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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Imperative programming / 3 / c / Solution 2026-10-25

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Represent the characteristic values of a bounded universe by one Boolean slot per possible member, a Boolean-array form of a bitset:

class BitMapSet(val MAXVALUE: Int) extends IntSet {
  require(MAXVALUE >= 0)
  private val bits = Array.fill(MAXVALUE)(false)
  private var count = 0

  def contains(elem: Int): Boolean =
    0 <= elem && elem < MAXVALUE && bits(elem)

  def add(elem: Int): Boolean = {
    require(0 <= elem && elem < MAXVALUE, "Outside bitmap universe")
    if (bits(elem)) false
    else {
      bits(elem) = true
      count += 1
      true
    }
  }

  def size: Int = count
}

The abstraction function is

α (bi t s, co u n t) = {x : 0 \leq x < M A X V A LU E, bi t s (x)}

. The representation invariant is that bits.length==MAXVALUE, its entries are Boolean, and count equals the number of true entries. Initialization gives the empty set; changing a false entry to true adds exactly its index and increments the count once. A repeated addition preserves both. All three operations have

O (1)

time complexity after initialization; storage and explicit array initialization are

Θ (M A X V A LU E)

The constructor requires $M A X V A LU E \geq 0$ , and add requires $0 \leq e l e m < M A X V A LU E$ . contains remains total by returning false outside the representable universe; short-circuit evaluation protects the index. This finite representation cannot implement unrestricted addition of every Int while preserving the abstract contract. Its strengthened add precondition therefore makes it a restricted implementation, not a behaviorally substitutable implementation of the unrestricted trait for arbitrary clients. The sequence client below proves that it stays within the restricted universe.

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Recamán's sequence 2026-10-25

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The integer sequence proposed by Bernardo Recamán Santos starts at zero. At step

n > 0

, subtract

n

if the result is positive and has not appeared before; otherwise add

n

. The commonly written nonnegative version gives the same sequence because zero is already present. The initial values are

0, 1, 3, 6, 2, 7, 13, 20, \dots

. The addition branch can repeat a value:

r (20) = r (24) = 42

. Avoiding a repeated subtraction is therefore not a global injectivity condition.^[ref]

Since

r (n) \leq r (n - 1) + n

, induction gives

0 \leq r (n) \leq n (n + 1) /2

. A prefix scan needs only constant extra storage but can take quadratic membership work. A bitset-style bounded membership table makes each lookup constant time, while its allocation cost depends on the chosen value universe. This separates the number of generated terms from the range of possible stored values.

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