounts. Currently I'm tackling returning the list of movements as I feel the total and avg's are trivial.
What I'm currently stuck on is saving the distance between the current head position and the current track so I can continue searching the list for the shortest seek time. What I did first was to sort the list and pass it into a separate helper function that only returns the the shortest-movement from the current head position so I can continually call this function.
Is the best way to do this to simply pass in another variable to hold the distance and keep checking based on this number? and then remove that disk from the list once I've satisfied the movement?
In Scheme having accumulators and state as extra parameters to a helper function and update using recursion to have state during the execution without actually mutating bindings.
The simplest example:
(define (reverse lst)
(let loop ((lst lst) (acc '()))
(if (null? lst)
acc
(loop (cdr lst) (cons (car lst) acc)))))
(reverse '(1 2 3)) ; ==> (3 2 1)
EDIT
Now since you said you're not used to named let here is the same with a helper procedure and keeping the same names:
(define (reverse lst)
;; a helper with an accumulator
(define (loop lst acc)
(if (null? lst)
acc
(loop (cdr lst) (cons (car lst) acc))))
;; call the helper
(loop lst '()))
Now the names loop, lst, acc are just variable names. However their names suggest loop is a tail recursive procedure. lst is a list and acc is short for accumulator which indicates on the base case it will be part of the result and for the default case it will be updated.
I need to write a good set function that checks whether its argument lst is a properly represented set, i.e. it is a list consisting only of integers, with no duplicates, and returns true #t or false #f. For example:
(good-set? (1 5 2)) => #t
(good-set? ()) => #t
(good-set? (1 5 5)) => #f
(good-set? (1 (5) 2)) => #f
so I have began writing the function as:
(define (good-set? lst)
so I don't know how to proceed after this. Can anybody help?
One option would be to use andmap and sets, as has been suggested by #soegaard:
(define (good-set? lst) ; it's a good set if:
(and (andmap integer? lst) ; all its elements are integers and
(= (length lst) ; the list's length equals the size
(set-count (list->set lst))))) ; of a set with the same elements
But if you can't use sets or other advanced procedures, then traverse the list and test if the current element is an integer and is not present somewhere else in the list (use member for this), repeating this test for each element until there are no more elements in the list. Here's the general idea, fill-in the blanks:
(define (good-set? lst)
(cond (<???> ; if the list is empty
<???>) ; then it's a good set
((or <???> ; if the 1st element is not an integer or
<???>) ; the 1st element is in the rest of the list
<???>) ; then it's NOT a good set
(else ; otherwise
(good-set? <???>)))) ; advance recursion
Sets are built into the Racket standard library: I would recommend not reimplementing them in terms of lists unless you really need to do something customized.
If we need to treat this as a homework assignment, I would recommend using a design methodology to systematically attack this problem. In this case, see something like How to Design Programs with regards to designing functions that work on lists. As a brief sketch, we'd systematically figure out:
What's the structure of the data I'm working with?
What tests cases do I consider? (including the base case)
What's the overall shape of the function?
What's the meaning of the natural recursion?
How do I combine the result of the natural recursion in order to compute a solution to the total?
For this, check if the first number is duplicated, if it is not, then recurse by checking the rest. As such:
(define (good-set? list)
(or (null? list) ; nothing left, good!
(let ((head (car list)))
(rest (cdr list)))
(and (number? head) ; a number
(not (member = head rest)) ; not in the rest
(good-set? rest))))) ; check the rest
If you need member, then
(define (member pred item list)
(and (not (null? list))
(or (pred item (car list))
(member pred item (cdr list)))))
I was wondering, how do you check if every element in a list is an integer or not? I can check the first element by using (integer? (car list), but if I do (integer? (cdr list), it always returns false (#f) because the whole of the last part of the list is not an integer as a group.
In this case let's say list is defined as.
(define list '(1 2 5 4 5 3))
(define get-integers
(lambda (x)
(if (null? x)
"All elements of list are integers"
(if (integer? (car x))
(get-integers (cdr x))
"Not all elements are an integer"))))
Practical Schemes provide functions for doing tests across whole sequences. An application of the andmap function, for example, would be appropriate. Racket provides a for/and to do something similar. If you really needed to write out the loop by hand, you'll be using recursion.
What you need to do is test each element in the list to see if it satisfies a condition (being an integer, in this case). When you evaluate (integer? (car list)) on a list of integers, you're checking if the first element in the list is an integer, and that's fine. But the expression (integer? (cdr list)) tests if a list is an integer (because cdr returns a list), and that won't work - you need to test the next element in the list, and the next, and so on until the list is empty.
There are several ways to do the above, the most straightforward would be to recur on the list testing each element in turn, returning false if a non-integer element was found or true if all the list was consumed without finding a non-integer element, like this:
(define (all-integers? lst)
(cond ((null? lst) #t)
((not (integer? (car lst))) #f)
(else (all-integers? (cdr lst)))))
A more practical approach would be to use a built-in procedure, like this:
(andmap integer? lst)
andmap will check if all the elements in lst evaluate to true for the given predicate. For example:
(andmap integer? '(1 2 3))
> #t
(andmap integer? '(1 "x" 3))
> #f
SRFI-1 uses the terms every and any rather than andmap and ormap. match can also be used:
(define list-of-integers?
(lambda (lst)
(match lst
(((? number?) ..1) #t)
(_ #f))))
Hi I am trying to write a program where given a list of lists check to see if they are equal in size and return #t if they are.
So for example if i were to write (list-counter? '((1 2 3) (4 5 6) (7 8 9))) the program would return #t, and (list-counter? '((1 2 3) (4 5 6) (7 8))) would return #f.
SO far this is what I have done:
(define list-counter?
(lambda (x)
(if (list? x)
(if (list?(car x))
(let (l (length (car x))))
(if (equal? l (length(car x))))
(list-counter?(cdr x))
) ) ) ) )
I think where I am going wrong is after I set the length of l to the length of the first list. Any help would be appreciated.
There are several ways to solve this problem. For instance, by hand and going step-by-step:
(define (all-lengths lists)
(if (null? lists)
'()
(cons (length (car lists))
(all-lengths (cdr lists)))))
(define (all-equal? head lengths)
(if (null? lengths)
true
(and (= head (car lengths))
(all-equal? head (cdr lengths)))))
(define (list-counter? lists)
(let ((lengths (all-lengths lists)))
(all-equal? (car lengths) (cdr lengths))))
Let me explain the above procedures. I'm dividing the problem in two steps, first create a new list with the lengths of each sublist - that's what all-lengths does. Then, compare the first element in a list with the rest of the elements, and see if they're all equal - that's what all-equal? does. Finally, list-counter? wraps it all together, calling both of the previous procedures with the right parameters.
Or even simpler (and shorter), by using list procedures (higher-order procedures):
(define (list-counter? lists)
(apply = (map length lists)))
For understanding the second solution, observe that all-lengths and all-equal? represent special cases of more general procedures. When we need to create a new list with the result of applying a procedure to each of the elements of another list, we use map. And when we need to apply a procedure (= in this case) to all of the elements of a list at the same time, we use apply. And that's exactly what the second version of list-counter? is doing.
You could write an all-equal? function like so:
(define (all-equal? list)
;; (all-equal? '()) -> #t
;; (all-equal? '(35)) -> #t
;; (all-equal? '(2 3 2)) -> #f
(if (or (null? list) (null? (cdr list)))
#t
(reduce equal? list)
))
then do:
(all-equal? (map length listOfLists))
Alternatively you can do:
(define (lists-same-size? list-of-lists)
(if (== (length listOfLists) 0)
#t
(let*
(( firstLength
(length (car listOfLists)) )
( length-equal-to-first?
(lambda (x) (== (length x) firstLength)) )
)
(reduce and #t (map length-equal-to-first? listOfLists))
)
)))
What this says is: if the list length is 0, our statement is vacuously true, otherwise we capture the first element of the list's length (in the 'else' part of the if-clause), put it in the closure defined by let's syntactic sugar (actually a lambda), and use that to define an length-equal-to-first? function.
Unfortunately reduce is not lazy. What we'd really like is to avoid calculating lengths of lists if we find that just one is not equal. Thus to be more efficient we could do:
...
(let*
...
( all-match? ;; lazy
(lambda (pred list)
(if (null? list)
#t
(and (pred (first list)) (all-match? (cdr list)))
;;^^^^^^^^^^^^^^^^^^^ stops recursion if this is false
)) )
)
(all-match? length-equal-to-first? listOfLists)
)
)))
Note that all-match? is already effectively defined for you with MIT scheme's (list-search-positive list pred) or (for-all? list pred), or in Racket as andmap.
Why does it take so long to write?
You are forced to write a base-case because your reduction has no canonical element since it relies on the first element, and list manipulation in most languages is not very powerful. You'd even have the same issue in other languages like Python. In case this helps:
second method:
if len(listOfLists)==0:
return True
else:
firstLength = len(listOfLists[0])
return all(len(x)==firstLength for x in listOfLists)
However the first method is much simpler to write in any language, because it skirts this issue by ignoring the base-cases.
first method:
if len(listOfLists)<2:
return True
else:
return reduce(lambda a,b: a==b, listOfLists)
This might sound a bit weird, but I think it is easy.
Run down the list, building a new list containing the length of each (contained) list, i.e. map length.
Run down the constructed list of lengths, comparing the head to the rest, return #t if they are all the same as the head. Return false as soon as it fails to match the head.