ruby - 石头，纸。剪刀锦标赛

Question

我需要实现一个石头剪刀布锦标赛模拟器，它将玩完所有回合并返回锦标赛。

这是我的锦标赛数组:

tournament = [
    [
        [ 
            ["Armando", "P"], ["Dave", "S"] 
        ],
        [ 
            ["Richard", "R"],  ["Michael", "S"] 

        ],
    ],
    [
        [ ["Allen", "S"], ["Omer", "P"] ],
        [ ["David E.", "R"], ["Richard X.", "P"] ]
    ]
]

例如 ["Richard", "R"], ["Michael", "S"] 表示理查德玩剪刀石头布的游戏。理查德应该是获胜者并晋级下一轮。

我正在尝试为此实现一个递归函数，但我无法正确设置数组

这是我的方法:

 def rps_tournament_winner(tournament)

  qualifying_round_winners = []
   round_winners = []
  # last round

  if tournament.length < 1
    return
  end

  tournament.each_with_index do |round,i|   
 puts round.inspect
     qualifying_round_winners[i] = []
     round_winners = []
     round.each_with_index do |game,j|
         winner = rps_game_winner(game)
         round_winners.push winner
     end
     qualifying_round_winners[i] = round_winners

  end

  rps_tournament_winner(qualifying_round_winners)
end

我发明了很多用于排位赛获胜者的数组，但无法找到解决方案。希望你能帮我。谢谢。

编辑:

这是我在该方法的每次迭代中应该得到的结果

第一轮之后

tournament = [
    [
        [ 
            ["Dave", "S"], ["Richard", "R"] 
        ],
        [ 
            ["Allen", "S"],  ["Richard X.", "P"] 

        ]
    ]
]

第二轮之后:

tournament = [
    [
        [ 
            ["Richard", "R"] , ["Allen", "S"]
        ]
    ]
]

获胜者

["Richard", "R"] 

编辑:2

完整代码如下: http://pastebin.com/gjKfiWLD

Answer 1

在递归时我们必须考虑两种情况:

要么 rps_tournament_winner 的参数是游戏，要么 rps_tournament_winner 的参数是锦标赛。请注意，锦标赛的一部分本身就是一场锦标赛(只是少了一轮)。如果这是一场比赛，我们会返回获胜者。如果它是一个小节/锦标赛，我们递归找到它的两个 child 的赢家。

def rps_tournament_winner(tournament)
    # Check if we're at a game
    if tournament[0][0].is_a? String
        return rps_game_winner(tournament)
    end
    # Otherwise keep going down the rabbit hole
    return rps_game_winner([rps_tournament_winner(tournament[0]),rps_tournament_winner(tournament[1])])
end

这是否有助于您理解？