CSE 105
Theory of
Computation
Alexander Tsiatas
Spring 2012
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At last we reach:
TURING MACHINES
2
Turing Machine Model
Draw a picture here
3
PDA’s vs. Turing Machines
PDA
• Processes input string left to
right
• Only memory access is the
top of the stack
• Non-deterministic
• Halts when all input is read
• No “blank” symbol
Both models: infinite memory
Turing Machine
• Processes input string in any
order; it starts out on the
tape
• Can move tape head to any
point on the tape
• Deterministic
• Halts whenever q_accept or
q_reject is reached
• “Blank” symbol included
Important Turing Machine detail
• There is ONE way for a Turing Machine to accept a
string.
– Visit state qaccept; the machine halts and accepts.
• There are TWO ways for a Turing Machine to not
accept a string.
– Visit state qreject; the machine halts and rejects.
– Infinite loop.
• The LANGUAGE of a Turing machine is: the set of
strings that are accepted by a Turing machine.
• All strings that are NOT accepted are NOT in the
language.
Turing Machine Formal Description
• In the TM transition function
δ: Q x Γ -> Q x Γ x {L,R},
a given input of the transition function is:
a) A current state, a character read, and whether we
came from the left or right
b) A current state, a character read
c) A current state, a character to write
d) A destination state, a character read
e) None of the above or more than one of the above
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Turing Machine Formal Description
• In the TM transition function
δ: Q x Γ -> Q x Γ x {L,R},
a given output of the transition function is:
a) A current state, a character read, and whether we came
from the left or right
b) A current state, a character to write, and whether we
should next go left or right
c) A destination state, a character read, and whether we
should write ‘L’ or ‘R’
d) A destination state, a character to write, and whether we
should next go left or right
e) None of the above or more than one of the above
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Turing Machine configurations
• Suppose we have at TM s.t. Γ={a,b,c,d,_}, Q = {qx |
1≤x≤10} U {qacc,qrej}. Suppose the current
configuration is xbq1ay. What does this mean?
a) The TM is in state q1, the tape contains “xbay”,
and the tape head is on the “b”.
b) The TM is in state q1, the tape contains “xbay”,
and the tape head is on the “a”.
c) The TM is in state q1, the tape contains “xbq1ay”,
and we don’t know where the tape head is.
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Executing a Transition, “yields”
• Suppose we have at TM s.t. Γ={a,b,c,d,_}, Q = {qx |
1≤x≤10} U {qacc,qrej}, and the transition function
includes rules δ(q1,a) = (q3,b,R) and δ(q1,b) = (q3,a,L).
We also have strings x,y in Γ*. Which configuration
does the current configuration, xbq1ay, yield?
a)
b)
c)
d)
e)
xq3bay
xbbq3y
xbaq3by
xq3aay
None of the above or more than one of the above
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Executing a Transition, “yields”
• Suppose we have at TM s.t. Γ={a,b,c,d,_}, Q = {q1,
q2, q3, q4,qacc,qrej}, and the transition function
includes rules δ(q2,c) = (q3,d,R) and δ(q2,d) =
(q3,c,L). We also have strings u,v,x,y in Γ*. Which
configuration does the current configuration,
xcq2dy, yield?
a)
b)
c)
d)
e)
xq3ddy
xcdq3y
xq3ccy
xq3ccdy
None of the above or more than one of the above
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More Transition Function
• (a) TRUE
• (b) FALSE
x0qrej11y yields x01q11y can never happen in any
Turing Machine
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Turing machines give you these:
DECIDABLE AND TURINGRECOGNIZABLE LANGUAGES
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So a language is not context-free.
What is it?
• L = {1^{2n} | n ≥ 0}.
• Not context-free.
– Exercise: prove it!
– So there is no PDA
or CFG with L as its
language.
– What about Turing
Machines?
Context-Free
Regular
What about this stuff???
L = {1^{2n} | n ≥ 0}
WHOA, that looks complicated. Let’s break this down.
L = {1^{2n} | n ≥ 0}
• How could you check to see if a number is a
power of 2?
a) Check to see if it’s even
b) Keep on dividing by 2; if you get an odd number,
it’s not a power of 2
c) Keep on dividing by 2; if you get an odd number
> 2, it’s not a power of 2
d) Keep on dividing by 2; if you get down to 0, it’s a
power of 2
L = {1^{2n} | n ≥ 0}
No 1’s
Even
1’s
One 1
Odd 1’s
STEP 1: Scan the tape left to right, cross off every other 1,
keep track if the number of 1’s is even, odd, or equal to 1.
L = {1^{2n} | n ≥ 0}
No 1’s
Even
1’s
One 1
Odd 1’s
STEP 2: If you’re left with just one 1, halt and accept!
L = {1^{2n} | n ≥ 0}
No 1’s
Even
1’s
One 1
Odd 1’s
STEP 3: If you’re left with an odd number of 1’s, but more
than one, halt and reject!
L = {1^{2n} | n ≥ 0}
No 1’s
Even
1’s
One 1
Odd 1’s
STEP 4: Send the tape head back to the left end of the tape.
L = {1^{2n} | n ≥ 0}
No 1’s
Even
1’s
One 1
Odd 1’s
STEP 5: Go back to STEP 1.
L = {1^{2n} | n ≥ 0}
• This is a “high-level” description of a Turing
Machine:
– STEP 1: Scan the tape left to right, cross off every other 1, keep track if
the number of 1’s is even, odd, or equal to 1.
– STEP 2: If you’re left with just one 1, halt and accept!
– STEP 3: If you’re left with an odd number of 1’s, but more than one,
halt and reject!
– STEP 4: Send the tape head back to the left end of the tape.
– STEP 5: Go back to STEP 1.
• On Homework and Exams: give high-level descriptions.
– This should be easier than drawing the diagram, specifying the
transition details, etc.
L = {1^{2n} | n ≥ 0}
• This language L is:
a)
b)
c)
d)
Decidable
Turing-recognizable
Decidable AND Turing-recognizable
Neither decidable nor Turing-recognizable
So a language is not context-free.
What is it?
• Decidable: there is a
TM that always halts
on all inputs.
• Turing-recognizable:
there is a TM that
always halts for
accepted strings, but
MAY NOT HALT on
rejected strings.
Regular
Context-Free
Decidable
Turing-recognizable
What’s out here…? After Exam 2.