I am trying to take a list of 16 numbers I have and make it into a list of 4, 4 element sublists to represent the game board of a magic square. I made a method that can take a list and return a single sublist, and now I am trying to recursively use this method to build the full board.
My problem however, is my initBoard returns nil no matter what and I know every other method is working as desired. Any clarification of my error would be greatly appreciated!
Also here is an example input list:
(4 5 15 10 14 11 1 8 9 16 6 3 7 2 12 13)
And what I want as the output would be:
((4 5 15 10) (14 11 1 8) (9 16 6 3) (7 2 12 13))
(defun smallList (lst cnt)
(cond ((>= cnt 4) nil)
(t (cons (car lst) (smallList (cdr lst) (+ 1 cnt))))))
(defun isEmpty (lst)
(if lst 1 -1))
(defun initBoard (lst)
(cond ((= (isEmpty lst) -1) nil)
(t (cons (smallList lst 0) (initBoard (cddddr lst))))))
Some remarks:
someList, lst, cnt is not idiomatic, use some-list, list, count
You don't need is-empty, just use endp or null, which returns a boolean (not -1 or 1). You could make an alias if you want (but why?):
(setf (symbol-function 'is-empty) #'endp)
You could use a loop for small-list:
(defun small-list (list)
(values (loop repeat 4 collect (pop list)) list))
The secondary value is the rest of the list, so that you don't need to cddddr.
But in fact, it might be better to initialize the whole board inside a single function:
(defun init-board (list)
(loop repeat 4 collect
(loop repeat 4 collect (pop list))))
The first LOOP collect lists of 4 elements, which are collected by the inner LOOP. The collected elements are popped from the input list.
Now, if I wanted to be extremely careful, I would add some checks and report errors on bad inputs:
(defun init-board (list)
(flet ((failure ()
(error "Input list should contain exactly 16 integers: ~S"
list)))
(loop
with current = list
repeat 4 collect
(loop
repeat 4
collect (if current
(let ((element (pop current)))
(check-type element integer)
element)
(failure)))
into board
finally (if list (failure) (return board)))))
Also, I would use a multi-dimensional array for boards.
(make-array '(4 4) :initial-contents (init-board list))
I just tested your three functions and it gave me the correct output, so perhaps your issue isn't where you think it is.
(initBoard '(4 5 15 10 14 11 1 8 9 16 6 3 7 2 12 13))
=> ((4 5 15 10) (14 11 1 8) (9 16 6 3) (7 2 12 13))
I would use the following recursive function:
(defun smalllist (l n)
(when l
(cons (subseq l 0 (min n (length l)))
(smalllist (nthcdr n l) n))))
Related
So im extremely new to Scheme, and i've been trying to do a multiplying table, if you put
(multiplying-table 10 3) should give a list (3 6 9 12 15 18 21 24 27 30)
Here is my code
(define (multiplying-table n value)
(cond ((= n 0) '())
(else (* n value)(Multiplying-table(- n 1)))))
You need to use cons to combine the first number 3 with the list of the rest.
Eg. (3 6 ...) is (cons 3 (cons 6 ...)). You are instead having two expressions where the (* n value) is only for effect since the result is never used. Because of that every call should return the empty list.
Usually there are 2 ways to do it. With recursion inside cons or with an accumulator. Since lists are created from end to beginning you could count n downwards and build the list from end to beginning. The base case would just return the accumulator which by default is the empty list. Here is an example:
(define (sequence from to)
(let loop ((n to) (acc '()))
(if (< n from)
acc
(loop (- n 1) (cons n acc)))))
(sequence 5 10) ; ==> (5 6 7 8 9 10)
I'm working with Scheme and want to do some simple programs. I created one in which the multiples of five are filtered out of a list. Now I want to write a boolean function in which it will tell me if a specific element is part of the filtered list! So if I write "8" it will say false, but "40" would say true. Thanks!
(define (multiple-of-5 some-integer)
(equal? (remainder some-integer 5) 0))
(filter multiple-of-5 '(2 5 8 20 25 27 32 40))
output so far:
(5 20 25 40)
How you produced the list (5 20 25 40) is irrelevant. Thus all your code is irrelevant to this question. Basically you want this:
(define some-list '(5 20 25 40)) ; doesn't matter how this came about
(find (lambda (e) (= e 8)) some-list) ; ==> #f
(find (lambda (e) (= e 40)) some-list) ; ==> 40 (thruthy)
Now find is defined in the SRFI-1 List library and it is voted into R7RS Large. I imagine you could make an abstraction more useful for your case like (exists? 40 some-list) that uses find or you can roll your own like in Torbio's answer.
You can implement a contains? recursive function which travels the list and checks if the input number is equal to the current list element. The condition to stop is if the list is empty or if it contains the number. Pretty self-explanatory.
(define (contains? l i)
(if (null? l) #f
(or (equal? (car l) i) (contains? (cdr l) i))))
(contains? (filter multiple-of-5 '(2 5 8 20 25 27 32 40)) 8) ; #f
(contains? (filter multiple-of-5 '(2 5 8 20 25 27 32 40)) 40) ; #t
I want to reverse only the continuous sequences, not all the elements of my original list.
Ex:
(reverseC '( 1 2 ( 4 5 ) 5 ) ) => ( 2 1 ( 5 4 ) 5 )
(reverseC '(1 4 2 (3 4) 9 6 (7 8)))) => (2 4 1 (4 3) 6 9 (8 7))
I was thinking of splitting it into 2 functions: one to reverse a simple list ( 1 2 3 ) -> ( 3 2 1 ) and one function
(main) to determine the continuous sequences, make a list out of them, apply reverse on that list and the remake the whole reversed list.
(defun reverse-list ( lista )
(if (eql lista () )
()
(append (reverse-list (cdr lista )) (list ( car lista)))
)
)
This is the reverse function but I have no idea how to do the other one. I'm new to Lisp and I come from Prolog so it's a pretty big change of scenery. Any idea is welcome.
(defun reverse-more (L)
(if (eql L nil)
nil
(let ( el (car L)) (aux (cdr L)))
(if (eql (listp el) nil)
...No idea on the rest of the code ...
There's already an accepted answer, but this seems like a fun challenge. I've tried to abstract some of the details away a bit, and produced a map-contig function that calls a function with each contiguous sublist of the input list, and determines what's a contiguous list via a predicate that's passed in.
(defun map-contig (function predicate list)
"Returns a new list obtained by calling FUNCTION on each sublist of
LIST consisting of monotonically non-decreasing elements, as determined
by PREDICATE. FUNCTION should return a list."
;; Initialize an empty RESULT, loop until LIST is empty (we'll be
;; popping elements off of it), and finally return the reversed RESULT
;; (since we'll build it in reverse order).
(do ((result '())) ((endp list) (nreverse result))
(if (listp (first list))
;; If the first element is a list, then call MAP-CONTIG on it
;; and push the result into RESULTS.
(push (map-contig function predicate (pop list)) result)
;; Otherwise, build up sublist (in reverse order) of contiguous
;; elements. The sublist is finished when either: (i) LIST is
;; empty; (ii) another list is encountered; or (iii) the next
;; element in LIST is non-contiguous. Once the sublist is
;; complete, reverse it (since it's in reverse order), call
;; FUNCTION on it, and add the resulting elements, in reverse
;; order, to RESULTS.
(do ((sub (list (pop list)) (list* (pop list) sub)))
((or (endp list)
(listp (first list))
(not (funcall predicate (first sub) (first list))))
(setf result (nreconc (funcall function (nreverse sub)) result)))))))
Here's your original example:
(map-contig 'reverse '< '(1 2 (4 5) 5))
;=> (2 1 (5 4) 5)
It's worth noting that this will detect discontinuities within a single sublist. For instance, if we only want continuous sequences of integers (e.g., where each successive difference is one), we can do that with a special predicate:
(map-contig 'reverse (lambda (x y) (eql y (1+ x))) '(1 2 3 5 6 8 9 10))
;=> (3 2 1 6 5 10 9 8)
If you only want to break when a sublist occurs, you can just use a predicate that always returns true:
(map-contig 'reverse (constantly t) '(1 2 5 (4 5) 6 8 9 10))
;=> (5 2 1 (5 4) 10 9 8 6)
Here's another example where "contiguous" means "has the same sign", and instead of reversing the contiguous sequences, we sort them:
;; Contiguous elements are those with the same sign (-1, 0, 1),
;; and the function to apply is SORT (with predicate <).
(map-contig (lambda (l) (sort l '<))
(lambda (x y)
(eql (signum x)
(signum y)))
'(-1 -4 -2 5 7 2 (-6 7) -2 -5))
;=> (-4 -2 -1 2 5 7 (-6 7) -5 -2)
A more Prolog-ish approach
(defun reverse-contig (list)
(labels ((reverse-until (list accumulator)
"Returns a list of two elements. The first element is the reversed
portion of the first section of the list. The second element is the
tail of the list after the initial portion of the list. For example:
(reverse-until '(1 2 3 (4 5) 6 7 8))
;=> ((3 2 1) ((4 5) 6 7 8))"
(if (or (endp list) (listp (first list)))
(list accumulator list)
(reverse-until (rest list) (list* (first list) accumulator)))))
(cond
;; If LIST is empty, return the empty list.
((endp list) '())
;; If the first element of LIST is a list, then REVERSE-CONTIG it,
;; REVERSE-CONTIG the rest of LIST, and put them back together.
((listp (first list))
(list* (reverse-contig (first list))
(reverse-contig (rest list))))
;; Otherwise, call REVERSE-UNTIL on LIST to get the reversed
;; initial portion and the tail after it. Combine the initial
;; portion with the REVERSE-CONTIG of the tail.
(t (let* ((parts (reverse-until list '()))
(head (first parts))
(tail (second parts)))
(nconc head (reverse-contig tail)))))))
(reverse-contig '(1 2 3 (4 5) 6 7 8))
;=> (3 2 1 (5 4) 8 7 6)
(reverse-contig '(1 3 (4) 6 7 nil 8 9))
;=> (3 1 (4) 7 6 nil 9 8)
Just two notes about this. First, list* is very much like cons, in that (list* 'a '(b c d)) returns (a b c d). list** can take more arguments though (e.g., **(list* 'a 'b '(c d e)) returns (a b c d e)), and, in my opinion, it makes the intent of lists (as opposed to arbitrary cons-cells) a bit clearer. Second, the other answer explained the use of destructuring-bind; this approach could be a little bit shorter if
(let* ((parts (reverse-until list '()))
(head (first parts))
(tail (second parts)))
were replaced with
(destructuring-bind (head tail) (reverse-until list '())
You can perform all at once with a single recursive function, with the usual warning that you should favor looping constructs over recursive approaches (see below):
(defun reverse-consecutive (list &optional acc)
(etypecase list
;; BASE CASE
;; return accumulated list
(null acc)
;; GENERAL CASE
(cons (destructuring-bind (head . tail) list
(typecase head
(list
;; HEAD is a list:
;;
;; - stop accumulating values
;; - reverse HEAD recursively (LH)
;; - reverse TAIL recursively (LT)
;;
;; Result is `(,#ACC ,LH ,#LT)
;;
(nconc acc
(list (reverse-consecutive head))
(reverse-consecutive tail)))
;; HEAD is not a list
;;
;; - recurse for the result on TAIL with HEAD
;; in front of ACC
;;
(t (reverse-consecutive tail (cons head acc))))))))
Exemples
(reverse-consecutive '(1 2 (3 4) 5 6 (7 8)))
=> (2 1 (4 3) 6 5 (8 7))
(mapcar #'reverse-consecutive
'((1 3 (8 3) 2 )
(1 4 2 (3 4) 9 6 (7 8))
(1 2 (4 5) 5)))
=> ((3 1 (3 8) 2)
(2 4 1 (4 3) 6 9 (8 7))
(2 1 (5 4) 5))
Remarks
#Melye77 The destructuring-bind expression does the same thing as [Head|Tail] = List in Prolog. I could have written this instead
(let ((head (first list))
(tail (rest list)))
...)
Likewise, I prefer to use (e)typecase over the generic cond expression whenever possible, because I think it is more precise.
I could have written:
(if acc
(if (listp (first list))
(nconc ...)
(reverse-consecutive ...))
acc)
... but I think it is less clear and not a good thing to teach beginners.
On the contrary, I think it is useful, even (especially) for beginners, to introduce the full range of available constructs.
For example, overusing recursive functions is actually not recommended: there are plenty of existing iteration constructs for sequences that do not depend on the availability of tail-call optimizations (which are not guaranteed to be implemented, though it is generally available with appropriate declarations).
Iterative version
Here is an iterative version which uses of the standard reverse and nreverse functions. Contrary to the above method, inner lists are simply reversed (contiguous chunks are only detected at the first level of depth):
(defun reverse-consecutive (list)
(let (stack result)
(dolist (e list (nreverse result))
(typecase e
(list
(dolist (s stack)
(push s result))
(push (reverse e) result)
(setf stack nil))
(t (push e stack))))))
I need to create a Scheme function that receives a list and a desired new size, the function then extends the list size by using the same list values. For example:
'(1 2 3) to size 6 will turn to '(1 2 3 1 2 3)
'(1 2) to size 5 will turn to '(1 2 1 2 1)
'(4 5 6 1) to size 7 will turn to '(4 5 6 1 4 5 6)
The new length function parameter can be equal or bigger than the current list size.
You can use SRFI 1 function circular-list (alongside Racket's built-in take) to do this:
(require srfi/1)
(define (take-circular lst n)
(take (apply circular-list lst) n))
If you want to avoid using SRFI 1, another method works like this:
(define (take-circular lst n)
(let ((size (length lst)))
(if (> n size)
(append lst (take-circular lst (- n size)))
(take lst n))))
If we had a list A holding (1 2 1 1 2 3 3 4 4 4), how could we get a new list B with ((1 . 30) (2 . 20) (3 . 20) (4 . 30)) in it, such that the number_after_dot is the percentage of the number_before_dot in the list A.
For example 1 is 30% of list A, 2 is 20% of list A, etc..
(1 . 30) is a pair, which could be made by (cons 1 30)
I think what you want to do is calculate the percentage of the list that is equal to each element. You used the word "unique" but that a bit confusing since your list has no unique elements. This is based on your sample input and output, where the list (1 2 1 1 2 3 3 4 4 4) is composed of "30% ones".
You can break this down roughly into a recursive algorithm consisting of these steps:
If the input list is empty, return the empty list.
Otherwise, get the first element. Calculate how many times it occurs in the list.
Calculate the percentage, and cons the element with this percentage.
Remove all the occurrences of the first item from the cdr of the list.
Recurse on this new list, and cons up a list of (element . percentage) pairs.
To do the first part, let's use filter:
> (filter (lambda (x) (eq? (car A) x)) A)
(1 1 1)
With your list A, this will return the list (1 1 1). We can then use length to get the number of times it occurs:
> (length (filter (lambda (x) (eq? (car A) x)) A))
3
To calculate the percentage, divide by the number of elements in the whole list, or (length A) and multiply by 100:
> (* 100 (/ (length (filter (lambda (x) (eq? (car A) x)) A)) (length A)))
30
It's easy to cons this with the element (car A) to get the pair for the final list.
To do the second step, we can use remove which is the inverse of filter: it will return a list of all elements of the original list which do not satisfy the predicate function:
> (remove (lambda (x) (eq? (car A) x)) A)
(2 2 3 3 4 4 4)
This is the list we want to recurse on. Note that at each step, you need to have the original list (or the length of the original list) and this new list. So you would need to somehow make this available to the recursive procedure, either by having an extra argument, or defining an internal definition.
There might be more efficient ways I'm sure, or just other ways, but this was the solution I came up with when I read the question. Hope it helps!
(define (percentages all)
(let ((len (length all))) ; pre-calculate the length
;; this is an internal definition which is called at ***
(define (p rest)
(if (null? rest)
rest
;; equal-to is a list of all the elements equal to the first
;; ie something like (1 1 1)
(let ((equal-to (filter (lambda (x) (eq? (car rest) x))
rest))
;; not-equal-to is the rest of the list
;; ie something like (2 2 3 3 4 4 4)
(not-equal-to (remove (lambda (x) (eq? (car rest) x))
rest)))
(cons (cons (car rest) (* 100 (/ (length equal-to) len)))
;; recurse on the rest of the list
(p not-equal-to)))))
(p all))) ; ***
The question formulation is very close to the idea of run-length encoding. In terms of run-length encoding, you can use a simple strategy:
Sort.
Run-length encode.
Scale the run lengths to get percentages.
You can implement run-length encoding like this:
(define (run-length-encode lst)
(define (rle val-lst cur-val cur-cnt acc)
(if (pair? val-lst)
(let ((new-val (car val-lst)))
(if (eq? new-val cur-val)
(rle (cdr val-lst) cur-val (+ cur-cnt 1) acc)
(rle (cdr val-lst) new-val 1 (cons (cons cur-val cur-cnt) acc))))
(cons (cons cur-val cur-cnt) acc)))
(if (pair? lst)
(reverse (rle (cdr lst) (car lst) 1 '()))
'()))
and scaling looks like:
(define (scale-cdr count-list total-count)
(define (normalize pr)
(cons (car pr) (/ (* 100 (cdr pr)) total-count)))
(map normalize count-list))
Now we need something to sort a list. I'll just use the sort function in racket (adapt as needed). The function to calculate the percentages for each number in the list is then:
(define (elem-percent lst)
(scale-cdr (run-length-encode (sort lst <)) (length lst)))
Some examples of use:
> (elem-percent '())
'()
> (elem-percent (list 1 2 3 4 5))
'((1 . 20) (2 . 20) (3 . 20) (4 . 20) (5 . 20))
> (elem-percent (list 1 2 1 1))
'((1 . 75) (2 . 25))
> (elem-percent (list 1 2 1 1 2 3 3 4 4 4))
'((1 . 30) (2 . 20) (3 . 20) (4 . 30))