NoteNextra-origin/content/CSE347/CSE347_L7.md

Lecture 7

Known NP-Complete Problems

• SAT and 3-SAT
• Vertex Cover
• Independent Set

How to show a problem L is NP-Complete

• Show L \in NP
  • Give a polynomial time certificate
  • Give a polynomial time verifier
• Show L is NP-Hard: for some known NP-Complete problem K, show K \leq_p L
  • Construct a mapping \phi from instance in K to instance in L, given an instance k\in K, \phi(k)\in L.
  • Show that you can compute \phi(k) in polynomial time.
  • Show that k \in K is true if and only if \phi(k) \in L is true.

Example 1: Subset Sum

Input: A set S of integers and a target positive integer t.

Problem: Determine if there exists a subset S' \subseteq S such that \sum_{a_i\in S'} a_i = t.

We claim that Subset Sum is NP-Complete.

Step 1: Subset Sum \in NP

• Certificate: S' \subseteq S
• Verifier: Check that \sum_{a_i\in S'} a_i = t

Step 2: Subset Sum is NP-Hard

We claim that 3-SAT \leq_p Subset Sum

Given any $3$-CNF formula \Psi, we will construct an instance (S, t) of Subset Sum such that \Psi is satisfiable if and only if there exists a subset S' \subseteq S such that \sum_{a_i\in S'} a_i = t.

How to construct \Psi?

Reduction construction:

Assumption: No clause contains both a literal and its negation.

3-SAT problem: \Psi has n variables and m clauses.

Need to: construct S of positive numbers and a target t

Ideas of construction:

For 3-SAT instance \Psi:

• At least one literal in each clause is true
• A variable and its negation cannot both be true

S contains integers with n+m digits (base 10)

p_1p_2\cdots p_n q_1 q_2 \cdots q_m

where p_i are representations of variables that are either 0 or 1 and q_j are representations of clauses.

For each variable x_i, we will have two integers in S, called v_i and \overline{v_i}.

• For each variable x_i, both v_i and \overline{v_i} have digits p_i=1. all other p positions are zero

• Each digit q_j in v_i is 1 if x_i appears in clause j; otherwise q_j=0

For example:

\Psi=(x_1\lor \neg x_2 \lor x_3) \land (\neg x_1 \lor x_2 \lor x_3)

	p_1	p_2	p_3	q_1	q_2
v_1	1	0	0	1	0
\overline{v_1}	1	0	0	0	1
v_2	0	1	0	0	1
\overline{v_2}	0	1	0	1	0
v_3	0	0	1	1	1
\overline{v_3}	0	0	1	0	0
t	1	1	1	1	1

Let's try to prove correctness of the reduction.

Direction 1: Say subset sum has a solution S'.

We must prove that there is a satisfying assignment for \Psi.

Set x_i=1 if v_i\in S'

Set x_i=0 if \overline{v_i}\in S'

1. We want set x_i to be both true and false, we will pick (in S') either v_i or \overline{v_i}
2. For each clause we have at least one literal that is true since q_j has a 1 in the clause.

Direction 2: Say \Psi has a satisfying assignment.

We must prove that there is a subset S' such that \sum_{a_i\in S'} a_i = t.

If x_i=1, then v_i\in S'

If x_i=0, then \overline{v_i}\in S'

Problem: 1,2 or 3 literals in every clause can be true.

Fix

Add 2 numbers to S for each clause j. We add y_j,z_j.

• All p digits are zero
• q_j of y_j is 1, q_j of z_j is 2, for all j, other digits are zero.
• Intuitively, these numbers account for the number of literals in clause j that are true.

New target are as follows:

	p_1	p_2	p_3	q_1	q_2
y_1	0	0	0	1	0
z_1	0	0	0	2	0
y_2	0	0	0	0	1
z_2	0	0	0	0	2
t	1	1	1	4	4

Time Complexity of construction for Subset Sum

• O(n+m)
• n is the number of variables
• m is the number of clauses

How many integers are in S?

• 2n for variables
• 2m for new numbers
• Total: 2n+2m integers

How many digits are in each integer?

• n+m digits
• Time complexity: O((n+m)^2)

Proof of reduction for Subset Sum

Claim 1: If Subset Sum has a solution, then \Psi is satisfiable.

Proof:

Say S' is a solution to Subset Sum. Then there exists a subset S' \subseteq S such that \sum_{a_i\in S'} a_i = t. Here is an assignment of truth values to variables in \Psi that satisfies \Psi:

• Set x_i=1 if v_i\in S'
• Set x_i=0 if \overline{v_i}\in S'

This is a valid assignment since:

• We pick either v_i or \overline{v_i}
• For each clause, at least one literal is true

QED

Claim 2: If \Psi is satisfiable, then Subset Sum has a solution.

Proof:

If A is a satisfiable assignment for \Psi, then we can construct a subset S' of S such that \sum_{a_i\in S'} a_i = t.

If x_i=1, then v_i\in S'

If x_i=0, then \overline{v_i}\in S'

Say t=\sum elements we picked from S.

• All p_i in t are 1
• All q_j in t are either 1 or 2 or 3.
  • If q_j=1, then y_j,z_j\in S'
  • If q_j=2, then z_j\in S'
  • If q_j=3, then y_j\in S'

QED

Example 2: 3 Color

Input: Graph G

Problem: Determine if G is 3-colorable.

We claim that 3-Color is NP-Complete.

Proof of NP for 3-Color

Homework

Proof of NP-Hard for 3-Color

We claim that 3-SAT \leq_p 3-Color

Given a 3-CNF formula \Psi, we will construct a graph G such that \Psi is satisfiable if and only if G is 3-colorable.

Construction:

1. Construct a core triangle (3 vertices for 3 colors)
2. 2 vertices for each variable x_i:v_i,\overline{v_i}
3. Clause widget

Clause widget:

• 3 vertices for each clause C_j:y_j,z_j,t_j (clause widget)
• 3 edges extended from clause widget
• variable vertex connected to extended edges

Key for dangler design:

Connect to all v_i with true to the same color. and connect to all v_i with false to another color.

''' TODO: Add dangler design image here. '''

Proof of reduction for 3-Color

Direction 1: If \Psi is satisfiable, then G is 3-colorable.

Proof:

Say \Psi is satisfiable. Then v_i and \overline{v_i} are in different colors.

For the color in central triangle, we can pick any color.

For each dangler color is connected to blue, all literals cannot be blue.

...

QED

Direction 2: If G is 3-colorable, then \Psi is satisfiable.

Proof:

QED

Example 3:Hamiltonian cycle problem (HAMCYCLE)

Input: G(V,E)

Output: Does G have a Hamiltonian cycle? (A cycle that visits each vertex exactly once.)

Proof is too hard.

but it is an existing NP-complete problem.

On lecture

Example 4: Scheduling problem (SCHED)

scheduling with release time, deadline and execution times.

Given n jobs, where job i has release time r_i, deadline d_i, and execution time t_i.

Example:

S=\{2,3,7,5,4\}. we created 5 jobs release time is 0, deadline is 26, execution time is 1.

Problem: Can you schedule these jobs so that each job starts after its release time and finishes before its deadline, and executed for t_i time units?

Proof of NP-completeness

Step 1: Show that the problem is in NP.

Certificate: \langle (h_i,j_i),(h_2,j_2),\cdots,(h_n,j_n)\rangle, where h_i is the start time of job i and j_i is the machine that job i is assigned to.

Verifier: Check that h_i + t_i \leq d_i for all i.

Step 2: Show that the problem is NP-hard.

We proceed by proving that SSS\leq_p Scheduling.

Consider an instance of SSS: \{ a_1,a_2,\cdots,a_n\} and sum b. We can create a scheduling instance with release time 0, deadline b, and execution time 1.

Then we prove that the scheduling instance is a "yes" instance if and only if the SSS instance is a "yes" instance.

Ideas of proof:

If there is a subset of \{a_1,a_2,\cdots,a_n\} that sums to b, then we can schedule the jobs in that order on one machine.

If there is a schedule where all jobs are finished by time b, then the sum of the scheduled jobs is exactly b.

Example 5: Component grouping problem (CG)

Given an undirected graph which is not necessarily connected. (A component is a subgraph that is connected.)

Problem: Component Grouping: Give a graph G that is not connected, and a positive integer k, is there a subset of its components that sums up to k?

Denoted as CG(G,k).

Proof of NP-completeness for Component Grouping

Step 1: Show that the problem is in NP.

Certificate: \langle S\rangle, where S is the subset of components that sums up to k.

Verifier: Check that the sum of the sizes of the components in S is k. This can be done in polynomial time using breadth-first search.

Step 2: Show that the problem is NP-hard.

We proceed by proving that SSS\leq_p CG. (Subset Sum \leq_p Component Grouping)

Consider an instance of SSS: \langle a_1,a_2,\cdots,a_n,b\rangle.

We construct an instance of CG as follows:

For each a_i\in S, we create a chain of a_i vertices.

WARNING: this is not a valid proof for NP hardness since the reduction is not polynomial for n, where n is the number of vertices in the SSS instance.