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In class we saw examples of reductions like from Independent Set (IS) to Longest common subsequence (arbitrary number of sequences) (LCS)

$V = \{v_1,\ldots,v_n\} E =\{ e_1,\ldots, e_m \}$ The reduction is $(G(V,E),k)$ to $\{S_0,S_1,\ldots, S_m\}$, where

$e_k = (v_i,v_j)$ gives $S_k = \langle v_1v_2 \ldots v_{i-1}v_{i+1}\ldots v_n \text{ } v_1v_2 \ldots v_{j-1}v_{j+1} \ldots v_n \rangle $

I understand the reduction for edge $(v_iv_j)$ we don't include $v_i$ in the first part of the sequence and don't include $v_j$ in the second part of the sequence.

They claim $G$ has an IS of size $k$ if and only if LCS has a solution of size k.

This reduction fails when $k$ is not a maximum independent size set. In that case we get longest common subsequence of size $< k$.

Even in another reduction from VC to a minimization problem Y, the same problem persists for me, their claim is size $k$ vertex cover gives a $k-1$ size solution for Y.

But if $k > \mathit{opt}$ vertex cover size then we don't get a minimum size for the other problem Y. I have verified this. Here optimal for Y $ < k-1$. I am happy to provide more details on this as well. Y is the Exemplar breakpoint distance between two genomes.

I am missing something in my understanding. I get that there is no contradiction and this does not prove Y, LCS are NPC, as we would still need to check LCS, Y could be verified polynomially.

Are we doing something wrong in class or am I missing something?

Yuval Filmus
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T.Harish
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  • Your reductions show NP-hardness. To prove that LCS and Y are NP-complete, you need to show that they are in NP. – Yuval Filmus Nov 20 '19 at 16:58
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    The first reduction shows that $(G,k)$ is a Yes instance of IS iff $(S_0,\ldots,S_m),k$ is a Yes instance of LCS. This is what you need to show in order to prove NP-hardness. – Yuval Filmus Nov 20 '19 at 16:59
  • Suppose k = 1 then $(G,1) is a yes instance always. But there might exist a longer common subsequence than 1. I get that still there always exists a size 1 common subsequence but that it is not the longest common subsequence. I understand that we still need to show that it is in NP to prove Np-complete. Could you please clarify that @YuvalFilmus ? Thanks! – T.Harish Nov 20 '19 at 20:14
  • I think you got the definition of LCS wrong. It asks whether there is a common sequence of size $k$. It doesn’t have to be longest, just as in maximum IS, we only ask whether there exists an IS of size $k$; it doesn’t have to be the largest. – Yuval Filmus Nov 20 '19 at 20:22
  • Sorry for not being clear. My questions was doing reductions from decision problems to min/max problems. So LCS and Y are not decision version problems. – T.Harish Nov 20 '19 at 20:28
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    That’s where I think you’re mistaken. When we say that an optimization problem is NP-hard, we really mean that the corresponding decision problem is NP-hard. NP-hardness is only defined directly for decision problems. – Yuval Filmus Nov 20 '19 at 20:40
  • Sorry could you elaborate more ? We say min vertex cover is Np-hard, but the decision version is Np complete. I assume decision problems are always Np-complete. I am assuming we can always verify given a certificate polynomially. But you can't verify polynomially that it is the optimal one. This was my understanding. – T.Harish Nov 20 '19 at 20:45
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    A problem like “the size of the maximum independent set is $k$” is probably not in NP, and we typically don’t consider such problems. We only consider ones of the form “the size of the maximum independent set is at least $k$”. You are just being confused by the name. In your undergraduate studies you will probably never encounter problems of the form “the optimum is exactly $k$”. – Yuval Filmus Nov 20 '19 at 20:49
  • Don’t think your material is special. Perhaps it hasn’t been explained very well, but trust me, you are using exactly the same convention as anybody else (or your professor is clueless about the material, and it’s best to ignore what they say). – Yuval Filmus Nov 20 '19 at 20:51
  • Thanks! If you want to formalize what you explained as an answer, please do. So this question can be resolved – T.Harish Nov 20 '19 at 20:55
  • I believe I already wrote such an answer somewhere... – Yuval Filmus Nov 20 '19 at 21:03

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