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Given a known strongly NP Complete problem $A$. If there is a polynomial time transformation from $A$ to another problem NP Complete problem $B$ does that imply anything about if $B$ is always/automatically strongly NP complete or not?

I don't think that it is necessary that $B$ is always/automatically strongly NP Complete but some other sources seem to suggest the opposite. Thus a bit confused.

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    $\begingroup$ Take any strongly NP Complete problem for A and any NP Complete B for which we know a pseudo-polynomial time algorithm. Is there a poly-time reduction from A to B? (remember that B is NPC). That should answer your question. $\endgroup$
    – Tassle
    Oct 11 at 10:38
  • $\begingroup$ @Tassle why don't you post your comment as an answer so we can upvote it and it can be accepted? $\endgroup$
    – Nathaniel
    Oct 11 at 10:39
  • $\begingroup$ @Nathaniel Yep I should have. Done $\endgroup$
    – Tassle
    Oct 11 at 10:41

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Take any strongly NP Complete problem for A and any NP Complete B for which we know a pseudo-polynomial time algorithm. Is there a poly-time reduction from A to B? (remember that B is NPC) That should answer your question.

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  • $\begingroup$ thanks. so I take a strongly NPC simply cannot be transformed to a weak NPC in Polynomial time? (just reconfirming) didn't see transformation of strongly NPC and this part mentioned anywhere..! thanks again. $\endgroup$ Oct 11 at 10:45
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    $\begingroup$ @TheoryQuest1 You got it kinda backwards. A weakly NPC problem is still an NPC problem, so any problem in NP (including strongly NPC problems) can be poly-time reduced to it. This is a counter example to the claim that such a reduction proves strong NP completeness. (You actually need the reduction to produce at most polynomially large numerical parameters if you want to prove strong NP completeness) $\endgroup$
    – Tassle
    Oct 11 at 11:07
  • $\begingroup$ aah.. I initially was sure of the same thing thus the "don't think that it is necessary that B is always/automatically strongly NP Complete.." but your comment threw me off :) $\endgroup$ Oct 11 at 11:28

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