NP Complexity

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NP Complexity

• By
• Mussie Araya

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What is NP Complexity?

• Formal Definition NP is the set of decision
  problems solvable in polynomial time by a
  non-deterministic 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.

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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.

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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.

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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

                      
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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.

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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 - A2 - B2)/2
• And For A B ? N we can compute the square
  A2 B2 from AB
• 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!

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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!

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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

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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.

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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.

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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?

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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.

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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!

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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

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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.

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Quantum Algorithm

• 3 bit register 23 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
  2n