I'm a coder that has just started to learn code, and I am weeks into my Racket journey. I am implementing a function called my-subset? that has two parameters lst1 and lst2 that returns true just when every element of lst1 is a member of lst2. In this scenario the empty list is considered a subset of all other lists. For this function, I don't want to use the built in subset function.

Here is some code I have written personally, I have been unable to finish the code, and would appreciate some help. The first function is a helper function that is comparing a certain value with every element in a list. the second function is the actual subset function, I am having troubles with the part after cond as I don't know how to write a case where there is no subset and I have to return false. I am comparing a value from lst 1 to every value in lst 2.

(define (member x lst)
  (cond [(empty? lst) #f]
        [(equal? x (first lst)) #t]
        [else (member x (rest lst))]))

(define (my-subset? lst1 lst2)
  (cond [(empty? lst1) #t]
        [(member (first lst1)(list lst2)) #t]
        [else (my-subset? (rest lst1) (list lst2))]))

CodePudding user response：

You are so close to the solution:

(define (member x lst)
  (cond 
    [(empty? lst)           #f]
    [(equal? x (first lst)) #t]
    [else                   (member x (rest lst))]))

(define (my-subset? lst1 lst2)
  (cond 
    [(empty? lst1)              #t]
    [(member (first lst1) lst2) (my-subset? (rest lst1) lst2)]
    [else                       #f]))

That is, if the first element of lst1 is a member of lst2, then lst1 is a subset of lst2 if the rest of the elements of lst1 also are part of lst2.