# 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.