# ch4

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```					4.1 Decidable languages
   Decidable problems concerning regular languages:

Let ADFA = {B,w | B is a DFA that accepts input
string w}.
The problem of testing whether a DFA B accepts
an input w is the same as the problem of testing
   Theorem 4.1
ADFA is a decidable language.

M = “On B,w, whether B is a DFA and w is a string:
1. Simulate B on input w.
2. If the simulation ends in an accept state,
accept. If it ends in a nonaccepting state,
reject.”
ANFA = {B,w | B is a NFA that accepts input string w}.

   Theorem 4.2
ANFA is a decidable language.

N = “On input B,w where B is an NFA ,and w is a
string:
1. Convert NFA B to an equivalent DFA C using
the procedure for this conversion given in
Theorem 1.19.
2. Run TM M from Theorem 4.1 on input C,w.
3. If M accepts, accept; otherwise reject.”
Let AREX = {R,w | R is a regular expression that
generates string w}.

   Theorem 4.3
AREX is a decidable language.

PROOF The following TM P decides AREX.
P =“On input R,w where R is a regular expression
and w is a string:
1. Convert regular expression R to an equivalent
DFA A by using the procedure for this
conversion given in Theorem 1.28.
2. Run TM M on input A,w.
3. If M accepts, accept; if M rejects, reject.”
EDFA = {A | A is a DFA and L(A)=}.

   Theorem 4.4
EDFA is a decidable language.

T = “On input A where A is a DFA:
1. Mark the start state of A.
2. Repeat until no new states get marked:
3. Mark any state that has a transition coming
into it from any state that is already marked.
4. If no accept state is marked, accept; otherwise
reject.”
EQDFA = {A,B | A and B are DFAs and L(A) = L(B)}.

   Theorem 4.5
EQDFA is a decidable language.
             
L(C )  L( A)  L( B )  L( A)  L( B )   

F = “On input A, B, where A and B are DFAs:
1. Construct DFA C as described.
2. Run TM T from Theorem 4.4 on input C.
3. If T accepts, accept. If T rejects, reject.”
DECIDABLE PROBLEMS CONCERNING
CONTEXT-FREE LANGUAGE
ACFG = {G,w | G is a CFG that generates string w}.

   Theorem 4.6
ACFG is a decidable language.

Proof The TM S for ACFG follows.
S = “On input G,w, where G is a CFG and w is a string:
1. Concert G to an equivalent grammar in Chomsky
normal form.
2. List all derivations with 2n-1 steps, where n is the
length of w.
3. If any of these derivations generate w, accept; if not,
reject.”
ECFG = {G | G is a CFG and L(G) = }.

   Theorem 4.7
ECFG is a decidable language.

PROOF
R = “On input G, where G is a CFG:
1. Mark all terminal symbols in G.
2. Repeat until no new variables get marked:
3. Mark any variable A where G has a rule
AU1U2…Uk and each symbol U1,…,Uk has
4. If the start symbol is not marked, accept;
otherwise reject.”
   Theorem 4.8
Every context-free language is decidable.

Proof Let G be a CFG for A and design a TM MG
that decides A. We build a copy of G into MG. It
works as follows.
MG = “On input w:
1. Run TM S on input G, w, where S is a
2. If this machine accepts, accept; if it rejects,
reject.”
FIGURE 4.2
The relationship among classes of languages.
4.2 The Halting Problem
1.   ATM = {M, w | M is a TM and M accepts w}
The following Turing machine U recognizes ATM.
U =“On input M, w, where M is a TM and w is a
string:
1. Simulate M on input w.
2. If M ever enter its accept state, accept; If
M ever enters its reject state, reject.”
ATM is sometimes called the halting problem.

2.   The proof of the undecidability of the halting
problem uses a technique called “diagonalization”,
discovered by mathematician Georg Cantor in 1873.
3.    Assume that we have two sets A and B and a
function f: AB
–   f is one-to-one if f(a)f(b) whenever ab.
–   f is onto if for every bB there is an aA s.t. f(a)=b.
–   A and B are the same size if there is a one-to-one, onto
function f: AB.
–   A function that is both one-to-one and onto is called a
correspondence.
Ex. Let N be the set of natural numbers {1, 2, 3, …}and let 
be the set of even natural numbers {2, 4, 6, …}. Using
Cantor’s definition, N and  have the same size.
f(n)=2n:
4.   A set S is countable if either it is finite or it has the
same size as N.
Ex.
Let Q be the set of positive rational numbers,
m
i.e. Q  { m, n  N }, Q is countable.
n
5.   For some infinite sets no correspondence with N
exists, such sets are called uncountable.
   Theorem 4.14     R is uncountable.
需證明 no correspondence exists between N and R.
The proof is by induction:
1. 假設存在一種 correspondence f between N and
R (每個 element in N 有唯一之 element in
R 與其對應).
2. 我們只要找出一 xR 使得 x 不包含在此對應
關係即可證明. 如何找？以下列範例說明：

永遠可以找到一 x 不存在 f 之中！
    Corollary 4.15
Some language are not Turing-recognizable.

1.   觀察一：the set of all string * is countable, for
any alphabet .
觀察二：the set of all Turing machines is countable
because each Turing machine M has an
encoding into a string M.
 The set of all Turing machines is countable.
2.   Let B be the set of all infinite binary sequences.
B is uncountable.
3.   Let L be the set of all languages over .
Let * = {s1, s2, s3, …}. Each language AL has
unique sequence A in B:
The ith bit of that sequence is a “1” if siA and is a
“0” if siA.
例: * = { , 0, 1, 00, 01, 10, 11, 000, 001, …};
A ={       0, 00, 01,           000, 001, …};
A = 0 1 0 1 1 0 0 1                   1 … .
The function f: LB, where f(A) equals the characteristic
sequence of A, is one-to-one and onto and hence a
correspondence.
B is uncountableL is uncountable.

Some languages are not recognized by any
Turing machine.
   The Halting Problem is Undecidable
Now we are ready to prove Theorem 4.9, the
undecidability of the language
ATM = {M,w| M is a TM and M accepts w}.

Proof:
Assume ATM is decidable and obtain a contradiction.
1. Suppose that H is a decider for ATM.
H(M,w) =    accept if M accepts w.
reject if M does not accept w.
2. Now we construct a new Turing machine D with H
as a subroutine.
D = “On input M, where M is a TM:
1. Run H on input M, M.
2. Output the opposite of what H outputs;
that is, if H accepts, reject and if
H                    rejects, accept.”
3. D(M) =      accept if M does not accept M.
reject if M accepts M.

D(D) =      accept if D does not accept D.
reject if D accepts D.
Figure 4.4    Entry i,j is accept if Mi accepts Mj

Figure 4.5 Entry i,j is the value of H on input Mi , Mj 
FIGURE 4.6
If D is in the figure, a contradiction occures at “?”
   Theorem 4.16
A language is decidable if and only if it is
both Turing-recognizable and co-Turing-
recognizable.
In other words, a language is decidable if and
only if both it and its complement are Turing-
recognizable.

Pf:
1. 証 Easy
2. 証 
2. 証 
If both A and A are Turing-recognizable,
we let M1 be the recognizer for A and M2
be the recognizer for A. The following
Turing machine M is a decider for A:
M = “On input w:
1. Run both M1 and M2 on input w in
parallel.
2. If M1 accepts, accept; if M2
accepts, reject.”
 Corollary 4.17
ATM is not Turing-recognizable.

4.3
4.5
4.8
4.11
4.12

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 views: 4 posted: 2/12/2012 language: pages: 24