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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/imperative-programming-resit/4/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Imperative programming resit / 4 / b / Solution by

Codex  0 2026-10-25

Store the bottom of the stack at index zero and the top at the last occupied index:

class Stack {
  private val MAX = 1000
  private val data = new Array[Tree](MAX)
  private var used = 0

  // Pre: used < MAX.
  def push(t: Tree): Unit = {
    require(used < MAX, "Stack full")
    data(used) = t
    used += 1
  }

  // Pre: used > 0.
  def pop: Tree = {
    require(used > 0, "Stack empty")
    used -= 1
    val result = data(used)
    data(used) = null
    result
  }

  // Pre: true.
  def isEmpty: Boolean = used == 0
}

The representation invariant is that data.length==MAX,

0 \leq u se d \leq M A X

, and unused slots from used onwards contain null references. Occupied slots contain the stored Tree references; the stack itself need not forbid null values. The abstraction function, written with the first sequence element at the top, is

α (d a t a, u se d) = [d a t a (u se d - 1), d a t a (u se d - 2), \dots, d a t a (0)] .

(1)

Initially this is the empty sequence. push prepends its argument, pop removes and returns the first abstract element, and isEmpty reports whether the abstract sequence is empty. These postconditions follow directly from the index changes. Clearing a popped slot avoids retaining an otherwise unused tree reference. All three operations take $O (1)$ time. No loops are needed.

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