1. CSC 4170
Theory of Computation
Decidable languages
Section 4.1

2. Examples of decidable languages
{1,3,5} 
{x | x is even}
{x | x is a perfect square}
{x | x2-10x = 0}
{x | x=y*z for some integers y,z>1 (i.e. x is not prime)}
{x | x is a prime (i.e. x is not divisible by anything except 1 and
itself)}
{<G> | G is a connected graph}
{<P> | P is a one-variable polynomial expression with an integral root}
Or the problem
\exists m,n,k 2^m+3^n=
Undecidable:
{<P> | P is a two-variable polynomial expression with an integral root}

3. The acceptance problem for DFAs is decidable
Let ADFA = {<B,w> | B is a DFA that accepts input string w}
Theorem 4.1: ADFA is a decidable language.
Proof idea: Here is a Turing machine M that decides ADFA:
M = “On input <B,w>, where 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 the simulation ends in a nonaccept state, reject.”

4. The acceptance problem for NFAs is decidable
Let ANFA = {<B,w> | B is an NFA that accepts input string w}
Theorem 4.2: ANFA is a decidable language.
Proof idea: Here is a Turing machine N that decides ANFA:
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 that we learned.
2. Run TM M from Theorem 4.1 on input <C,w>.
3. If M accepts, accept.
If M rejects, reject.”

5. The string generation problem for REs is decidable
Let AREX = {<R,w> | R is a regular expression that generates string w}
Theorem 4.3: AREX is a decidable language.
Proof idea: Here is a Turing machine P that decides AREX:
P = “On input <R,w>, where R is a regular expression and w is a string:
1. Convert R to an equivalent NFA B using the procedure for
this conversion that we learned.
2. Run TM N from Theorem 4.2 on input <B,w>.
3. If N accepts, accept.
If N rejects, reject.”

6. The emptiness problem for the language of a DFA is decidable
Let EDFA = {<A> | A is a DFA and L(A)=}
Theorem 4.4: EDFA is a decidable language.
Proof idea: Here is a Turing machine T that decides EDFA:
T = “On input <A>, where A is a DFA:
1. Mark the start state of A.
2. Repeat until no new states get marked:
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.”

7. Other decidable problems from the language theory
Each the following languages are also decidable:
EQDFA = {<A,B> | A and B are DFAs and L(A)=L(B)}
ACFG = {<G,w> | G is a CFG that generates string w}
Theorem 4.9: Every context-free language is decidable.
Proof idea: Suppose L is a context-free language. Let G be the CFG
that generates L, and S be a TM that decides ACFG.
Here is a Turing machine H that decides L:
H = “On input w:
1. Run S on input <G,w>.
2. If the machine accepts, accept; if it rejects, reject.”