问题
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.
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".
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.
Any help is highly appreciated.
回答1:
This could be done in three simple steps:
- 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
- Apply XOR operation to this new BDDs
- Q = R[x=0] xor R[x=1]
- Enumerate all models of Q
Applying this to your examples:
- R = {<0,0,1>, <1,0,1>, <0,0,0>} = (¬x ∧ ¬y ∧ z) ∨ (x ∧ ¬y ∧ z) ∨ (¬x ∧ ¬y ∧ ¬z)
- R[x=1] = {<0,1>} = (¬y ∧ z)
- R[x=0] = {<0,1>,<0,0>} = (¬y ∧ z) ∨ (¬y ∧ ¬z)
- Q = R[x=1] xor R[x=0] = (¬y ∧ ¬z)
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.
来源：https://stackoverflow.com/questions/14313049/find-unique-tuples-in-a-relation-represented-by-a-bdd