CSC 3130: Automata theory and formal languages

1
Fall 2009
The Chinese University of Hong Kong
CSC 3130 Automata theory and formal languages
NFA to DFA conversionand regular expressions
Andrej Bogdanov http//www.cse.cuhk.edu.hk/andrej
b/csc3130
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NFAs are as powerful as DFAs

• Obviously, an NFA can do everything a DFA can do
• But can it do more?
• Theorem

NO!
If a language L is accepted by some NFA, thenit
is also accepted by some DFA.
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Proof of theorem

• We will show a general way to convert every NFA
  into an equivalent DFA
• Step 1 Simplify NFA by eliminating e-transitions
• Step 2 Convert simplified NFA (without es)

We do this first
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NFA to DFA conversion intuition
0, 1
1
0
NFA
q0
q1
q2
0
0
1
0
DFA
q0 or q1
q0 or q2
q0
1
1
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NFA to DFA conversion intuition
0, 1
1
0
NFA
q0
q1
q2
0
0
1
0
DFA
q0, q1
q0, q2
q0
1
1
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General method
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Why the method works

• At the end, the DFA accepts when it is in a state
  that contains some accepting state of NFA
• So the DFA accepts only when the NFA accepts too

After reading n symbols, the DFA is in state
qi1,,qik if and only if the NFA is in one of
the states qi1,,qik
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Converting via the general method
0, 1
1
0
NFA
q0
q1
q2
DFA
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Converting via the general method
After eliminating the dead states and
transitions, we end up with the same picture
0
1
1
q0, q1
q0
1
0, 1
0
0
1
0
1
q0, q2
q0, q1, q2
Æ
q1
0
0, 1
0
q1, q2
q2
1
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Proof of theorem

• We will show a general way to convert every NFA
  into an equivalent DFA
• Step 1 Simplify NFA by eliminating e-transitions
• Step 2 Convert simplified NFA (without es)

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Eliminating e-transitions
0
?, 1
?
q0
q1
q2
eNFA
0
Transition table of corresponding NFA
inputs
0
1
q0
q1, q2
q0, q1, q2
q1
q0, q1, q2
Æ
states
q2
Æ
Æ
q0, q1, q2
Accepting states of NFA
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Eliminating e-transitions
0
?, 1
?
q0
q1
q2
NFA
0
0
0
0, 1
0
NFA without es
q0
q1
q2
0
0, 1
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Eliminating e-transitions General method

• For every state qi and every symbol a Î S,
  replace every path out of qi like
• For every accept state qf, make accepting all
  states connected to it via es

q4
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Regular expressions
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Operations on strings

• Given two strings s a1an and t b1bm, we
  define their concatenation st a1anb1bm
• We define sn as the concatenation sss n times

s abb, t cba st abbcba
s 011 s3 011011011
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Operations on languages

• The concatenation of languages L1 and L2 is
• Similarly, we write Ln for LLL (n times)
• The union of languages L1 ? L2 is the set of all
  strings that are in L1 or in L2
• Example L1 01, 0, L2 e, 1, 11, 111, .
  What is L1L2 and L1 ? L2?

L1L2 st s ? L1, t ? L2
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Operations on languages

• The star (Kleene closure) of L are all strings
  made up of zero or more chunks from L
• This is always infinite, and always contains e
• Example L1 01, 0, L2 e, 1, 11, 111, .
  What is L1 and L2?

L L0 ? L1 ? L2 ?
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Constructing languages with operations

• Lets fix an alphabet, say S 0, 1
• We can construct languages by starting with
  simple ones, like 0, 1 and combining them

0(0?1)all strings that start with 0
0(01)
(01)?(10)
0110
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Regular expressions

• A regular expression over S is an expression
  formed using the following rules
• The symbol Æ is a regular expression
• The symbol e is a regular expression
• For every a ? S, the symbol a is a regular
  expression
• If R and S are regular expressions, so are RS,
  RS and R.

A language is regular if it is represented by a
regular expression
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Examples
S 0, 1
01
0(1)
0, 01, 011, 0111,
0 followed by any number of 1s
(01)(01)
001, 0101, 01101, 011101,
0 followed by any number of 1s and then 01
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Examples
strings of length 1
01
0, 1
(01)
e, 0, 1, 00, 01, 10, 11,
any string
(01)010
any string that ends in 010
(01)01(01)
any string that contatins the pattern 01
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Examples
((01)(01))((01)(01)(01))
all strings whose length is even or divisible by
3 strings of length 0, 2, 3, 4, 6, 8, 9, 10,
12, ...
((01)(01))
strings of even length
(01)(01)
strings of length 2
((01)(01)(01))
strings of length divisible by 3
(01)(01)(01)
strings of length 3
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Examples
((01)(01)(01)(01)(01))
strings that can be broken in blocks, where each
block has length 2 or 3
(01)(01)(01)(01)(01)
strings of length 2 or 3
(01)(01)
strings of length 2
(01)(01)(01)
strings of length 3
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Examples
((01)(01)(01)(01)(01))
strings that can be broken in blocks, where each
block has length 2 or 3
00110
10
011
e
1
011010110
?
?
?
?
?
?
this includes all strings except those of length 1
((01)(01)(01)(01)(01))
all strings except 0 and 1
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Examples
(101001)(?000)
ends in at most two 0s
there can be at most two 0s betweenconsecutive 1s
there are never three consecutive 0s
Guess (101001)(?000) x x does not
contain 000
0110010110
0010010
00
e
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Examples

• Write a regular expression forall strings with
  two consecutive 0s.

S 0, 1
(anything) 00 (anything else)
(01)00(01)
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Examples

• Write a regular expression forall strings that
  do not contain two consecutive 0s.

S 0, 1
0110101101010
blocks ending in 1
last block
(1 01)
... at most one 0 in every block ending in 1
(e 0)
... and at most one 0 in the last block
(1 01)(e 0)
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Examples

• Write a regular expression forall strings with
  an even number of 0s.

S 0, 1
even number of zeros (two zeros)
two zeros 10101
(10101)
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Main theorem for regular languages

• Theorem

A language is regular if and only if it is the
language of some DFA
NFA
regularexpression
DFA
regular languages
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Road map
?
?
NFA
regularexpression
NFAwithout e
DFA
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Examples regular expression ? NFA

• R1 0
• R2 0 1
• R3 (0 1)

M2
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General method
NFA
regular expr
Æ
q0
q0
e
a
q0
q1
symbol a
?
?
?
RS
q0
q1
MS
MR
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General method continued
NFA
regular expr
MR
?
?
q0
q1
R S
?
?
MS
?
?
?
?
R
q0
q1
MR
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Road map
?
?
?
NFA
regularexpression
NFAwithout e
DFA