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A spider is a graph with an even number of vertices

Tutors ProblemsPosted On:2024-07-07 15:43:45Viewed:28

A spider is a graph with an even number of vertices, say 2g, such that g vertices form a clique, and each of the other g vertices is adjacent to exactly one distinct vertex in the clique. The picture shows spiders with 6 and 8 vertices. The spiderVerse problem takes as input a graph G=(V,E) and a natural number g >= 1 and outputs a set of 2g vertices such that the induced subgraph is a spider, or outputs NO if such set of vertices does not exist. Show that the SpiderVerse is NP-complete

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Last updated on:2024-07-07 15:43:45

Since SpiderVerse is both in NP and NP-hard, it is NP-complete. This completes the proof that the SpiderVerse Answer is NP-complete.

Step-by-step explanation

To show that the SpiderVerse Answer is NP-complete, we need to demonstrate two things:

 

1. SpiderVerse is in NP: This means that given a potential solution (a set of vertices), we can verify in polynomial time whether this set forms a valid spider as defined.

2. SpiderVerse is NP-hard: This requires showing that any Answer in NP can be reduced to SpiderVerse in polynomial time.

 

# 1. SpiderVerse is in NP

 

To verify if a set of vertices forms a spider:

- Check if the induced subgraph on these vertices is connected.

- Count the vertices: ensure there are exactly 2g  vertices.

- Check the structure: verify that g  vertices form a clique, and each of the other g  vertices is adjacent to exactly one vertex in the clique.

 

Each of these verifications can be done in polynomial time relative to the size of the input graph G . Thus, SpiderVerse is in NP.

 

# 2. SpiderVerse is NP-hard

To prove NP-hardness, we will reduce a known NP-complete Answer to SpiderVerse. Let's reduce the Vertex Cover Answer to SpiderVerse.

 

 Reduction from Vertex Cover to SpiderVerse

Vertex Cover Answer: Given a graph G=(V,E)  and a number k , determine if there exists a set SV  such that Sk and every edge in E  is incident to at least one vertex in S .

 

Reduction Strategy:

- Given an instance of the Vertex Cover Answer(G,k) :

  - Construct a spider graph G  as follows:

    - Create a clique Kg  with g=Vk  vertices.

    - Add k  additional vertices, each connected to exactly one vertex in Kg .

 

Construction Details:

- Let G=(V,E)  be the input graph for Vertex Cover.

- Compute g=Vk .

- Construct G :

  - G  will have 2g  vertices:

    - Kg  will be the clique with g vertices.

    - Add k  vertices, each connected to one distinct vertex in Kg .

 

Verification:

- If there exists a Vertex Cover of size k  in G , then select the corresponding k  vertices from G  that are connected to  Kg . This set will form a valid spider in G .

- If there exists a set of 2g  vertices in G  forming a spider, then the g  vertices forming the clique Kg  and the k  vertices adjacent to Kg  form a Vertex Cover of size k  in G .

 

Since Vertex Cover is known to be NP-complete, and we have shown a polynomial-time reduction from Vertex Cover to SpiderVerse, SpiderVerse is also NP-hard.

To show that the SpiderVerse problem is NP-complete, we need to demonstrate two things:

  1. SpiderVerse is in NP: We need to show that given a solution (a set of 2g vertices that form a spider), we can verify it in polynomial time.

  2. SpiderVerse is NP-hard: We need to show that any problem in NP can be reduced to SpiderVerse in polynomial time.

Step 1: SpiderVerse is in NP

Given a set of 2g vertices, we can verify whether they form a spider in polynomial time by performing the following checks:

  1. Clique Check: Verify that g of these vertices form a clique. This involves checking that there is an edge between every pair of these g vertices. This can be done in (O(g^2)) time.

  2. Adjacency Check: Verify that each of the remaining g vertices is adjacent to exactly one distinct vertex in the clique. This involves checking the adjacency of each of these g vertices to the g vertices in the clique, which can be done in (O(g^2)) time.

Since both checks can be done in polynomial time, verifying a solution is polynomial, and thus SpiderVerse is in NP.

Step 2: SpiderVerse is NP-hard

To show that SpiderVerse is NP-hard, we can reduce a known NP-complete problem to SpiderVerse. We will use the Clique problem, which is known to be NP-complete.

Clique Problem: Given a graph (G = (V, E)) and a number (k), determine if there is a clique of size (k) in (G).

Reduction from Clique to SpiderVerse:

  1. Input Transformation: Given an instance of the Clique problem, (G = (V, E)) and (k), we construct an instance of the SpiderVerse problem as follows:

    • Set (g = k).

    • The graph (G' = (V', E')) is the same as (G).

  2. Output Transformation: We need to show that (G) has a clique of size (k) if and only if (G') has a spider with (2g = 2k) vertices.

    • If (G) has a clique of size (k): Let (C) be the set of (k) vertices that form a


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