[SOLVED] ;; Below is my solution for sublist? from Exam 1. To remind you,

;; Below is my solution for sublist? from Exam 1.To remind you,
;; heres what I said about sublist?:

;; ;; Write a procedure that takes two lists of numbers, which I call big

Copyright By Assignmentchef assignmentchef

;; ;; and little.The function returns true if the little list occurs as
;; ;; a consecutive subsequence within the big list.Note that unlike an
;; ;; ordinary subset relationship, order matters the elements in the
;; ;; both lists must be in the the same order, not just have the same
;; ;; values somewhere in the list.
;; ;; Examples:
;; ;; > (sublist? (1 2 99 100 3 4) (99 100))
;; ;; > (sublist? (1 2 99 100 3 4) (100 99))
;; ;; > (sublist? (1 2 99 100 3 4) ())

(define (head-matches?-orig big little)
(cond [(null? little) #t]
[(null? big) #f]
[(equal? (car little) (car big)) (head-matches?-orig (cdr big) (cdr little))]
[else #f]))

(define (sublist?-orig big little)
(cond [(< (length big) (length little)) #f][(head-matches?-orig big little) #t][else (sublist?-orig (cdr big) little)]));; Here’s the same code, converted by me to datastructure cps.Note;; in this case I decided that length was a substantial procedure, so;; I have a cps-ized implementation for length (which yes, you will be;; required to convert to imperative form).(define-datatype continuation continuation?[length-cps-k (k continuation?)][len-big-k (big list?) (little list?) (k continuation?)][len-little-k (big list?)(little list?)(len-big number?)(k continuation?)][is-match-k (big list?)(little list?)(k continuation?)])(define (sublist?-ds-cps big little k)(length-ds-cps big (len-big-k big little k)))(define (head-matches?-ds-cps big little k)(cond [(null? little) (apply-k-ds k #t)][(null? big) (apply-k-ds k #f)][(equal? (car little) (car big)) (head-matches?-ds-cps (cdr big) (cdr little) k)][else (apply-k-ds k #f)]))(define (length-ds-cps lst k)(if (null? lst)(apply-k-ds k 0)(length-ds-cps (cdr lst) (length-cps-k k))));; hint for imperative form!Be careful about the way that the;; variables in this function may shadow global variable names if;; you’re not careful.(define apply-k-ds(lambda (k v)(cases continuation k [init-k () v] [length-cps-k (k) (apply-k-ds k (add1 v))] [len-big-k (big little k)(length-ds-cps little (len-little-k big little v k))] [len-little-k (big little len-big k) (if (< len-big v) (apply-k-ds k #f) (head-matches?-ds-cps big little (is-match-k big little k)))] [is-match-k (big little k) (apply-k-ds k #t) (sublist?-ds-cps (cdr big) little k))]))) ;; convert sublist?-ds-cps and the functions it calls to imperative;; form.Below is a test function which is how your imperative form;; function will be invoked – it does not need to be converted to;; imperative form.;; note you do not need to ellimate the use of cond – its obvious how;; cond could be replaced with ifs if desired.(define big)(define little)(define k);; add additional registers you need(define lst)(define (sublist?-impr-tester big-in little-in)(set! big big-in)(set! little little-in)(set! k (init-k));; you can edit sublist?-ds-cps if you wish, or you can make a new;; copy with a different name and call that here(sublist?-ds-cps)) CS: assignmentchef QQ: 1823890830 Email: [email protected]