CS 3343: Analysis of
Algorithms
Lecture 25: P and NP
Some slides courtesy of Carola Wenk
Have seen so far
• Algorithms for various problems
– Sorting, LCS, scheduling, MST, shortest path,
etc
– Running times O(n2) ,O(n log n), O(n), O(nm),
etc.
– i.e., polynomial in the input size
– Those are called tractable problems
• Can we solve all (or most of) interesting
problems in polynomial time?
Example difficult problem
• Traveling Salesperson
Problem (TSP)
– Input: undirected graph with
lengths on edges
– Output: shortest tour that
visits each vertex exactly
once
– An optimization problem
• Best known algorithm:
O(n 2n) time.
• Intractable
2
5 10
8
9
6
4 9
11 3
5
7
Another difficult problem
• Maximum Clique:
– Input: undirected graph
G=(V,E)
– Output: largest subset C of
V such that every pair of
vertices in C has an edge
between them
– An optimization problem
• Best known algorithm:
O(n 2n) time
What can we do ?
• Spend more time designing algorithms for those
problems
– People tried for a few decades, no luck
• Prove there is no polynomial time algorithm for
those problems
– Would be great
– Seems really difficult
What else can we do ?
• Show that those hard problems are
essentially equivalent. i.e., if we can solve
one of them in polynomial time, then all
others can be solved in polynomial time as
well.
• Works for at least 10 000 hard problems
The benefits of equivalence
• Combines research
efforts
• If one problem has
polynomial time
solution, then all of
them do
• More realistically:
Once an exponential
lower bound is
shown for one
problem, it holds for
all of them
P1
P2
P3
Summing up
• If we show that a problem ∏ is equivalent
to ten thousand other well studied
problems without efficient algorithms, then
we get a very strong evidence that ∏ is
hard.
• We need to:
– Identify the class of problems of interest
– Define the notion of equivalence
– Prove the equivalence(s)
Classes of problems: P and NP
• Implicitly, we mean decision problems: answer
YES or NO
– Other problems can be converted into decision
problems
– k-clique: is there a clique of size = k?
• P is set of problems that can be solved in
polynomial time
• NP (nondeterministic polynomial time) is the set
of problems that can be solved in polynomial
time by a nondeterministic computer
Nondeterminism
• Think of a non-deterministic computer as a
computer that magically “guesses” a solution,
then has to verify that it is correct
– If a solution exists, computer always guesses it
– One way to imagine it: a parallel computer that can
freely spawn an infinite number of processes
• Have one processor work on each possible solution
• All processors attempt to verify that their solution works
• If a processor finds it has a working solution
– So: NP = problems verifiable in polynomial time
P and NP
• P = problems that can be solved in polynomial time
• NP = problems for which a solution can be verified in
polynomial time
• K-clique problem is in NP:
– Easy to verify solution in polynomial time
• Is sorting in NP?
– Not a decision problem
• Is sortedness in NP?
– Yes. Easy to verify
P and NP
• Is P = NP?
– The biggest open problem in CS
– Most suspect not
– the Clay Mathematics Institute has offered a
$1 million prize for the first proof
NP
P
or
P = NP?
NP-Complete Problems
• We will see that NP-Complete problems are the
“hardest” problems in NP:
– If any one NP-Complete problem can be solved in
polynomial time…
– …then every NP-Complete problem can be solved in
polynomial time…
– …and in fact every problem in NP can be solved in
polynomial time (which would show P = NP)
– Thus: solve TSP in O(n100) time, you’ve proved that P
= NP. Retire rich & famous.
Reduction
• The crux of NP-Completeness is reducibility
– Informally, a problem ’ can be reduced to another
problem  if any instance of ’ can be “easily
rephrased” as an instance of , the solution to which
provides a solution to the instance of ’
• What do you suppose “easily” means?
• This rephrasing is called transformation
– Intuitively: If ’ reduces to , ’ is “no harder to solve”
than 
Reductions: ∏’ to ∏
x
YES
A for ∏
NO
x’
YES
A’ for ∏’
NO
Reductions: ∏’ to ∏
f
f(x’)= x
YES
A for ∏
NO
x’
YES
A’ for ∏’
NO
Reduction: an example
– ’: Given a set of Booleans, is at least one
TRUE?
– : Given a set of integers, is their sum
positive?
– Transformation: (x1, x2, …, xn) = (y1, y2, …, yn)
where yi = 1 if xi = TRUE, yi = 0 if xi = FALSE
Reductions
• ∏’ is polynomial time reducible to ∏ ( ∏’≤p ∏)
iff there is a polynomial time function f that
maps inputs x’ for ∏’ into inputs x for ∏, such
that for any x’
∏’(x’)=∏(f(x’))
• Fact 1: if ∏P and ∏’ ≤p ∏ then ∏’P
• Fact 2: if ∏NP and ∏’ ≤p ∏ then ∏’NP
• Fact 3 (transitivity):
if ∏’’ ≤p ∏’ and ∏’ ≤p ∏ then ∏” ≤p ∏
Recap
• We defined a large class of interesting
problems, namely NP
• We have a way of saying that one problem
is not harder than another (∏’ ≤p ∏)
• Our goal: show equivalence between hard
problems
Showing equivalence between
difficult problems
• Options:
– Show reductions between all pairs
of problems
– Reduce the number of reductions
using transitivity of “≤”
TSP
Clique
P3
∏’
P4
P5
Showing equivalence between
difficult problems
• Options:
– Show reductions between all pairs
of problems
– Reduce the number of reductions
using transitivity of “≤”
– Show that all problems in NP are
reducible to a fixed ∏.
To show that some
problem ∏’NP is equivalent to all
difficult problems, we
only show ∏ ≤p ∏’.
TSP
Clique
∏
P3
∏’
P4
P5
The first problem ∏
• Boolean satisfiability problem (SAT):
– Given: a formula φ with m clauses over n variables,
e.g., x1v x2 v x5 , x3 v ¬ x5
– Check if there exists TRUE/FALSE assignments to
the variables that makes the formula satisfiable
• Any SAT can be rewritten into conjunctive
normal form (CNF)
– a conjunction of clauses, where a clause is a
disjunction of literals
• A literal is a Boolean variable or its negation
– E.g.
(AB)  (BCD) is in CNF
(AB)  (ACD) is not in CNF
SAT is NP-complete
• Fact: SAT NP
• Theorem [Cook’71]: For any ∏’NP ,
we have ∏’ ≤ SAT.
• Definition: A problem ∏ such that for any
∏’NP we have ∏’ ≤p ∏, is called NP-hard
– Not necessarily decision problem
• Definition: An NP-hard problem that belongs
to NP is called NP-complete
• Corollary: SAT is NP-complete.
•SAT: the boolean satisfiability problem for formulas in conjunctive normal form
•0-1 INTEGER PROGRAMMING
•CLIQUE (see also independent set problem)
Karp's 21 NP-complete
•SET PACKING
problems, 1972
•VERTEX COVER
Since then thousands
•SET COVERING
problems have been
•FEEDBACK NODE SET
•FEEDBACK ARC SET
proved to be NP•DIRECTED HAMILTONIAN CIRCUIT
complete
•UNDIRECTED HAMILTONIAN CIRCUIT
•3-SAT
•CHROMATIC NUMBER (also called the Graph Coloring Problem)
•CLIQUE COVER
•EXACT COVER
•HITTING SET
•STEINER TREE
•3-dimensional MATCHING
•KNAPSACK (Karp's definition of Knapsack is closer to Subset sum)
•JOB SEQUENCING
•PARTITION
•MAX-CUT
Clique again
• Clique (decision variant):
– Input: undirected graph
G=(V,E), K
– Output: is there a subset C
of V, |C|≥K, such that every
pair of vertices in C has an
edge between them
SAT ≤p Clique
x’
• Given a SAT formula φ=C1,…,Cm over
x1,…,xn, we need to produce
G=(V,E) and K,
f(x’)=x
such that φ satisfiable iff G has a clique of
size ≥ K.
SAT ≤p Clique reduction
• For each literal t occurring in φ, create a
vertex vt
• Create an edge vt – vt’ iff:
– t and t’ are not in the same clause, and
– t is not the negation of t’
SAT ≤p Clique example
Edge vt – vt’  • t and t’ are not in the same clause, and
• t is not the negation of t’
• Formula: (x1 V x2 V x3 )  ( ¬ x2 v ¬ x3)  (¬ x1 v x2)
• Graph:
x1
¬x2
¬ x3
x2
x3
¬ x1
x2
• Claim: φ satisfiable iff G has a clique
of size ≥ m
Proof
Edge vt – vt’  • t and t’ are not in the same clause, and
• t is not the negation of t’
• “→” part:
– Take any assignment that
satisfies φ.
E.g., x1=F, x2=T, x3=F
– Let the set C contain one
satisfied literal per clause
– C is a clique
x1
¬x2
¬ x3
x2
x3
x2
¬ x1
Proof
Edge vt – vt’  • t and t’ are not in the same clause, and
• t is not the negation of t’
• “←” part:
– Take any clique C of size ≥ m
(i.e., = m)
– Create a set of equations that
satisfies selected literals.
E.g., x3=T, x2=F, x1=F
– The set of equations is consistent
and the solution satisfies φ
x1
¬x2
¬ x3
x2
x3
x2
¬ x1
Altogether
• We constructed a reduction that maps:
– YES inputs to SAT to YES inputs to Clique
– NO inputs to SAT to NO inputs to Clique
• The reduction works in polynomial time
• Therefore, SAT ≤p Clique →Clique NPhard
• Clique is in NP → Clique is NP-complete
Independent set (IS)
• Input: undirected graph
G=(V,E)
• Output: is there a subset
S of V, |S|≥K such that no
pair of vertices in S has
an edge between them
• Want to show: IS is NPcomplete
Clique ≤ IS
x’
• Given an input G=(V,E), K to
Clique, need to construct an
input G’=(V’,E’), K’ to IS,
f(x’)=x
such that G has clique of size
≥K iff G’ has IS of size ≥K’.
• Construction: K’=K,V’=V,E’=E
• Reason: C is a clique in G iff it
is an IS in G’s complement.
Classes of problems