Yes, by definition, every NP problem can be reduced to an NP-complete problem in polynomial time.
All NP-complete problems can be reduced to each other in polynomial time since an NP-complete problem is itself an NP problem.
An NP-complete problem lies in NP, the set of all decision problems whose solutions can be verified in polynomial time. Similarly, NP can be defined as a set of decision problems that can be solved in polynomial time by a non-deterministic Turing machine. A problem p in NP is NP-complete if every other problem in NP can be transformed (or reduced) to p in polynomial time.
Not sure if all NP issues can be resolved quickly. This is called the P-NP problem. However, if every NP-complete problem can be solved quickly, the definition of an NP-complete problem states that every problem in NP must be rapidly reducible to every NP-complete problem (i.e., it can be solved in polynomial time abbreviated by ). For this reason, NP-complete problems in general are often said to be harder or harder than NP problems.
A decision problem C is NP-complete if:
1. C is in NP and every problem in NP is reducible to C in polynomial time.
2. We can show that C is in NP by showing that the candidate solutions for C can be verified in polynomial time.
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