Keep the same player until an occupied, in-range cell is supplied:The outer loop invariant is that
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
}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, 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 legal moves occur. Articles by others on the same topic
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