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3.2 Pumping Lemma for Regular Languages Given a language L, how do we know whether it is regular or not? If we can construct an FA to accept the language or a regular expression denoting the language, then it is a regular. But if we can not, we still can not make sure that it is not regular. So, we have develop methods to show irregular languages. We have already mentioned one method in section 3.1 to show that a language L is not regular, if the number of equivalence classes of the relation R L is infinite. The second method we shall introduce is the pumping lemma for regular languages. If a language L is finite, then it is easy to figure out whether or not a given string is in L. L is finite, it is sure that we can list them all. So, L is a regular language. Theorem 2: Let L be a finite language. Then L is regular. Proof: It is easy to construct an NFA as follows to accept the finite set L={1, 2, …, n}. q1 q0 1 2 q2 qn n p1 p2 pn qf Lemma 4(The pumping lemma): Let L be an infinite regular language over an alphabet . Then there is a constant n such that for each string in L with | | n, can be written as = xyz, where |xy| n and |y| 1, and for all k 0, xy kz L. Proof : Since L is regular, by the Myhill-Nerode theorem there is a minimal DFA M = (Q, , , q 0, F) accepting L. Let n be |Q|. If L and || n. Let =a 1 a 2…a m, where m n. Let q i= ( q 0, a 1 a 2…a i), i = 1, 2, …, m. Since there are only n states in Q and m n, by the pigeon hole theorem there are two states of q 0 , q 1 , q 2 , …, and q n are same, say 0 i j n and q i = q j. q i = ( q 0, a 1 a 2…a i) = ( q 0, a 1 a 2…a j) = q j. We have that q i= ( q 0, a 1 a 2…a i) = ( q i, a (i+1) …a j). Hence, q i= ( q i, a (i+1) …a j ) = ( q i, (a (i+1) …a j) k) , for all k 0. Let x = a 1 a 2…a i, y = a (i+1) …a j, and z = a (i+1) …a m. We have that ( q 0, xy k z) = ( q i, y k z) = ( q i, z) = ( q 0, xyz) F, for all k 0. Therefore, = xy k z, where |xy| n and |y| 1, and xy k z L, for all k 0. For a true statement ” x P(x) n Q(n) ”, where P( ) and Q( ) are two predicates, suppose that we want to show that for some y P(y) is not true. We can first assume that P(y) were true. And show there is an n such that Q(n) is false. That the predicate Q(n) is false leads to a contradiction. Hence, P(y) is not true. The pumping lemma says that ( L)( L is regular and |L| is infinite ( n ) ( ( in L, | | n) ( x, y and z, |xy| n and |y| 1, = xyz)( i 0) (xy i z L))) Therefore, L is regular and |L| is infinite, we have that “ n, ” such that “ in L, | | n, x, y and z, = xyz, where |xy| n and |y| 1, and xy i z L, for all i 0.” To show that a set L is not regular by contradiction, we first assume that L were regular and n in the pumping lemma. And show that “ in L, | | n, x, y and z, = xyz, where |xy| n and |y| 1, i 0 such that xy i z L.” Example 6: Let L={0 i1 i | i>0}. Show that L is not regular. Solution : Assume that L were regular. Let n be the constant in the pumping lemma. Choose = 0 n 1 n. We have that L and | | > n. Let = xyz, where |xy| n and |y| 1. Since |xy| n and |y| 1, we have that x = 0 |x| and y = 0 |y|. It is obvious that xy i z L for i1, since the number of 0s is not equal to that of 1s in the string xy i z. Contradict to the assumption that L is regular. Example 7: Let L be the regular expressions over an alphabet ={0, 1}. Show that L is not regular. Solution : Assume that L were regular. Let n be the constant in the pumping lemma. Choose = ( n 0*)*…)*). We have that L and | | > n. Let = xyz, where |xy| n and |y| 1. Since |xy| n and |y| 1, we have that x = ( |x| and y = ( |y|. It is obvious that xy i z L for i1, since the number of the left parentheses is not equal to that of right parentheses in the string xy i z. Contradict to the assumption that L is regular. Example 8: Let L = { in {0, 1}* | i = the number of 0’s in , and j = the number of 1’s in , i , j 1 and gcd(i, j)=1}. Show that L is not regular. Solution : Assume that L were regular. Let n be the constant in the pumping lemma. Choose = 0 t 1 s, where s =2*3*5*…*p n, p n is the nth prime number and t = p n+1. We have that L and | | > n. Let = xyz, where |xy| n and |y| 1. Hence, x = 0 |x| and y = 0 |y|. Consider the case that i = 0, xy i z = xz = 0 t - |y| 1 s. Since t - |y| < p n+1, we have that gcd(t - |y| , s) > 1. Then xz L. Contradiction.