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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.
We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while
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
For every regular language there is a constant n depending only on L such that, for all strings x ∈ L if |x| ≥ n then there are strings u, v and w such that 1. x = uvw, 2. |u
Suppose G = (N, Σ, P, S) is a reduced grammar (we can certainly reduce G if we haven't already). Our algorithm is as follows: 1. Define maxrhs(G) to be the maximum length of the
The Emptiness Problem is the problem of deciding if a given regular language is empty (= ∅). Theorem 4 (Emptiness) The Emptiness Problem for Regular Languages is decidable. P
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
These assumptions hold for addition, for instance. Every instance of addition has a unique solution. Each instance is a pair of numbers and the possible solutions include any third
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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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