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CS360 A3

CS360 A3 - Assignment 3 solution set prepared by Jalaj...

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Assignment 3: solution set prepared by Jalaj Upadhyay Question 1 (a) We first give the grammar and then prove the correctness. S AB B aBb | C bCc | The correctness of the above grammar can be seen by making the following observation: the grammar produces a “a” or a “c” if and only if it produces a “b”; therefore, n a ( x ) + n c ( x ) = n b ( x ) for all x L ( G ) and no x / L ( G ) is produced by the grammar. It also produces all set of a * b * before producing b * c * because of S AB. (b) We first give the grammar and then prove the correctness. S aSc | B | C B bBc | C C cC | c Note that / G ; therefore, there is no production rule that gives just as output. Now we prove the correctness of the above grammar. Let us consider any arbitrary word w L . Since it is in L , the number of c is more than n a ( w )+ n b ( w ) . This can be produced by the grammar by first applying S aSc , n a ( w ) times; then S B once followed by B bBc , n b ( w ) times; and finally B C once followed by C cC to cover the extra c 0 s. Now consider any word w ¯ L ( w will have n a ( w ) + n b ( w ) n c ( w ) . However, it is simple to see that G cannot produce such a word because the production rule of S and B has at least one non-terminal output, c ; therefore, there is no way a c is not produced when an a or a b is generated. (c) Let A := { (0 , Z 0 , 0 Z 0 ) , (1 , Z 0 , 1 Z 0 ) , (0 , 0 , 00) , (0 , 1 , ) , (1 , 0 , ) , (1 , 1 , 11) } and B := { (1 , Z 0 , 1 Z 0 ) , (1 , 0 , ) , (1 , 1 , 11) } , where the tuple ( a, b, c ) means that if a is read and b is on the top of the stack, replace it by c . A PDA that accepts L is as follow: ( q s , A ) q s ( q s , B ) q 1 ( q 1 , B ) q 2 ( q 2 , A ) q 2 ( q 2 , ( , Z 0 , Z 0 )) q f , where ( a, b ) c means that move to state c when you are in state a and the tuple b is present. The start state is q s and the final state is q f . The automaton accepts by final state.

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CS360 A3 - Assignment 3 solution set prepared by Jalaj...

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