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3 B = {w#t ] s = a®b°#ab®. w is a substring of t, where w, t e {a,b}*}. Hint: consider %3D

3 B = {w#t ] s = a®b°#ab®. w is a substring of t, where w, t e {a,b}*}. Hint: consider %3D

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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3 {w#t ] wis a substring of t, where w,t e {a,b}*}. Hint: consider s = aPb°#aPb°.
Transcribed Image Text:3 {w#t ] wis a substring of t, where w,t e {a,b}*}. Hint: consider s = aPb°#aPb°.
Example Proof Prove that the language B = {0"1"|n > 0} is not a regular language. Proof. Assume that B is regular and let p be the pumping length for the language. Choose the string s = 0P1P so that |s| > p. By the pumping lemma (PL) for regular languages, we can partition s = ryz so that for any i> 0, s' = xy' z E B. Let's consider three cases for the partition y: 1. The string (partition) y consists only of 0's. Then, s' = xyYz = xy'z has more 0's than l's. Clearly s' 4 B and so condition number 1 of the PL is violated. This is a contradiction for our assumption that B is regular. 2. The string (partition) y consists only of l's. Using the same argument from the previous case, we obtain another contradiction. 3. The string (partition) y consists of 0's and l's. Then, s' = xyyz = xy?z may have the same number of 0's and l's, but some of the l's will come before some of the 0's and violate membership in the language B. Hence, we have another contradiction with our assumption of reg- ularity. Therefore, we cannot avoid a contradiction with any possible y partition and conclude that the language B cannot be a regular language. Instructions Use the Pumping Lemma for CFLS to prove that each of the following lan- guages is not context-free.
Transcribed Image Text:Example Proof Prove that the language B = {0"1"|n > 0} is not a regular language. Proof. Assume that B is regular and let p be the pumping length for the language. Choose the string s = 0P1P so that |s| > p. By the pumping lemma (PL) for regular languages, we can partition s = ryz so that for any i> 0, s' = xy' z E B. Let's consider three cases for the partition y: 1. The string (partition) y consists only of 0's. Then, s' = xyYz = xy'z has more 0's than l's. Clearly s' 4 B and so condition number 1 of the PL is violated. This is a contradiction for our assumption that B is regular. 2. The string (partition) y consists only of l's. Using the same argument from the previous case, we obtain another contradiction. 3. The string (partition) y consists of 0's and l's. Then, s' = xyyz = xy?z may have the same number of 0's and l's, but some of the l's will come before some of the 0's and violate membership in the language B. Hence, we have another contradiction with our assumption of reg- ularity. Therefore, we cannot avoid a contradiction with any possible y partition and conclude that the language B cannot be a regular language. Instructions Use the Pumping Lemma for CFLS to prove that each of the following lan- guages is not context-free.
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