找到独特的元组在重新$ P $的关系由BDD psented [英] Find unique tuples in a relation represented by a BDD

问题描述

让我们假设我用BDD重新present的元组的关系。为简单起见考虑的元组0或1的值。例如： R = {＆LT; 0,0,1>，＆LT; 1,0,1>，＆LT; 0,0,0>}重新present在BDD三元关系有三个变量，例如x，y和Z（一个用于每个元组的元素）。我要的是实现给定一个BDD像R和多维数据集C回到R，它将包含唯一的元组时，C中的变量被抽象的子集的操作。

Let's suppose I use a BDD to represent the set of tuples in a relation. For simplicity consider tuples with 0 or 1 values. For instance: R = {<0,0,1>, <1,0,1>, <0,0,0>} represent a ternary relation in a BDD with three variables, say x, y and z (one for each tuple's element). What I want is to implement an operation that given a bdd like for R and a cube C returns the subset of R that contains unique tuples when the variables in C are abstracted.

有关例如，如果C含有变量x（其中重新presents第一元件中的每个元组）的函数的结果必须是{＆所述; 0,0,0>}。请注意，当x被抽象出来的​​元组＆LT; 0,0,1>和＆lt; 1,0,1>成为相同的

For instance, if C contains the variable x (which represents the first element in each tuple) the result of the function must be {<0,0,0>}. Notice that when x is abstracted away tuples <0,0,1> and <1,0,1> become "the same".

现在假设R = {＆LT; 0,0,1>，＆LT; 1,0,1>，＆LT; 0,0,0>，＆LT; 1,0,0>}，我们要抽象x再次。在这种情况下，我会想到恒虚的结果，因为没有独特的元组在研发抽象后面的x。

Now suppose R = {<0,0,1>, <1,0,1>, <0,0,0>, <1,0,0>} and we want to abstract x again. In this case I would expect the constant false as result because there is no unique tuple in R after abstracting x.

任何帮助是非常AP preciated。

Any help is highly appreciated.

推荐答案

这可能是三个简单的步骤完成的：

This could be done in three simple steps:

• 请2 BDDS与要抽象变量的限制值：
  • 研究[X = 0] =限制R，其中X = 0
  • 研究[X = 1] =限制R，其中X = 1
  • Make two BDDs with restricted value of the variable you want to abstract:
    • R[x=0] = restrict R with x = 0
    • R[x=1] = restrict R with x = 1
    • 问：= R [X = 0] XOR - [R [X = 1]

    应用到您的例子：

    • 在R = {＆LT; 0,0,1>，＆LT; 1,0,1>，＆LT; 0,0,0>} =（¬x∧¬y∧Z）∨（X∧¬y∧ Z）∨（¬x∧¬y∧¬z）
    • 研究[X = 1] = {＆LT; 0,1>} =（¬y∧Z）
    • 研究[X = 0] = {＆LT; 0,1>，＆LT; 0,0>} =（¬y∧Z）∨（¬y∧¬z）
    • 问：= R [X = 1] XOR - [R [X = 0] =（¬y∧¬z）

    直觉这里要说的是异或将取消发生在两个BDDS条目。
    这是很容易的（但与指数的复杂性）推广到几个抽象的变量的情况。

    Intuition here is that XOR will cancel entries that occur in both BDDs.
    This is easily (but with exponential complexity) generalized to the case with several abstracted variables.

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