NFAs and DFAscpennycu/2019-2020/Spring...Deterministic vs. Nondeterministic Finite Automata...
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NFAs and DFAs
Transcript of NFAs and DFAscpennycu/2019-2020/Spring...Deterministic vs. Nondeterministic Finite Automata...
NFAs and DFAs
Reminder: Regular OperationsLet A and B be languages. We define the regular
operations union, concatenation, and star as
follows:
● Union: A ∪ B = {x | x ∈ A or x ∈ B }.
● Concatenation: A ⚬ B = {xy | x ∈ A and y ∈ B }.
● Star: A* = {x
1
x
2
...x
k
| k ≥ 0 and each x
i
∈ A }.
Proofs that Regular Languages Are Closed Under...● Union: A ∪ B = {x | x ∈ A or x ∈ B}.
○ Proof by Construction (last class)
But…
● Concatenation: A ⚬ B = {xy | x ∈ A and y ∈ B}.
● Star: A* = {x
1
x
2
...x
k
| k ≥ 0 and each x
i
∈ A}.
We Need Better Tools!
Deterministicvs.
NondeterministicFinite Automata
Everything discussed thus far was
deterministic, thus a DFA
Now, we will expand DFAs to allow
multiple potential paths.
Two types of FSMs
What is Determinism?
Are computers deterministic?
“If the current state is known, and the
current inputs are known, then the
future state is also known.”
● No Choices
● No Randomness
● No Oracles
● No Errors/Cheating
According to our previous definition,
FSM == DFA
What is Nondeterminism?
“If the current state is known, and the
current inputs are known, there may be
multiple possible future states.”
● Random choice?
● Parallel choice and simultaneous execution?
NFAs: Relaxing The Requirements
Multiple edges with the same label
ε (optional) edges
● ε is not in Σ● Does not consume input character
Only need one path to an accept state
What could go wrong?
How do we know which path to try?
● Try them all
● Always make the right choice
NFA Formal Definition
A Nondeterministic Finite
Automaton is a 5-tuple
N = (Q, Σ, δ, q
0
, F )
where:
Q Set of states (finite)
Σ Alphabet of symbols (finite)
δ The transition function
δ: Q x Σε ➝ 𝓟(Q)
q
0
The starting (initial) state
q
0
∈ Q
F The set of “Accept” states
F ⊆ Q
NFA Formal Definition Example
N = (Q, Σ, δ, q
0
, F )
Q {q
1
, q
2
, q
3
, q
4
}
Σ {0, 1}
δ
q
0
q
1
F {q
4
}N = ({q
1
, q
2
, q
3
, q
4
}, {0, 1}, δ, q
1
, {q
4
})
NFA Formal Definition Example
N = (Q, Σ, δ, q
0
, F )
Q {q
1
, q
2
, q
3
, q
4
}
Σ {0, 1}
δ
q
0
q
1
F {q
4
}N = ({q
1
, q
2
, q
3
, q
4
}, {0, 1}, δ, q
1
, {q
4
})
NFA Formal Definition of ComputationLet N = (Q, Σ, δ, q
0
, F )
Let y
1
y
2
...y
m
be a string w where y
i
∈ Σε
N accepts w if there is a sequence of
states r
0
, r
1
, r
2
...r
m
in Q such that:
1. r
0
= q
0
,
2. r
i+1
∈ δ(r
i
, y
i+1
) for 0 ≤ i < m, and
3. r
m
∈ F
N “recognizes”
language A if
A = {w | N accepts w}
We now have a
NONDETERMINISTIC tool that we
can use to understand Regular
Languages!
Wait… Can we say this?!?
Proof of Equivalency
NFAs and DFAs
Proof by Construction:
A method to convert
N = (Q, Σ, δ, q
0
, F )
into
M = (Q ’, Σ, δ’, q ’
0
, F ’)
What Were We Talking About Last Time???
http://cscrunch.com:8890
Convert NFA to DFAProof by Construction. (ignoring ε for now)
Let N = (Q, Σ, δ, q
0
, F ) be a NFA that recognizes
some language A.
Construct a DFA M = (Q ’, Σ, δ’, q ’
0
, F ’) that
recognizes the same language A.
For ε, define E(R) = {q | q can be reached from R
by traveling along 0 or more ε arrows} where
R ⊆ Q and q ∈ Q.
q’
0
= E({q
0
})
δ’(R, a) = {q ∈ Q | q ∈ E(δ(r, a)) for some r ∈ R}
Q’ = 𝓟(Q)
ex: {{p
0
}, {p
0
, p
1
}, {p
0
, p
2
}, {p
0
, p
1
, p
2
}, …}
q ’
0
= {q
0
}
F ’ = {R ∈ Q ’ | R contains an accept state of N }
or, F ’ = {R ∈ Q ′ | R ∩ F ≠ ∅}
δ’(R, a) = {q ∈ Q | q ∈ δ(r, a) for some r ∈ R }
where R ∈ Q’ and a ∈ Σ.
Practice: Convert this NFA to a DFA
Practice: Convert this NFA to a DFA
We defined DFAs.
We defined Computation using DFAs.
We defined Regular Languages using
DFAs.
We defined NFAs.
We defined Computation using NFAs.
We showed equivalence of NFAs to
DFAs.
Now What?
Regular OperationsThey eat their fiber.
Union, Concatenation, and Star:
3 Proofs by Construction
Regular Operations: UnionLet N
1
= (Q
1
, Σ, δ1
, q
1
, F
1
) recognize A
1
,
and N
2
= (Q
2
, Σ, δ2
, q
2
, F
2
) recognize A
2
.
Construct N = (Q, Σ, δ, q
0
, F ) to recognize A
1
∪A
2
.
Q = {q
0
} ∪ Q
1
∪ Q
2
F = F
1
∪ F
2
For q ∈ Q and any a ∈ Σε,
Regular Operations: ConcatenationLet N
1
= (Q
1
, Σ, δ1
, q
1
, F
1
) recognize A
1
,
and N
2
= (Q
2
, Σ, δ2
, q
2
, F
2
) recognize A
2
.
Construct N = (Q, Σ, δ, q
1
, F
2
) to recognize A
1
○ A
2
.
Q = Q
1
∪ Q
2
For q ∈ Q and any a ∈ Σε,
Regular Operations: StarLet N
1
= (Q
1
, Σ, δ1
, q
1
, F
1
) recognize A
1
.
Construct N = (Q, Σ, δ, q
0
, F ) to recognize A
1
*.
Q = {q
0
} ∪ Q
1
F = {q
0
} ∪ F
1
For q ∈ Q and any a ∈ Σε,
