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The path function δ : Q × Σ* → P(Q) is the extension of δ to strings:
This just says that the path labeled ε from any given state q goes only to q itself (or rather never leaves q) and that to ?nd the set of states reached by paths labeled wσ from q one ?rst ?nds all the states q′ reached by paths labeled w from q and then takes the set of all the states reached by an edge labeled σ from any of those q′.
We will still accept a string w i? there is a path labeled w leading from the initial state to a ?nal state, but now there may be many paths labeled w from the initial state, some of which reach ?nal states and some of which do not. When thinking in terms of the path function, we need to modify the de?nition of the language accepted by A so it includes every string for which at least one path ends at a ?nal state.
Let G be a graph with n > 2 vertices with (n2 - 3n + 4)/2 edges. Prove that G is connected.
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
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
value chain
spam messages h= 98%, m= 90%, l= 80% non spam h=12%, m = 8%, l= 5% The organization estimates that 75% of all messages it receives are spam messages. If the cost of not blocking a
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
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
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Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about
One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
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