So to demonstrate what I'm hoping to achieve I'll use a deck of cards.
为了演示我希望达到的效果，我将使用一副牌。
Let's say there are three people, each with their own randomly shuffled deck. The cards in the deck simply have values 1 through 13, and there are four of each card.
假设有三个人，每个人都有自己的随机洗牌。牌中的牌只值1到13，每张牌有4张。
When it comes time to draw a card, everyone takes their top card and shows its value to the other two players.
当需要抽牌的时候，每个人都拿着他们的最高卡，并向另外两名球员展示它的价值。
What I want now, is some way to map the values of each of these top cards to a single integer from 1 - 13. The goal being that this algorithm would generate something unique for each operation, and only allow for 4 of the same value (when the exact same inputs are calculated each of 4 times it can happen).
我现在想要的是，从1到13，将这些*牌的值映射到一个整数。这个算法的目标是为每个操作生成一个独特的东西，并且只允许4个相同的值(当相同的输入被计算出4倍的值时)。
I know I can use Cantor Pairing Function to generate a unique value, but again I want it to be in the range 1-13.
我知道我可以使用Cantor配对函数来生成一个唯一的值，但我还是希望它在1-13的范围内。
1 个解决方案
#1
0
The Cantor pairing function is only necessary if you want to map all positive integers. What's wrong with simply the following, with n = 13 and N = n^4?
如果你想映射所有正整数，那么康托配对函数是唯一必要的。只是下面的,有什么问题与n = 13 n = n ^ 4 ?
(val1 - 1) * n^3 + (val2 - 1) * n^2 + (val3 - 1) * n + (val4 - 1) + 1
The ordering can be given by the suit of the card, or the order in which they are drawn.
订购可以由卡片的套装，或他们所绘制的顺序。
#1
0
The Cantor pairing function is only necessary if you want to map all positive integers. What's wrong with simply the following, with n = 13 and N = n^4?
如果你想映射所有正整数，那么康托配对函数是唯一必要的。只是下面的,有什么问题与n = 13 n = n ^ 4 ?
(val1 - 1) * n^3 + (val2 - 1) * n^2 + (val3 - 1) * n + (val4 - 1) + 1
The ordering can be given by the suit of the card, or the order in which they are drawn.
订购可以由卡片的套装，或他们所绘制的顺序。