April 13th Class Notes - University of Central Florida
April 13th Class Notes Hw# 5 will be worth 50 points and it will be posted tonight or tomorrow 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 Hampath={|G 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 The graph below has a hampath One Hampath for the diagram is
A-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
Hampath={|Gdoesnt 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) Composites ={n|n exists Z+ and there exists a p and q such
that n=p*q} Can easily produce a certificate ie if n=143 p=11 and q=13 Primes={n|there is no way to represent n=p*q, where pa nd q exist in Z+ and P>1 and q>1}
This has been shown to be O(n**6) by AKS Clique={G,K|Graph G has a Clique of size k} The time to determine this would be n choose k A certificate would be all of the nodes in the clique
Subset-sum={(s,target| some subset in s adds up to the target) Brute force method: chack all subsets of S This would take 2**n where |s|=n A certificate could be generated which would include the
subet that adds to the target Example S={7,15,35,19,135,2} target =59 Certificate =7,15,35,2 It is not known if P is a subset of NP or they are equal
Section 7.4 SAT (satisfiability problem) Boolean formula (X1 V X2 V X3) ^(X3V X4) Cook Levin theorem SAT exists in P iff P=NP
Polynomial time mapping problem A< 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 A< pB, Than there exists a polynomial
solution to A To solve A A(w){ 1. 2.
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
Partition < p subset sum S= {3,16,19,25,14,201} F(w)=S= {3,16,19,25,14,201} Target = (3+16+19+25+14+201)/2
All problems in NP are poly time reducible to SAT X 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 prob
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