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This question already has an answer here:

The following variation on the vertex cover problem was given:

Given is an instance of graph $G = (V, E)$. Does $G$ have a vertex cover of size at most $\frac{|V|}{4}$?

I was asked to prove that this problem is also $\mathsf{NP}$-hard. I am pretty sure that this can be done by using Karp-reduction from the original vertex cover problem to this variation.

That means that I would need some transformation function $f$ that would transform an instance of the vertex cover problem, $I = (V, E, k)$, to an instance of the above described problem, $f(I) = (V', E')$.


I am not sure how to come up with the proper transformation function, but this is what I think have figured out:

  • Let $V = \{v_1, v_2, ..., v_n\}$.
  • Let $V' = V \cup V^+$, where $V^+ = \{v_{n+1}, v_{n+2}, ..., v_{4n}\}$, because this would mean that the total size $|V'| = 4 \cdot |V|$.
  • Let $E' = E \cup E^+$, where $E^+ = \{\{v_i, v_j\} \mid v_i \in V, v_j \in V^+, j \in \{n +i, 2n+i, 3n+i\}\}$. Basically, this means that each vertex in $V$ is connected to three different vertices in $V^+$ such that all vertices in $V^+$ are connected to exactly one vertex in $V$.

However, I am having a hard time proving the correctness of this reduction. I think I messed up the generation of $E'$, which would make the transformation function wrong. Anyone have an idea what the actual transformation function could be?

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marked as duplicate by xskxzr, David Richerby complexity-theory Feb 28 at 11:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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One note is that you'll have to use $k$ in your transformation.


Here's one approach that I think works:

We want some $G' = (V', E')$ that has a vertex cover of size $\le \frac{|V'|}{4}$ iff $G$ has a vertex cover of size $\le k$.

If $k = \frac{|V|}{4}$, just set $G' = G$.

When $k \neq \frac{|V|}{4}$, one idea is to add a disconnected graph $H$ to $G$, where we know the vertex cover size of $H$.

Case 1: $k > \frac{|V|}{4}$. Then we want to add a graph with a small vertex cover. A star graph should do it. A star graph $H$ with $m$ vertices has vertex cover size $1$.

Set $G' = G \cup H$, a graph with $|V| + m$ vertices. It has vertex cover of size $\le k+1$ iff $G$ has a vertex cover of size $\le k$.

What should $m$ be? Let's solve

$$\frac{k + 1}{|V| + m} = \frac{1}{4}$$

This gives $m = 4(k+1)-|V|$. So when $k > \frac{|V|}{4}$, add a disconnected $[4(k+1)-|V|]$-vertex star graph to $G$, and call the $\frac{|V|}{4}$ vertex cover algorithm on that. $\square$

Case 2: $k < \frac{|V|}{4}$, we can add a complete graph with some properly chosen number $m$ vertices*. Will leave it at that.

*In this case when you solve for $m$, $m$ may not always be an integer; this can be dealt with by adding a $\lceil m \rceil$-vertex complete graph, which "overshoots" a bit, and then adding a small star graph

[Let me know for further clarification/etc]

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