database 如何获取数据库中一组唯一对的总数？
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How do I get the total number of unique pairs of a set in the database?
提问by Kirk Ouimet
4 items:
4 项：
A
B
C
D
6 unique pairs possible:
可能有 6 对独特的配对：
AB
AC
AD
BC
BD
CD
What if I have 100 starting items? How many unique pairs are there? Is there a formula I can throw this into?
如果我有 100 个起始物品怎么办？有多少个独特的对？有没有我可以把它放进去的公式？
回答by Mike Christensen
What you're looking for is n choose k. Basically:
您正在寻找的是n 选择 k。基本上：
For every pair of 100 items, you'd have 4,950 combinations- provided order doesn't matter (AB and BA are considered a single combination) and you don't want to repeat (AA is not a valid pair).
对于每对 100 件商品，您将有 4,950 种组合- 前提是顺序无关紧要（AB 和 BA 被视为单一组合）并且您不想重复（AA 不是有效对）。
回答by audiomason
TLDR; The formula is n(n-1)/2where nis the number of items in the set.
TLDR；公式是n(n-1)/2哪里n是集合中的项目数。
Explanation:
解释：
To find the number of unique pairs in a set, where the pairs are subject to the commutative property(AB = BA), you can calculate the summationof 1 + 2 + ... + (n-1)where nis the number of items in the set.
要查找一组，其中对受独特的对数交换律(AB = BA)，就可以计算出总和的1 + 2 + ... + (n-1)地方n是在设定项目的数量。
The reasoning is as follows, say you have 4 items:
推理如下，假设您有 4 个项目：
A
B
C
D
The number of items that can be paired with Ais 3, or n-1:
可配对的项目数A为 3，或n-1：
AB
AC
AD
It follows that the number of items that can be paired with Bis n-2(because Bhas already been paired with A):
因此可以配对的项目数B为n-2（因为B已经配对A）：
BC
BD
and so on...
等等...
(n-1) + (n-2) + ... + (n-(n-1))
which is the same as
这与
1 + 2 + ... + (n-1)
or
或者
n(n-1)/2
回答by Joni
This is how you can approach these problems in general on your own:
这是您通常可以自己解决这些问题的方法：
The first of the pair can be picked in N (=100) ways. You don't want to pick this item again, so the second of the pair can be picked in N-1 (=99) ways. In total you can pick 2 items out of N in N(N-1) (= 100*99=9900) different ways.
可以通过 N (=100) 种方式选择一对中的第一个。您不想再次选择此项目，因此可以通过 N-1 (=99) 种方式选择该对中的第二个。总的来说，您可以用 N(N-1) (= 100*99=9900) 种不同的方式从 N 中挑选 2 个项目。
But hold on, this way you count also different orderings: AB and BA are both counted. Since every pair is counted twice you have to divide N(N-1) by two (the number of ways that you can order a list of two items). The number of subsets of two that you can make with a set of N is then N(N-1)/2 (= 9900/2 = 4950).
但请稍等，这样您也计算不同的排序：AB 和 BA 都计算在内。由于每对计算两次，因此您必须将 N(N-1) 除以 2（您可以对两个项目的列表进行排序的方式数）。你可以用一组 N 组成的两个子集的数量是 N(N-1)/2 (= 9900/2 = 4950)。
回答by atomsfat
I was solving this algorithmand get stuck with the pairs part.
我正在解决这个算法并陷入对部分的困境。
This explanation help me a lot https://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/
这个解释对我很有帮助 https://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/
So to calculate the sum of series of numbers:
所以要计算一系列数字的总和：
n(n+1)/2
But you need to calculate this
但是你需要计算这个
1 + 2 + ... + (n-1)
So in order to get this you can use
所以为了得到这个，你可以使用
n(n+1)/2 - n
that is equal to
这等于
n(n-1)/2
