;; 11.1

(define (answer-for-11.1)
  '(-1 11 16 16 16))

;; To see the derivations, evaluate the following expressions in the
;; interpreter from lecture 22:

;;  1. -(4,3)
;;  2. let x = 4 in +(let x = 7 in x, x)
;;  3. let x = 8 in ((proc(x)proc(y)*(y, 2) 10) x)
;;  4. let x = 9 in ((proc(x)proc(y)+(x, y) 7) x)
;;  5. let f = proc(g)*((g 3),4) in (f proc(w)if w then (f proc(x)1) else w)



;; 11.2

(define (answer-for-11.2)
  '((exn 1) (exn 3) (val 5) (val 6) (val 7) (val 7) (val 5)))

;; To see the derivations, evaluate the following expressions in the
;; interpreter from lecture 23:

;;  1. raise 1
;;  2. +(1, raise 3)
;;  3. if 0 then raise 1 else 5
;;  4. +(1, try 5 handle proc(x)+(x,1))
;;  5. +(1, try raise 5 handle proc(x)+(x,1))
;;  6. try try raise 5 handle proc(x)7 handle proc(y)8
;;  7. ((proc(f)proc(g)try (f 1) handle g proc(x)raise 1) proc(y)5)

;; 11.3

(define (answer-for-11.3)
  6)

;; To see the derivations, evaluate the following expression in the
;; interpreter from lecture 23:

;;   +(1, let f = letcc k in proc(y)continue k y in (f proc(x)5))