Recap lecture 29
Example of prefixes of a language,
Theorem: pref(Q in R) is regular, proof,
example, Decidablity, deciding whether two
languages are equivalent or not ?, method 1,
example, method 2, example
1
Example
Consider two languages L1 and L2,
expressed by the REs r1=a* and r2=+aa*
respectively. To determine whether L1 and
L2 are equivalent, let the corresponding FAs
a
be
a,b
a
FA1 p1
r3+
a
b
p2
FA2 r 
1
b
a,b
b
r2
2
Example continued …
As discussed earlier, the FA corresponding
to the language (L1L2c)U(L1cL2) helps
in this regard i.e. if this FA accepts any
word then L1 and L2 are not equivalent other
wise L1 and L2 are equivalent.
Following are the FAs corresponding to
c
c
L1 and L2
3
Example continued …
a
a,b
a
FA1
c
q1-
b
q2+
s3
a
FA2c
s1 -
a,b
b
b
s2 +
FAs corresponding to (FA1c U FA2)c and
(FA2c U FA1)c may be as follows
4
Example continued …
c
(FA1 UFA2)
c
a
q1 , r3
a
q1 , r1
c
(FA2 UFA1)
c
a
p 1 , s3
a
a,b
b
b
q2 , r2
-
p1 , s1
a,b
b
b
p2 , s2
Both the FAs have no final state, so these FAs
accept nothing. This implies that their union will
not also accept any string. Hence FA
corresponding to the language
(L1L2c)U(L1cL2) accepts nothing. Thus both
the languages are equivalent.
5
Example
Following is an FA, for which it is
required to decide whether it accepts
any string or not? Using steps
discussed in method 2, the following
FA can be checked whether it accepts
any word or not.
6
Example
1b
b
a
a
2
b
5
a
b
a,b
3
b
4+
a
a
6
7
Example continued …
1b
b
a
a
2
b
5
a
b
a,b
3
b
4+
a
a
6
8
Example continued …
1b
b
a
2
b
5
a
a,b
3
b
4+
a
a
6
9
Example continued …
1b
a
2
b
a
a,b
3
b
4+
a
5
6
10
Example continued …
1b
a
2
b
a
3
b
4+
a
5
6
11
Example continued …
As no final state of the previous FA is
marked, after applying the method, so the
given FA accepts no word.
12
Method 3
If the FA has N states, then test the words of
length less than N. If no word is accepted by
this FA, then it will accept no word.
Note: It is to be noted that all the methods
discussed above, to determine whether an FA
accepts certain word, are effective procedures.
Example: To determine whether the following
FA accepts certain word, using method 3, all
the strings of length less than 4 (i.e. less than
the number of states of the FA) are sufficient to
be tested
13
Example
2
a
a
a
1-
b
a
4+
b
b
3
b
i.e. , a, b, aa, ab, ba, bb, aaa, aab, aba, abb,
baa, bab, bba, bbb are required to be tested.
14
Example continued …
It can be observed that the strings aa, baa,
aaa are accepted by this FA, hence the
language accepted by this FA is not
empty. Following is another example, to
test whether an FA accepts certain word ?
15
Example
Consider the following FA. To determine
whether this FA accepts some word, all the
strings of length less than 4 (i.e. less than the
number of states of this FA) are to be tested
a
1-
a,b
a
b
b
2
b
3
a,b
4+
16
Example continued …
It can be observed that none of the strings ,
a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab,
bba, bbb, is accepted by this FA. Thus the
given FA cannot accept any word.
17
Observation
To find whether a regular expression defines
an infinite language or not ? The following
possibilities are required to be checked.
If a regular expression contains * then it may
define an infinite language, with the
exception * as *=  e.g.
(+a*)(*+)* defines finite language.
While (*+a*)* (*+)* defines an infinite
language.
There are so many ways to decide whether an
FA accepts a finite language or an infinite.
Following is a theorem regarding this
18
Theorem
Let F be an FA having N states
1. If F accepts a word w s.t.
N  length(w) < 2N
then F accepts infinite language.
2. If F accepts an infinite language then there are
some words w s.t.
N  length(w) < 2N
The first part can be proved, using the pumping
lemma version II.
Following is a remark regarding this theorem
19
Remark
There is an effective procedure to decide
whether an FA accepts finite or infinite
language. For a machine with N number of
states, the total number of strings to be tested,
defined over an alphabet of m letters, is
mN +mN+1 + mN+2 +… +m2N-1. After testing all
these strings on the machine, if any is accepted
then the machine accepts infinite language
other wise not. e.g. for machine of three states
and alphabet of two letters, 2 3 +2 4 +2 5 = 56
strings are to be tested.
20
Summing Up
Deciding whether two languages are
equivalent or not, example, deciding
whether an FA accept any string or not,
method 3, examples, finiteness of a
language
21