Transcript The Class NP

April
th
13
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 isn’t 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>|G has a Hamiltonian path}
• This doesn’t 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={<G>|Gdoesn’t 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<pSAT
• 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 prob<SAT <pX