Theory of Computation, Homework 3 Sample Solution. 3.8 b.) The following
machine M will do: M = "On input string : 1. Scan the tape and mark the first 1
which ...
Theory of Computation, Homework 3 Sample Solution 3.8 b.) The following machine M will do: M = "On input string : 1. Scan the tape and mark the first 1 which has not been marked. If no unmarked 1 is found, go to step 5. Otherwise, move the head back to the front of the tape. 2. Scan the tape and mark the first 0 which has not been marked. If no unmarked 0 is found, reject. 3. Move on to mark the next unmarked 0. If no such unmarked 0 is found, reject. 4. Move the head back to the front of the tape and go to step 1. 5. Move the head back to the front of the tape. Scan the tape to see if there is any unmarked 0 found. If yes, reject. Otherwise, accept. c.) This machine is identical to 3.8b) if we switch “reject” with “accept”. M = "On input string : 1. Scan the tape and mark the first 1 which has not been marked. If no unmarked 1 is found, go to step 5. Otherwise, move the head back to the front of the tape. 2. Scan the tape and mark the first 0 which has not been marked. If no unmarked 0 is found, accept. 3. Move on to mark the next unmarked 0. If no such unmarked 0 is found, accept. 4. Move the head back to the front of the tape and go to step 1. 5. Move the head back to the front of the tape. Scan the tape to see if there is any unmarked 0 found. If yes, accept. Otherwise, reject.
3.12 be a Turing machine with left reset and Let simulate all the operations that can perform, recognizable language. Obviously,
be an ordinary Turing machine. If recognizes the class of Turing-
can simulate
of
can
without
problems. How simulates of is described as follows. 1. overwrite with a marked 2. reset to the left-hand end 3. shift the whole tape one cell to the right but keep the mark in the same position 4. reset to the left-hand end 5. scan right to find the marked symbol; the next move will treat the marked symbol as a normal symbol. We assume the following during the shifting for Step 3: a. We use states to remember the symbol to be shifted right. b. The first symbol of the tape after the shift is a symbol not used by the original M. c. If the current symbol is c and the next symbol is the marked b, after the shifting, c becomes marked but b is not marked. d. The shifting stops when we see a blank symbol.
3.15 b.) Let and be two decidable languages and and be the corresponding TMs. The aim is to construct a TM based on and such that the concatenation is also decidable. Since a given input concatenation of strings and has finite possible partitions, a nondeterministic TM is chosen to simplifies the description. NTM ”On input 1. Nondeterministically split into and 2. run on and on 3. accept if both accepts and accepts ; reject otherwise” accepts an input
Obviously, the NTM and
accepts . Besides,
iff there exists a split of
eventually halts because
and
is decidable since there exists an NTM
Therefore,
such that
accepts
are both deciders.
which decides
.
c.) Let
be a decidable language and be the corresponding TM. The aim is to construct a TM based on such that is also decidable. Since a given input has finite possible combinations of strings where and , a nondeterministic TM is chosen to simplify the description. NTM ”On input 1. accept if 2. if 3. run
, nondeterministically split on for all i.
4. accept if
accepts all
,
into
, where
is not empty.
; reject otherwise”
Obviously, the NTM accepts an input iff or there exists a combination of such that accepts . Besides, eventually halts because is a decider. Therefore, is decidable since there exists an NTM which decides .
d.) Let
be a decidable language and based on such that is described as follows.
TM
be the corresponding TM. The aim is to construct a TM
, the complement of
, is also decidable. The resulting TM
”On input 1. run on rejects 2. accept if 3. reject if accepts
Obviously,
accepts an input
is a decider. Therefore,
iff
rejects
. Besides,
eventually halts because
is decidable since there exists a TM
which decides
.
e.) and be two decidable languages and and be the corresponding TMs. The Let aim is to construct a TM based on and such that the intersection is is described as follows. also decidable. The resulting TM ”On input 4. run and 1. accept if both Obviously, the NTM eventually halts because since there exists a TM
on and
accepts
accepts an input and
; reject otherwise” iff
is accepted by
are both deciders. Therefore,
which decides
.
and
. Besides, is decidable
3.16 b.) and be two Turing-recognizable languages and and be the corresponding Let TMs. The aim is to construct a TM based on and such that the concatenation is also Turing-recognizable. Since a given input concatenation of strings and has finite possible partitions, a nondeterministic TM is chosen to simplify the description. NTM 1. 2. 3. 4.
”On input Nondeterministically split into on input run reject if halts and rejects run on input
5. accept if
accepts ; reject if
and
halts and rejects
recognizes an input
Obviously, the NTM accepts
and
accepts . However,
are not deciders. Therefore, which recognizes
iff there exists a partition of
such that
may loop forever on some input because
is Turing-recognizable since there exists an NTM
.
c.) Let be a Turing-recognizable language and construct a TM based on such that
be the corresponding TM. The aim is to is also Turing-recognizable. Since a given input
has finite possible combinations of strings where nondeterministic TM is chosen to simplify the description.
,
,a
NTM ”On input 1. accept if 2. if 3. run
nondeterministically split on for all i.
4. accept if 5. reject if
such that
recognizes an input
accepts
is not a decider. Therefore, recognizes
.
into
accepts all , halts and rejects for any
Obviously, the NTM
and
. However,
, where
is not empty.
; ” iff
or there exists a combination of
may loop forever on some input because
is Turing-recognizable since there exists an NTM
which
d.) and be two Turing-recognizable languages and and be the corresponding Let TMs. The aim is to construct a TM based on and such that the intersection is also Turing-recognizable. The resulting TM is described as follows.
1. 2. 3. 4.
”On input run on rejects if run on accept if
Obviously, the TM However, Therefore, .
halts and rejects accepts
; reject if
recognizes an input
halts and rejects” iff
is accepted by
may loop forever on some input because
and
and then
.
both are not deciders.
is Turing-recognizable since there exists a TM
which recognizes
4.4 Since follows.
is just a special case of
, it is possible to adapt TM S for
as
, where is a CFG and is an empty string: TM S = "On input 1. Convert G to an equivalent grammar in Chomsky normal form. 2. If “S -> “ is a production rule in Chomsky normal form, accept; if not, reject." 4.10 To decide
is to determine if there exist strings generated by
at least the pumping length . Let The construction of TM TM
be a CFG for and design a TM is as follows, deciding
with lengths that decides
:
= “On input: < > 1. convert to Chomsky normal form from , where is the number of variables in 2. calculate the pumping length p = (in Chomsky normal form, each rule has 0 or two variables at the right, so the pumping length is that is computable) 3. accept if produces a string of length at least because the string can be pumped to generate infinitely other strings 4. otherwise, reject”
For step 3, it is decidable since it is possible to construct a DFA
such that the regular
language recognized by is the set of strings of lengths at least . Let which is a CFL from Problem 2.18a and be the CFG of Then TM decide . If TM accepts, produces no strings of lengths at least string of length at least p.
be in Theorem 4.8 can . If not, produces a
4.12 . Therefore, we can first construct two equivalent DFA
If and recognizing and decide if the two DFA are equivalent. TM
and then run TM
= "On input < >: 1. construct the equivalent NFA
and
for
2. construct the equivalent DFA
and
for the
in theorem 4.5 to
and and
3. construct a DFA , accepting . 4. run TM to decide if and are equivalent 5. accept if TM accepts; reject if TM rejects” 4.24 be a CFL containing all the palindromes,
Let
be
the regular language accepted by and be . The goal is that is decidable iff the emptiness of is also decidable. Since is a CFL from in theorem 4.8. Let be the CFG of Problem 2.18, its emptiness can be decided by TM The decider is constructed as follows. TM
= "On input 1. Let 2. 3. 4. 5.
⟨ ⟨where
is a DFA accepting some palindrome: be a CFL containing all the palindromes
Let be the regular language accepted by derive the CFG for run TM to decide accept if TM accepts; reject if TM rejects.
4.28 Let us prove it by contradiction. Suppose that every decider is in A. Since A is Turingrecognizable, A is also enumerable. Let be the ith decider in A. We may construct the following decider as follows: ”On input , (1) decide the order number of , i.e, (2) accept if rejects; rejects if
= , the index of accepts.”
be ;
is a decider as (1)-(2) halts. is different from any Apparently, contradiction to the assumption that every decider is in A.
in A, which is a
To show that is different from any in A, let be the list of all the strings in an canonical order of string (length than dictionary order). Then can be derived by applying the diagonalization method as illustrated by the following table. The table below demonstrates an example
. ...
...
Obviously,
reject
accept
accept
...
accept
accept
accept
...
reject
accept
reject
...
...
...
...
...
accept
reject
accept
...
is different from any language decided by
whose description appears in
.
