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if l1 is in NP-HARD, so for every L2!=empty set, l1*l2 is in np-hard.

when:

l1*l2={(w1,w2) , w1 in L1 and w2 in L2}

Is it true or false and why?

I can't approve it but I also don't find counter example.

Jason Sturges
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user1462787
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  • Vote to move to [cstheory.stackexchange.com](http://cstheory.stackexchange.com/)—this isn't a programming question! – Asherah Jun 25 '12 at 02:19
  • Oh, good point. I also think a move might be a good idea. Edit: On second thought, http://cstheory.stackexchange.com/ is for research questions... This questions seems a bit too simple... – Dino Jun 25 '12 at 08:25

1 Answers1

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L1 * L2 is NP-hard.

Proof: Let L be a language in NP, let f be a reduction of L to L1 and let w2 be in L2. Define g(x) = (f(x), w2). Now g is a polynomial time many-to-one reduction of L to L1*L2 because clearly:

x in L <==> (f(x), w(2)) in L1*L2

Dino
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