Question

How do we verify in scheme with the do cicle, if an element of the first list is in the second?

Answer 1

The do loop in racket has an interesting structure:

(do ([id init-expr step-expr-maybe] ...)
    (stop?-expr finish-expr ...)
  expr ...)

The documentation for r5rs provides an example:

(let ((x '(1 3 5 7 9)))
  (do ((x x (cdr x))
       (sum 0 (+ sum (car x))))
      ((null? x) sum)))

That statement returns 25, the sum of the elements of the loop. The x in the do loop is initialized to the x in the let, and then iteratively set to the cdr of itself each time through the loop. sum is initialized to 0, and accumulates the value of the car of x each time through. The stopping condition is when the iteration variable is empty, and the return value is the sum.

Ok, aside from the racket preference of square brackets, this looks good. There’s a do loop and a list. The loop does something over that list. We can use that to write a function that finds a specific atom in a list (using the racket brackets):

(define (find5 lst)
  (do ([x lst (rest x)]
       [found #f (or found (eq? 5 (first x)))])
    ((null? x) found)))

Instead of initializing and adding the value sum, I or into found. Also, I prefer first and rest over car and cdr and define them myself when they don’t exist. The way this function works should follow from the explanation of the example.

(find5 '(1 2 3 4 6))

Gives #f, as expected. Similarly:

(find5 '(1 2 3 4 5 6))

Gives #t.

Are you able to generalize finding a specific element in a list with a do loop into your specific question?