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Part VI NP-Hardness

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Lecture 23 Whats NP?

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Hard Problems

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Answer What is NP-complete? What is NP-hard? First, what is NP?

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Answer 1 Wrong! Did you take computer science before ?

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Answer 2 Still wrong! But, almost true. I can give you a counterexample!

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Integer Programming

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Decision version of Integer Programming

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How to prove a decision problem belonging to NP? How to design a polynomial-time nondeterministic algorithm?

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we need to study computation model-Turing Machine. To answer,

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Deterministic Turing Machine (DTM) Finite Control tape head

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The tape has the left end but infinite to the right. It is divided into cells. Each cell contains a symbol in an alphabet Γ. There exists a special symbol B which represents the empty cell. al ph a B e

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The head scans at a cell on the tape and can read, erase, and write a symbol on the cell. In each move, the head can move to the right cell or to the left cell (or stay in the same cell). a

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The finite control has finitely many states which form a set Q. For each move, the state is changed according to the evaluation of a transition function δ : Q x Γ Q x Γ x {R, L}.

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δ(q, a) = (p, b, L) means that if the head reads symbol a and the finite control is in the state q, then the next state should be p, the symbol a should be changed to b, and the head moves one cell to the left. q a b p

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δ(q, a) = (p, b, R) means that if the head reads symbol a and the finite control is in the state q, then the next state should be p, the symbol a should be changed to b, and the head moves one cell to the right. p q a b

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There are some special states: an initial state s and an final states h. Initially, the DTM is in the initial state and the head scans the leftmost cell. The tape holds an input string. s

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When the DTM is in the final state, the DTM stops. An input string x is accepted by the DTM if the DTM reaches the final state h. Otherwise, the input string is rejected. h x

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Multi-tape DTM Input tape (read only) Storage tapes Output tape (possibly, write only)

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Nondeterministic Turing Machine (NTM) Finite Control tape head

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The tape has the left end but infinite to the right. It is divided into cells. Each cell contains a symbol in an alphabet Γ. There exists a special symbol B which represents the empty cell. al ph a B e

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The finite control has finitely many states which form a set Q. For each move, the state is changed according to the evaluation of a transition function δ : Q x Γ 2^{Q x Γ x {R, L}}.

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Nondeterministic TM (NTM) There are multiple choices for each transition. For each input x, the NTM may have more than one computation paths. An input x is accepted if at least one computation path leads to the final state. L(M) is the set of all accepted inputs.

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Time of DTM Time M (x) = # of moves that DTM M takes on input x. Time M (x) < infinity iff x ε L(M).

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Space Space M (x) = maximum # of cells that M visits on each work (storage) tapes during the computation on input x. If M is a multitape DTM, then the work tapes do not include the input tape and the write-only output tape.

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Whats P? P is a class of decision problems that each can be solved by deterministic TM in polynomial time.

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Whats NP? NP is a class of decision problems that each can be solved by a nondeterministic TM in polynomial time. NP is a class of decision problems that each can be solved by a polynomial-time nondeterministic algorithm.

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Whats PSPACE? PSPACE is a class of decision problems that each can be solved by TM in polynomial space. Why didnt specify Deterministic or Nondeterministic? It doesnt matter due to Savitchs Theorem.

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P = ? NP = ? PSPACE They are central problems in computational complexity.

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If P = NP, then NP-complete P

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Ladner Theorem If NP P, then there exists a set A lying - between P and NP-complete class, i.e., A is in NP, but not in P and not being NP- compete.

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How to prove a decision problem belonging to NP? How to design a polynomial-time nondeterministic algorithm?

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Hamiltonian Cycle Given a graph G, does G contain a Hamiltonian cycle? Hamiltonian cycle is a cycle passing every vertex exactly once.

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Nondeterministic Algorithm Guess a permutation of all vertices. Check whether this permutation gives a cycle. If yes, then algorithm halts. What is the running time?

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Guessing Time Each guess can choose one from a constant number of choices. This is because that in NTM, the number of choices for each move is independent from input size. Guessing a permutation of n vertices needs (n long n) time.

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Minimum Spanning Tree Given an edge-weighted graph G, find a spanning tree with minimum total weight. Decision Version: Given an edge-weighted graph G and a positive integer k, does G contains a spanning tree with total weight < k.

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Nondeterministic Algorithm Guess a spanning tree T. Check whether the total weight of T < k. This is not clear!

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How to guess a spanning tree? Guess n-1 edges where n is the number of vertices of G. Check whether those n-1 edges form a connected spanning subgraph, i.e., there is a path between every pair of vertices.

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Co-decision version of MST Given an edge-weighted graph G and a positive integer k, does G contain no spanning tree with total weight < k?

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Algorithm Computer a minimum spanning tree. Check whether its weight < k. If yes, the algorithm halts.

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co-NP co-NP = {A | Σ* - A ε NP}

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NP co-NP So far, no natural problem has been found in NP co-NP, but not in P. P NP co-NP

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Linear Programming Decision version: Given a system of linear inequality, does the system have a solution? It was first proved in NP co-NP and later found in P (1979).

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Primality Test Given a natural number n, is n a prime? It was first proved in NP co-NP and later found in P (2004).

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Therefore A natural problem belonging to NP co- NP is a big sign for the problem belonging to P.

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Proving a problem in NP In many cases, it is not hard. In a few cases, it is not easy.

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Polynomial-time verification with polynomial-size certificate

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Integer Programming Decision version: Given A and b, does Ax > b contains an integer solution? The difficulty is that the domain of guess is too large, that is, it is not easy to know the existence of polynomial-size certificate.

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