### PPT - Mount Holyoke College

```Turing machines
Sipser 2.3 and 3.1
(pages 123-144)
A Context-free Grammar for
{anbncn| n ≥ 0}?
• Theorem 2.34 (Pumping lemma for CFLs):
If A is a CFL, then there is a number p
where,
if s is any string in A of length ≥ p,
then s = uvxyz such that:
1. For each i ≥ 0, uvixyiz ∈ A,
2. |vy| > 0, and
3. |vxy| ≤ p
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Proof idea
• Surgery on parse trees
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So…
• Is {anbncn| n ≥ 0} a CFL?
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Chomsky hierarchy
anbncn
Context-free languages
Regular
languages
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0n1n
5
Introducing… Turing machines
Infinite tape
a
b
a
b
⨆
⨆
⨆
Finite
control
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Formally…
• A Turing machine is a 7-tuple
(Q, Σ, Γ, δ, q0, qaccept, qreject), where
– Q is a finite set called the states
– Σ is a finite set not containing the blank symbol ⨆
called the input alphabet
– Γ is a finite set called the tape alphabet with ⨆∈Γ
and Σ⊆Γ
– δ:Q ×Γ → Q ×Γ ×{L,R}
– q0∈Q is the start state
– qaccept∈Q is the accept state
– qaccept∈Q is the reject state
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Recognizing {anbncn| n ≥ 0}
Infinite tape
a
a
b
b
c
c
⨆
⨆
⨆
Finite
control
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Configurations
• A configuration is
– Current state
– Current tape contents
– Current head location
• u q v means
– Current state is q
– Current tape contents is uv
– Current head points at first symbol of v
• Example
–
–
–
–
âaq1bbcc
In state q1
Tape contents are âabbcc
Tape head is on first b
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Yields
• A configuration C1 yields configuration C2 if
the Turing machine can legally go from C1
to C2 in a single step
•
• Written
yields
⊢
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Turing-recognizable languages
• A Turing machine accepts input w if a sequence
of configurations C1,C2,...,Ck exists where
1. C1 is the start configuration of M on input w
2. Each Ci yields Ci+1
3. Ck is an accepting configuration
• Defn 3.5: A language is Turing-recognizable if it
is accepted by some Turing machine.
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Recognizing
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```