纸牌游戏（在主要组中制作随机4组数字） [英] playing cards game (make Random 4 groups of Numbers in Main Group)

问题描述

我想制作纸牌游戏。我给每张卡一个1到52的数字

现在我需要给4个玩家的牌（每个玩家有13张牌）。那么我如何制作随机4组（每组包含13个数字）和数字从1-52并且没有组具有相同的数字？

Hi ,
I want to make playing cards game . I give every card a number from 1 to 52
now I need give cards to 4 players (every player has 13 cards) . so how I can make random 4 groups (every groups contain with 13 number ) and number from 1-52 and no group has the same number ?

推荐答案

最简单的方法是模拟一副真正的卡片：

将你的套牌作为整数列表：

The easiest way is to emulate a real deck of cards:
Take your "deck" as a list of integers:

private List<int> deck = new List<int>();
private const int cardsCount = 52;
private Random rand = new Random();

然后用所有数字填写：

Then fill it with all the numbers:

deck.AddRange(Enumerable.Range(1, cardsCount));

然后写一个shuffle方法：

Then write a shuffle method:

List<int> shuffled = new List<int>();
for (int i = 0; i < cardsCount; i++)
    {
    int index = rand.Next(deck.Count);
    shuffled.Add(deck[index]);
    deck.RemoveAt(index);
    }
deck = shuffled;

然后你可以从顶部处理卡片并获得保证随机和唯一的价值。



（我可能会创建一个Card类，以便将数字翻译成套装并且值更好，但这取决于你）

You can then deal the "cards" off the top and get guaranteed random and unique values.

(I'd probably create a Card class so that the number translated to a suit and value better, but that's up to you)

诀窍在于：从甲板上随机抽取5张卡片 [ ^ ]（ C ++ 代码，对不起）。



在你的情况下，它应该是这样的：



初始化

• 以这种方式分配数组： c（0）= 1，c（1） = 2，..，c（51）= 52 。

The trick is here: "Random extraction of 5 cards from a deck"[^] (C++ code, sorry).

In your case, it should work this:

Initalization

• Assign an array this way: c(0)=1, c(1)=2, .., c(51)=52.

• 在 0 和 51之间选择一个随机数（包含）说 r 。
• 使用 c（51）交换 c（r）。

• pick a random number between 0 and 51 (included) say r.
• swap c(r) with c(51)..

• 在 0 和之间选择一个随机数（ r ） 50 （已包含）。
• 使用 c（r） > c（50）。

• pick a random number (r) between 0 and 50 (included).
• swap c(r) with c(50).

它可能与游戏无关，但是OriginalGriff的解决方案1正在做很多额外的工作（在每次迭代时将列表内容移动到被删除元素上方）。使用数组而不是列表 Fisher-Yates shuffle [ ^ ]可以使用由内到外在O（n）中完成version。

It probably is irrelevant for a game, but OriginalGriff's Solution 1 is doing a lot of extra work (moving the list contents above the removed element at each iteration). Using an array instead of a list the Fisher-Yates shuffle[^] can be done in O(n) using the "inside-out" version.

private const int cardsCount = 52;
private Random rand = new Random();
private static int[] deck = Enumerable.Range(1, cardsCount).ToArray();

private static int[] Shuffle(int[] deck)
{
  int[] shuffled = new int[deck.Length];
  shuffled[0] = deck[0];
  for (int i = 1; i < deck.Length; i++)
  {
    int j = rand.Next(i + 1);
    if (j != i)
      shuffled[i] = shuffled[j];
    shuffled[j] = deck[i];
  }
  return shuffled;
}

使用类似：

Use like:

int[] shuffled = Shuffle(deck);

=============

编辑：2016-14-6 MTH

这有点旧，但值得注意的是：

如果避免产生shuffles中的偏差很重要，请特别注意Fisher-Yates shuffle：潜在的偏见来源。

（例如，为了有可能产生52卡片组的所有可能的改组，伪随机数发生器必须至少有226位内部状态！）

=============
2016-14-6 MTH
This is a bit old, but it's worth noting:
If avoiding bias in the resultant "shuffles" is important, pay particular attention to the Fisher-Yates shuffle: Potential sources of bias in the Wikipedia article.
(For example, to have the possibility of producing all possible shufflings of a 52 card deck the pseudo random number generator must have at least 226 bits of internal state!)

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