Non-decreasing list of lists in Scheme? - list

Question

We need a Scheme function called nondecreaselist, which takes in a list of numbers and outputs a list of lists, which overall has the same numbers in the same order, but grouped into lists that are non-decreasing.
For example, if we have input (1 2 3 4 1 2 3 4 1 1 1 2 1 1 0 4 3 2 1), the output should be:
((1 2 3 4) (1 2 3 4) (1 1 1 2) (1 1) (0 4) (3) (2) (1))
How would you implement this? I know we have to use recursion.
My attempt so far:
(define (nondecreaselist s)
(cond ((null? s) '())
((cons (cons (car s)
((if (and (not (null? (cadr s)))
(not (> (car s) (cadr s))))
((cadr s))
('()))))
(nondecreaselist (cdr s))))))
However, this gives me the error:
(int) is not callable:

Answer 1

(define decrease-list
(lambda (l)
((lambda (s) (s s l cons))
(lambda (s l col)
;; limitcase1: ()
(if (null? l)
(col '() '())
;; limitcase2: (a1)
(if (null? (cdr l))
(col l '())
(let ((a1 (car l)) (a2 (cadr l)))
;; limitcase3: (a1 a2)
(if (null? (cddr l))
(if (>= a2 a1)
(col l '())
(col (list a1) (list (cdr l))))
;; most usual case: (a1 a2 ...)
(s s (cdr l)
(lambda (g l*)
(if (>= a2 a1)
(col (cons a1 g) l*)
(col (list a1) (cons g l*)))))))))))))
1 ]=> (decrease-list '(1 2 3 4 1 2 3 4 1 1 1 2 1 1 0 4 3 2 1))
;Value: ((1 2 3 4) (1 2 3 4) (1 1 1 2) (1 1) (0 4) (3) (2) (1))
I did not comment it, if you have questions you can ask but I think you can also study yourself the code I wrote for you now.
Note also that one can consider the limit cases () and (a1) out of the loop and check these cases only once:
(define decrease-list
(lambda (l)
;; limitcase1: ()
(if (null? l)
'()
;; limitcase2: (a1)
(if (null? (cdr l))
(list l)
((lambda (s) (s s l cons))
(lambda (s l col)
(let ((a1 (car l)) (a2 (cadr l)))
;; limitcase3: (a1 a2)
(if (null? (cddr l))
(if (>= a2 a1)
(col l '())
(col (list a1) (list (cdr l))))
;; most usual case: (a1 a2 ...)
(s s (cdr l)
(lambda (g l*)
(if (>= a2 a1)
(col (cons a1 g) l*)
(col (list a1) (cons g l*)))))))))))))

Answer 2

There are a few problems with the posted code. There is no test expression in the second cond clause; there are too many parentheses around the if and its clauses. Perhaps the most significant problem is that the code is attempting to build a non-decreasing list, which is to be consed to the result of (nondecreaselist (cdr s)), but when the non-decreasing sequence is more than one number long this starts again too soon in the input list by going all the way back to (cdr s).
Fixing Up OP Code
The logic can be cleaned up. OP code already is returning an empty list when input is an empty list. Instead of testing (null? (cadr s)) (when (cdr s) is '(), cadr won't work on s), one could test (null? (cdr s)) before code attempts a (cadr s). But it is even better to move this logic; when the input list contains one element, just return a list containing the input list: ((null? (cdr s)) (list s)).
Instead of (and (not (> ;... the logic can be made more clear by testing for > and executing the appropriate action. In this case, when (> (car s) (cadr s)) a new sublist should be started, and consed onto the list of sublists that is the result returned from nondecreaselist.
Otherwise, (car s) should be added to the first sublist in the result returned from nondecreaselist. To accomplish this, we need to construct the return list by consing s onto the first sublist, and then consing that new sublist back onto the cdr of the list of sublists that is the result returned from nondecreaselist.
Here is some revised code:
(define (nondecreaselist s)
(cond ((null? s) '())
((null? (cdr s)) (list s))
((> (car s) (cadr s))
(cons (list (car s))
(nondecreaselist (cdr s))))
(else
(let ((next (nondecreaselist (cdr s))))
(cons (cons (car s)
(car next))
(cdr next))))))
Using a Helper Function
Another approach would be to define a helper function that takes an input list and an accumulation list as arguments, returning a list of lists. The helper function would take numbers from the front of the input list and either add them to the accumulator, creating a non-decreasing list, or it would cons the accumulated non-decreasing list to the result from operating on the rest of the input.
If the input lst to the helper function ndl-helper is empty, then a list containing the accumulated non-decreasing list sublst should be returned. Note that sublst will need to be reversed before it is returned because of the way it is constructed, as described below.
If the accumulator sublst is empty, or if the next number in the input list is greater-than-or-equal-to the largest number in the sublst, then the next number should simply be added to the sublst. By consing the number onto the front of sublst, only the car of sublst needs to be checked, since this will always be the largest (or equal to the largest) value in sublst. But, since sublst is in reverse order, it will need to be reversed before adding it to the growing list of lists.
Otherwise, lst is not empty, and sublst is not empty, and the next number in the input list is less than the largest number in sublst. Thus, a new sublist needs to be started, so the old sublst is reversed and consed onto the result of the remaining computation done by calling the helper function on the remaining lst with an empty accumulator sublst:
(define (nondecreaselist-2 lst)
(define (ndl-helper lst sublst)
(cond ((null? lst) (list (reverse sublst)))
((or (null? sublst)
(>= (car lst) (car sublst)))
(ndl-helper (cdr lst) (cons (car lst) sublst)))
(else
(cons (reverse sublst) (ndl-helper lst '())))))
(ndl-helper lst '()))
Both functions work:
> (nondecreaselist '(1 2 3 4 1 2 3 4 1 1 1 2 1 1 0 4 3 2 1))
((1 2 3 4) (1 2 3 4) (1 1 1 2) (1 1) (0 4) (3) (2) (1))
> (nondecreaselist-2 '(1 2 3 4 1 2 3 4 1 1 1 2 1 1 0 4 3 2 1))
((1 2 3 4) (1 2 3 4) (1 1 1 2) (1 1) (0 4) (3) (2) (1))