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CSE347 Analysis of Algorithms (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

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'

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.

Tip

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.

...

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

Proof

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.