CSE 130, Spring 2014 Name/ID ANSWER KEY Instructor: Ranjit Jhala
Final Exam
Instructions: read these first!
Do not open the exam, turn it over, or look inside until you are told to begin.
Switch off cell phones and other potentially noisy devices.
Write your full name on the line at the top of this page. Do not separate pages.
You may refer to any printed materials, but no computational devices (such as laptops, calculators, phones, iPads, friends, enemies, pets, lovers).
Read questions carefully. Show all work you can in the space provided. Where limits are given, write no more than the amount specified.
The rest will be ignored.
Avoid seeing anyone elses work or allowing yours to be seen.
Do not communicate with anyone but an exam proctor.
If you have a question, raise your hand.
When time is up, stop writing.
The points for each part are rough indication of the time that part should take.
Question
Points
Score
1
25
2
25
3
35
4
45
5
20
Total:
150
CSE 130, Spring 2014 Final Exam Page 1 of 12 1. [25 points] For each of the following OCaml or Scala programs, write down the value of ans.
(a) [5 points]
let rec range i j =
if i > j
then []
else i :: (range (i+1) j)
let ans = range 1 5
ans =
(b) [5 points]
let gelato g = let x = 10 in gx
[1;2;3;4;5]
let x = 100
let f y = x + y
let ans = gelato f
ans = 110
(c) [5 points]
val ans = for ( i <- 1 to 5; j <- i to 5; k <- j to 5; if (i*i + j*j == k*k))yield (i, j, k)ans = Vector((3,4,5))CSE 130, Spring 2014 Final Exam Page 2 of 12 (d) [6 points]class A () {def foo(x:Int) = 1 + this.bar(x)def bar(x:Int) = 1 + x}class B () extends A {override def bar(x:Int) = 100 + x}val x = (new A) foo 10val y = (new B) foo 10val ans = (x, y)ans = (12, 111)(e) [4 points]var m1 = Map(“tako” -> nom nom, uni -> blergh) varm2 =m1
m1 += (uni -> delicioso)
val ans = m2(uni)
ans = blergh
CSE 130, Spring 2014 Final Exam Page 3 of 12 2. [25 points] Consider the following Scala class and type definitions.
class A
class B extends A
type Point2A = {val x:A; val y:A}
type Point2B = {val x:B; val y:B}
type Point3A = {val x:A; val y:A; val z:A}
type Point3B = {val x:B; val y:B; val z:B}
Which of the below snippets of code typechecks? Circle the case that you believe holds.
(a) [5 points]
def ans = { def foo(p:Point2A) = error(ignore me)
val p3 : Point3A = error(ignore me)
foo(p3) }
Does Not Typecheck Typechecks [YES: width-subtyping]
(b) [5 points]
def ans = { def foo(p:Point3A) = error(ignore me)
val p2 : Point2A = error(ignore me)
foo(p2) }
Does Not Typecheck Typechecks [NO]
(c) [5 points]
def ans = { def foo(p: Point2A) = error(ignore me)
val p2 : Point2B= error(ignore me)
foo(p2) }
DoesNotTypecheck Typechecks [YES:depth-subtyping]
(d) [5 points]
def ans = { def foo(f:(Point2A) => Int) = error(ignore me)
def f2(p:Point2B): Int= error(ignore me)
foo(f2) }
Does Not Typecheck Typechecks[NO: CO-VARIANT inputs]
(e) [5 points]
def ans = { def foo(f:(Point3B) => Int) = error(ignore me)
def f2(p:Point2A): Int= error(ignore me)
foo(f2) }
Does Not Typecheck Typech[eYckEsS: CONTRA-VARIANT inputs]
CSE 130, Spring 2014 Final Exam Page 4 of 12
3. [35 points] A binary-search-ordered dictionary is a data structure that maps keys to values. We will represent dictio- naries using a polymorphic Ocaml datatype:
type (k, v) dict
= Empty
| Node of k * v * (k, v) dict * (k, v) dict
That is, a dictionary is represented as a tree, which is either empty, or a node with:
1. a binding from a k key to an v value, 2. a left sub-dictionary, and
3. a right sub-dictionary.
For example, consider the dictionary
fruit price
apple 2.25 banana 1.50 cherry 2.75 grape 2.65 kiwi 3.99 orange 0.75 peach 2.25
that represents the prices (per pound) of various fruits. This dictionary is represented by the tree (on the left) which in turnisrepresentedbytheOcamlvalue(oftype(string, float) dict)boundtofruitdontheright.
grape:
_____2.65_____
|| banana: orange:
___1.50___ ___0.75___ ||||
apple:cherry:kiwi:peach:
2.252.75 3.99 2.25
let fruitd =
Node (grape, 2.65,
Node (banana, 1.50,
Node (apple, 2.25, Empty, Empty),
Node (cherry, 2.75, Empty, Empty)),
Node (orange, 0.75,
Node (kiwi, 3.99, Empty, Empty),
Node (peach, 2.25, Empty, Empty)))
Notice the tree is Binary-Search-Ordered, meaning for each node with a key k,
keys in the left subtree are less than k, and keys in the right subtree are greater than k.
CSE 130, Spring 2014 Final Exam Page 5 of 12
(a) [5 points] Recall the type a option = None | Some of a. Write a function val find: k -> (k, v) dict -> v option
suchthatfind k devaluatestoSome vifvisthevalueassociatedwiththekeykinthedictionaryd,andNone otherwise. When you are done, you should get the following behavior:
# find cherry fruitd
: float option = Some 2.75
# find pomegranate fruitd
: float option = None
Fill in the blanks below to implement find as described.
let rec find k d =
match d with
| Empty ->
None
| Node (k, v, l, r) when k = k ->
Some v
| Node (k, v, l, r) when k < k ->
find k l
| Node (k, v, l, r) (* k < k *) ->
find k r
(b) [8 points] Next, write a function
val deleteMax : (k, v) dict -> (k * v * (k, v) dict)
such that deleteMax d returns a tuple of the largest key in d, the value corresponding to the key, and the dictionary without the corresponding key-value pair. When you are done you should get the following behavior: # let d0 = Node (banana, 1.50,
Node (apple, 2.25, Empty, Empty),
Node (cherry, 2.75, Empty, Empty)) ;;
# deleteMax d0 ;;
: (string, float, (string, float) dict) =
=(cherry, 2.75, Node (banana, 1.50,
Node (apple, 2.25, Empty, Empty),
Empty))
Fill in the blanks below to implement deleteMax as described. (It will only be called with non-Empty trees.)
CSE 130, Spring 2014
Final Exam
Page 6 of 12
let rec deleteMax d =
match d with
| Node (k, v, l, Empty) ->
(k, v, l)
|Node(k,v,l,r) ->
let (k, v, r) = deleteMax r in (k, v, Node (k, v, l, r))
(c) [8 points] Using deleteMax, write a function
val delete : k -> (k, v) dict -> (k, v) dict
suchthatdelete k dreturnsthedictionarywithallthekey-valuepairsofdexceptk.Ifkwasnotpresentind then the output should be the same as d. When you are done, you should get the following behavior:
#delete grape fruitd ;;
: (string, float) dict
= Node (cherry, 2.75,
Node (banana, 1.50,
Node (apple, 2.25, Empty, Empty),
Empty),
Node (orange, 0.75,
Node (kiwi, 3.99, Empty, Empty),
Node (peach, 2.25, Empty, Empty)))
Fill in the blanks below, using deleteMax, to implement delete:
let rec delete k d =
match d with
| Empty ->
Empty
| Node (k, v, Empty, r) when k = k ->
r
| Node (k, v, l, r) when k = k ->
let (k, v, l) = deleteMax l in Node (k, v, l, r)
| Node (k, v, l, r) when k < k ->
Node (k, v, delete k l, r)
| Node (k, v, l, r) (* when k < k *) ->
Node (k, v, l, delete k r)
CSE 130, Spring 2014 Final Exam Page 7 of 12
(d) [7 points] The following function implements a fold over the dictionaries. let rec fold f b t = match t with
| Empty -> b |Node(k,v,l,r)->letb0=foldfbr in
let b1 = f k v b0in
let b2 = fold f b1 l in
b2
What is the type of fold?
val fold : (k-v-a-a)-a-(k,v)tree-a
(e) [7 points] Fill in the blanks below to obtain a function
val keysWithValue : v -> (k, v) dict -> k list *) suchthatkeysWithValue v dthatreturnsthelistofkeysindwithvaluev.Whendone,youshouldget:
# keysWithValue 2.25 fruitd;;
: string list = [apple; peach]
let keysWithValue v d =
let f k v acc = ifv=vthenk:: accelseacc in letb = [] in fold f b d
CSE 130, Spring 2014 Final Exam Page 8 of 12
4. [45 points] Lets implement Scala-style for-loops in Ocaml, using the following functions: let skip = []
let yield x = [x]
let rec foreach xs f = match xs with
| []-> []
| x::xs -> f x @ foreach xs f
(a) [3 points] What is the type of skip?
val skip : a list
(b) [4 points] What is the type of yield?
val yield: a – a list
(c) [8 points] What is the type of foreach?
val foreach: a list – (a – b list) – b list
(d) [4 points] What is the value of ans?
let ans = foreach [1;2;3] (fun x ->
yield (x * x)
)
ans = [1;4;9]
(e) [6 points] What is the value of ans?
let ans = foreach [1; 2] (fun i ->
foreach [a; b] (fun c ->
ans =
[(1, a); (1, b); (2, a); (2, b)]
)
yield (i, c) )
CSE 130, Spring 2014 Final Exam Page 9 of 12 (f) [5 points] Recall the Scala code from the first question:
val ans = for ( i <- 1 to 5 ; j <- i to 5 ; k <- j to 5 ; if (i*i + j*j == k*k)) yield (i, j, k)Translate it to the equivalent Ocaml, by filling the blanks below using only the functions yield, skip. (The function range is from Question 1):let ans = foreach (range 1 5) (fun i ->
foreach (range i 5) (fun j ->
foreach (range j 5) (fun k ->
if then else
(i*i + j*j = k*k) yield (i,j,k) skip
) )
)
(g) [5 points] Rewrite the usual map function for lists using only foreach, skip and yield: (* val map : (a -> b) -> a list -> b list *)
let map f xs =
foreach xs (fun x – yield (f x)
)
(h) [5 points] Rewrite the usual filter function for lists using only foreach, skip and yield: (* val filter : (a -> bool) -> a list -> a list *)
let filter f xs =
foreach xs (fun x – if (f x)
then yield x else skip
)
CSE 130, Spring 2014 Final Exam Page 10 of 12 (i) [5 points] The function flatten of type:
val flatten : a list list -> a list
has the following behaviour:
# flatten [[1;2;3]; [4;5]; [6]] ;;
: int list = [1;2;3;4;5;6]
Write flatten using only foreach, skip and yield: let flatten xss =
foreach xss (fun xs – foreach xs (fun x – yield x
) )
CSE 130, Spring 2014 Final Exam Page 11 of 12 5. [20 points] Lets write a function to generate all permutations of a list.
(a) [5 points] Write a function insertAt with the following behavior:
scala> insertAt(0, cat, List(mouse, giraffe, hippo))
res0: List[String] = List(cat, mouse, giraffe, hippo)
scala> insertAt(1, cat, List(mouse, giraffe, hippo))
res1: List[String] = List(mouse, cat, giraffe, hippo)
scala> insertAt(2, cat, List(mouse, giraffe, hippo))
res2: List[String] = List(mouse, giraffe, cat, hippo)
scala> insertAt(3, cat, List(mouse, giraffe, hippo))
res3: List[String] = List(mouse, giraffe, hippo, cat)
Fill in the blanks to get a definition of insertAt
def insertAt[A](pos:Int, x:A, ys:List[A]): List[A] =
(pos, ys) match {
case (0, _)=>
case (n, y::ys_) =>
case (_, Nil)=>
x::ys
y :: insertAt(n-1, x, ys ) x :: Nil
}
(b) [5 points] Next, write a function spliceInto with the following behavior:
scala> spliceInto(cat, List(mouse, giraffe, hippo))
res4: List[List[String]] = List(List(cat, mouse, giraffe, hippo),
List(mouse, cat, giraffe, hippo),
List(mouse, giraffe, cat, hippo),
List(mouse, giraffe, hippo, cat))
Fill in the blanks to get a definition of spliceInto
def spliceInto[A](x:A, ys:List[A]) : List[List[A]] =
for (i <- ( 0 to ys.length ).toList) yield insertAt(i, x, ys)CSE 130, Spring 2014 Final Exam Page 12 of 12 (c) [10 points] Finally, use spliceInto to write a function permutations with the following behavior:scala> permutations(List(0,1,2))
res5: List[List[Int]] = List(List(0, 1, 2),
List(1, 0, 2),
List(1, 2, 0),
List(0, 2, 1),
List(2, 0, 1),
List(2, 1, 0))
Fill in the blanks below to obtain an implementation of permutations def permutations[A](xs:List[A]): List[List[A]] =
xs match {
case Nil=>
case x::xs_ =>
}
List(Nil)
for ( ys – permutations(xs ) ; zs – spliceInto(x,ys) ) yield zs
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