Solutions for Sample Questions for Midterm 2 (CS 421 Spring 2007)
On the actual midterm, you will have plenty of space to put your answers.
Some of these questions may be reused for the exam.
1. Write the definition of an OCaml variant type reg_exp to express abstract syntax
trees for regular expressions over a base character set of booleans. Thus, a
boolean is a reg_exp, epsilon is a reg_exp, the concatenation of two reg_exp’s is
a reg_exp, the “choice” of two reg_exp’s is a reg_exp, and the Kleene star of a
reg_exp is a reg_exp.
Solution:
type reg_exp =
Epsilon
| Var of bool
| Choice of (reg_exp * reg_exp)
| Concat of (reg_exp * reg_exp)
| Kleene_star of reg_exp
| Paren of reg_exp (* I would accept it with this case missing *)
2. Given the following OCAML datatype:
type int_seq = Null | Snoc of (int_seq * int)
write a tail-recursive function in OCAML all_pos : int_seq -> bool that returns true if
every integer in the input int_seq to which all_pos is applied is strictly greater than 0
and false otherwise. Thus all_pos (Snoc(Snoc(Snoc(Null,3),5),7)) should returns true,
but
all_pos (Snoc(Null,~1)) and all_pos (Snoc(Snoc(Null, 3),0)) should both return
false.
Solution:
let rec all_pos s =
(match s with Null -> true
| Snoc(seq, x) -> if x <= 0 then false else all_pos seq);;
3. Using the rules provided in class, derive a valid type judgment for
let rec fact = fun n -> if n = 0 then 1 else let r = fact (n - 1) in n * r in fact;;
(The rules will be provided for you on the exam, if this kind of question is asked.)
Solution: (I didn’t just write the tree because it wouldn’t fit.)
By the let_rec rule we have
(1) {fact : int -> int} |- (fun n -> if n = 0 then 1 else let r = fact (n – 1) in n * r) : int -> int
(2) {fact : int -> int} |- fact : int -> int
{} |- let rec fact = fun n -> if n = 0 then 1 else let r = fact (n – 1) in n * r in fact : int -> int
(2) is valid by the variable rule:
(2) {fact : int -> int} |- fact : int -> int
