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As far as my understanding goes, to show that a problem A is NP-hard, we use another NP-complete problem B. We reduce (in polynomial time) from B to A, i.e. use A to solve B. This shows that A is harder than or equal to B, which proves that it is NP-hard.

However, I don't understand why A is harder than B? I feel like I'm missing a part after the reduction step.

Apologies if my understanding is wrong, or if this is a duplicate question.

monadoboi
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  • As you said yourself, A is not necessarily harder than B -- it's just at least as hard as B. For this to be false, A would have to be easier than B -- but that can't possibly be, since the reduction describes a way to encode any B instance as an A instance -- meaning that literally every B-instance you can think of can be written as an A-instance. – j_random_hacker May 05 '18 at 19:33
  • Thank you! That makes sense :) Feel free to add it as an answer if you wish, though this question has been marked as duplicate now. – monadoboi May 05 '18 at 19:35

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