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95
(b)
It is possible to compute
x
n
in
log
n
time, and the rest of the formula
requires a constant number of multiplications. Thus, the number of
multiplications required is polynomial on the input size.
15.5
First, we should note that, for both problems, the decision problem is being
considered.
TRAVELING SALESMAN is
NP
complete if (1) it is in
NP
, and (2) it is
NP
hard. Proving (1) is easy, just provide a nondeterministic polynomial
time algorithm. To prove (2), we will reduce to TRAVELING SALESMAN
from the known
NP
complete problem HAMILTONIAN CYCLE. First, we
transform an input to HAMILTONIAN CYCLE into an input to TRAVEL
ING SALESMAN by giving each edge of the input graph an arbitrary dis
tance of 1, and then picking any arbitrary (large) number for the total dis
tance to beat. Then if TRAVELING SALESMAN returns “YES”, we know
that there exists a Hamiltonian cycle in the graph since a salesman’s circuit
is a Hamiltonian cycle. If TRAVELING SALESMAN returns “NO” then no
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This note was uploaded on 12/27/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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