7. 14 Points Pumping in regular languages. The pumping lemma states that every regular language has a pumping length p such that every string in the language can be pumped if it has length p or more. If p is a pumping length for language A, so is any length p'> p. The minimum pumping length for A is the smallest p that is pumping length for A. For example, if A = 01*, the minimum pumping length for A is 2. The reason is that the string s= 0 is in A and has length 1 yet s cannot be pumped; but any string in A of length 2 or more contains a 1 and hence can be pumped by dividing it so that x = 0, y =1, and z is the rest. For each of the following languages (over = {0,1)), give its minimum pumping length and justify your answer. Hint: See problem 1.55 page 91 and its solution page 99. (a) 001 U 0*1* (b) (01)* (c) (d) 1*01*01* (e) 10(11*0)*0 (f) 1011 (g) I* 7. 14 Points Pumping in regular languages. The pumping lemma states that every regular language has a pumping length p such that every string in the language can be pumped if it has length p or more. If p is a pumping length for language A, so is any length p'> p. The minimum pumping length for A is the smallest p that is pumping length for A. For example, if A = 01*, the minimum pumping length for A is 2. The reason is that the string s= 0 is in A and has length 1 yet s cannot be pumped; but any string in A of length 2 or more contains a 1 and hence can be pumped by dividing it so that x = 0, y =1, and z is the rest. For each of the following languages (over = {0,1)), give its minimum pumping length and justify your answer. Hint: See problem 1.55 page 91 and its solution page 99. (a) 001 U 0*1* (b) (01)* (c) (d) 1*01*01* (e) 10(11*0)*0 (f) 1011 (g) I*