We know that the INDEPENDENT-SET problem is NP Complete i.e $\langle G',k'\rangle$ means graph $G'$ has an Independent set of size $k'$. I am preparing for the finals an a sample question is to prove TRIANGLE-FREE-SET to be NP Complete. $$TRIANGLE-FREE-SET:=\{\langle G,k\rangle| \text{$G$ has a subset of vertices of size $k$ whose induced subgraph is triangle free}\}$$

To give a polynomial reduction of IS instance $\langle G',k'\rangle$ to TFS, I thought of adding an extra vertex and connecting this extra vertex to all the original vertex of the IS problem instance and then have $k=k'+1$. However, I am stuck trying to prove that if $\langle G,k\rangle \in TFS$ then $\langle G',k'\rangle \in IS$. Any leads appreciated.

  • $\begingroup$ Welcome to Computer Science! Let me direct you towards our reference questions which cover, among other things, commonly found reduction techniques. $\endgroup$
    – Raphael
    Dec 14, 2017 at 20:03
  • $\begingroup$ I do not see one for reduction from IS, for example. $\endgroup$ Dec 14, 2017 at 20:05
  • 2
    $\begingroup$ The problem with your reduction is that you can "cheat" by not selecting the special vertex. Try having enough special vertices to force at least one to be necessarily chosen. $\endgroup$ Dec 14, 2017 at 20:28

1 Answer 1


Probably you can get your answer from here: Reducing independent set to triangle-free subgraph

Basically in order to prove that $TFS$ is NP-complete you can just check the first proof in the link's answer. Given a graph $G$ with an independent set $k$, there exists a $G'$ with a $TFS$ of size $|V|+k+1$.


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