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For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable.
"Positiveness Problem".
Note that each instance of the Positiveness Problem is a regular language. (Each instance itself is, not the set of solved instances.) Clearly, we cannot take the set of strings in the language to be our instance, (since, in general, this is likely to be in?nite in size. But we have at least two means of specifying any regular language using ?nite objects: we can give a Finite State Automaton that recognizes the language as a ?ve-tuple, each component of which is ?nite, (or, equivalently, the transition graph in some other form) or we can give a regular expression. Since we have algorithms for converting back and forth between these two forms, we can choose whichever is convenient for us. In this case, lets assume we are given the ?ve-tuple. Since we have an algorithm for converting NFAs to DFAs as well, we can also assume, without loss of generality, that the automaton is a DFA.
A solution to the Positiveness Problem is just "True" or "False". It is a decision problem a problem of deciding whether the given instance exhibits a particular property. (We are familiar with this sort of problem. They are just our "checking problems"-all our automata are models of algorithms for decision problems.) So the Positiveness Problem, then, is just the problem of identifying the set of Finite State Automata that do not accept the empty string. Note that we are not asking if this set is regular, although we could. (What do you think the answer would be?) We are asking if there is any algorithm at all for solving it.
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State and Prove the Arden's theorem for Regular Expression
Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
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We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
Another striking aspect of LTk transition graphs is that they are generally extremely ine?cient. All we really care about is whether a path through the graph leads to an accepting
Differentiate between DFA and NFA. Convert the following Regular Expression into DFA. (0+1)*(01*+10*)*(0+1)*. Also write a regular grammar for this DFA.
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
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