CSCI 2670
HW 8
11/2/05
Solutions
1.
Show T = {<M>  M is a TM with L(M)
R
= L(M)} is an undecidable language.
Proof:
Assume T is decidable and let D be a decider for T.
Consider the following
Turing machine S:
S = “On input <M,w>, where M is a Turing machine
1. Create the following Turing machine M
1
M
1
= “On input x
1.
If x = 0
n
1
n
for some n,
accept
2.
If M accepts w,
accept.
”
2. Run D on input <M
1
>
3. If D accepts,
accept
4. If D rejects,
reject
”
S is a decider since step 1 creates a Turing machine and moves on, step 2 calls a
decider, and steps 3 and 4 halt.
Note that if M accepts w, then L(M
1
) = Σ*, which is
in T.
If M does not accept w, then L(M
1
) = 0
n
1
n
, which is not in T.
Therefore, D
accepts <M
1
> iff M accepts w, and reject otherwise.
By construction, S accepts if D
accepts (i.e., if M accepts w) and rejects if D rejects (i.e., if M does not accept w).
Therefore, S decides A
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 sm
 Halting problem

Click to edit the document details