1. NP-complete problems The Chinese University of Hong Kong Fall 2008
CSC 3130: Automata theory and formal languages
NP-complete problems
Andrej Bogdanov

2. Polynomial-time reductions
Language L polynomial-time reduces to L’ if there exists a polynomial-time computable map R that takes an instance x of L into instance y of L’ s.t.
x ∈ L if and only if y ∈ L’ L
(IS) L’
(CLIQUE) x
(G, k) R y
(G’, k’)
x ∈ L
y ∈ L’
(G has IS of size k)
(G’ has clique of size k’)

3. The Cook-Levin Theorem
Every L ∈ NP reduces to SAT
SAT = {f: f is a satisfiable Boolean formula}
f =
(x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1)
f =
(x1∨x2 ) ∧ (x1) ∧ (x2)
is satisfiable
is not satisfiable
x1 = T x2 = F x3 = T x4 = T

4. NP-hardness Language L is NP-hard if every L’ in NP reduces to L
Intuitively, NP-hard means “harder than all of NP”
The Cook-Levin Theorem says
SAT NP
SAT is NP-hard P

5. NP-complete L is NP-complete if L is NP-hard and L ∈ NP
Intuitively, NP-complete means “hardest in NP”
Recall that SAT ∈ NP, so SAT is NP-complete
SAT NP
If SAT ∈ P, then P = NP P

6. Proof of Cook-Levin Theorem
Every L ∈ NP reduces to SAT
To prove it, we have to describe a reduction R: w R
Boolean formula f
f is satisfiable
w ∈ L

7. Proof of Cook-Levin Theorem
All we know about L: It has a poly-time NTM M
Let’s look at computation tableau of M on input w S w M q0 w1 w2
S-th configuration symbol at time T T
Since M is nondeterministic,
there may be many possible tableaus
qacc …

8. Proof of Cook-Levin TheoremS
n = length of input w q0 w1 w2
height of tableau is p(n)
for some polynomial p T u
width is at most p(n)
k possible tableau symbols
qacc …
true,
if the (T, S) cell of tableau contains u
1 ≤ S ≤ p(n)
xT, S, u =
1 ≤ T ≤ p(n)
false,
if not
1 ≤ u ≤ k

9. Proof of Cook-Levin Theoremw R
Boolean formula f
f is satisfiable
w ∈ L
We will design a formula f such that:
variables of f :
xT, S, u
assignment to xT, S, u
way to fill up the tableau
satisfying assignment
accepting computation tableau
f is satisfiable
M accepts w

10. Proof of Cook-Levin Theorem
true,
if the (T, S) cell of tableau contains u
1 ≤ S ≤ p(n)
xT, S, u =
1 ≤ T ≤ p(n)
false,
if not
1 ≤ u ≤ k
We want to construct (in time poly(n)) a formula f :
true,
if the xs represent a valid accepting tableau
f(x1, 1, 1, ..., xp(n), p(n), k) =
false,
if not

11. Proof of Cook-Levin TheoremS
f = fcell ∧ f0 ∧ fmove ∧ facc q0 w1 w2 T
fcell:
“Every cell contains
exactly one symbol” u
f0:
“The first row is
q0w1w2...wk☐...☐”
qacc …
fmove:
“The moves between rows follow the transitions of M”
facc:
“qacc appears somewhere in the last row”

12. Proof of Cook-Levin Theorem
Desired meaning
Implementation:
fcell:
“Every cell contains exactly one symbol”
or: “Exactly one of xS, T, 1 ∨ ... ∨ xS, T, k is true”
fcell = fcell 1, 1 ∧ ... ∧ fcell p(n), p(n)
where
fcell T, S = (xT, S, 1 ∨ ... ∨ xT, S, k)
∧ (xT, S, 1 ∨ xT, S, 2)
∧ (xT, S, 1 ∨ xT, S, 3)
∧ ∧ (xT, S, k-1 ∨ xT, S, k)
at least one symbol
no two symbols

13. Proof of Cook-Levin Theorem
Desired meaning
Implementation:
f0:
“The first row is q0w1w2...wk☐...☐”
facc:
“qacc appears somewhere in the last row”
f0 =
x1, 1, q0 ∧ x1, 1, w1 ∧ x1, 1, w2 ∧ ... ∧ x1, p(n),☐
facc =
xp(n), 1, qacc ∨ xp(n), 2, qacc ∨ ... ∨ xp(n), p(n), qacc

14. Valid and invalid windows
… 6q3t0u0 …
… 0k6t0q0 …0
… 6t3t0u0 …
… 0t6t0u0 …0
invalid window
valid window
… 6c3a0t0 …
… 0c6a0p0 …0
… 6t3q3u0 …
… 0t6a0q7 …0
valid if d(q3, u) = (q7, a, R)
invalid window
… 6t3t0u0 …
… 0t6t0q3 …0
… 6c3a0t0 …
… 0b6a0t0 …0
valid window
valid window

15. Proof of Cook-Levin Theorem
Desired meaning
Implementation: q0 a a b b q3 a b
fmove:
“The moves between rows follow transitions of M” b c q7 b
fmove 2, 2
qacc …
fmove = fmove 1, 1 ∧ ... ∧ fmove p(n)-3, p(n)-3 ∨
fmove T, S =
(xT, S, a1 ∧ xT, S+1, a2 ∧ xT, S+2, a3
∧ xT+1, S, a4 ∧ xT+1, S+1, a5 ∧ xT+2, S+1, a6)
over all
valid windows a1 a2 a3 a4 a5 a6

16. Other NP-complete problems
CLIQUE = {(G, k): G is a graph with a clique of k vertices}
IS = {(G, k): G is a graph with an independent set of k vertices}
VC = {(G, k): G is a graph with a vertex cover of k vertices}
CLIQUE, IS and VC are NP-complete
CLIQUE IS
SAT VC NP

17. Proving NP-hardness
To show L is NP-hard, it is enough to reduce from some L’ we already know is NP-hard
For now we can take L’ = SAT
To show L is NP-complete, we also need to argue that L is in NP
This is usually the easy part
roadmap:
CLIQUE VC IS
3SAT
SAT

18. 3SAT SAT = {f: f is a satisfiable Boolean formula}
3SAT = {f: f is a satisfiable Boolean formula in conjunctive normal form with at most 3 distinct literals per clause}
(x2∨(x1∧x2 ))∧(x1∧(x1∨x2 ))
literal: xi or xi
gates
CNF: AND of ORs of literals
(conjunctive normal form)
(x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1)
literals
clause
3CNF: CNF with ≤3 lit/clause

19. NP-hardness of 3SAT Theorem 3SAT is NP-hard
Proof: We describe a reduction R from SAT
3SAT is NP-hard
Boolean formula f R
3CNF formula f’
f’ is satisfiable
f is satisfiable

20. Reducing SAT to 3SAT Example: f = (x2∨(x1∧x2 ))∧(x1∧(x1∨x2 )) x4 x5 x7
AND OR
NOT x2 x1
T T T T T T F F T F T F T F F T F T T F F T F T F F T F F F F T
(x4∨x5∨x7)
(x4∨x5∨x7)
(x4∨x5∨x7)
∧(x4∨x5∨x7)
We give extra variables to every gate (“wire”)

21. Reducing SAT to 3SAT Step 1: Add variable xn+j for each gate Gj in f
Boolean formula f R
3CNF formula f’
Step 1: Add variable xn+j for each gate Gj in f
Step 2: Write 3CNF fj for each gate Gj, j = {1, ..., t}
Step 3: Output
requires that output of f is true
f’ = fn+1 ∧ fn+2 ∧ ... ∧ ft ∧ xn+t

22. Reducing SAT to 3SAT
Boolean formula f R
3CNF formula f’
f’ is satisfiable
f is satisfiable
Every satisfying assignment of f extends uniquely to a satisfying assignment of f’
Conversely, every satisfying assignment of f’ must contain a satisfying assignment of f

23. Independent set Theorem IS is NP-hard
IS = {(G, k): G is a graph with an independent set of k vertices}
Theorem
IS is NP-hard
CLIQUE VC
An independent set is a subset of vertices so that no pair is connected 1 2 IS
3SAT
{1, 2}, {1, 3}, {4} are independent sets ✓
SAT 3 4

24. Reducing 3SAT to IS Proof: We describe a reduction from 3SAT to IS
3SAT = {f: f is a satisfiable Boolean formula in 3CNF}
IS = {(G, k): G is a graph with an independent set of k vertices}
3CNF formula f R
(G, k)
G has an independent set of size k
f is satisfiable

25. Reducing 3SAT to IS Example: f = (x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1) etc.
TTT
all interconnected
TTF TT x1
TFT T TF
TFF
FTT F FT
etc.
FTF FF
FFT
FFF
Put an edge for every inconsistency

26. Reducing 3SAT to IS Example: f = (x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1)
satisfying assignment of f
IS of size 3 in G
x1x2
TTT
TTF TT
TFT x1 TF
TFF T
x1 = T x2 = F x3 = T x4 = T
FTT
any IS of size 3 in G
satisfying assignment of f FF
FFT G
FFF
edges = inconsistencies

27. Reducing 3SAT to IS
3CNF formula f R
(G, k)
G has a vertex for every clause Ci of f and every satisfying assignment of Ci
G has an edge between any two vertices that represent inconsistent assignments
k is the number of clauses in f

28. Reducing 3SAT to IS
3CNF formula f R
(G, k)
f is satisfiable
G has an IS of size k
Every satisfying assignment of f gives an independent set of size k in G
Conversely, from every IS of size k in G we can “extract” a consistent and satisfying assignment of f

29. Vertex cover Theorem VC is NP-hard
VC = {(G, k): G is a graph with a vertex cover of size k}
Theorem
VC is NP-hard
CLIQUE VC 1 2
A vertex cover is a set of vertices that touches (covers) all edges ✓ IS ✓
3SAT ✓
{2, 4}, {3, 4}, {1, 2, 3}
are vertex covers
SAT 3 4

30. Reducing IS to VC Proof: We describe a reduction from IS to VC Example
(G, k)
(G’, k’)
G has an IS of size k
G’ has a VC of size k’ 1 2
vertex covers
independent sets
{2, 4}, {3, 4},
{1, 2, 3}, {1, 2, 4},
{1, 3, 4}, {2, 3, 4},
{1, 2, 3, 4}
∅, {1}, {2}, {3}, {4},
{1, 3}, {2, 3} 3 4

31. Reducing IS to VC
Claim
Proof
S is an independent set of G if and only if S is a vertex cover of G
S is an independent set of G
no edge has both endpoints in S
every edge has an endpoint in S
S is a vertex cover of G

32. Reducing IS to VC R
(G, k)
(G’, k’)
Set G’ = G, k’ = n – k (n = number of vertices) by previous Claim.
G has an IS of size k
G’ has a VC of size k’

33. The ubiquity of NP-complete problems
We saw a few examples of NP-complete problems, but there are many more
A surprising fact of life is that most CS problems are either in P or NP-complete
A 1979 book by Garey and Johnson lists 100+ NP-complete problems