算法设计与分析P-NP(2014)

polynomial time algorithm does not exist for problem A or we are not smart.

Open problem: PNP? $1 million prize.
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相关概念: P,NP,NPC,NP-hard
P: 能在多项式时间内解决的问题 NP: 不能在多项式时间内解决或不确定能不能在多项式 时间内解决，但能在多项式时间验证的问题 NPC: NP完全问题，所有NP问题在多项式时间内都能约 化(Reducibility)到它的NP问题，即解决了此NPC问题，所有 NP问题也都得到解决。 NP hard:NP难问题，所有NP问题在多项式时间内都能约 化(Reducibility)到它的问题(不一定是NP问题)。
NP-complete: A subset of NP that seems to be hardest to solve. The most-likely relationship of the three problem classes
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Class P and Class NP
Based on definitions, PNP. If we can design a polynomial time algorithm for problem A, then problem A is in P. However, if we have not been able to design a polynomial time algorithm for problem A, then there are two possibilities:
Question: Do there exist polynomial time algorithms for those famous problems？
Answer: No body knows.
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Story
Research topic: Prove that polynomial time algorithms do not exist for those famous problems, e.g., Hamilton circuit problem.
The first NPC problem is Satisfiability probelm
Proved by Cook in 1971 and obtains the Turing Award for this work
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Boolean formula
A boolean formula f(x1, x2, …xn), where xi are boolean variables (either 0 or 1), contains boolean variables and boolean operations AND, OR and NOT .
Satisfiability problem
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The First NP-complete Problem
Theorem: Satisfiability problem is NP-complete.
It is the first NP-complete problem. S. A. Cook in 1971 /wiki/Stephen_Cook Won Turing prize for his work.
To answer if PNP, a rich area, NP-completeness theory is developed.
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Definitions (informally)


P: polynomial time solvable(( in O(nk) time)
NP:consists of those problem that are verifiable in
P and NP
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NP-Co time vs exponential time（多项式时间和指数 时间)
Polynomial O(nk), where n is the input size (e.g., number of nodes in a graph, the length of strings , etc) of our problem and k is a constant (e.g., k=1, 2, 3, etc). Exponential time: 2n or nn. n= 2, 10, 20,
polynomial time.
What we mean here is that if we were somehow given a “solution”, then we can verify that the solution is correct in time polynomial in the input size to the problem. Example: Hamilton Circuit: given an order of the n distinct vertices (v1, v2, …, vn), we can test if (vi, v i+1) is an edge in G for i=1, 2, …, n-1 and (vn, v1) is an edge in G in time O(n) (polynomial in the input size).
Clause: variables and their negations are connected with OR operation, e.g., (x1 OR NOTx2 OR x5) Conjunctive normal form of boolean formula: contains m clauses connected with AND operation. Example: (x1 OR NOT x2) AND (x1 OR NOT x3 OR x6) AND (x2 OR x6) AND (NOT x3 OR x5). Here we have four clauses.
Significance:
If Satisfiability problem can be solved in polynomial time, then ALL problems in class NP can be solved in polynomial time. If you want to solve PNP, then you should work on NPC problems such as satisfiability problem. We can use the first NPC problem, Satisfiability problem, to show that other problems are also NPcomplete.
The formula is true for assignment x1=1, x2=0, x3=1.
Note: for n Boolean variables, there are 2n assignments. Testing if formula=1 can be done in polynomial time for any given assignment.Given an assignment that satisfies formula=1 is hard.


Can arbitrary instances of problem Y be solved using polynomial number of standard computational steps, plus a polynomial number of calls to a black box that solves problem X? If yes, then Y is polynomial-time reducible to X.
Story
All algorithms we have studied so far are polynomial time algorithms (unless specified).
Facts: people have not yet found any polynomial time algorithms for some famous problems, (e.g., Hamilton Circuit, longest simple path, Steiner trees).
You can get Turing award if you can give the proof.
In order to answer the above question, people define two classes of problems, P class and NP class.
Suppose we have a black box (an algorithm) that could solve instances of a problem X; If we give the input of an instance of X, then in a single step, the black box will return the correct answer. Question:
Y p X
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NP-Complete(NPC)
A problem X is NP-complete if it is in NP and any problem Y in NP has a polynomial time reduction to X.
it is the hardest problem in NP If an NP-complete problem can be solved in polynomial time, then any problem in class NP can be solved in polynomial time.