Lecture 36: NP, Completed

$ \def\verify{ {\mathrm{verify}} } $

COSC 311 Algorithms, Fall 2022

Announcements

1. Final Exam: Friday, Dec. 16 9:00–12:00
   • same format as midterms
   • ~8 questions
2. Final Guide:
   • posted this weekend
3. Grading:
   • assignments 2, 3 this weekend
   • assignments 4, 5 next week

Previously

Two Classes of Problems:

P: decision problems solvable in polynomial time

NP: decision problems with a polynomial time verifier

A decision problem $A$ is NP complete if

1. $A \in $ NP
2. For every $B \in $ NP, $B \leq_P A$.

Theorem [Cook, Levin]. Boolean Satisfiability (SAT) is NP complete.

Today

1. More NP Complete Problems
2. Coping with NP Completeness

Simpler Boolean Formulae

Terminology:

• a literal is a variable or its negation: $x, \bar{x}$
• a clause is an expression of the from
  1. $(z_1 \wedge z_2 \wedge \cdots \wedge z_k)$ (conjuctive clause) where each $z_i$ is a literal, or
  2. $(z_1 \vee z_2 \vee \cdots \vee z_k)$ (disjunctive clause) where each $z_i$ is a literal
• a conjunctive normal form (CNF) expression is an expression of the form $C_1 \wedge C_2 \wedge \cdots \wedge C_\ell$ where each $C_i$ is a disjunctive clause

Observation: a CNF formula evaluates to true $\iff$ all clauses evaluate to true

3-SAT

Definition. A 3-CNF formula is a Boolean formula in conjunctive normal form such that every clause contains 3 literals.

Example.

$\varphi(w, x, y, z) = (x \vee y \vee z) \wedge (y \vee \bar z \vee w) \wedge (\bar x \vee \bar y \vee \bar w)$

3-SAT:

• Input: a 3-CNF formula $\varphi$
• Output: “yes” $\iff \varphi$ is satisfiable

3-SAT is NP-Complete

Theorem (Tseytin 1970). Any Boolean formula $\varphi$ can be efficiently (in polynomial time) transformed into a 3-CNF formula $\psi$ such that:

1. if $\varphi$ is satisfiable, then so is $\psi$
2. if $\varphi$ is not satisfiable, then neither is $\psi$

Relationships

Showing NP Completeness

In order to show a problem $A$ is NP complete, show:

1. $A \in NP$
   • describe a polynomial time verifier for $A$
2. $B \leq_P A$ for any NP complete problem $B$
   • describe a polynomial time reduction from $B$ to $A$

IS is NP Complete

Theorem. IS in NP Complete.

Question. What do we need to show?

Example

$\varphi(w, x, y, z) = (x \vee y \vee z) \wedge (y \vee \bar z \vee w) \wedge (\bar x \vee \bar y \vee \bar w)$

Construction, Formalized

Input:

• 3-SAT formula $\varphi = C_1 \wedge C_2 \wedge \cdots \wedge C_k$
  • clause $C_i = (x_i \vee y_i \vee z_i)$ with $x_i, y_i, z_i$ literals (variables or negated variables)

Output:

• graph $G = (V, E)$ on $n = 3 k$ vertices
  • $V = \{ x_1, y_1, z_1, x_2, y_2, z_2, \ldots, x_k, y_k, z_k\}$
  • edges:
    • for each $i$, $x_i, y_i, z_i$ form a triangle
    • if $x_i = \neg x_j$, add edge $(x_i, x_j)$ (sim. for other variables)

Claim 1

Suppose $\varphi$ a 3-SAT formula with $k$ clauses, $G$ corresponding graph. If $\varphi$ is satisfiable, then $G$ has an independent set of size $k$.

Claim 2

Suppose $\varphi$ a 3-SAT formula with $k$ clauses, $G$ corresponding graph. If $G$ has an independent set of size $k$, then $\varphi$ is satisfiable.

Conclusion

The correspondence $\varphi \to G$ is a polynomial time reduction from 3-SAT to IS.

• $\implies \text{3-SAT} \leq_P \text{IS}$.
• $\implies$ IS is NP complete

Previously. Showed Vertex Cover (VC) satisfies IS $\leq_P$ VC

• $\implies$ VC is NP complete

More Relationships

NP Hard Problems

A problem $A$ is NP Hard if $B \leq_P A$ for some NP-complete problem $B$.

Coping with NP Hardness

Fact of Life. Many important practical problems are NP-Hard.

Question. So what do we do about it?

Coping Strategies

What if we need to solve an NP hard problem?

Where to go from Here?

1. More algorithms!
   • parallel & distributed algorithms (COSC 273, 373)
   • computational geometry (COSC 225)
   • randomized algorithms
   • streaming and sublinear algorithms
   • approximation algorithms
2. More complexity!
   • automata/computability theory (COSC 401)
   • computational complexity
   • cryptography
   • models of computation

Thank You!