April 13th Class Notes

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April 13th Class Notes

• Hw 5 will be worth 50 points and it will be
  posted tonight or tomorrow

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Definition of the Class NP

• 1)Language that can be accepted on a
  nondeterministic TM in polynomial time. If it
  isnt in the language it may not reject in
  polynomial time
• 2)Given a certificate, one can verify in
  polynomial time whether or not an input is in the
  language

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HampathltGgtG has a Hamiltonian path

• This doesnt have a Hampath since there are three
  vertices of degree one and we would need a way in
  and a way out

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The graph below has a hampath
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One Hampath for the diagram isA-B-D-F-E-C

• This path can be used to represent a certificate.
  Verifying this path would take less than
  polynomial time
• We can also prove that this can NP by checking
  all path options at once

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HampathltGgtGdoesnt have a hamiltonion path

• There is not an easy certificate that can be
  generated to prove that no path exists
• This is Co-NP (compliment NP)

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Composites nn exists Z and there exists a p
and q such that npq

• Can easily produce a certificate ie if n143 p11
  and q13

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Primesnthere is no way to represent npq,
where pa nd q exist in Z and Pgt1 and qgt1

• This has been shown to be O(n6) by AKS

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CliqueG,KGraph G has a Clique of size kThe
time to determine this would be n choose kA
certificate would be all of the nodes in the
clique
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Subset-sum(s,target some subset in s adds up
to the target)

• Brute force method chack all subsets of S
• This would take 2n where sn
• A certificate could be generated which would
  include the subet that adds to the target
• Example S7,15,35,19,135,2 target 59
• Certificate 7,15,35,2

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It is not known if P is a subset of NP or they
are equal
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Section 7.4

• SAT (satisfiability problem)
• Boolean formula (X1 V X2 V X3) (X3V X4)
• Cook Levin theorem SAT exists in P iff PNP

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Polynomial time mapping problem

• Alt pB iff W exists in A iff f(w) exists in B
• And f is computable in polynomial time
• If a polynomial solution to B was known and Alt
  pB, Than there exists a polynomial solution to A

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To solve A

• A(w)
• Calculate f(w)
• Run the solution for B with input f(w)
• Given a solution to B we solve A by callin
  the soln to B a polynomial number of times

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Partition lt p subset sum

• S 3,16,19,25,14,201
• F(w)S 3,16,19,25,14,201
• Target (316192514201)/2

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All problems in NP are poly time reducible to SAT
XltpSAT

• If all problems in the NP reduce (in polytime) to
  a problem Y. Then Y is known as NP-Complete and Y
  has to be in the NP. Also SAT is NP-Complete
• To prove a problem is NP_Complete reduce it from
  a known NP-Complete problem
• Any probltSAT ltpX