Last updated on Sunday, October 12, 2025

2021 CS-210 Midterm

The exam without solutions is available as PDF: 2021-midterm-empty.pdf.

Exercise 1

trait Tree[T]:
    def height: Int = this match
        case EmptyTree(_) => 0
        case Node(left, _, right, _) => Math.max(left.height, right.height) + 1

    def isBalanced: Boolean = this match
        case EmptyTree(_) => true
        case Node(left, _, right, _) => left.isBalanced && right.isBalanced && Math.abs(left.height - right.height) <= 1

case class EmptyTree[T](leq: (T, T) => Boolean) extends Tree[T]
case class Node[T](left: Tree[T], elem: T, right: Tree[T], leq: (T, T) => Boolean)
extends Tree[T]

Tests

def intLeq(a: Int, b: Int) = a <= b

val exTree1 = Node(
    Node(
        Node(EmptyTree(intLeq), 1, EmptyTree(intLeq), intLeq),
        2,
        Node(EmptyTree(intLeq), 3, EmptyTree(intLeq), intLeq),
        intLeq
    ),
    5,
    Node(
        Node(EmptyTree(intLeq), 7, EmptyTree(intLeq), intLeq),
        9,
        Node(
            Node(EmptyTree(intLeq), 10, EmptyTree(intLeq), intLeq),
            15,
            Node(EmptyTree(intLeq), 25, EmptyTree(intLeq), intLeq),
            intLeq
        ),
        intLeq
    ),
    intLeq
)

val exTree2 = Node(
    Node(
        Node(
            Node(EmptyTree(intLeq), 25, EmptyTree(intLeq), intLeq),
            15,
            Node(EmptyTree(intLeq), 10, EmptyTree(intLeq), intLeq),
            intLeq
        ),
        9,
        Node(EmptyTree(intLeq), 7, EmptyTree(intLeq), intLeq),
        intLeq
    ),
    5,
    Node(
        Node(EmptyTree(intLeq), 3, EmptyTree(intLeq), intLeq),
        2,
        Node(EmptyTree(intLeq), 1, EmptyTree(intLeq), intLeq),
        intLeq
    ),
    intLeq
)

val tree3 = Node(
    Node(
        Node(
            Node(EmptyTree(intLeq), 25, EmptyTree(intLeq), intLeq),
            15,
            Node(EmptyTree(intLeq), 10, EmptyTree(intLeq), intLeq),
            intLeq
        ),
        9,
        Node(EmptyTree(intLeq), 7, EmptyTree(intLeq), intLeq),
        intLeq
    ),
    5,
    EmptyTree(intLeq),
    intLeq
)

assert(exTree1.height == 4)
assert(exTree2.height == 4)
assert(EmptyTree(intLeq).height == 0)
assert(exTree1.isBalanced)
assert(exTree2.isBalanced)
assert(!tree3.isBalanced)

Exercise 2

enum Expr:
  case Var(name: String)
  case Op(name: String, args: List[Expr])

import Expr._

final case class UnknownVarException(name: String) extends Exception
final case class UnknownOpException(name: String) extends Exception

There was multiple correct solutions. Here two possible solutions:

def eval1(e: Expr, context: Map[Var, Int]): Int = e match 
    case Var(name) => context.get(Var(name)) match
        case Some(n) => n
        case _ => throw UnknownVarException(name)
    case Op(name, args) =>
        val nargs = args.map(eval1(_, context))
        name match
            case "*" => nargs.foldLeft(1)(_ * _)
            case "+" => nargs.foldLeft(0)(_ + _)
            case _ => throw UnknownOpException(name)

def eval2(e: Expr, context: Map[Var, Int]): Int = e match 
    case sym: Var if context.contains(sym) => context(sym)
    case sym: Var => throw UnknownVarException(sym.name)
    case Op("*", args) => args.foldLeft(1)(_ * eval2(_, context))
    case Op("+", args) => args.foldLeft(0)(_ + eval2(_, context))
    case Op(name, _) => throw UnknownOpException(name)

Comments

A lot of you assumed that the arguments of Ops can only be Vars, but these can be Ops as well as in Op("*", List(Op("+", List(Var("x"), Var("x"))), Var("y"))), so the function needs to call itself recursively for each operation argument.
Minor: Map[Var, Int] maps Var to Ints, so the argument passed to context.get must be a Var, not a String.

Tests

for eval <- Seq(eval1, eval2) do
    assert(eval(Op("+", List()), Map()) == 0)
    assert(eval(Op("+", List(Var("x"))), Map(Var("x") -> 2)) == 2)
    assert(eval(Op("+", List(Var("x"), Var("y"))), Map(Var("x") -> 2, Var("y") -> 3)) == 5)
    assert(eval(Op("*", List()), Map()) == 1)
    assert(eval(Op("*", List(Var("x"))), Map(Var("x") -> 2)) == 2)
    assert(eval(Op("*", List(Var("x"), Var("y"))), Map(Var("x") -> 2, Var("y") -> 3)) == 6)
    assert(eval(Op("*", List(Op("+", List(Var("x"), Var("x"))), Var("y"))), Map(Var("x") -> 2, Var("y") -> 3)) == 12)

Exercise 3

Exercise 3.1

The contravariant type parameter A in map is used in a covariant context:

scala> trait Transform[-A, +B]:
     |     def apply(x: A): B
     |     def map[C](f: A => C): Transform[A, C]
     |     def followedBy[C](t: Transform[B, C]): Transform[A, C]
     |
-- Error: ----------------------------------------------------------------------
3 |    def map[C](f: A => C): Transform[A, C]
  |               ^^^^^^^^^
  |contravariant type A occurs in covariant position in type A => C of parameter f
1 error found

There are many possible fixes:

trait Transform[-A, +B]:
  def mapMinimal[a <: A, B](f: a => B): Transform[A, B]
  def mapWithReturn[a <: A, B](f: a => B): Transform[a, B]
  def mapFullyGeneralized[a <: A, b >: B](f: a => b): Transform[a, b]
  def mapAfter[C](f: B => C): Transform[A, C]
  def mapBefore[C](f: C => A): Transform[C, B]

As a general explanation, here is one way to think of these variance problems: think of what may go wrong.

Variance annotations tell you what casts are safe on a type. Scala checks for you that your annotations are correct. Here we have a - annotation on A, telling us that it’s safe to cast Transform[A] to Transform[a] where a is smaller, more restrictive than A (a subtype).

Now consider map. map takes, as originally written, a function from A, and could e.g. apply it to an A.

So we can do the following:

Create an instance of Transform[Any, String] containing an arbitrary object.
Cast it to Transform[Int, String]
Call map and pass it an Int => String function.
Have map call that function with the previously stored Any, which would be invalid and unsound.

So clearly the annotation is wrong, or otherwise said map isn’t compatible with the annotation. Now consider possible fixes:

trait Transform[-A, +B]:
  def map[a <: A, b >: B](f: a => b): Transform[a, b]
  def mapAfter[C](f: B => C): Transform[A, C]
  def mapBefore[C](f: C => A): Transform[C, B]

The first restricts what map can do: it tells us that the function accepts something that is a subtype of A (but not necessarily A), so map can’t call it with an A. (I made a similar change on the B side).

The second and third ones put the A and B parameters on the “right” side: either to pre-process the input of the transformer or to post-process its output.

Here’s a concrete implementation of this trait to help with experiments:

case class T[-A, +B](f0: A => B) extends Transform[A, B]:
  def map[a <: A, b >: B](f: a => b): Transform[a, b] =
    T(f)
  def mapAfter[C](f: B => C): Transform[A, C] =
    T(f0 andThen f)
  def mapBefore[C](f: C => A): Transform[C, B] =
    T(f andThen f0)

val t: Transform[Any, String] = T((o: Any) => o.toString())
t.map[Int, String]((x: Int) => x.toString)

Exercise 3.2

Is the following subtype assertion true?                yes    no

List[Triangle]            <:  List[AnyRef]              [X]    [ ]
List[Shape]               <:  List[AnyVal]              [ ]    [X]
List[String]              <:  List[Shape]               [ ]    [X]
Transform[Point, Circle]  <:  Transform[Point, Any]     [X]    [ ]
Transform[Point, Shape]   <:  Transform[Shape, Point]   [ ]    [X]
Transform[Shape, Point]   <:  Transform[Point, Shape]   [X]    [ ]

Exercise 4

Axioms:

(1) Nil.reverse === Nil

(2) (x::xs).reverse === xs.reverse ++ (x::Nil)

(3) (xs++ys)++zs === xs++(ys++zs)

(4) Nil map f === Nil

(5) (x::xs) map f === f(x) :: (xs map f)

(6) Nil ++ ys === ys

(7) xs ++ Nil === xs

(8) x::xs ++ ys === x::(xs ++ ys)

(9) (l1 ++ l2).map(f) === l1.map(f) ++ l2.map(f)

Exercise 4.1

We must prove:

(10) (l1 ++ l2).reverse === l2.reverse ++ l1.reverse


By induction on l1:

- l1 is Nil:

        (Nil ++ l2).reverse ===
    (6) l2.reverse

        l2.reverse ++ Nil.reverse ===
    (1) l2.reverse ++ Nil ===
    (7) l2.reverse

- l1 is x::xs

    IH: (xs ++ l2).reverse === l2.reverse ++ xs.reverse

         (x::xs ++ l2).reverse ===
    (8)  (x::(xs ++ l2)).reverse ===
    (2)  (xs ++ l2).reverse ++ (x::Nil) ===
    (IH) l2.reverse ++ xs.reverse ++ x::Nil ===
    (3)  l2.reverse ++ (xs.reverse ++ x::Nil) ===
    (2)  l2.reverse ++ (x::xs).reverse ===
         l2.reverse ++ l1.reverse

Exercise 4.2

We must prove:

(l1 ++ l2).reverse.map(f) === l2.reverse.map(f) ++ l1.reverse.map(f)


Direct proof using axiom 10 proved in 4.1:

     (l1 ++ l2).reverse.map(f) =
(10) (l2.reverse ++ l1.reverse).map(f) =
(9)  l2.reverse.map(f) ++ l1.reverse.map(f)


Or without using axiom 10:

By induction on l1:

- l1 is Nil:

        ((l1 ++ l2).reverse).map(f) ===
    (6) l2.reverse.map(f)

        l2.reverse.map(f) ++ l1.reverse.map(f) ===
    (1) l2.reverse.map(f) ++ Nil.map(f) ===
    (4) l2.reverse.map(f) ++ Nil ===
    (7) l2.reverse.map(f)


- l1 is x::xs

    IH: ((xs ++ l2).reverse).map(f) === l2.reverse.map(f) ++ xs.reverse.map(f)

         ((x::xs) ++ l2).reverse.map(f) ===
    (8)  (x::(xs ++ l2)).reverse.map(f) ===
    (2)  ((xs ++ l2).reverse ++ (x::Nil)).map(f) ===
    (9)  ((xs ++ l2).reverse.map(f)) ++ ((x::Nil).map(f)) ===
    (IH) l2.reverse.map(f) ++ xs.reverse.map(f) ++ (x::Nil).map(f) ===
    (3)  l2.reverse.map(f) ++ (xs.reverse.map(f) ++ (x::Nil).map(f)) ===
    (9)  l2.reverse.map(f) ++ (xs.reverse ++ x::Nil).map(f)) ===
    (2)  l2.reverse.map(f) ++ ((x::xs).reverse.map(f)) ===

Exercise 5

Exercise 5.1

There was multiple correct solutions:

def takeWhileStrictlyIncreasing1(list: List[Int]): List[Int] = list match
    case Nil => Nil
    case x :: Nil => list
    case x1 :: x2 :: xs => if x1 < x2 then x1 :: takeWhileStrictlyIncreasing1(x2 :: xs) else List(x1)

def takeWhileStrictlyIncreasing2(list: List[Int]): List[Int] = {
    def fun(ls: List[Int], last: Int): List[Int] = ls match
        case x :: xs if x > last => x :: fun(xs, x)
        case _ => Nil

    list match
        case x :: xs => x :: fun(xs, x)
        case Nil => Nil
}

def takeWhileStrictlyIncreasing3(list: List[Int]): List[Int] = {
    def fun(ls: List[Int], last: Option[Int]): List[Int] = (ls, last) match
        case (x :: xs, Some(last)) if x > last => x :: fun(xs, Some(x))
        case (x :: xs, None) => x :: fun(xs, Some(x))
        case _ => Nil

    fun(list, None)
}

Comments

It was possible to write a correct implementation using foldLeft. Note that foldLeft traverses the whole list and cannot “early return” like a recursive call. This possible solution was not very clean as it required a flag or a condition to check that foldLeft should do nothing once it passed the first decreasing step. Avoid using foldLeft when you do not need to traverse the whole list every time.
When we do not ask for a tail recursive implementation, the solution can be more readable without an accumulator. You should favor a simple and readable version.
The solution takeWhileStrictlyIncreasing3 uses Option and is elaborate. We include this solution for a good example of using Option but takeWhileStrictlyIncreasing2 is simpler and more readable.

Exercise 5.2

def increasingSequences(list: List[Int]): List[List[Int]] = list match
    case Nil => Nil
    case _ =>
        val prefix = takeWhileStrictlyIncreasing(list)
        prefix :: increasingSequences(list.drop(prefix.length))

Comments

It is important to store prefix to avoid calling twice takeWhileStrictlyIncreasing.
In the empty list case, some students made the confusion of returning a list containing the empty list. In this case, we have: List[List[Int]]() == Nil != List(Nil) == List(List[Int]()).