1. CS154, Lecture 16: More NP-Complete Problems; PCPs

2. Is 3SAT solvable in O(n) time on a multitape TM?
Are there logic circuits of size 6n for 3SAT?
If yes, then not only is P=NP, but there would be a “dream machine” that could crank out short proofs of theorems, quickly optimize all aspects of life… recognizing quality work is all you need to produce

3. There are thousands of NP-complete problems
Your favorite topic certainly has an NP-complete problem somewhere in it
Even the other sciences are not safe: biology, chemistry, physics have NP-complete problems too!

4. Given a favorite problem   NP, how can we prove it is NP-hard?
Generic Recipe:
Take a problem  that you know to be NP-hard (3-SAT)
Prove that  P 
Then for all A  NP, A P  and  P 
We conclude that A P , and  is NP-hard

5.  is NP-Complete NP P  

6. The Clique Problem
Given a graph G and positive k, does G contain a complete subgraph on k nodes?
CLIQUE = { (G,k) | G is an undirected graph with a k-clique }
Theorem (Karp): CLIQUE is NP-complete

7. Proof Idea: 3SAT P CLIQUE
Transform a 3-cnf formula  into (G,k) such that
  3SAT  (G,k)  CLIQUE
Want transformation that can be done in time that is polynomial in the length of 
How can we encode a logic problem as a graph problem?

8. 3SAT P CLIQUE
We transform a 3-cnf formula  into (G,k) such that
  3SAT  (G,k)  CLIQUE
Let C1, C2, …, Cm be clauses of . Assign k := m. Make a graph G with m groups of 3 nodes each.
Group i corresponds to clause Ci of . Each node in group i is labeled with a literal of Ci
Put edges between all pairs of nodes in different groups, except pairs of nodes with labels 𝑥 𝑖 and  𝑥 𝑖
Put no edges between nodes in the same group
When done putting in all the edges, erase the labels

9. (x1  x1  x2)  (x1  x2  x2)  (x1  x2  x2)
|V| = 9
k = 3

10. (x1  x1  x1)  (x1  x1  x2)  (x2  x2  x2)  (x2  x2  x1)

11. Claim:   3SAT  (G,m)  CLIQUE
Claim: If   3SAT then (G,m)  CLIQUE
Proof: Given a SAT assignment A of , for every clause C there is at least one literal in C that’s set true by A
For each clause C, let vC be a vertex from group C whose label is a literal that is set true by A
Claim: S = {vC : C  } is an m-clique
Proof: Let vC,vC’ be in S. Suppose (vC,vC’) ∉ E.
Then vC and vC’ must label inconsistent literals, call them x and x
But assignment A cannot satisfy both x and x
Therefore (vC,vC’)  E, for all vC, vC’  S.
Hence S is an m-clique, and (G,m)  CLIQUE

12. Claim:   3SAT  (G,m)  CLIQUE
Claim: If (G,m)  CLIQUE then   3SAT
Proof: Let S be an m-clique of G. We construct a satisfying assignment A of .
Claim: S contains exactly one node from each group.
Now for each variable x of , make assignment A: Assign x to 1  There is a vertex v ϵ S with label x
For all i = 1,…,m, at least one vertex from group i is in S.
Therefore, for all i = 1,…,m. A satisfies at least one literal in the ith clause of . Therefore A is a satisfying assignment to 

13. (G, k)  CLIQUE iff (G’, k)  IS
Independent Set
IS: Given a graph G = (V, E) and integer k, is there S  V such that |S|=k and no two vertices in S have an edge?
IS = {(G,k) | G is an undirected graph with an IS of size k}
CLIQUE: Given G = (V, E) and integer k, is there S  V such that |S|=k and every pair of vertices in S have an edge?
CLIQUE ≤P IS:
Given G = (V, E), output G’ = (V, E’) where E’ = {(u,v) | (u,v) ∉ E}.
(G, k)  CLIQUE iff (G’, k)  IS

14. The Vertex Cover Problema e c d b a e c d
vertex cover = set of nodes C that cover all edges:
For all edges, at least one endpoint is in C

15. VERTEX-COVER = { (G,k) | G is a graph with a vertex cover of size at most k}
Theorem: VERTEX-COVER is NP-Complete
(1) VERTEX-COVER  NP
(2) IS P VERTEX-COVER

16. IS P VERTEX-COVER
Want to transform a graph G and integer k into G’ and k’ such that
(G,k)  IS  (G’,k’)  VERTEX-COVER

17. IS P VERTEX-COVER
Claim: For every graph G = (V,E), and subset S  V,
S is an independent set if and only if (V – S) is a vertex cover
Proof: S is an independent set
(8 u, v  V)[ (u  S and v  S) ) (u,v) ∉ E ]
(8 u, v  V)[ (u,v)  E ) (u ∉ S or v ∉ S) ]
(V – S) is a vertex cover
Therefore (G,k)  IS  (G,|V| – k)  VERTEX-COVER
Our polynomial time reduction: f(G,k) := (G, |V| – k)

18. The Subset Sum Problem
Given: Set S = {a1,…, an} of positive integers and a positive integer t
Is there an A µ {1, … ,n} such that t = i 2 A ai ?
SUBSET-SUM = {(S, t) | 9 A µ S s.t. t = i 2 A ai }
A simple number-theoretic problem
Theorem: SUBSET-SUM is NP-complete
Note: There is an O(n ⋅ t) time algorithm for Subset Sum.
Does this prove P=NP?

19. VC P SUBSET-SUM
Want to reduce a graph to a set of numbers
Given (G, k), let E = {e0,…,em-1} and V = {1,…,n}
Our subset sum instance (S, t) will have |S| = n+m
“Edge numbers”: For every ej 2 E, put bj = 4j in S
“Node numbers”: For every i 2 V, put ai = 4m + j : i 2 ej 4j in S
Set the target number: t = k ¢ 4m + j=0m-1 (2 ¢ 4j)

20. For every ej 2 E, put bj = 4j in S
For every i 2 V, put ai = 4m + j : i 2 ej 4j in S
Set the target number: t = k ¢ 4m + j=0m-1 (2 ¢ 4j)
Claim: If (G,k)  VC then (S,t)  SUBSET-SUM
Suppose C µ V is a VC with k vertices.
Let S’ = {ai : i 2 C}  {bj : |ej \ C| = 1}
S’ = (node numbers corresponding to nodes in C) plus (edge numbers corresponding to edges covered only once by C)
Claim: The sum of all numbers in S’ equals t
Think of the numbers as being in “base 4”… as vectors with m+1 components

21. i 2 C ai + ej 2 F bj = t = k ¢ 4m + j=0m-1 (2 ¢ 4j)
For every ej 2 E, put bj = 4j in S
For every i 2 V, put ai = 4m + j : i 2 ej 4j in S
Set the target number: t = k ¢ 4m + j=0m-1 (2 ¢ 4j)
Claim: If (S,t)  SUBSET-SUM then (G,k)  VC
Suppose C µ V and F µ E satisfy
i 2 C ai + ej 2 F bj = t = k ¢ 4m + j=0m-1 (2 ¢ 4j)
Claim: C is a vertex cover of size k.
Proof: Subtract out the bj numbers from the above sum. What remains is a sum of the form: i 2 C ai = k ¢ 4m + j=0m-1 (cj ¢ 4j)
where each cj > 0. But cj = number of nodes in C covering ej
This implies C is a vertex cover!

22. Finding Paths - Two Problems
Let G denote a graph, and s and t denote nodes.
SHORTEST PATH
= {(G, s, t, k) | G has a simple path of length < k from s to t }
LONGEST PATH
= {(G, s, t, k) | G has a simple path of length > k from s to t }
Are either of these in P? Are both of them?

23. HAMPATH = { (G,s,t) | G is an directed graph with a Hamiltonian path from s to t}
Theorem: HAMPATH is NP-Complete
(1) HAMPATH  NP
(2) 3SAT P HAMPATH
See Sipser for the proof

24. HAMPATH P LONGEST-PATH
= {(G, s, t, k) | G has a simple path of length > k from s to t }
Can reduce HAMPATH to LONGEST-PATH by observing:
(G, s, t) 2 HAMPATH  (G, s, t, |V|) 2 LONGEST-PATH
Therefore LONGEST-PATH is NP-hard.

25. Coping with NP-Completeness
There are thousands of NP-complete problems
Many are solved all the time !?!
Average Case vs. Worst Case; Heuristics vs. Algorithms [Beyond Worst-Case Analysis (CS264)]
Special cases/parameters that make a problem easy
Approximation Algorithms