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NP Complexity
By
Mussie Araya
What is NP Complexity?
Formal Definition: NP is the set of decision
problems solvable in polynomial time by a nondeterministic Turing machine
Informal Definition: NP contains all decision
problems for which the 'yes'-answers have
simple proofs of the fact that the answer is
indeed 'yes' : More precisely, these proofs have
to be verifiable in polynomial time by a
deterministic machine.
Sub Classes
• P class and NP complete class
– Definition of class P: all decision problems
which can be solved by a deterministic Turing
machine using in polynomial time.
– Definition of NP complete
– Let C be a decision problem:
– C is in NP
– C is NP hard.
NP Hardness
C is NP hard iff every problem that is NP is
reducible to C.
How do I know C is in NP.
Verify possible solution in polynomial time.
How do I know C is NP hard?
Show a known NP Complete problem can be
reduced to C.
Reduction
• Reduction is a a relation that is transitive
and reflexive on P(N), where P(N) is the
power set of N and N is any set.
• Given two elements of P(N), A & B and a
set of functions F from N  N that is
closed under composition:
• A is reducible to B under f iff
– There exist f in F such that for all x in N, x is in
A iff f(x) is in B
More On Reduction
• There exist many reductions
• Interesting one is Turing reduction
• Definition: Given two sets of natural
numbers, we say A is Turing reducible to
B : if there is an oracle that computes the
characteristic function of A when run with
oracle B.
Example of Turing reduction
• Reduction from multiplication to squaring
• Assume we want to compute multiplication
knowing just addition and the following
multiplication function
A × B = ((A + B)^2 − A^2 − B^2)/2
– And For A & B є N we can compute the square A^2
+ B^2 from A*B
– Notice assuming we know multiplication We can
compute the square, thus this type of reduction is
Turing reduction
– Further note that knowing squaring we can compute
multiplication thus multiplication and squaring are
equally hard!
Many One Reduction
• Assuming we have a restriction that we
can only call square once at end of
computation step
– The problem of going from multiplication to
squaring is not possible in such as sqrt(2)
– However under same restriction: we can
compute the square of any number from
single multiplication
– This reduction implies the two problems are
not equally hard!
More On reduction
• In general reduction must be easy. relative
to the complexity of typical problems in the
class [...]
• If the reduction itself were difficult to
compute, an easy solution to the complete
problem wouldn't necessarily yield an easy
solution to the problems reducing to it."
Michael Sipser
SAT: Boolean Satisfiability Problem
• Proven NP by Steven Cook in 1971.
• Decision problem in a form of a Boolean
formula consisting and , or , negation,
variable. Also known as conjunctive
normal form.
• Problem : is there on interpretation that
satisfies the given formula.
• To See if a problem is NP hard reduce to
SAT.
Independent Set Problem
• Given a graph G, an independent set is a subset
of its vertices that are pair wise not adjacent.
• Problem given a graph G, does it have an
Independent set of at least size K?
• Proof of NP completeness.
– Is it in NP?
• Yes because we can check in any sub vertices if an edge is
present between any two vertices.
Is it NP hard? Can SAT be reduced to SAT? If so every
problem can be reduce to SAT hence, every problem can be
reduced to ISP.
Reduction of SAT to ISP
• Create a Boolean formula in CNF.
• Create vertices for every lateral in the
formula. Place an edge between laterals
which are
– 1. negations of each other.
– 2. In the same clause.
Does the resulting graph have an independent
set of size k where k is the number of clauses
iff the original Boolean formula is Satisfiable?
Proof of an equivalence
• Find an interpretation I that satisfies the
Boolean formula.
• Take one clause from each lateral which is
true under I.
• Each clause forms an independent set
because edges only exist between
laterals in the same clause and no
interpretation makes a lateral and its
negation true.
Proof continued
• Assume we have an independent set of size k or
grater.
• It can not contain any laterals in the same clause
because are pair wise adjacent.
• It can not contain lateral and its negation
because there exist an edge between them.
• Therefore it is easy to find an interpretation that
satisfies the corresponding conjunction. Assign
true to their interpretation. QED!
Current Conjecture
• P != NP
– NP complete problems can not be solved in
polynomial time using deterministic Turing
machine
Solution:
Computer scientist use exhaustive, greedy
algorithms and heuristics.
Some Physicist have current research on
developing Quantum Computers
Solutions
• Quantum computers
. More on quantum computing
Assuming classical computer based on a 3
bit register. Bits in register are in single
definite state such as 000.
Assume quantum computer on a register
described by a wave function. Each bit can
exist as superposition of all allowed states.
Quantum Algorithm
• 3 bit register 2^3 states
• Initialize all eight states
• In each step of algorithm, each vector or
state is modified by a unitary operator.
• On termination: value is read from register
via quantum measurement.
• Note the number of classical registers
required to estimate state of n bit quantum
computer is 2^n