are all np complete problems equivalent?
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What is the point of classifying the two if they are the same? In other words, if we have an NP problem then through (2) this problem can transform into an NP-complete problem. Therefore, the NP problem is now NP-complete, and NP = NP-complete. Both classes are equivalent.
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If P equals NP, every NP problem would contain a hidden shortcut, allowing computers to quickly find perfect solutions to them. But if P does not equal NP, then no such shortcuts exist, and computers' problem-solving powers will remain fundamentally and permanently limited.
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The corresponding sides may or may not be equal. But, the ratios of the corresponding parts of one object will be equal to that of the other objects. Now, all squares are always similar. Their size may not be equal but their ratios of corresponding parts will always be equal.
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Consistent and Admissible Heuristics. Any consistent heuristic is also admissible.
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Its decision version, the vertex cover problem, was one of Karp's 21 NP-complete problems and is therefore a classical NP-complete problem in computational complexity theory.
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The choice made by a greedy algorithm may depend on choices made so far but not on future choices or all the solutions to the subproblem. ... In other words, a greedy algorithm never reconsiders its choices. This is the main difference from dynamic programming, which is exhaustive and is guaranteed to find the solution.
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In computational complexity theory, NP-hardness (non-deterministic polynomial-time hardness) is the defining property of a class of problems that are informally "at least as hard as the hardest problems in NP". A simple example of an NP-hard problem is the subset sum problem.
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5 Answers. According to Godel's incompleteness theorem, ZFC cannot prove its own consistency. Therefore, it is relatively consistent with ZFC that there are not any set models of ZFC.
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The set of natural numbers satisfies the supremum property and hence can be claimed to be complete. But the set of natural numbers is not dense. It is actually discrete. There are neighbourhoods of every natural number such that they contain no others.
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Examples. An example of an NP-hard problem is the decision subset sum problem: given a set of integers, does any non-empty subset of them add up to zero? That is a decision problem and happens to be NP-complete.
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NP is the class of languages with polynomial-size membership proofs. P ⊆ NP still holds: If I can decide membership in polynomial time, I do not need a certificate. NP is closed under union, intersection, and concatenation; but is not known to be closed under complement.
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Polynomial-Time Algorithms. A polynomial-time algorithm is an algorithm whose execution time is either given by a polynomial on the size of the input, or can be bounded by such a polynomial. ... Although n log n is not, strictly speaking, a polynomial, the size of n log n is bounded by n2, which is a polynomial.
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It is easy to prove that the halting problem is NP-hard but not NP-complete. ... There are also NP-hard problems that are neither NP-complete nor Undecidable. For instance, the language of true quantified Boolean formulas is decidable in polynomial space, but not in non-deterministic polynomial time (unless NP = PSPACE).
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The running time is O(NW) for an unbounded knapsack problem with N items and knapsack of size W. W is not polynomial in the length of the input though, which is what makes it pseudo-polynomial. ... So the runtime is more accurately said to be O(N. 2bits in W), which is exponential.
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The complexity class NP itself is the class of all decision problems with efficiently verifiable solutions, where ”efficiently” = ”in time polynomial in the length of the input in binary”. A problem is a decision problem if its output is a yes/no answer.
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8 Answers. Divide and Conquer works by dividing the problem into sub-problems, conquer each sub-problem recursively and combine these solutions. Dynamic Programming is a technique for solving problems with overlapping subproblems.
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['Optimal Substructure: If an optimal solution contains optimal sub solutions then a problem exhibits optimal substructure.', 'Overlapping subproblems: When a recursive algorithm would visit the same subproblems repeatedly, then a problem has overlapping subproblems.']