Finite Automata
(Finite State Machine)
Lecture 4-5
Ref. Handout p18-24
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2
Controlling Things ...
If x has happened then
If in state y then
do something
do something
change to new state
If x happened while in state y
do something
change to new state
3
State Transition Table
State
Event
happy
work due
next week
anxious
work due
tomorrow
Work
cancelled
anxious
work done
happy
Action
New State
none
start
work
happy
none
hand in
work
anxious
happy
happy
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State Diagrams
a state transition
A
x
B
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State Transition Diagram
cancelled
start
Happy
tomorrow
Anxious
done
next week
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A Finite Automata
cancelled
Happy
next week
tomorrow
Anxious
done
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Components
starting state (one per machine)
d
input
symbol
transition
halting
state
(one or
more)
non-halting
state
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Sequence of Input Symbols
What input sequences leave this FA in its halt state?
a
S1
S2
b
 ac
yes
 ab
no
 abc
no
 abac yes
c
S3
{ (ab)nac | n>=0 }
{ ac, abac, ababac, abababac, ... }
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The Original Example ...
Some
acceptable
inputs
cancelled
Happy
next week
tomorrow
done
Anxious
nnnntc
ntd
Λ
nntcntd
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Recognising Strings
the set of all strings accepted/recognised by an FA
=
the language accepted/recognised by the FA
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Rejecting Strings
An FA rejects a string if
with some symbols remaining
there is no transition for the next symbol
or
with no symbols remaining
the FA is not in a halt state
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Rejected / Accepted Strings
tdnnn
tdnnt
n
ntc
ntcd
c
Happy
n
t
yes/no
yes/no
yes/no
yes/no
yes/no
Anxious
d
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Recall some simple languages ...
Can you draw FA to recognise them?
{ Λ, a, aa, aaa, aaaa, ...}
= { an | n >= 0 }
{ a, aa, aaa, aaaa, ...}
= { an | n > 0 }
{ 00, 01, 10, 11 }
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An Assignment Related Example
Draw an FA to recognise non-negative integers
Extend it to all integers (i.e. allowing +,- signs)
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Memo for In-class test 4 [
questions
my
answers
correct
answers
/5]
comments
1
2
3
4
5
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Non-Deterministic Automata
a
1
b
c
cd
cdacd
2
What happens
here?
d
3
d
4
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Non-Deterministic Automata
 In deterministic automata, there can
never be two transitions to different
state for the same input symbol.
 If there are more than one transition to
different states for the same input
symbol, we call it a non-deterministic
automata.
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Deterministic FAs and
Non-Deterministic FAs
For every
non-deterministic FA (NFA)
there is a
deterministic FA (DFA)
which accepts
the same language
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A Simple Example
2
a
1
NFA
b
3
a
DFA
a
b
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Step by Step (1)
2
a
1
NFA
b
3
a
{1}
a
{ 2,3 }
after ‘a’, must be in
state 2 or 3
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Step by Step (2)
2
a
1
NFA
b
3
a
from state 2, read
‘b’, move to state 3
{1}
a
{ 2,3 }
b
{3}
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Last Step
2
a
1
NFA
b
3
a
{1}
a
{ 2,3 }
b
{3}
3 is a halt state
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A General Algorithm
1. Write down {1} as the start state of the DFA
2. Choose a state in the DFA which hasn’t
been processed
3. Work out all transitions from it adding new
states
4. Repeat from step 2 until all states have been
processed
5. Mark halting states
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Building DFAs
start
Machine A
Language L
halt
start
Machine B
Language K
halt
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What does this do?
Machine A
Language L
halt
Machine B
Language K
halt
start
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What does this do?
start
Machine A
Language L
??
Machine B
Language K
halt
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What does this do?
start
Machine A
Language L
halt
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Using Building Blocks
For example, you are asked to define a format
for a collection of pictures’ file names
G = { a1.gif, a2.gif, a3.gif, ... a33.gif, .... }
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Using Building Blocks
Now you are asked to define a format for
another collection of pictures’ file names
P = { b1.jpg, b2.jpg, b3.jpg, ... b33.jpg, .... }
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What if we need an FA for the union of G and P
i.e.
GUP
a
1-9
1-9
.
i
g
b
f
1-9
1-9
.
j
p
g
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Using Building Blocks
Let’s change the language P to
P = { a1.jpg, a2.jpg, a3.jpg, ... a33.jpg, .... }
(replace ‘b’ by ‘a’)
a
1-9
1-9
.
j
p
g
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What happens now? If we need an FA for the union
of G and P.
i.e. G U P
a
1-9
1-9
.
i
g
a
f
1-9
1-9
.
It’s a NFA!
j
p
g
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Limitations of FAs
Can you draw an FA to accept this?
{ a nb n | n > 0 }
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Memo for In-class test 5 [
questions
my
answers
correct
answers
/5]
comments
1
2
3
4
5
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