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Question 5. What set is generated by the following grammar, G₁? S→ aS | aSbS | E Give a proof (in the same style and with the same strategy as Question 4). This is a little trickier.

Question 5. What set is generated by the following grammar, G₁? S→ aS | aSbS | E Give a proof (in the same style and with the same strategy as Question 4). This is a little trickier.

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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Please answer Question 5, Question 4 is used as a reference Pls send me solution fast within 20 min and i will rate instantly for sure!! Solution must be in typed form!!
Question 4. Let Σ = [a, b]. Prove that the CFG G4 with rules SaSb | bsa | SS | E generates the set of all strings in Σ* with an equal number of a's and b's. Do this by (1) defining two recursive functions a,b: (EU {S})→→N such that a(w) is the number of a's in w and b(w) is the number of b's in w, and then (2) proving • for all wE (ZU {S})*, if S→w then a(w) = b(w), by induction on the definition of →* for all w E Σ*, if a(w) = b(w), then S→w, by strong induction on wl, as we did with the balanced-parentheses grammar (see this post). This proof is easier than the balanced-parentheses proof, since we don't have to worry about prefixes at all. Question 5. What set is generated by the following grammar, G₁? SaS | aSbS | E Give a proof (in the same style and with the same strategy as Question 4). This is a little trickier.
Transcribed Image Text:Question 4. Let Σ = [a, b]. Prove that the CFG G4 with rules SaSb | bsa | SS | E generates the set of all strings in Σ* with an equal number of a's and b's. Do this by (1) defining two recursive functions a,b: (EU {S})→→N such that a(w) is the number of a's in w and b(w) is the number of b's in w, and then (2) proving • for all wE (ZU {S})*, if S→w then a(w) = b(w), by induction on the definition of →* for all w E Σ*, if a(w) = b(w), then S→w, by strong induction on wl, as we did with the balanced-parentheses grammar (see this post). This proof is easier than the balanced-parentheses proof, since we don't have to worry about prefixes at all. Question 5. What set is generated by the following grammar, G₁? SaS | aSbS | E Give a proof (in the same style and with the same strategy as Question 4). This is a little trickier.
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