NP (1% extra credit on final exam) Let ONE-SHORT = {F | F is a 3cnf-formula wher
ID: 3827122 • Letter: N
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NP (1% extra credit on final exam) Let ONE-SHORT = {F | F is a 3cnf-formula where some assignment satisfies all but one clause}. • Show that ONE-SHORT is in NP by sketching a polynomial time verifier for it. • Show that ONE-SHORT is NP-hard by giving a polynomial time reduction from 3SAT. Hint: Remember that the input for 3SAT F needs to be converted to an input for ALMOST-3SAT F such that F 3SAT F ONE-SHORT NP (1% extra credit on final exam) Let ONE-SHORT = {F | F is a 3cnf-formula where some assignment satisfies all but one clause}. • Show that ONE-SHORT is in NP by sketching a polynomial time verifier for it. • Show that ONE-SHORT is NP-hard by giving a polynomial time reduction from 3SAT. Hint: Remember that the input for 3SAT F needs to be converted to an input for ALMOST-3SAT F such that F 3SAT F ONE-SHORT NP (1% extra credit on final exam) Let ONE-SHORT = {F | F is a 3cnf-formula where some assignment satisfies all but one clause}. • Show that ONE-SHORT is in NP by sketching a polynomial time verifier for it. • Show that ONE-SHORT is NP-hard by giving a polynomial time reduction from 3SAT. Hint: Remember that the input for 3SAT F needs to be converted to an input for ALMOST-3SAT F such that F 3SAT F ONE-SHORTExplanation / Answer
To convert a 3SAT instance to an INDSET instance, we need a graph G and number n such that an independent set of size at least n in G
gives us a way to choose which literal in each clause of should be true,
doesn't simultaneously choose a literal and its negation, and
has size polynomially large in the length of the formula .
The complexity class NP (nondeterministic polynomial time) contains all problems that can be solved in polynomial time by a single-tape NTM. Formally:
NP = k = 0 NTIME(nk)
Equivalently: A language is in NP iff there is a polynomial-time verifier for it.
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