Question

Write a function sublist that takes two lists as arguments, and returns true if the first list appears as a contiguous sublist somewhere within the second list, and false otherwise.

> (sublist ’(c d e) ’(a b c d e f g))

#t

> (sublist ’(a c e) ’(a b c d e f g))

#f

Write a function lgrep that returns the “lines” within a list that contain a given sublist. Use the sublist function implemented in previous exercise to build lgrep.

> (lgrep ’(c d e) ’((a b c d e f g)

(c d c d e)

(a b c d)

(h i c d e k)

(x y z)))

((a b c d e f g) (c d c d e) (h i c d e k))

You may assume that all elements of (all lines of) all argument lists are atoms.

using the programming language scheme

i need the question to be answered using the programming language Scheme

Answer 1

Code: (both "Sublist" and "lgrep" functions included)

;;EX: (sublist '(c d e) '(c d c d e)) returns #t
(define (sublist ls1 ls2)
(cond ((null? ls2) #f) ;;if second list empty returns false
((null? ls1) #t) ;;if first list is empty returns true
(else (sublist1 ls1 (check (car ls1) ls2)))))

;;Ex: (check '(c d e) '(c d c d e))returns ((c d c d e) (c d e))
(define (check n list) ;;returns list strting with first element of given sublist
(cond ((null? list) '())
((equal? n (car list)) (cons list (check n (cdr list))))
(else (check n (cdr list)))))

;;Ex:(sublist '(c d e) '((c d c d e) (c d e))) returns #t
(define (sublist1 ls1 ls2)
(cond ((null? ls2) #f)
((null? ls1) #t)
((equal? (sublist2 ls1 (car ls2)) #t) #t)
(else (sublist1 ls1 (cdr ls2)))))

;;Ex:(sublist '(c d e) '(c d c d e)) returns #t
(define (sublist2 ls1 ls2)
(cond ((null? ls2) #f) ;;if second list empty returns false
((null? ls1) #t) ;;if first list is empty returns true
((equal? (car ls1) (car ls2)) (sublist3 (cdr ls1) (cdr ls2))) ;;if first element of list 1 equal to first element
;;check rest of the elements of list1 are contigeous subset of list2
(else (sublist2 ls1 (cdr ls2))))) ;;else continue with rest of list2

(define (sublist3 ls1 ls2) ;;HELPER_FUNCTION
(cond ((null? ls1) #t) ;;if second list empty returns false
((null? ls2) #f) ;;if first list is empty returns true
((equal? (car ls1) (car ls2)) (sublist3 (cdr ls1) (cdr ls2))) ;;if first element of list 1 equal to first element check rest elements
(else #f))) ;;else sublist is not contigeous, so returns false.
  
;;MAIN_FUNCTION
(define (lgrep list1 list2)
(cond ((null? list2) '()) ;;If list is empty returns '()
((null? list1) list2);;if first list is empty returns 2nd list
((equal? (sublist list1 (car list2)) #t) (cons (car list2) (lgrep list1 (cdr list2))));;if sublist is equal then add it to it
(else (lgrep list1 (cdr list2))))) ;;else continue in rest of the list

Snapshot of Code:

Output 1:

Output 2: