Quiz 1: Continuations and Tail Recursion

1. Write a tail-recursive version of factorial in Scheme.

;;; Compute n!
(define (factorial n)
  (trfact n 1))
;;; Compute a*(n!)
(define (trfact n a)
  ; if n is 0, we're done, a*n! = a*0! = a*1 = a
  (if (zero? n) a
  ; otherwise, a*n! = a*n*n-1*...*2*1 = a*n*(n-1)!
      (trfact (- n 1) (* n a))))

2. Convert (+ e1 e2 e3) to continuation-passing style. You must ensure that the expressions are executed in the order e2, e3, e1.

After computing e2, we

• compute e3
• compute e1
• add the results

Here is the continuation for e2.

(lambda (E2) (+ e1 E2 e3))

Our expression is now

((lambda (E2) (+ e1 E2 e3)) 
 e2)

What is the continuation for e3?

• compute e1
• add the three results

Which may be expresed as

(lambda (E3) (+ e1 E2 E3))

Plugging in, our expression is now

((lambda (E2)
  ((lambda (E3) (+ e1 E2 E3))
   e3))
 e2)

Finally, we want to say evaluate e1 and add the three results. The continuation is

(lambda (E1) (+ E1 E2 E3))

Putting it all together, we get

((lambda (E2)
    ((lambda (E3) 
       ((lambda (E1)
          (+ E1 E2 E3))
        e2))
     e3))
 e2)

Now, let's say that there were alternate versions of e1, e2, and e3 that took continuations as arguments. We'll call them e1-cps, e2-cps, e3-cps.

Switching stuff around to use e2-cps, we get

(e2-cps ((lambda (E2)
  ((lambda (E3)
    ((lambda (E1)
       (+ E1 E2 E3))
     e1))
   e3))))

Switching stuff around to use e3-cps, we get

(e2-cps ((lambda (E2)
  (e3-cps ((lambda (E3)
    ((lambda (E1)
       (+ E1 E2 E3))
     e1)))))))

Finally, we switch stuff around to use e1-cps.

(e2-cps ((lambda (E2)
  (e3-cps ((lambda (E3)
    (e1-cps ((lambda (E1)
       (+ E1 E2 E3)))))))))))

Or, in English,

• Evaluate e2, call the result E2
• Evaluate e3, call the result E3
• Evaluate e1, call the result E1
• Sum E1, E2, and E3.

We now have a precise order.