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--What does the following Scheme function do? Explain the answer provided. --Wha

ID: 3546838 • Letter: #

Question


                    


--What does the following Scheme function do? Explain the answer provided.                 

                

                    --What is the result of (guess '(A B C))? Show the derivation process.                 

                

                    
                

                

                    
                

                

                    (DEFINE(guess lis)                 

                

                    (COND                 

                

                    (NULL? lis) '())                 

                

                    (ELSE(append(guess(CDR lis)) (LIST (CAR lis)) ) )                 

                

                    ))                 

                                     

Explanation / Answer

result is (C B A) reversed the list......solution is given at the ending




Let me start with the basics....


1)NULL:


NULL? function tests its parameters to determine whether it is the empty list and returns #T if it is.

eg.,

(NULL? '(A B)) returns ( )

(NULL? '( )) returns #T

(NULL? 'A) returns ( )

(NULL? '( ( ) ) ) returns ( ) (parameter is not empty list, but list containing an empty list)

(NULL? NIL) returns #T


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2)CAR:


CAR function returns the first element of a given list.

e.g.,

(CAR '(ABC)) returns A

(CAR '((A B) C D)) returns (A B)

(CAR 'A) is an error (A is not a list)

(CAR '(A)) returns A


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3)CDR:


CDR returns the remainder of a given list after its CAR is removed.


(CDR '(A B C)) returns (B C)

(CDR '((A B) C D)) returns (C D)

(CDR 'A) is an error

(CDR '(A)) returns ( ) (empty list, also equivalent to NIL)


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now coming to the question


given lis = (A B C)


code runs ,

now as lis is not NULL, else executes i.e, append(guess(CDR lis)) (LIST (CAR lis))


==> append (guess((B C)) (LIST (A)))


==> append ( append (guess (C) LIST (B)) LIST (A))


==> append (append ( append () LIST (C))) LIST (B) ) LIST (A))


==> append (append LIST(C) LIST (B)) LIST (A)


==> append (LIST (C B) LIST (A))


==> LIST (C B A)


==> ( C B A)



I reversed the list given i.e, reversed (A B C ) to (C B A)