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Conversion algorithm from a NFA to a regular expression.
Start with the transition table for the NFA with the following
state naming conventions:
the first state is 1 or q1 or s1 which is the starting state.
states are numbered consecutively, 1, 2, 3, ... n
The transition table is a typical NFA where the table entries
are sets of states and phi the empty set is allowed.
The set F of final states must be known.
We call the variable r a regular expression.
We can talk about r being the regular expression with i,j subscripts
ij
Note r is just a (possibly) different regular expression from r
12 53
Because we need multiple columns in a table we are going to build, we
also use a superscript in the naming of regular expression.
1 3 k k-1
r r r r are just names of different regular expressions
12 64 1k ij
2
We are going to build a table with n rows and n+1 columns labeled
| k=0 | k=1 | k=2 | ... | k=n
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | | n
r | r | r | r | ... | r Only build column n
11 | 11 | 11 | 11 | | 11 for 1,final state
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | | n The final regular expression
r | r | r | r | ... | r is then the union, +, of
12 | 12 | 12 | 12 | | 12 the column n
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | | n
r | r | r | r | ... | r
13 | 13 | 13 | 13 | | 13
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | |
r | r | r | r | ... |
21 | 21 | 21 | 21 | |
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | |
r | r | r | r | ... |
22 | 22 | 22 | 22 | |
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | |
r | r | r | r | ... |
23 | 23 | 23 | 23 | |
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | |
r | r | r | r | ... |
31 | 31 | 31 | 31 | |
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | |
r | r | r | r | ... |
32 | 32 | 32 | 32 | |
----+--------+-------+-------+-----+------
| 0 | 1 | 2 | |
r | r | r | r | ... |
33 | 33 | 33 | 33 | |
^ 2
|- Note n rows, all pairs of numbers from 1 to n
Now, build the table entries for the k=0 column:
/
0 / +{ x | delta(q ,x) = q } i /= j
r = / i j
ij \
\ +{ x | delta(q ,x) = q } + epsilon i = j
\ i j
where delta is the transition table function,
x is some symbol from sigma
the q's are states
0
r could be phi, epsilon, a, 0+1, or a+b+d+epsilon
ij
notice there are no Kleene Star or concatenation in this column
Next, build the k=1 column:
1 0 0 * 0 0
r = r ( r ) r + r note: all items are from the previous column
ij i1 11 1j ij
Next, build the k=2 column:
2 1 1 * 1 1
r = r ( r ) r + r note: all items are from the previous column
ij i2 22 2j ij
Then, build the rest of the k=k columns:
k k-1 k-1 * k-1 k-1
r = r ( r ) r + r note: all items are from previous column
ij ik kk kj ij
Finally, for final states p, q, r the regular expression is
n n n
r + r + r
1p 1q 1r
Note that this is from a constructive proof that every NFA has a language
for which there is a corresponding regular expression.
Some minimization rules for regular expressions
These can be applied at every step.
Note: phi is the empty set
epsilon is the zero length string
0, 1, a, b, c, are symbols in sigma
x is a variable or regular expression
( ... )( ... ) is concatenation
( ... ) + ( ... ) is union
( ... )* is the Kleene Closure = Kleene Star
(phi)(x) = (x)(phi) = phi
(epsilon)(x) = (x)(epsilon) = x
(phi) + (x) = (x) + (phi) = x
x + x = x
(epsilon)* = (epsilon)(epsilon) = epsilon
(x)* + (epsilon) = (x)* = x*
(x + epsilon)* = x*
x* (a+b) + (a+b) = x* (a+b)
x* y + y = x* y
(x + epsilon)x* = x* (x + epsilon) = x*
(x+epsilon)(x+epsilon)* (x+epsilon) = x*
Now for an example:
Given M=(Q, sigma, delta, q0, F) as
delta | a | b | c Q = { q1, q2}
--------+------+------+----- sigma = { a, b, c }
q1 | {q2} | {q2} | {q1} q0 = q1
--------+------+------+----- F = { q2}
q2 | phi | phi | phi
--------+------+------+-----
| k=0 | k=1 (using e for epsilon)
-----+-------------+------------------------------------
r | c + epsilon | (c+e)(c+e)* (c+e) + (c+e) = c*
11 | |
-----+-------------+------------------------------------
r | a + b | (c+e)(c+e)* (a+b) + (a+b) = c* (a+b)
12 | |
-----+-------------+------------------------------------
r | phi | phi (c+e)* (c+e) + phi = phi
21 | |
-----+-------------+------------------------------------
r | epsilon | phi (c+e)* (a+b) + e = e
22 | |
-----+-------------+------------------------------------
| k=0 | k=1 | k=2 (using e for epsilon)
-----+-------------+----------+-------------------------
r | c + epsilon | c* |
11 | | |
-----+-------------+----------+-------------------------
r | a + b | c* (a+b) | c* (a+b)(e)* (e) + c* (a+b) only final
12 | | | state
-----+-------------+----------+-------------------------
r | phi | phi |
21 | | |
-----+-------------+----------+-------------------------
r | epsilon | e |
22 | | |
-----+-------------+----------+-------------------------
the final regular expression minimizes to c* (a+b)
Additional topics include Moore Machines and
Mealy Machines
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