NP-complete

Posted 2022-01-09 mb61caa1c74a413

目录

• 1.1 NP-complete
• 1.1.1 The “first” NP-complete problem
  • 1.1.2 Establishing NP-completeness
  • 1.1.3 Implications of Karp
  • 1.1.4 Implications of Cook–Levin
  • 1.1.5 Implications of Karp + Cook–Levin
  • 1.1.6 Some NP-complete problems
• 2.1 Asymmetry of NP
  • 2.1.1 NP = co-NP ？

1.1 NP-complete

NP-complete: A problem Y ∈ NP with the property that for every problem X ∈ NP, X ≤ P Y

Proposition. Suppose Y ∈ NP-complete. Then, Y ∈ P iff P = NP.
Pf. ⇐ If P = NP, then Y ∈ P because Y ∈ NP.
Pf. ⇒ Suppose Y ∈ P.
・Consider any problem X ∈ NP. Since X ≤ P Y, we have X ∈ P.
・This implies NP ⊆ P.
・We already know P ⊆ NP. Thus P = NP

其实定义一个问题是否是NP问题是很简单的，但是要满足所有的NP问题都可以规约到该问题看起来很难实现，因为NP太多了。。。。
那么是否真正存在我们定义中的NPC问题呢？答案是肯定的！

1.1.1 The “first” NP-complete problem

1.1.2 Establishing NP-completeness

Recipe. To prove that Y ∈ NP-complete:
・Step 1. Show that Y ∈ NP.
・Step 2. Choose an NP-complete problem X.
・Step 3. Prove that X ≤ P Y.

Proposition. If X ∈ NP-complete, Y ∈ NP, and X ≤ P Y, then Y ∈ NP-complete.
Pf. Consider any problem W ∈ NP. Then, both W ≤ P X and X ≤ P Y
・By transitivity, W ≤ P Y.
・Hence Y ∈ NP-complete.

1.1.3 Implications of Karp

1.1.4 Implications of Cook–Levin

1.1.5 Implications of Karp + Cook–Levin

1.1.6 Some NP-complete problems

Basic genres of NP-complete problems and paradigmatic examples.
・Packing/covering problems: SET-COVER, VERTEX-COVER,INDEPENDENT-SET.
・Constraint satisfaction problems: CIRCUIT-SAT, SAT, 3-SAT.
・Sequencing problems: HAMILTON-CYCLE, TSP.
・Partitioning problems: 3D-MATCHING, 3-COLOR.
・Numerical problems: SUBSET-SUM, KNAPSACK.

2.1 Asymmetry of NP

Asymmetry of NP： We need short certificates only for yes instances.

2.1.1 NP = co-NP ？

Fundamental open question. Does NP = co-NP?
・Do yes instances have succinct certificates iff no instances do?
・Consensus opinion: no.

关于NP问题是否和CO-np问是完全对等的？
业界公共的意识是：不等的！

Theorem. If NP ≠ co-NP, then P ≠ NP.
Pf idea.
・P is closed under complementation.
・If P = NP, then NP is closed under complementation.
・In other words, NP = co-NP.
・This is the contrapositive of the theorem