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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.
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
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
Computer has a single FIFO queue of ?xed precision unsigned integers with the length of the queue unbounded. You can use access methods similar to those in the third model. In this
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
The Myhill-Nerode Theorem provided us with an algorithm for minimizing DFAs. Moreover, the DFA the algorithm produces is unique up to isomorphism: every minimal DFA that recognizes
proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .
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write grammer to produce all mathematical expressions in c.
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
In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
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