Algorithms Q&A - Recent questions and answers in NP-Completeness

Answered: Long chain of friends

Wed, 13 Dec 2023 05:23:48 +0000

A chain of friends that is of length n is essentially a Hamiltonian Path.

Therefore, the “long chain of friends” problem is a generalization of the Hamiltonian Path problem.

The reduction diagram can be drawn from the Hamiltonian Path to a Long chain of friends problem by instantiating the value of k = n.

Answered: Prove that Sub Graph Isomorphism is Hard

Wed, 13 Dec 2023 01:27:58 +0000

To prove the decision version of whether a graph has a subgraph we will show a reduction from Clique to this Subgraph problem, I will call this problem SbG.

Hence for a NP-complete reduction we would first have to prove that SbG is in NP. We can verify in polynomial time that H is indeed a subgraph in G, this is trivial as if we have the <K,E`> we can check whether Graph G indeed has those vertices and edge pairs by traversing those specific edges and vertices in G.

Given the <K,E`> in graph H, and <V,E> in graph G. Let us assume H is complete graph with K vertices if and only if G has a Clique sized K. We can see that since H is a complete graph on K vertices a Clique with size K in the overall graph G would be this complete graph with exactly K vertices. Hence SbG is at least as hard as the Clique problem, Clique <_p SbG

*I can not add images otherwise I would draw the box as well

Answered: k-degree constrained spanning tree

Tue, 12 Dec 2023 21:51:45 +0000

If in a tree, the maximum degree is 2 , then the tree is simply a path. This allows us to observe that the problem of finding spanning tree where no vertex has degree more than 2 is equivalent to finding a Hamiltonian path.

Therefore, the k -degree constrained spanning tree problem is simply a generalization of the Hamiltonian Path problem. Specifically, Hamiltonian Path problem can be directly reduced to the k -degree constrained spanning tree problem by leaving the original graph unchanged and simply using the value of k = 2 .

Answered: Snow White and 7 Conference Rooms

Tue, 20 Dec 2022 03:54:07 +0000

Set this question as F.

Firstly, there are n people here and we need to divide the n people into 7 rooms. If there is a solution and we need to check whether it qualifies. There is a maximum of n people per room, we just need to check if they know anyone. If we use the adjacency matrix to store the relationships between people, it takes only O(n^2) time complexity to traverse through. Therefore, question F belongs to NP.

Secondly, Coloring question is known as a NP-complete problem.

Thirdly, there is an example show that we can reduce Coloring to F.

F: We set each person as a node and connect them if they know each other. In this situation, the relationships of all people can be stored in a graph.

In question Coloring, any two nodes of the same color cannot be adjacent to each other.

In F, any two people who know each other can be in a same room.

We create a graph like the relationships of F, and we tried to use 7 colors to paint the whole graph. If there is a solution the F can be solved, else it is not possible.

Above all, F reduces to Coloring question in polynomial time.

Answered: What's the time complexity for solving Sudoku with backtrack method?

Thu, 02 May 2019 05:52:44 +0000

Can be optimized, the goal is to choose the location that is most prioritized, and choose the number that can retain the maximum success rate.

Answered: Prove that 3-Coloring is NP Hard (starting with SAT as known NP hard problem)

Sun, 03 Dec 2017 01:51:20 +0000

Step 1: Prove 3-Coloring is in NP
We need to check whether each edge in graph G(V,E) connects two vertexes with two different colors. If we represent the graph using Adjacency Matrix A[1..n, 1..n] where A[i,j]=1 if (i,j) is an edge. Then we can check all of the edges in O(n), which is absolutely a polynomial time complexity. Thus, 3-Coloring is in NP.

Step 2: Reduce SAT as known NP hard problem to 3-Coloring

I. Since there are 3 colors, saying a, b and c. There are only 3 situations for each edge to have different colors, the two vertexes an edge can be (a,b)(a,c)(b,c).

II. If we can prove there exists assignments of two vertexes x i,y i for edge i (i:1..n)which make the boolean formula true (x 1 or y 1) and (x 2 or y 2) and...(x n or y n) be true. ( according to (2) I, we can know that (a or b) , (a or c), (b or c) are the only three true situations )

By now, we reduce 3-Coloring to SAT problem, which is a known NP hard problem. So we can get the conclusion that 3-Coloring is NP-Hard.

Answered: Cook-Levin Theorem

Tue, 06 Dec 2016 18:03:55 +0000

The construction of the CNF formula is the essence of the Cook Levin theorem. The theorem itself is quite involved, but in the past people have presented that as their P4 project. You can do a quick google search for that and find it.

The fact that the formula is polynomial in terms of f(n), is the key aspect of that. The theorem shows that the solving a given problem instance using its NTM is equivalent to the formula being satisfiable. Thus, CSAT is NP-hard.

Answered: Prove that Vertex Cover is NP-hard problem

Wed, 23 Nov 2016 03:10:15 +0000

Because in a graph G, if we are given a vertex cover, we can find the remaining set of vertices to be a set of independent set. Thus we can do the following reduction in polynomial time:

That means, if we want to find the minimum VC, we can do so by finding out the maximum IS.

As IS is proven to be NP-hard, VC is NP-hard too.

The full chain starting SAT can be written as:

SAT <=p CLIQUE <=p Independent Set <=p Vertex Cover