STINGY SAT is the following problem: given a set of clauses (each a disjunction of literals) and an integer K , find a satisfying assignment in which at most K variables are true, if such an assignment exists. Prove that isNP -complete.

Short Answer

Expert verified

STINGY SAT is the required assignment, it has been proved that STINGY SAT is NP complete.

Step by step solution

01

Statement of  is -complete

There are steps to show the given hassle as NP (Non-deterministic polynomial) Completeness.

Take a look at whether the STINGY SAT is in “NP”

To test whether the STINGY SAT is in “NP”, first test whether or not the answer incorporates an enjoyable challenge via evaluating the system.

Moreover, check that fewer than “okay” literals are assigned with “actual” cost by means of examining the literals as soon as.

Reduce a recognized NP-whole hassle to STINGY SAT

To prove the NP-completeness, allow us to reduce SAT to STINGY SAT.

It's far supplied that in a SAT method if there is “f” with “k” variables, then it virtually chooses the for instance of STINGY SAT.

Now, its miles had to show that: “f” is a “sure-example” of SAT if and best if is a “yes-example” of STINGY SAT.

02

Prove

Proof:

Anticipate that “f” is a “sure-example” of SAT. There are absolutely “k” variables. Consequently, no more than “k” variables are “genuine”.

Therefore, any gratifying venture of instance “f” for SAT will also be the enjoyable venture of example for STINGY SAT.

Subsequently, is a “yes-instance” of STINGY SAT.

Count on that is a “sure-example” of STINGY SAT, then any fulfilling assignment of that example will also be gratifying undertaking of instance “f” for SAT.

Therefore, “f” is a “yes-instance” of SAT.

Because SAT is NP-Complete, the STINGY SAT is NP-Complete.

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Most popular questions from this chapter

Optimization versus search.Recall the traveling salesman problem:

TSP

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Output: A tour which passes through all the cities and has lengthb, if such a tour exists.

The optimization version of this problem asks directly for the shortest tour.

TSP-OPT

Input:A matrix of distances

Output:The shortest tour which passes through all the cities.

Show that if TSP can be solved in polynomial time, then so can TSP-OPT.

Search versus decision. Suppose you have a procedure which runs in polynomial time and tells you whether or not a graph has a Rudrata path. Show that you can use it to develop a polynomial-time algorithm for RUDRATA PATH (which returns the actual path, if it exists).

Give a simple reduction from 3D MATCHING to SAT, and another from RUDRATA CYCLE to SAT.

(Hint: In the latter case you may use variables xijwhose intuitive meaning is “vertex i is the j th vertex of the Hamilton cycle”; you then need to write clauses that express the constraints of the problem.)

Prove that the following problem is NP-complete: given an undirected graph

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In task scheduling, it is common to use a graph representation with a node for each task and a directed edge from task i to j task if i is a precondition for j. This directed graph depicts the precedence constraints in the scheduling problem. Clearly, a schedule is possibe if and only if the graph is acyclic; if it isn’t, we’d like to identify the smallest number of constraints that must be dropped so as to make it acyclic.

Given a directed graph G=(V,E), a subset E'Eis called a feedback arc set if the removal of edges E' renders G acyclic.

FEEDBACK ARC SET (FAS): Given a directed graph G=(V,E)and a budget , find a feedback arc set of role="math" localid="1658907144825" bedges, if one exists.

(a)Show that FAS is in NP.

FAS can be shown to be NP-complete by a reduction from VERTEX COVER. Given an instance (G,b)of VERTEX COVER, where G is an undirected graph and we want a vertex cover of size b, we construct a instance (G',b)of FAS as follows. If G=(V,E)has vertices v1,K,vnthen make G'=(V',E')a directed graph with 2n verticesw1,w1',k,wn,wn',andn+2|E|(directed) edges:

  • (wi,wi')foralli=1,2,k,n
  • (wi',wj)and(wj',wi)forevery(vi,vj)E.
  • Show that if G contains a vertex cover of size b, then G' contains a feedback arc set of size b .
  • Show that if G' contains a feedback arc set of size b, then G contains a vertex cover of size (at most) b. (Hint: Given a feedback arc set of size b in G', you may need to first modify it slightly to obtain another one which is of a more convenient form, but is of the same size or smaller. Then, argue that G must contain a vertex cover of the same size as the modified feedback arc set.)
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