Np complexity pdf

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Neralla Manikanta. Pablo Recalde. B Claim 33 Assumption. I Remark. Lemma 31 becomes void due to Claim 33, which is addressed by Theorem The following tools are introduced for Theorem I Lemma Successive reductions due to ri0, ri1,.

Lemmas 26 and Follows from Lemma J I Lemma 36 Any conditional scope is derivable from its scope. Follows from Lemmas 34 and It is a generalization of Lemma I Theorem 39 Satisfiability. Remark 32 and Claim See also Lemmas 34— The following steps specify this assignment construction process. Thus, the reductions see Lemma 23 due to ri0 partition L into L ri0 and L0 ri0. Then, the reductions due to ri1 partition L ri0 into L ri1 ri0 and L0 ri1 ri0.

Then, ri2 partitions L0 ri1 ri0 into L ri2 ri1 and L0 ri2 ri1 , i. Salum 9 n. Note that Vn ri0, ri1,. Thus, the interruption-termination duality provides an efficiently characterizable structure see [2] pg. I Proposition Thus, Reduce runs in time O m , or in time O mn see L Hence, Scope runs in time O mn , or in time O mn2.

Note that the number of the scans is at most n. I Definition 42 Quantified Boolean Formula. I Note. I Note W I Note J W m I Remark. Note also that a prime clause Ck denotes a formula in Disjunctive Normal Form. Salum 11 I Note A proof sketch is as follows. This result denotes a duality theorem. References 1 Thomas J.

The complexity of satisfiability problems. Princeton University Press, The Graph Isomorphism problem is tackled via an example see Figure 3. Thus, the following properties hold. Related Papers. Here, by using the relationship between incidence coloring of graphs and independent dominating sets, we improve the previous complexity results and show the following theorem.

We say that a set of vertices are independent if there is no edge between these vertices. A dominating set of a graph G is a subset D of V G such that every vertex not in D is joined to at least one vertex of D. Proof of Theorem 1. We prove the two parts of the theorem together. Our proof consists of five steps. Step 1. Step 2. Call them old variables. For each old variable x, consider the new variables x0 , x1.

Without loss of generality suppose that the old variable x appears negated in c0 , c1 ,. Step 3. Consider any union of k of the sets Ai. Since each Ai contains at least 2 distinct elements and no literal is contained in more than 2 sets, the union contains at least k distinct elements.

For each clause c denote its representative literal by z c. Note that there is a polynomial-time algorithm which finds an SDR, when ever it exists. Step 4. Join the vertex c to one of the vertices x1 or x2. Next, for every clause c join the vertex c to the vertices z3c , z4c ,. Call the resulting graph G.

Step 5. Consider two copies of G. In the following we introduce the members of maximal independent set T. Members of T. Therefore, if D contains a vertex from X , then it does not have any vertex from Y and vice versa. It is easy to extend this set into an independent dominating set for T.

This completes the proof. Proof of Theorem 2. It is easy to see that theses sets are disjoint independent dominating sets for H and a partition for the vertices of H. Now, assume that G is not 3-colorable.

To the contrary suppose that T1 , T2 ,. But this is a contradiction. So the vertices of a given 3-regular graph G can be partitioned into l-perfect codes if and only if the vertices of G can be assigned 4 different colors in such a way that closed neighborhood of each vertex is assigned all 4 colors, i.

Let G be a 3-regular graph. Then the vertices of G can be partitioned into l-perfect codes if and only if the vertices of H can be partitioned into independent dominating sets. Proof of Theorem 3. Also, Tovey in [23] showed that instances of 3-SAT in which every variable occurs three times are always satisfiable.

Regarding this result, solving the following question can be interesting. We proved that determine whether the vertices of a given 3-regular graph can be partitioned into a number of independent dominating sets is NP-complete. However, one further step does not seem trivial. The complexity of that problem for the family of planar 3-regular graphs can be interesting. First, we introduce the construction of Jk. Lemma 1. Let G be a k-regular graph. Consider the bipartite graph G 2 G 2 is obtained from G by replacing each edge with a path with exactly one inner vertex.

References [1] A. Ahadi, S. Akbari, A.