Exercise 1.20

With applicative order, it performs 4 remainder operations.

(gcd 206 40)
=> (gcd 40 (remainder 206 40))
=> (gcd 40 6)
=> (gcd 6 (remainder 40 6))
=> (gcd 6 4)
=> (gcd 4 (remainder 6 4))
=> (gcd 4 2)
=> (gcd 2 (remainder 4 2))
=> (gcd 2 0)
=> 2

With normal order, it performs 18 remainder operations. Each b gets evaluated once in the (= b 0) predicate (14 operations). The final a gets evaluated in the end (4 operations). Together, that makes 18.

(gcd 206 40)
=> (gcd 40 (remainder 206 40))
=> (gcd (remainder 206 40)
        (remainder 40 (remainder 206 40)))
=> (gcd (remainder 40 (remainder 206 40))
        (remainder (remainder 206 40)
                   (remainder 40 (remainder 206 40))))
=> (gcd (remainder (remainder 206 40)
                   (remainder 40 (remainder 206 40)))
        (remainder (remainder 40 (remainder 206 40))
                   (remainder (remainder 206 40)
                              (remainder 40 (remainder 206 40)))))
=> (remainder (remainder 206 40)
              (remainder 40 (remainder 206 40)))