Proof question-Algorithm introduction 8.3 proves stingy sat problem

Stingy sat question: given a set of clauses (where each clause is a disjunction of the text) and an integer k, an assignment with a maximum of k variables of true is satisfied-if the assignment exists. Proving stingy is a complete problem for np-.

Thinking of solving problems
1. What is sat problem
Suppose we have such a set of clauses:
(a⋃b⋃c) ⋂ (A⋃b¯) ⋂ (B⋃c¯) (a¯⋂c) ⋂ (A¯⋃b¯⋃c¯)
All we need to do is find the value of A,b,c (TRUE or false) so that the result of the expression is true, assuming that A,b,c is true, then the value of the entire expression is false, it is not difficult to find, The expression does not have an appropriate ABC value so that the whole equation is true, and for this problem we need to search all combinations of values to determine whether a combination can be determined to make the equation true, and if it does not exist we return it, and if it exists we return the combination, The method of searching with exhaustive is O (2n), if we can find a simple algorithm to reduce the complexity of the algorithm to polynomial time, then we think this problem is a P problem, but we think this problem is a NP problem for the time being, because the NP problem can not reduce its complexity, The variant is a polynomial solution. 、

2. The understanding of the stingy sat question
Stingy sat problem is actually a variant of the SAT problem, such as still a problem, and now a variable, this variable is k, I
What we need to make sure is that we can't be more than K-a-true variables.

3. Basic Steps
To prove that the stingy SAT is np-complete problem, first prove that the stingy SAT is NP-problem, and then prove that the SAT is np-complete problem, and then the SAT into a stingy sat.

4. Certification Process
(1) Pre-condition
We assume that (F,K) is an example of a stingy sat, where X is a set of assignments, that is, F is an instance of the SAT (including K variables), because X can make (f,k) a true solution can be validated in polynomial time, so stingy sat is NP problem
(2) Target
The SAT protocol to the stingy sat, that is, the certificate:
X is the solution of F and only if X is (f,k)
(Sat) (stingy Sat)
(3) Proof of adequacy
If x is the solution of F, then at most k variables are true, X assigns (F,K) is also true, so X is the solution of (F,K)
(4) Proof of necessity
Assuming X is the solution of (f,k), it is clear that X is also the solution of F
(5) Conclusion

Stingy sat is also a np-complete problem