Farley Lai, 00764474
Theory of Computation, Homework 5
7.6
i.) union
Without loss of generality, let
and
be the TMs that decide two languages
and
in
M
1
M
2
L
1
L
2
P. Then we construct a TM
that decides the union of
and
in polynomial time.
M
L
1
L
2
”on input
M
=
w
1.
Run
on
. Accept if it accepts.
M
1
w
2.
Run
on
. Accept if it accepts.
M
2
w
3.
Otherwise, reject.”
Since stage1 and stage2 both run in polynomial time, TM
can decide the union of
and
M
L
1
L
2
in polynomial time. Hence,
is closed under union.
P
ii.) concatenation
Without loss of generality, let
and
be the TMs that decide two languages
and
in
M
1
M
2
L
1
L
2
P. Then we construct a TM
that decides the concatenation of
and
in polynomial time.
M
L
1
L
2
”on input
M
=
w
1.
For each possible split of
w
=
w w
1
2
a.
Run
on
and run
on
. Accept if both accept.
M
1
w
1
M
2
w
2
2.
Otherwise, reject.”
Assume the
. There are
possible splits of
at most. Besides, the sum of the time
w
j
j
=
n
(
n
)
O
w
complexities of
and
is also polynomial. Thus, step1 run in polynomial time and TM
M
1
M
2
M
can decide the concatenation of
and
in polynomial time. Hence,
is closed under
L
1
L
2
P
concatenation.
iii.) complement
Without loss of generality, let
be the TM that decides a language
in P. Then we construct
M
L
a TM
that decides the complement of
in polynomial time.
M
0
L
”on input
M
0
=
w
1.
Run
on
M
w
2.
Accept if it rejects. Otherwise, reject if it accepts.”
Since step1 run in polynomial time, TM
can decide the complement of
in polynomial time.
M
0
L
Hence,
is closed under complement.
P
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View Full Document7.7
i.) union
Without loss of generality, let
and
be the NTMs that decide two languages
and
N
1
N
2
L
1
L
2
in NP. Then we construct an NTM
that decides the union of
and
in polynomial time.
N
L
1
L
2
”on input
N
=
w
1.
Run
on
nondeterministically. Accept if it accepts.
N
1
w
2.
Run
on
nondeterministically. Accept if it accepts.
N
2
w
3.
Otherwise, reject.”
Since stage1 and stage2 both run in polynomial time nondeterministically, NTM
can decide
N
the union of
and
in polynomial time. Hence,
is closed under union.
L
1
L
2
P
N
ii.) concatenation
Without loss of generality, let
and
be the NTMs that decide two languages
and
N
1
N
2
L
1
L
2
in NP. Then we construct an NTM
that decides the concatenation of
and
in polynomial
N
L
1
L
2
time.
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 Spring '13
 zhang
 poly nomial time

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