NPCompleteness Nondeterministic Polynomial Completeness

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NP-Completeness (Nondeterministic Polynomial
Completeness)

• Sushanth Sivaram Vallath
• Z. Joseph

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Overview

• NP-Completeness
• Longest Path Problem
• Hamiltonian Cycle Problem
• Reduction
• Decision Problems
• More Problems
• HCP and Traveling Salesman
• 3SAT and Clique Problem

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Polynomial (P) Problems

• Are solvable in polynomial time
• Are solvable in O(nk), where k is some constant.
• Most of the algorithms we have covered are in P

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Nondeterministic Polynomial (NP) Problems

• This class of problems has solutions that are
  verifiable in polynomial time.
• Thus any problem in P is also NP, since we would
  be able to solve it in polynomial time, we can
  also verify it in polynomial time

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NP-Complete Problems

• Is an NP-Problem
• Is at least as difficult as an NP problem (is
  reducible to it)
• More formally, a decision problem C is
  NP-Complete if
• C is in NP
• Any known NP-hard (or complete) problem p C
• Thus a proof must show these two being satisfied

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Exponential Time Algorithms
NP-Complete
Exponential-time (Hard)
22
2n
22
n2
Polynomial-time (Easy)
nlogn
F(n)
vn
Log(n)
a(n)
n
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Examples

• Longest path problem (similar to Shortest path
  problem, which requires polynomial time)
  suspected to require exponential time, since
  there is no known polynomial algorithm.
• Hamiltonian Cycle problem Traverses all vertices
  exactly once and form a cycle.

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Reduction

• P1 is an unknown problem (easy/hard ?)
• P2 is known to be difficult
• If we can easily solve P2 using P1 as a
  subroutine then P1 is difficult
• Must create the inputs for P1 in polynomial time.
• P1 is definitely difficult because you know
  you cannot solve P2 in polynomial time unless you
  use a component that is also difficult (it cannot
  be the mapping since the mapping is known to be
  polynomial)

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• Longest Path problem
• Input Weighted graph, s, t
• Output The longest path between s and t
• Hamiltonian Cycle
• Input Unweighted graph
• Output Yes/No

Mapping is done by taking any edge. The nodes it
connects are used to call Longest Possible Path.
This will give Hamiltonian cycle if LPP
traverses same Number of nodes as number in graph.
Hamiltonian Cycle Problem
Unweighed graph
m times
Weighted graph(weight 1)
Longest Path Problem
Weight
Longest path
s,t
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Where does NP Complete lie?
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Decision Problems

• Represent problem as a decision with a boolean
  output
• Easier to solve when comparing to other problems
• Hence all problems are converted to decision
  problems.
• P all decision problems that can be solved in
  polynomial time
• NP all decision problems where a solution is
  proposed, can be verified in polynomial time
• NP-complete the subset of NP which are the
  hardest problems

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Alternative Representation

• Every element p in P1 can map to an element q in
  P2 such that p is true (decision problem) if and
  only if q is also true.
• Must find a mapping for such true elements in P1
  and P2, as well as for false elements.
• Ensure that mapping can be done in polynomial
  time.
• Note P1 is unknown, P2 is difficult

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Polynomial Reductions
Elements that are true
Elements that are true
Polynomial-time Matching-algorithm
Thus solution can be expressed in terms of
solutions of P2, which is known to be
NP-Complete. Thus P1 is NP Complete as well.
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Traveling Salesman problem

• Input
• Weighted graph G
• Length l
• Output
• Yes if a circuit exists of length l
• No otherwise
• TSP can be reduced from Hamiltonian cycle. TSP
  can be represented as a subroutine of HC, so as
  to represent TSP as NPC.

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Cooks Theorem

• Stephen Cook (Turing award winner) found the
  first NP-Complete problem, 3SAT.
• Basically a problem from Logic.
• Generally described using Boolean formula.
• A Boolean formula involves AND, OR, NOT operators
  and some variables.
• Ex (x or y) and (x or z), where x, y,
  z are boolean variables.
• Problem Definition Given a boolean formula of m
  clauses, each containing n boolean variables,
  can you assign some values to these variables so
  that the formula can be true?
• Boolean formula (x v y v ?) ? (x v y v ?)
• Try all sets of solutions. Thus we have
  exponential set of possible solutions. So it is a
  NPC problem.
• Having one definite NP-Complete problem means
  others can also be proven NP-Complete, using
  reduction.

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3SAT applied to Clique Problem

• Clique Problem It is a subgraph of a graph which
  contains all possible edges between each pair of
  vertices in the subgraph.
• Is NP complete if it is reducible to 3SAT.
• 1) Represent as decision tree problem
• Input Graph G,K
• output Yes, if clique with at least K nodes
  exists,
• No, otherwise

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Clique (Complete graph) in terms of 3SAT

• For example (x1 OR x2 OR x3) AND (x1 OR x2
  OR X3) AND (x1 OR x2 OR x3). This is a 3SAT
  problem and we will create a graph from it. We
  will put all possible edges, except edges in the
  same clause and between a variable and its
  negation.
• Does there exist a subgraph of l nodes which is
  a clique?
• Graph showing
  Clique

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A node and its negations are not connected
(x1 v x2 v x3) ? (x1 v x2 v x3) ? (x1 v x2 v x3)
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Reduction of clique from 3SAT

• 3SAT is satisfiable, if one variable from each
  clause should be true. Suppose there are m
  clauses, then it is satisfiable if the equivalent
  graph has a clique of size atleast m.
• If it is given that 3SAT problem is satisfiable,
  then we can select one variable from each clause
  to form a clique of size atleast m, as there
  will be atleast m inter-connected nodes (one
  true node from each clause).
• Thus clique problem can be reduced to a
  NP-complete problem from 3SAT.

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• References
• 1) NP-Completeness - TUSHAR KUMAR J.
• RITESH BAGGA
• 2) Introduction to Algorithms (Cormen et al)
• 3) Wikipedia.com