最小的检查，以找到一个列表重复 [英] minimal checks to find repeats in a list

问题描述

给定一个序列（D1，D2，...，DN），我要评估产品，谢谢！（1 - DIJ）对于所有的i = j的，其中DIJ = 1如果DI = DJ，否则为0。

在code，我只有检查的时候DIJ我

  PROD = 1;
的for（int i = 1;我n种; ++ I）{
    对于（INT J =; J＆LT; = N; ++ j）条{
        督促* =（1  -  DIJ）;
    }
}
 

我知道当我得到DIJ = 1我管不了，但我想要做的就是的DIJ的最小离pression检查。这样，我有一个EX pression，然后我就可以使用不同的序列，并对其进行评估。所以，我知道我能做到 I＆LT;！第j ，而不是 I = j的。所以，我想扩大了该产品，并得到这样的事情对于n = 3：

 （1  -  D12）（1  -  D13）（1  -  D23）= 1  -  D12  -  D13  -  D23 + D12 * D13 + D12 * D23 + D13 * D23  -  D12 * D13 * D23
 

但是，还有更多，我可以做。这当然pression实际上总是等于

  1  -  D12  -  D13  -  D23 + 3 * D12 * D13  -  D12 * D13 * D23
 

我的问题是：

1. 为什么D12 * D13 = D12 * D23？这始终是真实的（意思是无所谓对D顺序是什么），但我真的不知道为什么，因为在我看来，这意味着D13 = D23这并不总是正确的（这取决于在D序列） 。这是有助于使EX pression小的关系。

2. 我如何才能找到像这样的所有关系，并得到一个最小的前pression？是前pression上面最小？我甚至不知道。

解决方案

您正试图确定是否D包含任何重复或没有。最后，需要你每个条目与对方，这仅仅是列举两种元素的所有独特的组合进行比较。这最终被 N *（N-1）/ 2 。你可以做的更好，首先分拣D和再寻找重复的相邻的一对（一点点 O（N *日志（N）），或者，假设你坚持有界整数的范围内，就可以将其降低到线性时间有位向量，或者如果你喜欢冒险，一个基数排序。

Given a sequence (d1, d2, ..., dn), I want to evaluate the product (1 - Dij) for all i != j where Dij = 1 if di = dj and 0 otherwise.

The code I have only checks Dij when i

prod = 1;
for (int i=1; i<n; ++i) {
    for (int j=i; j<=n; ++j) {
        prod *= (1 - Dij);
    }
}

I know I can stop when I get Dij=1, but what I'm trying to do is get a minimal expression of the Dij's to check. This way I have one expression and then I can use difference sequences and evaluate it. So I know that I can do i<j instead of i != j. So I want to expand out this product and get something like this for n=3:

(1 - D12) (1 - D13) (1 - D23) = 1 - D12 - D13 - D23 + D12*D13 + D12*D23 + D13*D23 - D12*D13*D23

But there is more that I can do. This expression is actually always equal to

1 - D12 - D13 - D23 + 3 * D12*D13 - D12*D13*D23

My questions are:

1. Why is D12 * D13 = D12 * D23? This is always true (meaning it doesn't matter what the d sequence is) but I don't really get why because it seems to me that this means D13 = D23 which isn't always true (it depends on the d sequence). This is the relation that helps make the expression smaller.

2. How can I find all the relations like this and get a minimal expression? Is the expression above minimal? I don't even know.

解决方案

You are trying to determine if D contains any duplicates or not. Ultimately, that requires you to compare every entry with each other, which is simply enumerating all unique combinations of two elements. That ends up being N*(N-1)/2. You can do a little better by sorting D first and then searching for duplicate adjacent pairs (O(N*log(N)), or, assuming you are sticking to a bounded range of integers, you can reduce it to linear time with a bit vector, or if you are feeling adventurous, a radix sort.

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