Consider the language of odd-length palindromes with a special center symbol'c', L = { xcxR: x is in (a+b)* and x² is just the reverse of x }. One example is thestring "baacaab". What can be said about a PDA for deciding L?A. We can build a deterministic PDA for L with just the stack alphabet [ = {Z, A}.B. We can build a deterministic PDA for L with just the stack alphabet [ = {Z, A, B }.C. We need a nondeterministic PDA for L, with just the stack alphabet [ = {Z, A}.D. We need a nondeterministic PDA for L, with just the stack alphabet [ = {Z, A, B }.

construct push down automata for string baaCaab which option is correct.

Answered: Consider the language of odd-length…

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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Implement a queue Q with two stacks S1 and S2, so that the stacks that operate together as a queue may perform the operations ENQUEUE(Q, X) and DEQUEUE(Q, Y), where Q appears to be a queue and X is the element to be inserted into Q and Y is the element to be removed from Q. Assume that the ADT operations PUSH, POP, and EMPTYSTACK are only available for stacks.Show how your technique works on a list a, b, c, which will be utilised as a queue by the stacks S1 and S2.
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Consider the equals() method from the java.util.AbstractList class. (a) Draw a control flow graph for this method. Several possiblevalues can be used for the node number in the graph. Choosesomething reasonable.(b) Label edges and nodes in the graph with the corresponding codefragments. You may abbreviate predicates as follows whenlabeling your graph:A: o == thisB: !(o instanceof List)C: e1.hasNext() && e2.hasNext()C: e1.hasNext() && e2.hasNext()D: !(o1 == null ? o2 == null :o1.equals(o2))E: !(e1.hasNext() $\parallel$ e2.hasNext())(c) Node coverage requires (at least) four tests on this graph.Explain why.

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