Multi-level recursion in lisp - recursion

I have a problem with multilevel recursion. I have two numbers n1 and n2. When counter is between n1 and n2 i want thoose numbers to be in the result, keeping the list internal structure.
(defun izdvoj (lista n1 n2 counter)
(cond ((null lista) counter)
((and (atom (car lista))
(< counter n1)
(izdvoj (cdr lista) n1 n2 (+ counter 1))))
((and (atom (car lista))
(> counter n2)
(izdvoj (cdr lista) n1 n2 (+ counter 1))))
((atom (car lista))
(cons (car lista) (izdvoj (cdr lista) n1 n2 (+ counter 1))))
(t
(cons (izdvoj (car lista) n1 n2 counter) (izdvoj (cdr lista) n1 n2 counter)))))
(izdvoj '(1 2 (3 (4) 5 (6 (7)))(8 (9 (10 ((11))) 12)) (13 ((14) (15)))) 7 13 0)
The result should be ((((7))) (8 (9 (10 ((11))) 12)) (13))
And I get (((4) ((7))) (((((11))) 12)) (((14) (15))))
Any suggestions please?

The easiest way to do this is by returning the index so far as well as the result in the helper. That way you know how to continue when you have done car and are ready to do the cdr:
(defun get-range (list from-index to-index)
(labels ((aux (list cur acc)
(cond ((null list)
(values (nreverse acc) cur))
((not (atom (car list)))
(multiple-value-bind (res cur)
(aux (car list) cur '())
(aux (cdr list) cur (cons res acc))))
((<= from-index cur to-index)
(aux (cdr list) (1+ cur) (cons (car list) acc)))
(t (aux (cdr list) (1+ cur) acc)))))
(nth-value 0 (aux list 1 '()))))
(get-range '(a b (c d (e (f))) (g h)) 3 5)
; ==> ((c d (e ())) ())

Related

Implementing last-non-zero without continuations

last-non-zero takes a list of numbers and return the last cdr whose car is 0.
So, I can implement it using continuations, but how do I do this with natural recursion.
(define last-non-zero
(lambda (ls)
(let/cc return
(letrec
((lnz
(lambda (ls)
(cond
((null? ls) '())
((zero? (car ls)) ;; jump out when we get to last 0.
(return (lnz (cdr ls))))
(else
(cons (car ls) (lnz (cdr ls))))))))
(lnz ls)))))
Here's an obvious version which is not tail-recursive:
(define (last-non-zero l)
;; Return the last cdr of l which does not contain zero
;; or #f if there is none
(cond
((null? l)
#f)
((zero? (car l))
(let ((lnzc (last-non-zero (cdr l))))
;; This is (or lnzc (cdr l)) but that makes me feel bad
(if lnzc
lnzc
(cdr l))))
(else
(last-non-zero (cdr l)))))
Here is that version turned into a tail-recursive equivalent with also the zero test moved around a bit.
(define (last-non-zero l)
(let lnzl ([lt l]
[r #f])
(if (null? lt)
r
(lnzl (cdr lt) (if (zero? (car lt)) (cdr lt) r)))))
It's much clearer in this last version that the list is traversed exactly once.
Please indicate if I have correctly understood the problem:
#lang scheme
; returns cdr after last zero in lst
(define (last-non-zero lst)
; a helper function with 'saved' holding progress
(define (lnz-iter lst saved)
(if (null? lst)
saved
(if (zero? (car lst))
(lnz-iter (cdr lst) (cdr lst))
(lnz-iter (cdr lst) saved))))
(lnz-iter lst '()))
(last-non-zero '(1 2 3 0 7 9)) ; result (7 9)
Racket's takef-right can do it:
> (takef-right '(1 2 0 3 4 0 5 6 7) (lambda (n) (not (zero? n))))
'(5 6 7)
But assuming you have an assignment where you're supposed to write the logic yourself instead of just using a built in function, one easy if not very efficient approach is to reverse the list, build a new list out of everything up to the first zero, and return that. Something like:
(define (last-non-zero ls)
(let loop ([res '()]
[ls (reverse ls)])
(if (or (null? ls) (zero? (car ls)))
res
(loop (cons (car ls) res) (cdr ls)))))
Using your implementation where you return the argument in the event there are no zero you can just have a variable to keep the value you think has no zero values until you hit it and then update both:
(define (last-non-zero lst)
(let loop ((lst lst) (result lst))
(cond ((null? lst) result)
((zero? (car lst)) (loop (cdr lst) (cdr lst)))
(else (loop (cdr lst) result)))))
(last-non-zero '()) ; ==> ()
(last-non-zero '(2 3)) ; ==> (2 3)
(last-non-zero '(2 3 0)) ; ==> ()
(last-non-zero '(2 3 0 1 2)) ; ==> (1 2)
(define last-non-zero
(lambda (l)
((lambda (s) (s s l (lambda (x) x)))
(lambda (s l* ret)
(if (null? l*)
(ret '())
(let ((a (car l*))
(r (cdr l*)))
(if (zero? a)
(s s r (lambda (x) x))
(s s r
(lambda (r)
(ret (cons a r)))))))))))
Also possible, to use foldr:
(define (last-non-zero l)
(reverse (foldl (lambda (e res) (if (zero? e) '() (cons e res))) 0 l)))
Or use recursion:
(define (last-non-zero l (res '()))
(cond ((empty? l) res)
((zero? (car l)) (last-non-zero (cdr l) (cdr l)))
(else (last-non-zero (cdr l) res))))

Return the longest sequence of consecutive numbers from list in lisp

I am a lisp newbie.
I'm trying to create a function in lisp that receives an unsorted list and the function has to sort de list and return a list with the longest sequence of numbers.
Example: (2 1 8 9 3 11 10 20 12 21)(1 2 3 8 9 10 11 12 20 21) -> return (8 9 10 11 12)
I don't want to use the sort function and I have created 2 functions (With some help) to sort, but now I have no idea how I could find and return the longest sequence of numbers.
I could go through the list but, how I can store the numbers and check if a list of consecutive numbers is longer than another?
This are my functions to sort
(defun sortOne (list)
(let ((ca1 (car list)) (cd1 (cdr list)))
(if (null cd1)
list
(let ((cd (sortOne cd1))) ; cd = sorted tail
(let ((ca2 (car cd)) (cd2 (cdr cd)))
(if (<= ca1 ca2)
(cons ca1 cd)
(cons ca2 (cons ca1 cd2))))))))
(defun sortAll (list)
(if (null list)
nil
(let ((s (sortOne list)))
(cons (car s) (sortAll (cdr s))))))
Hope someone can help me.
¡Thanks!
Tonight I managed to do it, but surely it is not the best solution, I would like to know how to use a lambda function or recursion to do it better.
(defun listilla (lista)
(setq lista (sort lista #'<))
(setq lista1 (list (car lista)))
(setq lista2 '())
(loop for i from 0 to (- (length lista) 2) do
(cond ((= (nth i lista) (- (nth (+ i 1) lista) 1))
(push (nth (+ i 1) lista) (cdr (last lista1))))
(t (push lista1 lista2)
(setq lista1 (list (nth (+ i 1) lista)))
)
)
)
(push lista1 lista2)
(setq masLargo (car lista2))
(loop for i from 1 to (- (length lista2) 2) do
(if (< (length (nth i lista2)) (length (nth (+ i 1) lista2)))
(setq masLargo (nth (+ i 1) lista2))
)
)
masLargo
)
(print (listilla '(23 15 6 5 78 4 77)))
(defun group-consecutives (l &optional (acc '()))
(cond ((null l) (nreverse acc))
((and acc (= 1 (- (car l) (caar acc)))) (consecutives (cdr l) (cons (cons (car l) (car acc)) (cdr acc))))
(t (consecutives (cdr l) (cons (list (car l)) (when acc (cons (nreverse (car acc)) (cdr acc))))))))
(defun longest-consecutive (l)
(car (sort (consecutives (sort l #'<)) #'> :key #'length)))
(longest-consecutive '(2 1 8 9 3 11 10 20 12 21))
;;=> (8 9 10 11 12)
Probably the second function is easier to understand like this:
(defun sort-increasing (l)
(sort l #'<))
(defun sort-groups-by-length (groups)
(sort groups #'> #'length))
(defun longest-consecutive (l)
(car (sort-groups-by-length (group-consecutives (sort-increasing l))))))))

recursively count sublists beginning with a number

I am trying to write code in Lisp counting sublists beginning with number, recursively. I ve trying to use numberp but my code, when arrived to an atom, doesn't count the rest of the list.
With my code here,
(defun nombres (liste)
(cond
((atom liste) 0)((atom (car liste)) 0)
((and (numberp (caar liste)) (+ (nombres (cdr liste)) 1)))
(t (nombres (cdr liste))) ) )
I can get a count of sublists but when arrived to an atom , it doesn't count the rest.
[67]> (nombres '((a b d) (5 g) (7 m)))
2
[68]> (nombres '((a b d) (5 g) g (7 m)))
1
When I test the sublist with (listp (car list), it gives me nil.
[69]> (defun nombres (liste)
(cond
((atom liste) 0)((atom (car liste)) 0)
((listp (car liste))(and (numberp (caar liste)) (+ (nombres (cdr liste)) 1))) (t (nombres (cdr liste))) ) )
NOMBRES
[70]> (nombres '((a b d) (5 g) g (7 m) m))
NIL
I want to get something like :
(nombres '((a b d) a (5 g) (b) (7 m) j (8 h l g)))
3
Thanks for your help
You need to think about the cases you need to handle.
The end of the list => return the result
A sublist that has a number in front => add one to the result
Anything else => continue to the next element
These will quite easily translate to a COND:
(cond ((endp list) ...) ; 1
((and (listp (car list)) ; 2
(numberp (caar list)))
...)
(t ...) ; 3
Using an accumulator as an optional parameter, the counting is easy to fill in:
(defun count-sublists (list &optional (acc 0))
(cond ((endp list) acc)
((and (listp (car list))
(numberp (caar list)))
(count-sublists (cdr list) (1+ acc)))
(t (count-sublists (cdr list) acc))))
(count-sublists '((a b d) a (5 g) (b) (7 m) j (8 h l g)))
;=> 3
The standard Common Lisp function count-if is easier to use:
CL-USER > (count-if (lambda (item)
(and (consp item)
(numberp (first item))))
'((a b d) a (5 g) (b) (7 m) j (8 h l g)))
3

"Multiplication of Arbitrary Precision Numbers" in Scheme

The following is code to a problem that I have been working on for a few days. The problem I encountered is that for some reason when I call:
(apa-multi '(7 3 1 2) '(6 1 4))
the return is:
'(4 8 9 5 6 8)
The answer that it should output is
'(4 4 8 9 5 6 8)
When I call:
(apa-multi '(3 1 2) '(6 1 4))
The output is:
'(1 9 1 5 6 8)
which is correct.
I have debugged my code multiple times, and I can't seem to find out what the problem is (by the way, I know that the "remove-empty" function that I wrote is most likely unnecessary). Can anyone tell me where I am going wrong here? (My goal for this problem is to keep the arbitrary precision numbers in list format, and I can not create a function that converts numbers from list->num or num->list.) I believe that I have provided all of the necessary code for someone to work out what I was going for, but if not please let me know. The hint that I have for this is that " Multiplication of d = dndn−1 ...d1 by e = emem−1 ...e1 can be carried out by the rule de=d∗e1 +10∗(d∗em em−1...e2).)"
(define (remove-empty L)
(define (remove-empty-h L accum)
(cond ((null? L) accum)
((null? (car L))
(remove-empty (cdr L)))
(else (cons (car L) (remove-empty-h (cdr L) accum)))))
(remove-empty-h L '()))
(define (apa-add lst1 lst2)
(define (apa-add-h lst1 lst2 carry)
(cond ((and (null? lst1) (null? lst2))
(if (not (= 0 carry))
(list carry)
'()))
((null? lst1) (append (apa-add-h lst1 '() carry)
(list (+ (car (reverse-l lst2)) carry))
(reverse-l(cdr (reverse-l lst2)))))
((null? lst2) (append (apa-add-h '() lst2 carry)
(list (+ (car (reverse-l lst1)) carry)))
(reverse-l(cdr (reverse-l lst1))))
(else
(append (apa-add-h (cdr lst1) (cdr lst2) (quotient (+ (car lst1) (car lst2) carry) 10))
(list (modulo (+ (car lst1) (car lst2) carry) 10))))))
(apa-add-h (reverse-l lst1) (reverse-l lst2) 0))
(define (d-multiply lst factor)
(define (d-multiply-h lst factor carry)
(cond ((null? lst) (if (= carry 0)
'()
(list carry)))
((>= (+ (* (car lst) factor) carry) 10)
(append ;(list (check-null-and-carry-mult lst carry))
(d-multiply-h (cdr lst) factor (quotient (+ (* (car lst) factor) carry) 10))
(list (modulo (+ (* (car lst) factor) carry) 10))))
(else (append ;(list (check-null-and-carry-mult lst carry))
(d-multiply-h (cdr lst) factor (quotient(+ (* (car lst) factor) carry) 10))
(list (+ (* (car lst) factor) carry))))))
(remove-empty (d-multiply-h (reverse-l lst) factor 0)))
(define (nlength l)
(if (null? l)
0
(+ 1 (nlength (cdr l)))))
(define (apa-multi d e)
(define temp '())
(cond ((= (max (nlength e) (nlength d)) (nlength e))
(set! temp e)
(set! e d)
(set! d temp))
(else
(set! temp d)
(set! d e)
(set! e temp)))
(define (apa-multi-h d e)
(cond ((null? e) (list 0))
(else (append (apa-add (d-multiply d (car e))
(append (apa-multi-h d (cdr e)) (list 0)))))))
(apa-multi-h d (reverse-l e)))
The reason your code does not work is because your apa-add is broken. For example:
> (apa-add '(7 3 1 2) '(6 1 4))
'(9 2 6)
> (+ 7312 614)
7926
The rest of your code seems to work, at least for your 2 examples, if you use a working apa-add.
I admit I did not try to understand your code fully; the poor formatting and the set! procedures at the end made me want to start from scratch. So even if you could simple correct your apa-add, maybe have a look at my version anyway, because it is way shorter and probably easier to understand.
Building on my previous answer for apa-add multiplication is a matter of apa-adding, multiplying one list by a digit at a time, and adding zeroes at the end of the intermediary multiplications just as you'd do it manually:
(define (car0 lst) (if (empty? lst) 0 (car lst)))
(define (cdr0 lst) (if (empty? lst) empty (cdr lst)))
(define (apa-add l1 l2) ; apa-add (see https://stackoverflow.com/a/19597007/1193075)
(let loop ((l1 (reverse l1)) (l2 (reverse l2)) (carry 0) (res '()))
(if (and (null? l1) (null? l2) (= 0 carry))
res
(let* ((d1 (car0 l1)) (d2 (car0 l2)) (ad (+ d1 d2 carry)) (dn (modulo ad 10)))
(loop (cdr0 l1) (cdr0 l2) (quotient (- ad dn) 10) (cons dn res))))))
(define (mult1 n lst) ; multiply a list by one digit
(let loop ((lst (reverse lst)) (carry 0) (res '()))
(if (and (null? lst) (= 0 carry))
res
(let* ((c (car0 lst)) (m (+ (* n c) carry)) (m0 (modulo m 10)))
(loop (cdr0 lst) (quotient (- m m0) 10) (cons m0 res))))))
(define (apa-multi l1 l2) ; full multiplication
(let loop ((l2 (reverse l2)) (app '()) (res '()))
(if (null? l2)
res
(let* ((d2 (car l2)) (m (mult1 d2 l1)) (r (append m app)))
(loop (cdr l2) (cons '0 app) (apa-add r res))))))
Not sure why it doesn't work, all those appends and reverses are hard to follow, and not sure what's going on with all that set! stuff. Putting the state into a tail call is a lot easier to follow and usually more efficient to boot.
(define (apa-add . Lists)
(define (cdrs-no-null L)
(cond ((null? L) '())
((null? (cdar l)) (cdrs-no-null (cdr L)))
(else (cons (cdar l) (cdrs-no-null (cdr l))))))
(let loop ((carry 0) (Lists (map reverse Lists)) (sum '()))
(if (null? Lists)
(if (zero? carry) sum (cons carry sum))
(loop (quotient (fold + carry (map car Lists)) 10)
(cdrs-no-null Lists)
(cons (modulo (fold + carry (map car Lists)) 10) sum)))))
(define (apa-mult . Lists)
(define (mult-by-factor n order L)
(let loop ((order order) (L (reverse L)) (carry 0) (sum '()))
(cond ((> order 0) (loop (- order 1) L carry (cons 0 sum)))
((null? L) (if (zero? carry)
sum
(cons carry sum))) ;;bug here if carry > 9
(else (loop 0
(cdr L)
(quotient (+ carry (* n (car L))) 10)
(cons (modulo (+ carry (* n (car L))) 10) sum))))))
(define (apa-mult2 L1 L2)
(let ((rL1 (reverse L1))
(rL2 (reverse L2))
(zip-with-order
(lambda (L)
(let loop ((order 0) (L L) (accum '()))
(if (null? L)
accum
(loop (+ 1 order)
(cdr L)
(cons (cons (car L) order) accum)))))))
(fold apa-add '(0) (map (lambda (x)
(mult-by-factor (car x) (cdr x) L2))
(zip-with-order rl1)))))
(fold apa-mult2 '(1) Lists)))
(apa-mult '(3 1 2) '(6 1 4)))
;Value 7: (1 9 1 5 6 8)
(apa-mult '(2 0 0) '(3 1 2) '(6 1 4))
;Value 8: (3 8 3 1 3 6 0 0)
(apa-mult '(7 3 1 2) '(6 1 4))
;Value 9: (4 4 8 9 5 6 8)

"Addition of arbitrary precision numbers" in Scheme

I am trying to add two numbers (from lists) together in scheme while preserving list format. However, when the numbers being used for calculations add to something greater than or equal to 10 I get an undesired result. For example:
(define (reverse lst)
(if (null? lst)
'()
(append (reverse (cdr lst))
(list (car lst)))))
(define (apa-add lst1 lst2)
(cond ((null? lst1) lst2)
((null? lst2) lst1)
((>= (+ (car lst1) (car lst2)) 10)
(append (apa-add (cdr lst1) (cdr lst2))
(list (quotient(+ (car lst1) (car lst2)) 10))
(list (modulo (+ (car lst1) (car lst2)) 10))))
(else
(append (apa-add (cdr lst1) (cdr lst2))
(list (+ (car lst1) (car lst2)))))))
(apa-add (reverse '(4 4 5)) (reverse'(3 5 8)))
returns
'(7 9 1 3)
How can I modify my code to fix this error? I wanted to use a let statement so I could add the evaluation of (quotient (+ (car lst1) (car lst2)) 10) to (list (+ (car lst1) (car lst2))) of the next call, but I couldn't think of how to do this.
Making this into a tail call simplifies things a bit.
(define (apa-add lst1 lst2)
(let loop ((carry 0) (L1 (reverse lst1)) (L2 (reverse lst2)) (sum '()))
(cond ((and (null? l1) (null? l2))
(if (zero? carry) sum (cons carry sum)))
((null? L1)
(loop (quotient (+ carry (car l2)) 10)
'()
(cdr L2)
(cons (modulo (+ carry (car l2)) 10) sum)))
((null? L2)
(loop (quotient (+ carry (car l1)) 10)
(cdr l1)
'()
(cons (modulo (+ carry (car l1)) 10) sum)))
(else
(loop (quotient (+ carry (car l1) (car l2)) 10)
(cdr l1)
(cdr l2)
(cons (modulo (+ carry (car l1) (car l2)) 10) sum))))))
(apa-add (list 4 4 5) (list 3 5 8))
;Value 4: (8 0 3)
probably wouldnt be too hard to convert to an n-arity function.
(define (apa-add . Lists)
(define (cdrs-no-null L)
(cond ((null? L) '())
((null? (cdar l)) (cdrs-no-null (cdr L)))
(else (cons (cdar l) (cdrs-no-null (cdr l))))))
(let loop ((carry 0) (Lists (map reverse Lists)) (sum '()))
(if (null? Lists)
(if (zero? carry) sum (cons carry sum))
(loop (quotient (fold + carry (map car Lists)) 10)
(cdrs-no-null Lists)
(cons (modulo (fold + carry (map car Lists)) 10) sum)))))
(apa-add (list 4 4 5) (list 3 5 8) (list 1 0 2 7))
;Value 11: (1 8 3 0)
(apa-add (list 4 4 5) (list 3 5 8))
;Value 12: (8 0 3)

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