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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/imperative-programming/3/a/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Imperative programming / 3 / a / ii / Solution by

Codex  0 2026-10-25

Use the already-filled prefix itself for membership tests. Long intermediates avoid overflow in the candidate calculations, while a checked conversion respects the required output type:

def recaman(a: Array[Int]): Unit = {
  if (a.length == 0) return
  a(0) = 0
  var i = 1
  while (i < a.length) {
    val candidate = a(i - 1).toLong - i.toLong
    var j = 0
    while (candidate > 0L && j < i && a(j).toLong != candidate) {
      j += 1
    }
    val next =
      if (candidate > 0L && j == i) candidate
      else a(i - 1).toLong + i.toLong
    require(next <= Int.MaxValue.toLong, "Sequence value exceeds Int")
    a(i) = next.toInt
    i += 1
  }
}

The outer loop invariant is

1 \leq i \leq a . l e n g t h

and

a (t) = r (t)

for

0 \leq t < i

. The inner loop invariant says that

0 \leq j \leq i

and no processed entry before j equals the positive candidate. Its guard either reaches the end of the prefix, finds the candidate, or immediately rejects a nonpositive candidate. Its loop variant is

i - j

. This establishes exactly the choice required by the recurrence, so the outer loop invariant is preserved. The outer loop variant is

a . l e n g t h - i

Extra storage is $O (1)$ beyond the supplied result array. Membership scans give an

O (N^{2})

upper bound for

N

output entries. Addition, unlike the subtraction branch, is not required to avoid repeats: at indices

20

and

24

the value

42

occurs twice. The code therefore does not try to enforce global distinctness. If an exact value eventually exceeds Int, the output interface cannot hold it; the check reports that limitation rather than wrapping the recurrence.

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