# SICP Exercise 1.20 - GCD

Euclid’s algorithm for computing the GCD (Greatest Common Divisor) of two numbers is based on a very neat idea: The GCD of two numbers $a$ and $b$, with $a \gt b$ is the same as the GCD of $a-b$ and $b$. A little thought would convince us that this means the GCD of $a$ and $b$ is then equal to the GCD of $remainder \left( a, b \right)$ and $b$, which in turn is equal to the GCD of $b$ and $remainder \left( b, remainder \left( a, b \right) \right)$ and so on. In Scheme code,

```
(define (gcd a b)
(if (= b 0)
a
(gcd b (remainder a b))))
```

Note that the terminating clause for the recursion is when $b=0$, in which case we just return $a$, since any number is a divisor of $0$, and $0 \times 0 = 0$.

The exercise asks us to compare the applicative and normal orders of evaluation
as applied to the procedure `gcd`

, under the substitution model.

Under applicative-order of evaluation (which is the default in most languages
today), the arguments to a procedure are evaluated first and the procedure is
then called with these values. For example, the call `(gcd (* 4 3) (* 2 3))`

will cause the argument expressions `(* 4 3)`

and `(* 2 3)`

to be evaluated to
`12`

and `6`

, respectively, followed by the actual call `(gcd 12 6)`

.

In contrast, the normal-order of evaluation (a characteristic of some “lazy”
functional languages like Haskell) delays the evaluation of argument
expressions until their values are absolutely required. Let us trace the
process generated in the normal-order evaluation of `(gcd 206 40)`

.

```
(gcd 206 40)
-> (if (= 40 0)
206
(gcd 40 (remainder 206 40))))
;; At this point, the antecedent for the `if` clause must be evaluated:
-> (if false
206
(gcd 40 (remainder 206 40))))
-> (gcd 40 (remainder 206 40))
-> (if (= (remainder 206 40) 0)
40
(gcd (remainder 206 40)
(remainder 40 (remainder 206 40))))
;; Eval the if condition now, resulting in one call to remainder
-> (if (= 6 0)
40
(gcd (remainder 206 40)
(remainder 40 (remainder 206 40))))
-> (gcd (remainder 206 40)
(remainder 40 (remainder 206 40))))
-> (if (= (remainder 40 (remainder 206 40)) 0)
(remainder 206 40)
(gcd (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))))
;; This step calls remainder twice
-> (if (= 4 0)
(remainder 206 40)
(gcd (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))))
-> (gcd (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))))
-> (if (= (remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))
0)
(remainder 40 (remainder 206 40))
;; else
(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))))))
;; This step calls remainder 4 times
-> (if (= 2 0)
(remainder 40 (remainder 206 40))
;; else
(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))))))
-> (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)))))
-> (if (= (remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40))))
0)
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))
(gcd (remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40))))
(remainder (remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))
(remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))))))
;; This step calls remainder 7 times and the recursion terminates here.
-> (if (= 0 0)
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))
(gcd (remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40))))
(remainder (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)))
;; The remaining 4 remainder calls are now evaluated
-> 2
```

In all, the number of remainder calls in the normal-order evaluation of
`(gcd 206 40)`

is therefore $4 + 7 + 4 + 2 + 1 = 18$

Now, here’s the applicative-order evaluation of the same expression:

```
(gcd 206 40)
-> (if (= 40 0)
206
(gcd 40 (remainder 206 40)))
-> (gcd 40 (remainder 206 40))
;; 1 call to remainder
-> (gcd 40 6)
-> (if (= 6 0)
40
(gcd 6 (remainder 40 6)))
-> (gcd 6 (remainder 40 6))
;; 1 call to remainder; total 2
-> (gcd 6 4)
-> (if (= 4 0)
6
(gcd 4 (remainder 6 4)))
-> (gcd 4 (remainder 6 4))
;; 1 call to remainder; total 3
-> (gcd 4 2)
-> (if (= 2 0)
4
(gcd 2 (remainder 4 2)))
-> (gcd 2 (remainder 4 2))
;; 1 call to remainder; total 4
-> (gcd 2 0)
-> (if (= 0 0)
2
(gcd 0 (remainder 2 0)))
-> 2
```

As we can see, under the applicative order, we needed only 4 calls to
`remainder`

, whereas in the normal-order evaluation of the same expression, it
took 18 calls to `remainder`

to arrive at the same result. Also, one can
readily see that the normal order of evaluation generally uses more space than
applicative order, since it needs to keep all the unevaluated expressions
somewhere in memory.