完美最小散列数学组合 [英] Perfect minimal hash for mathematical combinations

问题描述

首先，定义两个整数 N 和 K ，其中 N'GT; = K ，无论是在编译时已知的。例如： N = 8 和 K = 3

First, define two integers N and K, where N >= K, both known at compile time. For example: N = 8 and K = 3.

接下来，定义一组整数 [0，N）（或 [1，N] 如果说的使答案更简单），并将其命名为取值。例如： {0，1，2，3，4，5，6，7}

Next, define a set of integers [0, N) (or [1, N] if that makes the answer simpler) and call it S. For example: {0, 1, 2, 3, 4, 5, 6, 7}

取值的子集与 K 元素的数量由下式给出Ç （N，K）。示例

The number of subsets of S with K elements is given by the formula C(N, K). Example

我的问题是这样的：创建一个完美的哈希值最小为这些子集。这个例子哈希表的大小将是 C（8,3）或 56 。

My problem is this: Create a perfect minimal hash for those subsets. The size of the example hash table will be C(8, 3) or 56.

我不关心排序，只是有是哈希表中的56项，而且我可以从一组 K 整数迅速确定哈希。我也不在乎可逆性。

I don't care about ordering, only that there be 56 entries in the hash table, and that I can determine the hash quickly from a set of K integers. I also don't care about reversibility.

例如哈希：哈希（{5，2，3}）= 42 。 （数目42是不重要的，至少不是这里）

Example hash: hash({5, 2, 3}) = 42. (The number 42 isn't important, at least not here)

有没有一个通用的算法，这种将工作与 N 和 K 的任何值？我没能找到一个通过搜索谷歌，还是我自己天真的努力。

Is there a generic algorithm for this that will work with any values of N and K? I wasn't able to find one by searching Google, or my own naive efforts.

推荐答案

有一个算法，以code和德codeA结合到其在给定的固定<$ C所有组合的字典顺序号$ C> K 。该算法是线性的，以 N 为code和组合去code。什么语言你有兴趣吗？

There is an algorithm to code and decode a combination into its number in the lexicographical order of all combinations with a given fixed K. The algorithm is linear to N for both code and decode of the combination. What language are you interested in?

编辑：我这里是code C ++中（它开创的组合的序列中的辞书号所有，而不是那些有n个元素的组合 K 元素，但确实是很好的起点）：

here is example code in c++(it founds the lexicographical number of a combination in the sequence of all combinations of n elements as opposed to the ones with k elements but is really good starting point):

typedef long long ll;

// Returns the number in the lexicographical order of all combinations of n numbers
// of the provided combination. 
ll code(vector<int> a,int n)
{
    sort(a.begin(),a.end());
    int cur = 0;
    int m = a.size();

    ll res =0;
    for(int i=0;i<a.size();i++)
    {
        if(a[i] == cur+1)
        {
            res++;
            cur = a[i];
            continue;
        }
        else
        {
            res++;
            int number_of_greater_nums = n - a[i];
            for(int j = a[i]-1,increment=1;j>cur;j--,increment++)
                res += 1LL << (number_of_greater_nums+increment);
            cur = a[i];
        }
    }
    return res;
}
// Takes the lexicographical code of a combination of n numbers and returns the 
// combination
vector<int> decode(ll kod, int n)
{
    vector<int> res;
    int cur = 0;

    int left = n; // Out of how many numbers are we left to choose.
    while(kod)
    {
        ll all = 1LL << left;// how many are the total combinations
        for(int i=n;i>=0;i--)
        {
            if(all - (1LL << (n-i+1)) +1 <= kod)
            {
                res.push_back(i);
                left = n-i;
                kod -= all - (1LL << (n-i+1)) +1;
                break;
            }
        }
    }
    return res;
}

我很抱歉，我有一个算法，你所要求的，现在的问题，但我相信这将是一个很好的锻炼，试图为明白我做什么上面。事实是这是我教的课程算法设计与分析的算法之一，这就是为什么我是有pre-写的。

I am sorry I have an algorithm for the problem you are asking for right now, but I believe it will be a good exercise to try to understand what I do above. Truth is this is one of the algorithms I teach in the course "Design and analysis of algorithms" and that is why I had it pre-written.

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