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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/imperative-programming-resit/7/b/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Imperative programming (resit) / 7 / b / ii / Solution by

Codex  0 2026-10-25

Use the heap-numbered binary tree with root label zero and children

2 k + 1, 2 k + 2

. Its strict descendants in breadth-first search order are $1, 2, 3, \dots$ . A machine-safe Node[Int] version stops at the largest representable label:

class NumberNode(val label: Int) extends Node[Int] {
  require(label >= 0)
  def children(): Iter[Node[Int]] = new Iter[Node[Int]] {
    private var candidate = 2L * label.toLong + 1L
    private val last = candidate + 1L
    def hasNext: Boolean = candidate <= last && candidate <= Int.MaxValue
    def next(): Node[Int] = {
      require(hasNext)
      val answer = new NumberNode(candidate.toInt)
      candidate += 1L
      answer
    }
  }
}

val traversal = breadthFirst(new NumberNode(0))
var expected = 1
while (expected <= 1000) {
  assert(traversal.hasNext)
  assert(traversal.next().label == expected)
  expected += 1
}

The arithmetic is widened before computing child labels, so no negative wrapped labels occur. Repeated hasNext calls should not skip nodes; a leaf root should produce immediate exhaustion, and uneven finite trees should retain sibling order across parents. A fresh second traversal should start again at one.

Literal unbounded increase is impossible with the bounded Int type. For a genuinely infinite test, change the label and child arithmetic to BigInt; the same construction then has no representational endpoint. The shown finite-prefix test already exercises lazy traversal without allocating the entire very large Int-labelled tree.

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