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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/5/c/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 / 5 / c / Solution by

Codex  0 2026-10-25

Keep the same player until an occupied, in-range cell is supplied:

val board = new Chomps(6)
var player = 1
while (board.Count() > 0) {
  board.Display()
  var x = -1
  var y = -1
  while (!board.Legal(x, y)) {
    println("Player " + player + ": enter x, then y")
    try {
      x = scala.io.StdIn.readInt()
      y = scala.io.StdIn.readInt()
    } catch {
      case _: NumberFormatException =>
        x = -1
        y = -1
    }
    if (!board.Legal(x, y)) println("Choose an occupied cell")
  }
  board.Move(x, y)
  if (board.Count() == 0)
    println("Player " + player + " loses; player " + (3 - player) + " wins")
  else player = 3 - player
}

The outer loop invariant is that player is the next player's number, either one or two, and the board is the result of the accepted moves. The inner loop invariant says that the board and player have not changed; rejected inputs do not consume a turn. Assuming the input source continues and eventually supplies a legal coordinate pair, the inner loop finishes. No program can guarantee that a user stops entering illegal moves, so that interaction assumption is necessary.

Every accepted move removes at least its selected token, so the outer loop variant is Count(), a decreasing nonnegative integer. In a nonempty order ideal,

(0, 0)

is occupied. Removing it removes the entire board, whereas a move at any other cell leaves it. Therefore the player making the final move is exactly the loser; the code reports that player before changing the turn. At most

36

legal moves occur.

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