scala tictactoe避免可变变量

Question

我在scala中写了一个简单的tic tac toe游戏实现。

package tiktaktoe

/**
  * Created by donbeo on 28/12/15.
  */
class Board {

  val size = 3
  // size of the board
  val goal = 3
  // number in line.  These values are assumed to be always 3 and 3
  val board = Array.fill(size, size)(0) // the board

  def isValidPosition(pos: Pos): Boolean = {
    if (pos._1 >= 0 && pos._2 >= 0 && pos._1 < size && pos._2 < size && board(pos._1)(pos._2) == 0) true
    else false
  }

  def setPosition(pos: Pos, number: Int): Unit = {
    if (isValidPosition(pos)) {
      board(pos._1)(pos._2) = number
      println(this)
      val winner = winningPlayer
      if (winner != 0) println("The winner is: Player " + winner)
    }
    else throw new Exception("Not Valid Position")
  }

  def winningPlayer: Int = {
    if (isLine(0) != 0) isLine(0)
    if (isLine(1) != 0) isLine(1)
    if (isLine(2) != 0) isLine(2)
    if (isColumn(0) != 0) isColumn(0)
    if (isColumn(1) != 0) isColumn(1)
    if (isColumn(2) != 0) isColumn(2)
    if (isDiagonal != 0) isDiagonal
    else 0
  }

  private def isLine(line: Int): Int = {
    // this works only if goal = size
    val number = board(line)(0)
    if (number == 0) return 0
    else if (board(line) == Array.fill(size)(number)) return number
    else 0
  }

  private def isColumn(column: Int): Int = {
    val number = board(0)(column)
    if (number == 0) 0
    else if (board(1)(column) == number && board(2)(column) == number) number
    else 0
  }

  private def isDiagonal: Int = {
    val number = board(1)(1)
    if (board(2)(2) == number && board(1)(1) == number) number
    else if (board(0)(2) == number && board(2)(0) == number) number
    else 0
  }

  override def toString = "Board=\n" + board.deep.mkString("\n") + "\n"
}


class Player(number: Int) {

  private def computeNextMove(board: Board): Pos = {
    (2, 3)
  }

  def nextMove(board: Board, pos: Pos): Unit = {
    board.setPosition(pos, number)
  }
}

object PlayGame {

  def main(args: Array[String]) {
    val board = new Board
    val players = List(new Player(1), new Player(2))
    println("Start the game \n\n" + board)

    players(0).nextMove(board, (1, 0))
    players(1).nextMove(board, (1, 1))
    players(0).nextMove(board, (1, 2))
    players(1).nextMove(board, (0, 0))
    players(0).nextMove(board, (2, 1))
    players(1).nextMove(board, (2, 2))

  }

}

游戏基于修改棋盘状态的功能Player.nextMove。据我所知，最好避免scala中的可变变量。 我想知道如何在不使用可变board

的情况下实现游戏