Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with regular expressions rather than just symbols in Σ∪{ε}. We will explain the algorithm using the example of Figure 1.
We begin by adding a new start state s and ?nal state f to the automaton and by extending it to include an edge between every state in Q∪{s} to every state in Q ∪ {f}, including self edges on states in Q. We then consolidate all the edges from a state i to a state j into a single edge, labeled with a regular expression that denotes the set of strings of length 1 or less leading directly from state i to state j in the original automaton. If there was no path directly from i to j in the original automaton the label is ∅. If there were multiple edges (or edges labeled with multiple symbols) the label is the ‘+' of the symbols on those edges (as in the edge from 2 to 1 in the example). There will be an edge from s labeled ε to the original start state and one labeled ∅ to every other state other than f. Similarly, there will be an edge labeled ε from each state in F in the original automaton to state f and one labeled ∅ from those in Q-F to f. The expression graph for the example automaton is given in the right hand side of the ?gure.
The idea, now, is to systematically eliminate the nodes of the transition graph, one at a time, by adding new edges that are equivalent to the paths through that state and then deleting the state and all its incident edges. In general, suppose we are working on eliminating node k. For each pair of states i and j (where i is neither k nor f and j is neither k nor s) there will be a path from i to j through k that looks like:
A.(A+C)=A
Sketch an algorithm for the universal recognition problem for SL 2 . This takes an automaton and a string and returns TRUE if the string is accepted by the automaton, FALSE otherwi
The project 2 involves completing and modifying the C++ program that evaluates statements of an expression language contained in the Expression Interpreter that interprets fully pa
What are the issues in computer design?
We represented SLk automata as Myhill graphs, directed graphs in which the nodes were labeled with (k-1)-factors of alphabet symbols (along with a node labeled ‘?' and one labeled
Computer has a single unbounded precision counter which you can only increment, decrement and test for zero. (You may assume that it is initially zero or you may include an explici
The initial ID of the automaton given in Figure 3, running on input ‘aabbba' is (A, aabbba) The ID after the ?rst three transitions of the computation is (F, bba) The p
Let L 3 = {a i bc j | i, j ≥ 0}. Give a strictly 2-local automaton that recognizes L 3 . Use the construction of the proof to extend the automaton to one that recognizes L 3 . Gi
It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ v) directly computes another (p, v) via
wwwwwwwwwwwwwwwwwwww
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd