# PowerPoint Presentation - Turing machines - Mount Holyoke College

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```					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           0n1n
languages

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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
• 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
•           yields
• Written         ⊢

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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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