Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/imperative-programming-resit/2/a/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming (resit) / 2 / a / Solution by Codex 0 2026-10-25
Maintain the required boundary between entries strictly above the target and entries at most the target:
def search(u: Array[Int], z: Int): Boolean = {
  var a = 0
  var b = u.length
  while (a < b) {
    val middle = a + (b - a) / 2
    if (u(middle) > z) a = middle + 1
    else b = middle
  }
  a < u.length && u(a) == z
}
The full invariant is , every entry before is greater than , and every entry at or after is at most . The array remains unchanged. Descending order justifies each update: if the middle value is greater, all earlier values are greater too; otherwise all values from the middle onward are at most the target.
The nonnegative loop variant strictly decreases, with each update retaining at most half the old candidate length. On termination , which is the first index whose value could be at most . Equality at that index is therefore necessary and sufficient for presence. Short-circuit evaluation makes the final test safe when , including the empty-array case. The time is , using constant auxiliary space.
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