Prerequisite question: (This should not be answered only for reference) Let B = { 1ky| y E {0, 1)' and y contains at least k 1s, for k Show that B is a regular language. Let C = {1ky| y E {0, 1)" and y contains at most k 1s, for k Show that C isn't a regular language 1.49 a. -1) b. l) Prerequisite answer: (Only for reference) Problem 1.49 Solution: a. b. Lets assume that C is regular and P be the pumping length given by the pumping lemma. Lets consider the string S-1P0p1P e B. The pumping lemma tells that s can be split into 3 pieces s= abz for som1. By pumping lemma ac1P-i0P1P, but this not possible and where labl p. As b is a contradiction. Therefore, C isn't a regular language. Based on the above answer please answer this question: (This is the Main question to be answered) 8. Consider the language C in Problem 1.49(b. You can prove C is non-regular by (1) assuming (to the contrary) that C is regular; (2) selecting a sufficiently long string w that (according to the Pumping Lemma) should be pumpable, (3) selecting a "pumping factor" i, and finally (4) showing a contradiction: the resuting string zy'z is not n C. The key is finding the right" w and the right" i. (a) Give a "wrong" choice for w that is sufficiently long (according to the Pumping Lemma) and that does not lead to a contradiction. Justify your answer [2 points] (b) Using your right" choice in your answer to 1.49(b), give a "wrong" choice fori. Justify your answer [2 points] Prerequisite question: (This should not be answered only for reference) Let B = { 1ky| y E {0, 1)' and y contains at least k 1s, for k Show that B is a regular language. Let C = {1ky| y E {0, 1)" and y contains at most k 1s, for k Show that C isn't a regular language 1.49 a. -1) b. l) Prerequisite answer: (Only for reference) Problem 1.49 Solution: a. b. Lets assume that C is regular and P be the pumping length given by the pumping lemma. Lets consider the string S-1P0p1P e B. The pumping lemma tells that s can be split into 3 pieces s= abz for som1. By pumping lemma ac1P-i0P1P, but this not possible and where labl p. As b is a contradiction. Therefore, C isn't a regular language. Based on the above answer please answer this question: (This is the Main question to be answered) 8. Consider the language C in Problem 1.49(b. You can prove C is non-regular by (1) assuming (to the contrary) that C is regular; (2) selecting a sufficiently long string w that (according to the Pumping Lemma) should be pumpable, (3) selecting a "pumping factor" i, and finally (4) showing a contradiction: the resuting string zy'z is not n C. The key is finding the right" w and the right" i. (a) Give a "wrong" choice for w that is sufficiently long (according to the Pumping Lemma) and that does not lead to a contradiction. Justify your answer [2 points] (b) Using your right" choice in your answer to 1.49(b), give a "wrong" choice fori. Justify your answer [2 points]
