np

Table of Contents

1 P & NP & NPC

  • `class P`: consists of problems that are solvable is polynomial time.
  • `class NP`: consists of problems that are verifiable in polynomial time.
  • `varifiable`: given a certificate of a solution,

we could verify that the certificate is correct in time polynomial in the size of the input to the problem.

\(P \subseteq NP\)

  • NPC: NP-complete. it is in NP and is as hard as any problem in NP.

If any NP-complete problem can be solved in polynomial time, every problemm in NP has a polynomial algorithm.

  • co-NP: the set of languages L such that \(\overline{L} \in NP\)

1.1 3 key concepts in showing a problem to be NP-complete

1.1.1 Decision problem and optimization problem

Optimization problem: each feasible solution has an associated value, we wish to find one with the best value.

Decision problem: Answer is yes or no

Optimization problem 可以加一个 threshold 让它变成 decision problem. 因此,只要OP是简单的,DP也是简单的。

1.1.2 Reductions

已经知道problem B是P, 如果有一个procedure that transform any instance \(\alpha\) of A into some instance \(\beta\) of B with the following characteristics:

  • the transformation takes polynomial time
  • the answers are the same.

That is, the answer for \(\alpha\) is yes if and only if answer for \(\beta\) is yes

the procedure is called a polynomial-time reduction algorithm.

显而易见,解决A的方法是:

  1. given an instance \(\alpha\) of A, use procedure to transform it into \(\beta\) of B.
  2. run P on B
  3. use answer of \(\beta\) for \(\alpha\).

对于NPC,我们使用其inverse: Assume No P algorithm for A. Assume A is NPC, we can prove B is also NPC.

1.1.3 A first NPC problem

2 Intro problems

  • an instance of a problem: the input to a problem. E.g. Graph G for PATH problem.

2.1 Shortest simple path

一个有向图中,找到从一个起点出发的最短路径。

complexity: 即使包含负边,也可以在$O(VE)$找到。

2.2 Longest simple path

仅仅决定一个图中是否含有一条path,包括一个给定数字以上的路径.

complexity: 都是NP-Complete。

2.3 Euler tour

在一个连同有向图中,一个cycle,遍历所有边有且仅有一次。

complexity: 决定一个图是否含有Euler tour,\(O(E)\).

2.4 hamiltonian cycle

在一个有向图中,a simple cycle,包括所有顶点。

complexity: 决定一个有向图是否含有hamiltonian cycle是NP-Complete

2.5 2-CNF/3-CNF satisfiability

  • k-CNF: k-Conjuntive normal form
  • A boolean formula is satisfiable if 它包括的变量去一种0和1的组合,可以使其值为1.
  • A boolean formula is in k-CNF if it is AND of clauses of ORs of exactly k variables or their negations.

\((x_1 \vee \neg x_2) \wedge (\neg x_2 \vee x_3) \wedge (\neg x_2 \vee \neg x_3)\) is 2-CNF.

complexity: determine whether 2-CNF is satisfiable is P, determine whether 3-CNF is satisfiable is NP-complete

3 NPC and reducitility

3.1 reducitility

  • language \(L_1\) is polynomial-time reducible to \(L_2\), written \(L_1 \le_p L_2\),

if 存在P时间的函数f, s.t. \(x \in L_1\) if and only if \(f(x) \in L_2\)

意味着问题$L1$可以转化成问题$L2$,所以只要$L2$有解,$L1$同样有解。 \(L_1\) is not more than a polynomial factor harder than \(L_2\).

Lemma 34.3: if \(L_1 \le_p L_2\), then \(L_2 \in P\) implies \(L_1 \in P\).

3.2 NPC & NPH

a language \(L \subseteq {0,1}^*\) is NP-complete if:

  1. \(L \in NP\)
  2. \(L' \le_p L\) for every \(L' \in NP\)

NP-hard: L satisfies 2 but not 1.

Theorem 34.4: 如果任意一个NPC problem都是P solvable的,那么\(P=NP\). 如果任意NP problem都不是P solvable的,那么所有NPC problem都不是P solvable的。

4 NPC Problems

4.1 clique

图中选出一些顶点,两两相连。

5 Homeworks

5.1 34.1-6

把P作为语言,证明其对并,交,连,补,*是封闭的。

If \(L_1,L_2 \in P\), then

  • \(L_1 \bigcup L_2 \in P\)
  • \(L_1 \bigcap L_2 \in P\)
  • \(L_1 L_2 \in P\)
  • \(\overline{L_1} \in P\)
  • \(L_1^* \in P\)

5.1.1 Solution

Assume \(L,L_1,L_2 \in P\):

  • \(L_1 \bigcup L_2\): decide if \(x \in L_1\) and \(x \in L_2\). 只要有一个成立,就会有\(x \in L_1 \bigcup L_2\). Otherwise not.
  • \(L_1 \bigcap L_2\): 同上,只要两个都成立,就有\(x \in L_1 \bigcap L_2\). Otherwise not.
  • \(L_1L_2\): $L1,L2$是两个字符串,$L1L2$是把两个连接起来。用$xij$来表示一个字符串的子串。

对于所有n种k,decide\(x_{1k} \in L_1\), \(x_{(k+1)n} \in L_2\).

  • \(\overline{L}\): \(x \in \overline{L}\) 等价于 \(x \notin L\)
  • \(L^*\):

Kleene star: \(L^*=\bigcup_{i\in N}L^k\) where \(L^(k+1) = LL^k\).

我们证明对于所有k,有$\bigcupi=0-k$成立。用induction.

当$k=0$,空集,显然。

假设k成立,对于k+1,有$Lk+1=LLk$。

5.2 34.2-9

Prove that \(P \subseteq co-NP\)

5.2.1 Solution

\(L \in P\) => \(\overline{L} \in P \subseteq NP\)

5.3 34-1

5.3.1 Solution

a).

INDEPENDENT-SET = { $ < G,k > $:G is a graph containing a independent-set of size k }

图中选出一些顶点,两两不连。

Prove NPC:

证明是NP. 构造一个验证。$V((G=(V,E), K), C)$,其中C是certificate,也就是一个顶点集,用来判断是不是Independent set。 需要做两件事:1. \(|C| \le K\) 2. C构成independent set。 两个都可以在P内完成。

证明是NP-hard。用clique来推导。根据G,构造一个$G'$,顶点不变,所有边去掉,所有没有边的加上边。 这时候,有: 如果nodes are independent set in \(G\), then they form a clique in \(G'\); 如果nodes form a clique in \(G'\), then they are independent set in \(G\).

b).

  1. find the maximum size independent set. Since \(K \le |V|\), \(O(V)\).
  2. Once the maximum K is found, we need to 决定哪些v在里面。

在G中选一个v,将其和所有其他顶点相连。黑箱测试$(G',K)$。 如果成功,说明v不在independent set中。\(G=G'\).继续测试下一个v。 如果失败,说明v在set中。用G做下一个v的测试。

复杂度分析:共有v个点,每个点要连其余v-1个点,共\(O(V^2)\).

c).

每个节点的degree都是2,说明,是一个cycle。显然,every other vertex. 直接选出来就可以了,所以\(O(V)\).

d).

maximum independent set is the side with the larger number of vertices. 直接比较一下多少就可以,所以\(O(V)\).

Author: Hebi Li

Created: 2017-06-14 Wed 22:43

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